\documentclass[11pt,a4paper]{article} \usepackage[T1]{fontenc} \usepackage[utf8]{inputenc} \usepackage{lmodern,amsmath,amssymb,amsthm,mathtools,microtype} \usepackage[margin=26mm]{geometry} \usepackage{booktabs,array,longtable,enumitem,needspace,multicol,hyperref} \hypersetup{hidelinks,pdftitle={Serre modularity over real quadratic fields},pdfauthor={Luis V. Dieulefait, Nuno Freitas and Ariel Pacetti},pdfkeywords={Serre's conjecture; real quadratic fields; Hilbert modular forms; Galois representations; compatible systems; finite flat group schemes; Selmer groups},bookmarksdepth=2} \makeatletter \newcommand{\compactcontents}{% \begingroup \setcounter{tocdepth}{1}% \setlength{\parskip}{0pt}% \small \renewcommand{\@pnumwidth}{2em}% \renewcommand{\@tocrmarg}{3em}% \section*{Contents}% \renewcommand{\l@section}{\@dottedtocline{1}{0em}{1.9em}}% \renewcommand{\l@part}[2]{\addpenalty{-\@highpenalty}% \addvspace{5pt}\begingroup\bfseries \@dottedtocline{0}{0em}{0em}{##1}{##2}\endgroup}% \begin{multicols}{2}\raggedcolumns \@starttoc{toc}% \end{multicols} \endgroup} \makeatother \setlength{\emergencystretch}{3em} \setlength{\parskip}{3pt} \setlength{\parindent}{14pt} \setlist{nosep,leftmargin=*,topsep=3pt} \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{hypothesis}[theorem]{Hypothesis} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \newtheorem{definition}[theorem]{Definition} \providecommand{\Q}{\mathbf Q} \providecommand{\Z}{\mathbf Z} \providecommand{\F}{\mathbf F} \providecommand{\OO}{\mathcal O} \providecommand{\cR}{\mathcal R} \providecommand{\GL}{\operatorname{GL}} \providecommand{\SL}{\operatorname{SL}} \providecommand{\PGL}{\operatorname{PGL}} \providecommand{\PSL}{\operatorname{PSL}} \providecommand{\Gal}{\operatorname{Gal}} \providecommand{\Ind}{\operatorname{Ind}} \providecommand{\Sym}{\operatorname{Sym}} \providecommand{\HT}{\operatorname{HT}} \providecommand{\WD}{\operatorname{WD}} \providecommand{\rd}{\operatorname{rd}} \providecommand{\ssm}{\mathrm{ss}} \providecommand{\br}{\overline r} \providecommand{\brho}{\overline\rho} \providecommand{\bchi}{\overline\chi} \providecommand{\Frob}{\operatorname{Frob}} \providecommand{\C}{\mathbf C} \providecommand{\om}{\overline\chi_3} \providecommand{\tr}{\operatorname{tr}} \providecommand{\Aut}{\operatorname{Aut}} \providecommand{\Out}{\operatorname{Out}} \providecommand{\ad}{\operatorname{ad}} \providecommand{\SC}{\mathrm{SC}} \providecommand{\SP}{\mathrm{SP}} \providecommand{\Ext}{\operatorname{Ext}} \providecommand{\Cl}{\operatorname{Cl}} \providecommand{\ur}{\mathrm{ur}} \providecommand{\Hom}{\operatorname{Hom}} \providecommand{\diag}{\operatorname{diag}} \providecommand{\St}{\operatorname{St}} \providecommand{\cO}{\mathcal O} \providecommand{\End}{\operatorname{End}} \providecommand{\Nm}{\operatorname N} \providecommand{\A}{\mathbb A} \providecommand{\eps}{\varepsilon} \begin{document} \begin{center} {\Large\bfseries Serre modularity over real quadratic fields}\\[8pt] {\large Luis V. Dieulefait$^{1}$,\quad Nuno Freitas$^{2}$, \quad Ariel Pacetti$^{3}$}\\[6pt] {\small $^{1}$Departament de Matem\`atiques i Inform\`atica, Universitat de Barcelona, 08007 Barcelona, Spain\\ \hspace*{1em}and Centre de Recerca Matem\`atica (CRM), 08193 Bellaterra, Spain\\[1pt] $^{2}$Instituto de Ciencias Matem\'aticas (ICMAT--CSIC), 28049 Madrid, Spain\\[1pt] $^{3}$CIDMA, Departamento de Matem\'atica, Universidade de Aveiro, 3810-193 Aveiro, Portugal\\[3pt] \texttt{apacetti@ua.pt}\quad \texttt{ldieulefait@ub.edu}\quad \texttt{nuno.freitas@icmat.es}}\\[5pt] {\small 3 November 2026} \end{center} \begin{abstract} Assuming GRH for Dedekind zeta functions, we prove Serre’s conjecture over the real quadratic fields \(\mathbf Q(\sqrt d)\) for \[ d\in\{2,3,5,6,7,10,11,13,17,21,33,37,41,57,65,73,89\}. \] More precisely, every continuous, absolutely irreducible, totally odd two-dimensional Galois representation over these fields in odd residual characteristic is modular. The result for \(d=17\) is unconditional; for \(d=33\), GRH can be replaced by the assertion that Dedekind zeta functions have no zeros with real part greater than \(31/32\). The proof combines conductor reduction with local ramification estimates and global discriminant bounds. Its principal new ingredients are a modularity criterion for \(2\)-adic representations with scalar reduction and integral trace arguments that force quadratic descent or bound coefficient fields. Uniform local lifting constructions allow these methods to treat arbitrary conductor. \end{abstract} \smallskip {\small\noindent\textbf{2020 Mathematics Subject Classification.} Primary 11F80; Secondary 11F33, 11F41, 11R42.\par \noindent\textbf{Keywords.} Serre's conjecture; real quadratic fields; Hilbert modular forms; Galois representations; compatible systems; finite flat group schemes; Selmer groups.\par} \compactcontents \clearpage \section{Introduction} Let $K$ be a real quadratic field. We study Serre's conjecture for continuous absolutely irreducible totally odd representations $G_K\to\GL_2(\overline\F_p)$, with $p$ odd and with arbitrary conductor. Here modularity means occurrence in the reduction of the Galois representation of a regular algebraic cuspidal Hilbert modular form. We do not prescribe its weight or level. Over $\Q$, the conjecture originated in Serre's questions about modular eigenvalues modulo a prime and was subsequently refined by his predictions for weight, level and character \cite[Section~3, Problem~1]{h:Serre75}, \cite{h:Serre87}. The lifting theorems of Wiles, Taylor--Wiles and Skinner--Wiles, together with potential modularity, laid the foundations for its proof \cite{h:Wiles,h:TW,r:t:SW99,h:Taylor}. An essential step in this development was the construction of compatible systems containing a given representation. Dieulefait proved such an existence theorem over $\Q$, in its original crystalline range, using potential modularity and descent \cite[Theorem~1.1]{h:Dieulefait04}. This permits a change of residual characteristic while retaining a single characteristic-zero compatible system. The method was used to establish cases of the Fontaine--Mazur conjecture and small-level cases of Serre's conjecture \cite{h:Dieulefait07,h:KWsmall}. Khare and Wintenberger subsequently proved Serre's conjecture in full generality \cite{h:KWI,h:KWII}, with Kisin's two-adic modularity lifting theorem providing the final dyadic input \cite{h:Kisin2}. Later lifting theorems permit shorter versions of this proof, as in Dieulefait--Pacetti \cite{h:DP23}. Over totally real fields, the corresponding strategy of connecting residual representations through compatible systems is formulated in the Rayuela conjecture \cite{h:Rayuela}. Over a real quadratic field, discriminant bounds need not exclude all small-characteristic representations of small conductor. For example, the elliptic curve \[ y^2+\sqrt6xy+(1+\sqrt6)y =x^3+(1-\sqrt6)x^2 -(1+3\sqrt6)x-1-2\sqrt6 \] has everywhere good reduction over $\Q(\sqrt6)$; its label is \href{https://www.lmfdb.org/EllipticCurve/2.2.24.1/1.1/a/1}{\texttt{2.2.24.1-1.1-a1}} in~\cite{h:LMFDB}. Thus the base cases must establish modularity of surviving representations, rather than only nonexistence. In addition, residual reducibility and ordinarity of a characteristic-zero source are different conditions. A congruence is useful only when modularity can be transferred back to that actual source. Our starting point is the characteristic-three method of Dieulefait--Pacetti over $\Q(\sqrt{13})$ \cite{h:DP13}. Successive discriminant comparisons restrict the residual images, and a conductor reduction relates arbitrary representations to the remaining base case. We retain this distinction between the residual base and the conductor argument, remove the restriction at two, and supply new source theorems where residual modularity alone is insufficient. Fontaine's local bounds, refined local ramification calculations, and global discriminant estimates provide the arithmetic input \cite{c:FontaineBT,c:MT,s:Odlyzko}; finite-flat groups and integral trace orders control the scalar and dihedral cases. The results below use the lifting and automorphic inputs specified in Sections~\ref{shared:systems} and~\ref{external:dyadic}. In particular, statements involving the split-dyadic scalar argument are relative to the dyadic structural and propagation results listed there, including the cited preprints. These inputs are distinct from the analytic hypothesis GRH. Write $d$ for a squarefree radicand. \begin{theorem}\label{main:grh}\label{main:dyadic} Let \[ d\in\{2,3,5,6,7,10,11,13,17,21,33,37,41,57,65,73,89\}, \qquad K=\mathbf Q(\sqrt d). \] Assume GRH for Dedekind zeta functions. For every odd prime $p$, every continuous absolutely irreducible totally odd representation \[ \bar\rho:G_K\longrightarrow \mathrm{GL}_2(\overline{\mathbf F}_p) \] is modular. \end{theorem} For $d=17$ the theorem holds without GRH. For $d=33$ the proof also works with the zero-free hypothesis stated in Corollary~\ref{s:cor:no-grh}. The analytic arguments for both variants are included in Section~\ref{s:sec:residual}. Further nonordinary source arguments and individual residual-image exclusions hold without GRH; these refinements are recorded in the corresponding proofs. \subsection{Structural theorems} The field list is an application of three mechanisms: conductor reduction to a specified terminal characteristic, scalar modularity from a quadratic Selmer condition, and coefficient-field rigidity from simultaneous integral trace orders. We state these mechanisms separately, since their hypotheses have different roles. \paragraph{Conductor reduction.} A strongly compatible system, and the conventions for weight two, are specified in Section~\ref{shared:systems}. The following theorem concerns a system already constructed in characteristic zero; it does not supply an initial lift of an arbitrary residual representation. \begin{theorem}[Conductor reduction at a split terminal prime] \label{shared:terminal-criterion} Let $K$ be real quadratic, suppose two does not split in $K$, and let $\ell$ be an odd rational prime that splits in $K$. Assume: \begin{enumerate}[label=(\roman*)] \item every absolutely irreducible totally odd representation $G_K\to\GL_2(\overline\F_\ell)$ of determinant $\bar\chi_\ell$ and unramified outside $\ell$ is modular; \item for every reducible representation with that determinant and ramification support, its constituent quotient is totally odd and extends to $G_\Q$. \end{enumerate} Then every nondihedral potentially modular weight-two strongly compatible system over $K$, whose determinant is cyclotomic times a finite totally even character, is modular. \end{theorem} The proof is given in Section~\ref{struct:odd-conductor}. The quotient in hypothesis (ii) is automatically totally odd, since its two constituents have product $\bar\chi_\ell$. \paragraph{A scalar Selmer condition.} Suppose two splits in $K$, with dyadic places $w_0,w_1$. Put \begin{equation}\label{struct:selmer-space} V^+_{w_0}(K)= \left\{[a]\in K^\times/K^{\times2}: \begin{array}{l} a\text{ is totally positive},\\ v_{\mathfrak q}(a)\equiv0\pmod2\quad(\mathfrak q\nmid2),\\ a\in K_{w_1}^{\times2} \end{array}\right\}. \end{equation} This finite squareclass space classifies totally real quadratic extensions unramified outside $w_0$ and split at $w_1$. Quadratic conjugation exchanges the two choices of dyadic place. The condition we use is \begin{equation}\tag{SV}\label{struct:SV} V^+_{w_0}(K)=0. \end{equation} It is not an odd-narrow-class-number condition. \begin{theorem}[Scalar dyadic modularity]\label{s:thm:scalarFM} Let $K$ be a real quadratic field in which two splits and \eqref{struct:SV} holds. Under the dyadic structural, propagation and classicality inputs of Section~\ref{external:dyadic}, let $E/\Q_2$ be finite and let $r:G_K\to\GL_2(E)$ be continuous and absolutely irreducible, unramified outside finitely many places, with $\det r=\varepsilon_2$ and $\bar r^{\mathrm{ss}}=\mathbf1\oplus\mathbf1$. Suppose that at both dyadic places $r$ is absolutely irreducible, de Rham and has distinct Hodge--Tate weights. Then $r$ is automorphic. \end{theorem} Sections~\ref{s:sec:field-preparation}--\ref{s:sec:scalarModularity} construct the auxiliary fields and prove the theorem. Together with a residual base at two and a two-power ordinary class number, it gives the split-dyadic Serre criterion of Theorem~\ref{struct:split-criterion}. The seven split-dyadic fields in the main theorem satisfy these arithmetic conditions. The Selmer vanishing is also valid when the narrow class number is even. \paragraph{Integral trace orders and symmetry.} Let $\mathcal R$ be a strongly compatible system of cyclotomic determinant over a real quadratic field $K$, with trace field $D$. For finitely many coefficient orbits, form their simultaneous integral matrix order $E_0$, including the trace scalars and every dyadic coefficient place, as in Section~\ref{r:orders}. Its rank is four times the sum of the trace-field degrees. The integral order, rather than a single residual two-dimensional realization, is essential here. \begin{theorem}[A trace-order bound and quadratic symmetry] \label{struct:packet} Let $K/\Q$ be real quadratic with conjugation $\sigma$, and let $\mathcal R$ be an absolutely irreducible strongly compatible rank-two system with cyclotomic determinant and trace field $D$. Let $h,\eta$ be restrictions of quadratic characters of $G_\Q$. Form the coefficient orbits generated by \[ H(\mathcal R)=\mathcal R\otimes h, \qquad G(\mathcal R)=\mathcal R^\sigma\otimes\eta. \] Suppose $\mathcal R\not\simeq\mathcal R\otimes h$ and their simultaneous integral order satisfies $\dim_{\F_2}(E_0/2E_0)<32$. Then either \[ \mathcal R^\sigma\simeq\mathcal R\otimes\eta \quad\text{or}\quad \mathcal R^\sigma\simeq\mathcal R\otimes\eta h, \] or $[D:\Q]\le2$ or $D/\Q$ is biquadratic. In the latter alternatives every coefficient prime has residue degree at most two. In the former alternatives, a totally odd regular system is modular by quadratic descent. \end{theorem} The arithmetic work in applying Theorem~\ref{struct:packet} lies in bounding the actual order. The finite-flat ramification theorem \ref{r:break} and the regular-order construction in Section~\ref{r:orders} supply different parts of this argument. The bound of twenty-six over $\Q(\sqrt{11})$ then gives the coefficient-field alternative directly; the bound of ten over $\Q(\sqrt3)$ uses Corollary~\ref{struct:two-orbit}. Two further source theorems have a separate role. Proposition~\ref{t:5:prop:initial-exceptional-lift} constructs the initial lift in the exceptional characteristic-five case over $\Q(\sqrt5)$, which the general prescribed-lift theorem excludes. Theorem~\ref{t:21:red3:source-theorem} proves modularity of the nonordinary reducible characteristic-three source over $\Q(\sqrt{21})$. Its proof kills the anti-invariant part of the full pseudodeformation cotangent space and applies Lemma~\ref{t:involution}; no finiteness theorem for the whole pseudodeformation ring is needed. These are source theorems, not consequences of residual-image bounds alone. \subsection{Field applications} The terminal characteristics of the outer conductor reductions are listed below. A source proof may use other coefficient characteristics internally. At every backward step the lifting theorem is applied to the actual old source, with its weight and local ordering retained. \begin{center}\small \begin{tabular}{@{}p{0.24\textwidth}p{0.18\textwidth}p{0.46\textwidth}@{}} \toprule Radicands $d$ & Terminal prime & Source arguments\\\midrule $7,13,37$ & $3$ & Supported residual bases and auxiliary composita.\\[3pt] $2,3,6,10,11$ & $7,13,5,3,5$ & Finite-flat extensions, trace orders and descent.\\[3pt] $5$ & $11$ & Exceptional initial lift and split-eleven sources.\\[3pt] $21$ & $5$ & Mixed local types and scalar or dihedral rigidity.\\[3pt] $17,33,41,57,65,73,89$ & $2$ & The scalar Selmer criterion and a residual base at two.\\ \bottomrule \end{tabular} \end{center} Part~\ref{part:structural} develops the lifting constructions, conductor criteria, scalar theorem and integral-order methods. Part~\ref{part:sources} proves the required source theorems. Part~\ref{part:applications} verifies the remaining residual bases and assembles the field results. The appendices specify the longer finite calculations; their programs and outputs are provided in~\cite{code}. Buzzard--Diamond--Jarvis formulated the weight conjecture at unramified residual primes \cite{h:BDJ}; see also \cite{h:Gee}. Our modularity results yield the weight part of Serre's conjecture under the image hypotheses of Gee--Liu--Savitt \cite[Theorem~6.2.1]{h:GLS}. Namely, suppose the residual representation remains absolutely irreducible after adjoining the roots of unity of order the residual characteristic and, in characteristic five, that the projective image of this restriction is not the alternating group on five letters. Then the occurring Serre weights are exactly those prescribed by the local crystalline-lift recipes, including at ramified residual primes and for all predicted companion weights. These consequences inherit the assumptions of the corresponding modularity theorem; the weight theorem introduces no further analytic hypothesis. %\subsection*{Acknowledgments} %We thank Toby Gee for helpful comments on the construction of %lifts. We are especially grateful to Lassina Demb\'el\'e for many %valuable discussions and for his deep contribution to the %proof of the base cases in the first examples considered. % %\Needspace{10\baselineskip} \part[General criteria and methods]{General criteria and local methods}\label{part:structural} \section{Compatible systems and congruences}\label{shared:systems} We use $d$ for the squarefree radicand, $D$ for a fundamental discriminant, and $q$ for the residual coefficient-field cardinality. The finite arithmetic and numerical checks are supplied in \cite{code}; the three longer certificate descriptions appear in the appendices. Throughout, $G_F=\operatorname{Gal}(\overline F/F)$ and the cyclotomic character $\chi_p$ has labelled Hodge--Tate weight one. Irreducibility means absolute irreducibility. A representation is totally odd if every real complex conjugation has determinant $-1$. A system over a totally real field is a pure, regular, totally odd, absolutely irreducible, strongly compatible rank-two system. We require compatibility of the Frobenius-semisimplified Weil--Deligne representations at every finite place, including a place above the coefficient characteristic, and compatibility of labelled weights. Its conductor support consists of the places where the common Weil--Deligne representation is ramified or has nonzero monodromy. Outside this support its $p$-adic member is crystalline above $p$ and unramified at the other places. Weight two means labelled weights $\{0,1\}$. Regular weights may include negative integers. \label{shared:pure-special} On the closure of a special local component the Frobenius eigenvalues have ratio $\mathrm N(v)$. For a pure system, monodromy cannot vanish at such a place: otherwise both eigenvalues would have the same complex absolute value, contradicting that ratio. We write $\mathcal R\rightsquigarrow\mathcal R'$ when a congruence, possibly followed by a finite character twist, has the property that modularity of $\mathcal R'$ implies modularity of $\mathcal R$. Every such implication below is an application of a lifting theorem to the actual old characteristic-zero member. Finite twists preserve modularity. Equality of good Frobenius polynomials and Chebotarev then propagate modularity from one member to the whole system. \subsection{Residual witnesses and lifting} \begin{lemma}\label{shared:weightzero} Let $F$ be totally real of even degree. An irreducible residual representation with a regular algebraic cuspidal Hilbert modular witness has a parallel-weight-two cuspidal witness, after allowing arbitrary level. \end{lemma} \begin{proof} Transfer the witness to the quaternion algebra over $F$ ramified at every real place and split at every finite place. At sufficiently small level $U=U^pU_p$, reduction of a primitive integral eigenvector is nonzero, with coefficient module $V$ over a finite field $k$. Its $U_p$-action factors through a finite quotient $G$; write $U'_p$ for the kernel. The map $V\hookrightarrow\operatorname{Map}(G,V)$ given by $v\mapsto(g\mapsto gv)$ is equivariant for $(hf)(g)=f(gh)$. The target is a sum of regular representations. Pointwise application and induction in stages give the same away-from-$p$ Hecke eigencharacter in scalar forms of level $U^pU'_p$. Their integral space is $\mathcal O^X$ for a finite double-coset set $X$, so reduction is surjective. Deligne--Serre's lifting lemma~\cite[Lemme~6.11]{g:DS} lifts the eigencharacter to characteristic zero. A norm character would have reducible residue; inverse Jacquet--Langlands therefore gives a parallel-weight-two cuspidal form. Retaining both operators $T_v,S_v$, with polynomial $X^2-T_vX+\mathrm N(v)S_v$, preserves the residual trace and determinant. Chebotarev and Brauer--Nesbitt identify the residue. No exact characteristic-zero central character or ordinary ordering is prescribed by this argument. \end{proof} Totally odd solvable residual representations have solvable Artin witnesses, followed by regular weight raising, in the scopes of automorphic induction and Langlands--Tunnell; see \cite[Theorem~1.2(1), Lemma~2.1 and proof of Theorem~2.5]{h:AKT}. Lemma~\ref{shared:weightzero} gives weight two whenever it is needed. We use Dickson's classification and the rank-two adequacy criterion \cite[Proposition~A.2.1]{g:BLGG} throughout. \begin{lemma}\label{shared:oddMLT} Let $K$ be real quadratic, $p>2$, and let $r:G_K\to\mathrm{GL}_2(\overline{\mathbf Q}_p)$ be finitely ramified and totally odd, with modular residue whose restriction to $G_{K(\zeta_p)}$ is irreducible. Then $r$ is modular in either case: \begin{enumerate}[label=(\roman*)] \item $p$ splits in $K$, and $r$ is potentially semistable with distinct labelled weights at the places above $p$; \item $r$ is de Rham of labelled weights $\{0,1\}$ above $p$, and its residue has a parallel-weight-two modular witness. \end{enumerate} \end{lemma} \begin{proof} Apply respectively \cite[Theorem~4.7]{g:Tung} and \cite[Theorem~2.3]{h:AKT}, dualizing to their inverse-cyclotomic normalization when required. In (ii), the parallel-weight-two witness gives the required weight-zero automorphic convention. \end{proof} \begin{lemma}\label{shared:dyadicMLT} Let $F$ be a totally real extension of a real quadratic field, and let $r:G_F\to\mathrm{GL}_2(\overline{\mathbf Q}_2)$ be finitely ramified and totally odd, with determinant $\chi_2\psi$, where $\psi$ has finite order. Suppose its residue is nonsolvable. Then $r$ is modular in either case: \begin{enumerate}[label=(\roman*)] \item $r$ is nearly ordinary of regular weights at every dyadic place and its residue has a nearly ordinary regular modular witness with the required residual ordinary ordering; \item $r$ is potentially Barsotti--Tate and potentially nonordinary at every dyadic place and its residue has a regular modular witness. \end{enumerate} \end{lemma} \begin{proof} The first case is \cite[Theorem~5.2.1]{g:Allen}. If the line and quotient have weights $b_\tau>a_\tau$ with sum one, its parameters are $w_\tau=a_\tau$ and $k_\tau=b_\tau-a_\tau+1$, so $k_\tau+2w_\tau=2$. Nonsolvability excludes the CM-dihedral exception; residual distinctness is not imposed. In the second case choose finite Galois extensions of the dyadic completions over which the source is Barsotti--Tate. The solvable local realization theorem~\cite[Chapter~X, Section~2]{g:AT}, in the form~\cite[Theorem~3.1 and Example~3.3]{g:Conrad}, realizes them in a solvable totally real Galois extension. No degree is prescribed. The nonsolvable residual projective image is $\mathrm{PSL}_2(\mathbf F_{2^a})$, $a\ge2$; its image after solvable Galois restriction is normal with solvable quotient and is unchanged. Solvable base change retains a cuspidal modular witness, put in weight two by Lemma~\ref{shared:weightzero}. Apply \cite[Theorem~3.3.5]{h:Kisin2}. Its condition that a dyadic completion equal $\mathbf Q_2$ applies only at potentially ordinary places, and there are none. For descent take a cyclic series. Strong multiplicity one makes the automorphic representation invariant at each step because the Galois representation extends downstairs. The descended Galois representation differs from the given one by a character of the cyclic quotient. Twisting and iterating proves modularity over $F$; see~\cite[Lemma~2.2.2]{g:BLGGT}. \end{proof} \subsection{Globalization and compatibility} \begin{lemma}\label{shared:global} Let $K$ be real quadratic, fix a finite set $S$ of places containing the coefficient places and all residual ramification, and let $\bar r$ be totally odd with irreducible cyclotomic restriction. At coefficient characteristic two assume nonsolvable image. Fix a totally odd global de Rham determinant lifting its determinant and unramified outside $S$, and nonempty compatible local components with regular weights at the coefficient places. Outside the exceptional characteristic-five case of \cite[Definition~4.1.1]{h:EGPZ}, these components admit a global lift unramified outside $S$. Once such a lift has been constructed, it is potentially modular and belongs to a strongly compatible system in the following cases: \begin{enumerate}[label=(\roman*)] \item odd coefficient characteristic and parallel weight two, with irreducible cyclotomic residual restriction; \item odd coefficient characteristic, adequate cyclotomic residue, and regular potentially diagonalizable crystalline local components; \item coefficient characteristic two, determinant $\chi_2$, and either ordinary crystalline components of regular weights at all dyadic places, or potentially nonordinary crystalline components of weight two at all dyadic places. \end{enumerate} \end{lemma} \begin{proof} Existence is \cite[Theorem~4.2.1 and Remark~1.1.2]{h:EGPZ}. Away from the coefficient prime, inertial conditions and conductor bounds specify unions of components. At that prime, take components through the supplied points; potential diagonalizability persists after a fixed sufficiently large local extension. For weight two first use the avoiding weight-two residual witness of \cite[Theorem~5.1.1]{g:Snowden}, followed by the actual-source theorem~\cite[Theorem~2.3]{h:AKT}. The potential-modularity field is chosen Galois and disjoint from the residual kernel and cyclotomic fields. Thus the actual source becomes modular after this extension. This argument starts with the weight-two point already constructed; it does not assume the existence of an exceptional initial characteristic-five point. For regular potentially diagonalizable sources with $p\ge7$, use \cite[Theorem~C, Theorem~4.5.1 and Corollary~4.5.2]{g:BLGGT}, with avoidance imposed before the CM quadratic base change. For the smaller odd characteristics, Snowden's avoiding residual theorem gives an ordinary weight-two witness over a totally real field, disjoint from the residual and cyclotomic fields. After a further disjoint CM quadratic extension, its local representations are potentially diagonalizable by~\cite[Lemma~1.4.3(1)]{g:BLGGT}. Apply the potentially diagonalizably automorphic alternative of \cite[Theorem~1]{g:DG} to the actual source, with polarization multiplier its determinant; adequacy and exclusion of $\zeta_p$ are retained. Solvable descent returns potential Hilbert modularity. At two, first make the finite local preparation which trivializes the residual dyadic restrictions, inside a totally real extension avoiding the global residual kernel. The ordinary weight-two residual witness of \cite[Proposition~6.7]{g:Thorne2} can then be chosen over a further avoiding field, with trivial finite central character. Its residual ordinary ordering agrees with that of the source because the prepared residual characters are trivial. Apply Lemma~\ref{shared:dyadicMLT} to the actual source. This proves potential modularity before a compatible system is invoked. The construction of compatible systems by potential modularity and descent originates in Dieulefait's work over $\mathbf Q$ \cite[Theorem~1.1]{h:Dieulefait04}. For the totally real fields and local conditions here, use the Brauer-induction construction of \cite[Proposition~9.2.1 and Remark~9.2.2]{g:Snowden}, retaining the finite characters in the Brauer identity. Mackey's formula and Frobenius reciprocity express the self-pairing of each virtual member in terms of pairings of restricted modular representations with finite character twists. Compatibility and Chebotarev make these pairings independent of the coefficient prime. At the original prime the self-pairing is one, so each virtual member is an irreducible or its negative. Its dimension two excludes the negative. Local--global compatibility for the modular systems \cite[Theorem~1 and equation~(1)]{g:Skinner} includes the coefficient prime and arbitrary cohomological weights. Restriction--induction identities give the common Weil representations; purity determines monodromy~\cite[Lemma~1.3.6(2)--(3)]{g:BLGGT}. Labelled weights are compatible by the same restriction argument. \end{proof} For the quadratic fields here other than $\mathbf Q(\sqrt5)$, the characteristic-five exception is excluded uniformly: an exceptional projective $S_5$ extension would contain $K(\zeta_5)$ and therefore give an abelian quotient of order four, whereas $S_5^{\mathrm{ab}}=C_2$. The exceptional initial lift over $\mathbf Q(\sqrt5)$ is treated separately below. \section{An auxiliary prime for congruences}\label{shared:protection} In this section $K$ is real quadratic. The finite set $S$ may include any prescribed terminal reset primes. \begin{proposition}\label{shared:protect} Let $\cR$ be a nondihedral weight-two system of determinant $\chi\psi$, where $\psi$ has finite two-power order. Fix a finite set $S$ of rational primes containing $2,3$, all primes dividing the discriminant of $K$, and every conductor prime. There are a prime $N>\max(S\cup\{5\})$, a degree-one prime $\lambda$ of $K$ of norm $l>N$, and a congruent weight-two system $\cR'$ such that $\cR\rightsquigarrow\cR'$ in characteristic $N$, its determinant remains $\chi\psi$, and its only new conductor prime is $\lambda$. At $\lambda$ its type is induced from the unramified quadratic extension with inverse inertia characters of order $N$. While this type is retained and conductor is supported at places over $S\cup\{l\}$, the residual representations at every rational prime in $S$ have nonsolvable image. At odd such characteristics their cyclotomic restrictions are adequate. \end{proposition} \begin{proof} Let $K_S/K$ be the maximal elementary quadratic extension unramified outside the places over $S$ and infinity. It is finite. Write $2^a$ for the order of $\psi$. By \cite[Proposition~4.6]{h:Rayuela}, choose $N>\max S$, $N>5$, split in $K$ and the coefficient field, with $N\equiv1\pmod{2^{\max\{2,a+1\}}}$, such that a residual member contains $\SL_2(\F_N)$. Its determinant is surjective on inertia above $N$, so its image is $\GL_2(\F_N)$. Let $R/K$ be the projective kernel field. Prescribe Frobenius to act trivially on $K_S$, by inversion on $K(\zeta_N)$, and as projective complex conjugation on $R$. The first two prescriptions agree on the intersection, since $K(\zeta_N)/K$ is totally ramified above $N$, whereas $K_S/K$ is unramified there. The only nontrivial solvable quotient of $\PGL_2(\F_N)$ is its determinant-squareclass quotient. Our congruence makes $\bar\psi$ square-valued, so that quotient is cut out by $\bchi_N^{(N-1)/2}$ and lies in $K(\zeta_N)$. Inversion fixes it, and projective complex conjugation belongs to $\PSL_2(\F_N)$, since $-1$ is square modulo $N$. The prescriptions therefore agree on $R\cap K_SK(\zeta_N)$. Chebotarev gives a positive-density set of such prime ideals. Those of residue degree greater than one have zero prime-ideal density. We may consequently choose $\lambda$ of degree one and arbitrarily large norm $l$, outside the rational primes already used. Thus $l$ splits in $K$, $l\equiv-1\pmod N$, and the residual Frobenius eigenvalues are $a,-a$. Put $\delta=\chi_N\psi$. Choose $\alpha$ lifting $a$ with $\alpha^2=-\delta(\Frob_\lambda)$, and define local inertia and Frobenius by \[ T=\begin{pmatrix}\zeta_N&0\\0&\zeta_N^{-1}\end{pmatrix}, \qquad \Phi=\begin{pmatrix}0&\alpha\\\alpha&0\end{pmatrix}. \] They satisfy $\Phi T\Phi^{-1}=T^l$ and have determinant $\delta$; reduction gives the original unramified local representation. Retain the old crystalline weight-two lifts above $N$ and compatible conditions elsewhere. Lemma~\ref{shared:global} gives a global weight-two lift and its system. Lemma~\ref{shared:oddMLT} at the split prime $N$ gives backward modularity. For a later coefficient characteristic $t\in S$, the two order-$N$ inertia characters at $\lambda$ remain distinct and are exchanged by Frobenius. The residual representation is therefore irreducible and its projective image contains an element of order $N>5$, excluding $A_4,S_4,A_5$. If it were dihedral, the inducing quadratic extension would be unramified outside $S$: at $\lambda$ its odd-order inertia lies in the rotations. But Frobenius at $\lambda$ exchanges the rotation eigenspaces, making $\lambda$ inert in that quadratic extension, contrary to its complete splitting in $K_S$. Dickson's classification forces nonsolvable projective image. Cyclotomic restriction retains its perfect derived subgroup, hence irreducibility. The order-$N$ element excludes the inadequate characteristic-three and characteristic-five cases; for $t>5$, the usual rank-two criterion applies. In characteristic two, nonsolvability is the required conclusion. \end{proof} \section{Local lifts and persistent inertial types}\label{shared:local} \begin{lemma}\label{c:crystalline} Let $F/\Q_p$ be finite, with $p>2$, and let $\br:G_F\to\GL_2(\overline\F_p)$ have determinant $\bchi_p$. Then $\br$ has a regular crystalline, potentially diagonalizable lift with determinant exactly $\chi_p$. \end{lemma} \begin{proof} Apply \cite[Proposition~2.7]{c:BIP} with prescribed determinant $\chi_p$. In rank two, its construction also gives potential diagonalizability. An irreducible residual representation is induced from a character of the unramified quadratic extension; the constructed lift is induced from a crystalline character, so becomes a sum of crystalline characters after restriction. A reducible representation is lifted as a crystalline extension of crystalline characters, to which \cite[Lemma~1.4.3(1)]{g:BLGGT} applies. The unramified scalar correction in the fixed-determinant construction preserves both properties. \end{proof} \begin{lemma}\label{c:monodromy} Let $F/\Q_p$ be finite, and let $r:G_F\to\GL_2(\overline\Q_p)$ be potentially semistable with labelled Hodge--Tate weights $\{0,1\}$ and determinant $\chi_p$ times a finite-order character. If its monodromy is nonzero, then $r$ is potentially ordinary. Thus a potentially nonordinary representation with these weights is potentially Barsotti--Tate. If $F=\Q_p$ and $r$ is attached to a weight-two newform which is Steinberg at $p$, with determinant $\chi_p$ and $a_p=s\in\{1,-1\}$, its ordinary quotient is unramified with Frobenius value $s$. \end{lemma} \begin{proof} Pass to a finite extension over which $r$ is semistable and the finite determinant factor is trivial. Apply the filtered-module functor to $r^\vee$, with filtration jumps zero and one. If the residue degree is $f$ and $N\ne0$, the relation $N\varphi=p\varphi N$ gives eigenvalues $\alpha,p^f\alpha$ for $\varphi^f$, with the first eigenline equal to $\ker N$. Equality of the total Hodge and Newton numbers gives $v_p(\alpha)=0$. Every labelled Hodge number on this line is zero or one. Weak admissibility bounds their average by the Newton slope zero, so all are zero. The line is weakly admissible and corresponds to a crystalline weight-zero, hence unramified, subrepresentation of $r^\vee$. Its dual is an ordinary quotient of $r$. Consequently potential nonordinarity forces $N=0$, and the resulting potentially crystalline representation with these weights is potentially Barsotti--Tate. In the last assertion, local--global compatibility identifies the Frobenius eigenvalues on the dual module as $s,ps$ \cite[equation~(1) and Theorem~1]{g:Skinner}. The same argument gives its unramified line with unit eigenvalue $s$; dualizing replaces the Frobenius value by $s^{-1}=s$. \end{proof} \begin{lemma}\label{c:dyadiclocal} Let $F/\Q_2$ be quadratic, and let $\br:G_F\to\GL_2(\overline\F_2)$ have determinant one. \begin{enumerate}[label=(\roman*)] \item If $\br$ is reducible, it has a crystalline ordinary lift of determinant $\chi_2$ with regular labelled weights. It may retain any specified invariant residual line. Its invariant line has the larger weight at every embedding, and the lift lies on an ordinary component of the fixed-determinant crystalline space. If $F/\Q_2$ is ramified, the weights may be chosen to be $\{1-n,n\}$ at both embeddings for some $n\ge2$. \item If $\br$ is irreducible, it has a crystalline lift of determinant $\chi_2$ and weights $\{0,1\}$. Every potentially Barsotti--Tate lift with these weights is potentially nonordinary. \end{enumerate} \end{lemma} \begin{proof} In the reducible case use the proof of \cite[Proposition~2.7]{c:BIP}, starting with the given residual line and quotient. Choose crystalline character lifts with product $\chi_2$. Multiplying the first by a sufficiently large positive power of $\chi_2$ and the second by the inverse power preserves their reductions and product, since $\bchi_2=1$. Their weights are respectively positive and negative at every embedding. The crystalline extension constructed there lifts the given extension and has those labelled weights. Its unramified scalar determinant correction reduces to one and preserves the filtration. We obtain weights $a_\tau2$, with residue field $k$ of size $q$. We give the cyclotomic character $\chi_p$ Hodge--Tate weight one. Put $K_0=\GL_2(\OO_F)$, choose a uniformizer $\pi$, put $K_1=1+\pi M_2(\OO_F)$, and write \[ I^+=\left\{\begin{pmatrix}1+\pi a&b\\\pi c&1+\pi d\end{pmatrix}:a,b,c,d\in\OO_F\right\}, \qquad J=\OO_F^\times I^+. \] Fix a nontrivial additive character $\psi:k\to\overline\Q_p^\times$. Define an affine generic character of $J$ by \[ \theta\left(z\begin{pmatrix}1+\pi a&b\\\pi c&1+\pi d\end{pmatrix}\right) =\psi(\bar b+\bar c), \qquad z\in\OO_F^\times. \] The ambiguity in $z$ does not affect the two displayed residue coordinates. Let $Z$ and $U$ denote the scalar subgroup and the upper unipotent subgroup of $\GL_2(k)$. \begin{lemma}\label{a:struct:lem:finite} Every irreducible $\overline\F_p$-representation of $\GL_2(k)$ with trivial central character is a quotient of $\Ind_{ZU}^{\GL_2(k)}1$. \end{lemma} \begin{proof} Let $V$ be such a representation. A finite $p$-group acting on a nonzero vector space in characteristic $p$ has a nonzero fixed vector. Thus $V^U\ne0$. Since $Z$ acts trivially, $V^{ZU}\ne0$. Frobenius reciprocity gives a nonzero map $\Ind_{ZU}^{\GL_2(k)}1\to V$, which is surjective by irreducibility. \end{proof} \begin{lemma}\label{a:struct:lem:type} The representation $\sigma=\Ind_J^{K_0}\theta$ is irreducible of dimension $q^2-1$. It is the compact type of a simple-supercuspidal inertial type $\tau$ with determinant one. Its natural integral lattice has reduction equal to the inflation of $\Ind_{ZU}^{\GL_2(k)}1$. \end{lemma} \begin{proof} On $K_1=1+\pi M_2(\OO_F)$, the character is $1+\pi A\mapsto\psi(\bar A_{21})$. Under conjugation by $\GL_2(k)$, its stabilizer is $ZU$. Indeed the trace pairing on $M_2(k)$ is nondegenerate, including after composing with the nontrivial additive character of $k$, and the centralizer of a nonzero nilpotent matrix is $ZU$. The restriction of $\sigma$ to $K_1$ is therefore a sum of $q^2-1$ distinct characters permuted transitively by $K_0$. This proves irreducibility and the dimension assertion. The simple-supercuspidal construction identifies $\sigma$ as the compact type of the representations obtained by extending $\theta$ to $\langle g\rangle F^\times I^+$, where $g=\left(\begin{smallmatrix}0&1\\\pi&0\end{smallmatrix}\right)$; see \cite[Section~3.1]{c:AL}. To check the occurrence condition, let an irreducible smooth representation contain $\theta$ on $J$, and use its central character to extend $\theta$ to $F^\times I^+$. By Frobenius reciprocity the compact induction from this latter group surjects onto the representation. That compact induction is the direct sum of the two irreducible simple-supercuspidal representations in \cite[Section~3.1, p.~6]{c:AL}. Changing the extension on $g$, or its central character on a uniformizer, gives an unramified twist. Conversely each such twist contains $\sigma$. The occurrence characterization of inertial types therefore identifies $\sigma$ with $\sigma(\tau)$, up to duality; compare \cite[Theorem~1.1.3]{a:GK}. The Langlands parameter is the induced parameter of \cite[Section~3.4 and Theorem~3.18]{c:AL}. The central character on $\OO_F^\times$ is trivial, so $\det\tau=1$. All values of $\theta$ are $p$-power roots of unity and reduce to one. The coset basis of the induced lattice therefore reduces to the permutation representation on $K_0/J$. Since $K_1\subset J$ and $J/K_1=ZU$, this is the asserted inflation. Taking the dual does not change its Jordan--H\"older support. \end{proof} \begin{theorem}[A universal wild inertial type]\label{shared:wild-reset}\label{a:struct:prop:universal} Let $\bar r:G_F\to\GL_2(\overline\F_p)$ be continuous with determinant $\bar\chi_p$. Then $\bar r$ admits a potentially crystalline lift of Hodge--Tate weights $\{0,1\}$ at every embedding, determinant exactly $\chi_p$, and inertial type $\tau$ from Lemma~\ref{a:struct:lem:type}. For $F=\Q_p$, write this type as $\tau_{p,w}$. Its inertia remains absolutely irreducible after reduction in every coefficient characteristic greater than $p$, for every stable lattice. \end{theorem} \begin{proof} We use the potentially Barsotti--Tate Breuil--M\'ezard theorem of Gee--Kisin, \cite[Corollaries~3.5.3 and~3.5.6]{a:GK}. To match their conventions, apply their theorem to the dual residual representation, and dualize the resulting lift at the end. Their cyclotomic character has Hodge--Tate weight $-1$. The central-character restriction in Corollary~3.5.3 is therefore trivial in the present determinant situation. At least one of the nonnegative multiplicities $\mu_V$ is positive. To see this, use the existence of some potentially Barsotti--Tate lift of any rank-two residual representation. This existence is explicitly justified in the proof of \cite[Theorem~A.2]{a:GK}: the irreducible case is contained in \cite[Proposition~7.8.1]{g:Snowden}, and the reducible case follows from the crystalline-extension construction cited there. The nonzero potentially Barsotti--Tate ring for that lift has positive Hilbert--Samuel multiplicity; the Breuil--M\'ezard formula makes at least one $\mu_V$ positive. Every $V$ with positive multiplicity has trivial central character. By Lemmas~\ref{a:struct:lem:finite} and~\ref{a:struct:lem:type}, it occurs in the reduction of the simple-supercuspidal type. The Breuil--M\'ezard formula for Hodge type zero and type $\tau$ consequently has positive right side. Its potentially crystalline lifting ring is nonzero and $p$-torsion free, hence has nonempty generic fibre. Dualizing supplies a lift $r$ with the desired Hodge--Tate weights and type: a two-dimensional determinant-one type is self-dual. Finally $\det r/\chi_p$ is unramified and reduces to one. It is unramified because it is potentially crystalline of weight zero with trivial inertial type. As $p$ is odd, its inverse has an unramified square root reducing to one. Twisting $r$ by this square root fixes its determinant without changing its reduction, weights or inertial type. For $F=\Q_p$, the parameter is induced from a ramified quadratic extension. Its restriction to wild inertia has two distinct characters of order $p$, exchanged by a quadratic inertia element \cite[Section~3.4 and Theorem~3.18]{c:AL}. In coefficient characteristic greater than $p$ the two characters remain distinct, and the exchanging element prevents either eigenline from being stable. The finite inertial image has order divisible only by $2$ and $p$, so semisimplicity makes this conclusion independent of the stable lattice. \end{proof} \begin{remark} For $p=3$ this theorem applies to both unramified and ramified quadratic completions. It gives a local lift of weight two. It makes no assertion that an arbitrary-weight global source is modular whenever such a replacement is modular. A backward modularity implication must be verified separately for the actual global source. \end{remark} \begin{lemma}\label{t:triadic-switch} Let $\ell\ne3$ be an odd prime with $\ell\equiv-1\pmod3$. Let $\bar s$ be the reduction, with any stable lattice, of a determinant-$\chi_3$ representation of $G_{\Q_\ell}$ which is either a quadratic twist of special type or tame dihedral with inertia of order three or six. After a finite coefficient extension, $\bar s$ has lifts of both kinds with determinant exactly $\chi_3$. The scalar quadratic inertial character determines whether the cuspidal lift has inertia of order three or six, and the special lift may be chosen with nonzero monodromy. \end{lemma} \begin{proof} Remove the scalar quadratic character. Wild inertia is trivial; tame inertia is trivial or unipotent of order three. Frobenius has characteristic polynomial $X^2-1$ modulo three: in the special case its eigenvalues are $\ell\alpha,\alpha$, $\alpha=\pm1$, and in the cuspidal case its trace is zero and determinant is $\ell$. The two eigenvalues are distinct. A nonzero inertial extension has one Frobenius eigenline as its invariant line and is normalized by diagonal conjugation. There are two nonsplit orientations and the split case. For a cuspidal lift take \[ T=\begin{pmatrix}0&-1\\1&-1\end{pmatrix},\qquad S=\begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad \Phi=bS,\quad b^2=-\ell,\quad b\equiv1\pmod3. \] Hensel's lemma supplies $b$. We have $T^3=1$, $STS=T^{-1}$ and $\det\Phi=\ell$. The standard lattice gives nontrivial unipotent inertia with eigenvalue $-1$ on its invariant line; replacing $\Phi$ by $-\Phi$ gives the other orientation. For trivial residual inertia use $T=\operatorname{diag}(\zeta_3,\zeta_3^{-1})$ over $\Q_3(\zeta_3)$ and the same exchanging Frobenius. For a special lift use \[ \Phi=\operatorname{diag}(\ell,1),\qquad U(t)=\begin{pmatrix}1&t\\0&1\end{pmatrix},\qquad \Phi U(t)\Phi^{-1}=U(\ell t). \] Choose its three-primary tame parameter to be a unit multiple when residual inertia is nontrivial, and three times a unit multiple otherwise. Both choices have nonzero monodromy. An unramified quadratic sign selects either orientation. Restoring the removed quadratic character preserves the determinant and gives the required inertial order. \end{proof} \begin{lemma}\label{t:21:local:ordinary3} Let $p\in\{2,3\}$, let $F/\Q_p$ have absolute ramification index $p-1$, and let $r:G_F\to\GL_2(E)$ be crystalline of labelled weights $\{0,1\}$ and determinant $\chi_p$. If both residual constituents are unramified, then $r$ is ordinary. \end{lemma} \begin{proof} After a finite unramified extension the residual image is a $p$-group, so its underlying $\F_p$-representation has a fixed nonzero vector. Let $G$ be a Barsotti--Tate group with coefficient action realizing a lattice. A nonzero etale quotient has coefficient rank one, by the weights, and proves ordinarity. Otherwise $G$ is connected. Its Cartier dual is also connected, since $r^\vee(1)\simeq r$ and connectedness is invariant under isogeny. Coefficient-uniformizer torsion is therefore biconnected. The schematic closure of the fixed $\F_p$-line has order $p$ and unramified generic character. Its Oort--Tate parameter valuation lies between zero and $p-1$. The endpoints give etale and multiplicative models. For $p=3$ the only other valuation is one, which gives a ramified quadratic generic character. Thus the closure is etale or multiplicative, both impossible in a biconnected group: the latter would give an etale quotient of its Cartier dual. The canonical maximal etale quotient descends through the unramified extension, proving ordinarity over $F$. \end{proof} \section{Dyadic automorphy and pro-modularity inputs}\label{external:dyadic} \label{s:hyp:dyadic} The local weight-two lifts and their support-preserving global construction are proved in Section~\ref{s:sec:lifts}, using the fixed-determinant Breuil--M\'ezard theorem \cite[Corollary~8.0.2]{s:Tung2} and the prescribed-lift theorem \cite[Theorem~4.2.1]{h:EGPZ}. Potential modularity is established separately there. For backward modularity we use Kisin's and Tung's theorems with their local-source and residual-image hypotheses \cite[Theorem~3.3.5]{h:Kisin2}, \cite[Theorem~8.0.3]{s:Tung2}. The scalar argument uses the representation spaces, relative presentations and local geometry of \cite[Section~3]{s:OAI}, and the completed-forms and local block comparisons of \cite[Propositions~4.1, 4.2 and~4.4, Lemma~4.6]{s:OAI}. The localized propagation theorem \cite[Theorem~5.1]{s:OAI} applies over totally real solvable fields of even degree completely split at two, to a potentially pro-modular characteristic-two curve whose generic representation is not virtually solvable and whose allowed odd local images are finite. In these conventions the completed-form central character is $(\det r)\varepsilon_2$ and the local block parameter is $r\otimes\varepsilon_2$. The regular de Rham locally irreducible parameters are treated by the local correspondence \cite[Theorems~1.1 and~1.3]{s:CDP}. For irreducible solvable residue we use \cite[Proposition~10.3 and Theorem~10.6]{s:FM}. The application is over an abelian totally real field completely split at two, with every dyadic restriction of the actual source absolutely irreducible, so there are no locally reducible places. The auxiliary fields and scalar-source deduction needed here are proved in Sections~\ref{s:sec:field-preparation} and~\ref{s:sec:scalarModularity}. \section{Conductor reduction at a split odd prime}\label{struct:odd-conductor} We prove Theorem~\ref{shared:terminal-criterion}. The local constructions above preserve the actual residual extension; the lifting implications in Section~\ref{shared:systems} concern the old characteristic-zero source at each congruence. \begin{proof}[Proof of Theorem~\ref{shared:terminal-criterion}] Twist by the inverse square root of the odd-order determinant factor, leaving determinant $\chi\psi$ with $\psi$ of two-power order. Let $S$ contain $2,3,5,\ell$, all primes dividing the discriminant of $K$, and all rational conductor primes. Apply Proposition~\ref{shared:protect}; write $\mathfrak q$ for the auxiliary place, of split rational norm $q>N>\max S$. The order-$N$ type gives nonsolvable residual images and adequate cyclotomic restrictions at all later coefficient characteristics in $S$. In characteristic two, use Lemma~\ref{c:dyadiclocal} at the unique quadratic dyadic completion to choose a crystalline target of exact determinant $\chi_2$. For a reducible local residue take an ordinary regular target, retaining the old ordinary filtration when present; for an irreducible local residue take a weight-two target, which is potentially nonordinary. Lemma~\ref{c:awaytwo} supplies determinant- $\chi_2$ lifts at odd conductor places and keeps unramified places unramified. Retain the protective type at $\mathfrak q$, correcting its unramified scalar. Lemma~\ref{shared:global} gives the new system, good at two. For the old weight-two source, Lemma~\ref{c:monodromy} and descent of the unique potentially ordinary line give the two cases of Lemma~\ref{shared:dyadicMLT}; in the ordinary case the target has matching residual ordering. Thus modularity returns to the old system. The new weights may be higher than two. Reset in characteristic $\ell$, imposing the type of Theorem~\ref{shared:wild-reset} at both $\ell$-places and retaining $\mathfrak q$ and all cleared conditions. Lemma~\ref{shared:global} gives a weight-two system, with backward modularity by Lemma~\ref{shared:oddMLT}(i). For each remaining odd conductor characteristic $t\ne\ell$, use the crystalline potentially diagonalizable components of Lemma~\ref{c:crystalline}, retaining the two wild $\ell$-types, the protective place and all cleared conditions. If $t$ is nonsplit, first reset at $\ell$ whenever the source is not of weight two; then apply Lemma~\ref{shared:oddMLT}(ii) backwards, using Lemma~\ref{shared:weightzero}. If $t$ splits, use Lemma~\ref{shared:oddMLT}(i) without a reset. In either case Lemma~\ref{shared:global} supplies the replacement system. The protective order-$N$ element ensures adequacy and excludes the exceptional characteristic-five image. Each step clears $t$, may change the weights, and introduces no conductor prime outside $S\cup\{q\}$. Finally work in coefficient characteristic $q>\max\{5,\ell\}$. The retained wild inertia at an $\ell$-place remains irreducible modulo $q$, by Theorem~\ref{shared:wild-reset}. It lies in the cyclotomic subgroup, since $K(\zeta_q)/K$ is unramified there. Thus the residual cyclotomic restriction is absolutely irreducible and adequate. If the residual image is solvable, an odd solvable Artin witness and split-$q$ lifting already prove modularity of the old system. Otherwise the crystalline replacements above $q$, Lemma~\ref{shared:global} and split-$q$ lifting remove $\mathfrak q$. The resulting system has exact cyclotomic determinant, arbitrary regular weights and conductor supported at $\ell$. Its actual $\ell$-adic member is absolutely irreducible at both $\ell$-places: a Galois line would give a Weil--Deligne line, contradicting the retained irreducible inertia type. If its residue is irreducible, (i) and \cite[Theorem~6.1.1]{c:Zhang} give modularity. If it is reducible, apply (ii) and \cite[Theorem~5.1.1]{c:Zhang}. Here $K$ is totally real abelian, $\ell$ splits completely, and the constituent quotient is also nontrivial at both $\ell$-places: a residual constituent has tame inertia $\omega_\ell^a$, so the quotient has inertia $\omega_\ell^{1-2a}$, an odd power modulo the even integer $\ell-1$. Propagate modularity backwards and undo the finite twist. \end{proof} \section{Auxiliary fields with scalar Selmer vanishing}\label{s:sec:field-preparation} Let $K=\Q(\sqrt d)$ be real quadratic, with $d>1$ squarefree and $d\equiv1\pmod8$. The space $V^+_{w_0}(K)$ and condition \eqref{struct:SV} are defined in the introduction. No class-number hypothesis is imposed in this section. \begin{lemma}[Cyclic towers with bounded odd splitting]\label{s:lem:cyclic-tower} Let $K=\mathbf Q(\sqrt d)$ be real quadratic with two split, let $\Sigma$ be a finite set of odd rational primes, and let $n\geq2$. There is an odd prime $r\notin\Sigma$ and a totally real cyclic field $L/\mathbf Q$ of degree $2^n$ such that: \begin{enumerate} \item $L$ is ramified only at $r$, where it is totally ramified; \item $r$ is inert in $K$; \item $2$ splits completely in $L$; \item every $\ell\in\Sigma$ has at most two primes above it in $KL$. \end{enumerate} The prime $r$ can avoid any additional prescribed finite set. \end{lemma} \begin{proof} Let $P$ be the union of $\Sigma$ and the prime divisors of $d$, and put \[ B_n=\mathbf Q(\zeta_{2^{n+1}},2^{1/2^n}),\qquad C_n=B_n\bigl(t^{1/2^n}:t\in P\bigr). \] The field $B_n$ is Galois over $\mathbf Q$ and ramified only at two. At a prime of $B_n$ above an odd rational prime $t$, the normalized valuation of $t$ is one. These valuations show that the classes of the $t\in P$ generate a direct sum of copies of $\mathbf Z/2^n\mathbf Z$ in $B_n^\times/(B_n^\times)^{2^n}$: a product $\prod_{t\in P}t^{b_t}$ that is a $2^n$-th power has $2^n\mid b_t$ for each $t$. Kummer theory consequently gives \[ \operatorname{Gal}(C_n/B_n)\simeq (\mathbf Z/2^n\mathbf Z)^P, \] where a vector $(a_t)$ acts by $t^{1/2^n}\mapsto\zeta_{2^n}^{a_t}t^{1/2^n}$. Let $j$ be the number of prime factors of $d$. If $j$ is odd, choose $a_t=1$ for every $t\in P$. If $j$ is even, choose $a_t=2$ for exactly one prime dividing $d$, and $a_t=1$ for every other $t\in P$. Then $\sum_{t\mid d}a_t$ is odd and each $a_t$ has two-adic valuation at most one. The field $C_n$ is Galois over $\mathbf Q$. Apply Chebotarev to the conjugacy class of the prescribed automorphism $h\in\operatorname{Gal}(C_n/B_n)$. We obtain a prime $r$ split in $B_n$ for which the Kummer exponents have the stated $2$-adic valuations. Conjugating $h$ can multiply all exponents by an odd integer, which does not change those valuations or their parity. Since $h(\sqrt d)=-\sqrt d$, the prime $r$ is inert in $K$. Since $r$ splits in $B_n$, it satisfies \[ r\equiv1\pmod{2^{n+1}},\qquad 2\in(\mathbf F_r^\times)^{2^n}. \] Take $L$ to be the unique subfield of $\mathbf Q(\zeta_r)$ of degree $2^n$. The first congruence makes $L$ totally real; the second makes two split completely. For $t\in P$, its Frobenius in $\operatorname{Gal}(L/\mathbf Q)$ has order $2^n$ when $a_t$ is odd, and order $2^{n-1}$ when $a_t=2$. This follows from the $2^n$-th power residue symbol at $r$. If $\ell\nmid d$, we chose $a_\ell=1$, so $L$ has one prime above $\ell$ and the quadratic extension $KL/L$ yields at most two. If $\ell\mid d$, it ramifies in $K$ and is unramified in $L$. The unramified local extension supplied by $L$ and the ramified quadratic extension supplied by $K$ are disjoint. Thus the number of primes over $\ell$ in $KL$ equals their number in $L$, which is at most two. This proves all assertions. \end{proof} \begin{lemma}[Propagation of one-dyadic scalar Selmer vanishing]\label{s:lem:selmer-propagation} Let $K$ be a totally real field with two places $w_1,w_2$ above two, and suppose that the only totally real quadratic extension of $K$ unramified outside $w_1$ and split at $w_2$ is the trivial extension. Let $F/K$ be cyclic of degree $2^n$, split at $w_1,w_2$, unramified outside a single finite prime $\mathfrak r\nmid2$, and totally ramified there. Then the only totally real quadratic extension of $F$ unramified outside the full fibre above $w_1$ and split at every place above $w_2$ is trivial. \end{lemma} \begin{proof} Let $V$ be the corresponding subspace of $H^1(G_F,\mathbf F_2)$. It is stable under the $2$-group $\Gamma=\operatorname{Gal}(F/K)$. If $V\ne0$, the elementary invariant-vector theorem for representations of a finite $2$-group in characteristic two gives a nonzero $c\in V^\Gamma$. Let $H/F$ be the quadratic extension cut out by $c$. Then $H/K$ is Galois, and its group $G$ is a central extension of the cyclic group $\Gamma$ by $C_2$. Every such central extension is abelian: a lift of a generator of $\Gamma$, together with the central kernel, generates $G$. Because $H/F$ is unramified at $\mathfrak r$ and $F/K$ is totally ramified there, the inertia subgroup $I\subset G$ maps isomorphically onto $\Gamma$ and intersects $\operatorname{Gal}(H/F)$ trivially. Consequently $G=I\times\operatorname{Gal}(H/F)$. The fixed field $E=H^I$ is quadratic over $K$, totally real, and unramified at $\mathfrak r$. At every finite place away from $w_1$ and $\mathfrak r$, both steps $H/F$ and $F/K$ are unramified, so $E/K$ is unramified there. At $w_2$, both steps split completely, so $E/K$ splits. This contradicts the assumed vanishing over $K$. \end{proof} \begin{theorem}[Uniform scalar field preparation]\label{s:thm:fieldprep} Let $K$ be real quadratic with two split and satisfying \eqref{struct:SV}. Let $\Sigma$ be a fixed finite set of odd rational primes, and fix integers $n>a\ge1$. There are totally real abelian two-extensions $K\subset A\subset F$, split completely at two, and a dyadic place $v_0$ of $A$, with the following properties. Put \[ D=[F:\mathbf Q],\qquad M=\{v\mid v_0\text{ in }F\},\qquad m=|M|, \] and let $S_A,S_F$ consist of the real places, the dyadic places, and the places above $\Sigma$. \begin{enumerate} \item $D=2^{n+1}$, $[F:A]=2^{n-a}$, and $m/D=2^{-a-1}$. The numbers $u_A,u_F$ of allowed odd places are at most $2|\Sigma|$. \item For every finite field $k$ of characteristic two, \[ \{c\in H^1(G_{F,S_F},k):\operatorname{loc}_v c=0 \text{ for all }v\in S_F\setminus M\}=0. \] Here and below $k$ has trivial Galois action. \item Restriction $H^1(G_{A,S_A},k)\longrightarrow H^1(G_{F,S_F},k)$ is injective. \item Let $H^1_{\mathrm{str}}(F,k)$ have zero local condition at $M$, full local condition at $S_F\setminus M$, including the real places, and unramified condition outside $S_F$. Then \[ \dim_k H^1_{\mathrm{str}}(F,k)\le 2D-3m+u_F. \] \end{enumerate} The degrees $[A:\mathbf Q]$ and $[F:A]$ can both be made arbitrarily large, while the allowed odd-place bound remains fixed. \end{theorem} \begin{proof} Apply Lemma~\ref{s:lem:cyclic-tower}, take the fields $A=KL_a$ and $F=KL$ in Lemma~\ref{s:lem:selmer-propagation}, using \eqref{struct:SV} as its base vanishing hypothesis. The first three assertions follow from the two preceding lemmas and complete splitting at two. For injectivity, a class killed under restriction factors through the cyclic group $\Gal(F/A)$. Its only nonzero quadratic character is ramified at the fresh conductor prime $r$, which is outside $S_A$. Such a class is excluded from $H^1(G_{A,S_A},k)$, proving injectivity. For the last assertion use the Selmer dimension formula from local and global Tate duality, with the real local conditions included \cite[Theorem~4.2]{s:Paterson}. The scalar module $k$ is self-dual with Tate twist, since the mod-two cyclotomic character is trivial. The dual local conditions are full at $M$, zero at $S_F\setminus M$, and unramified elsewhere. Their global Selmer group vanishes by the second assertion. The global invariant terms therefore cancel. Each unmarked dyadic place contributes $\dim H^1(G_{\mathbf Q_2},k)-1=2$, and each marked place contributes $-1$. A real place with full local condition contributes $1-1=0$; an allowed odd finite place contributes at most one. Unramified local conditions contribute zero outside $S_F$. Consequently the strict Selmer dimension is at most \[ 2(D-m)-m+u_F=2D-3m+u_F, \] as claimed. Taking $a$ and $n-a$ large proves the final assertion. \end{proof} When $\Sigma=\varnothing$, taking $n=a+2$ in Theorem~\ref{s:thm:fieldprep} gives $[F:A]=4$ with $[A:\Q]$ arbitrarily large. There are no odd local Euler terms, so the same Selmer dimension formula gives $\dim_k H^1_{\mathrm{str}}(F,k)=2[F:\Q]-12$. \begin{remark}\label{struct:selmer-limitation} Condition~\eqref{struct:SV} is a genuine restriction. For $K=\Q(\sqrt{113})$, the totally positive element $\alpha=(11+\sqrt{113})/2$ has norm two and valuation one at one dyadic prime. At the other its unit value $u$ satisfies $u+2/u=11$, so $u\equiv1\pmod8$ and is a square in $\Q_2$. Thus $[\alpha]$ is a nonzero element of the corresponding $V^+_{w_0}(K)$. This is a limitation of the criterion, not an obstruction to modularity over that field. \end{remark} \section{The scalar dyadic endpoint}\label{s:sec:scalarModularity} We prove Theorem~\ref{s:thm:scalarFM} using the field preparation of Section~\ref{s:sec:field-preparation} and the dyadic inputs listed in Section~\ref{external:dyadic}. The argument treats arbitrary finite ramification and regular weights. \begin{proof}[Proof of Theorem~\ref{s:thm:scalarFM}] First suppose $r$ is virtually solvable, in the sense that the identity component of its algebraic monodromy group is solvable. An irreducible regular two-dimensional geometric representation with this property is induced from a character of a quadratic extension of $K$. Indeed, a nontrivial unipotent radical would have a unique invariant line preserved by the whole monodromy group, while scalar identity component would give finite projective image and, after finite extension, equal Hodge--Tate weights. The remaining torus case has two lines permuted by a quadratic quotient. The inducing quadratic field must be CM. The inducing character is algebraic. If its field had a real place, the absolute-value relation on units and Dirichlet's unit theorem would force all real algebraic exponents to agree. At the two real embeddings above any split real place of $K$, the induced weights would therefore coincide. This contradicts regularity and also excludes mixed signature. Its inducing character is de Rham and finitely ramified, hence is the $2$-adic avatar of an algebraic Hecke character. Each dyadic place is nonsplit in the inducing extension: splitting would make the local induction a sum of characters, contrary to local absolute irreducibility. Quadratic automorphic induction gives a cuspidal automorphic representation with Galois representation $r$; its regular algebraic infinity type follows from the given regular Hodge--Tate weights. We henceforth assume that $r$ is not virtually solvable. Its restriction to every finite extension is then absolutely irreducible. \smallskip \noindent\emph{The seed and the auxiliary fields.} A cuspidal modular seed with scalar semisimple reduction is available over $\mathbb Q$: take the compatible system of Ramanujan's form $\Delta$, twisted by $\varepsilon_2^{-5}$. Its determinant is $\varepsilon_2$, it is non-CM, and its residual semisimplification at $2$ is $\mathbf1\oplus\mathbf1$. Indeed Jacobi's identity gives $\Delta\equiv\sum_{j\ge0}q^{(2j+1)^2}\pmod2$, so the trace at every odd prime is zero and the residual determinant is one. Base change supplies a cuspidal seed over $K$ and every abelian auxiliary field used below. Its determinant agrees exactly with the target; hence its quaternionic transfer has the required central character. It has tame level one, and the small uniform subgroups for completed forms are chosen at two. Choose a finite set $\Sigma$ of odd rational primes supporting the ramification of $r$ and the seed, and enlarge it when necessary to include the prime factors of $d$. Theorem~\ref{s:thm:fieldprep} supplies \[ K\subset A\subset F, \qquad D_A=[A:\mathbb Q],\quad D=[F:\mathbb Q], \qquad m=[F:A], \] with $A/\mathbb Q$ and $F/\mathbb Q$ abelian, $2$ completely split, and uniformly bounded numbers $u_A,u_F$ of allowed odd places. The marked set $M$ consists of the $m$ places of $F$ over one selected dyadic place of $A$. Both $D_A$ and $m$ may be made arbitrarily large, with $m/D=1/D_A$. The same theorem gives injectivity of scalar residual extension classes under $A\subset F$, and gives \[ \{c\in H^1(G_{F,S_F},k): \operatorname{loc}_v c=0\text{ for all }v\in S_F\setminus M\}=0. \tag{S1}\label{s:eq:scalarDualVanish} \] Here $S_F$ consists of the real and dyadic places and the allowed odd places. The coefficient field $k$ can be enlarged without changing the vanishing. The real local conditions in \eqref{s:eq:scalarDualVanish} are zero conditions, so the assertion uses the totally positive squareclass condition~\eqref{struct:SV}, not only an ordinary class-number condition. For the strict Selmer group, Theorem~\ref{s:thm:fieldprep}(iv) gives \[ \dim_k H^1_{\mathrm{str}}(F,k)\le 2D-3m+u_F. \tag{S2}\label{s:eq:scalarStrictBound} \] Let $c$ be a fixed generator bound for the augmentation ideal of the characteristic-two universal local pseudodeformation ring of $G_{\Q_2}$ with residual pseudorepresentation $\mathbf1\oplus\mathbf1$ and fixed determinant one. This bound is independent of $A$: every selected dyadic completion is $\Q_2$, and extending the residue field preserves a fixed bound. Choose \[ D_A>\max\{4,u_A+c+3\},\qquad m>u_F+4. \tag{S3}\label{s:eq:scalarDegreeChoices} \] The uniform bound on $u_A,u_F$ and independent growth of $D_A,m$ make these choices possible. Since $D=mD_A$, they imply $D>u_F+4$ and \[ (2D-2m-4-u_F)-D=m(D_A-2)-(u_F+4)>0. \] They also give the dimension bound needed for entry from $A$ below. \smallskip \noindent\emph{Connected transverse charts.} Over the characteristic-two fixed-determinant pseudodeformation base for $G_F$, take the finite-type framed representation scheme and record an invariant projective line at each mark. In its residual fibre retain the open locus where no marked line is globally invariant. At such a point the residual representation at a mark is scalar: a nonzero unipotent extension has only one invariant line, whereas the global invariant line and the chosen transverse line are different. An upper extension with diagonal characters frozen must therefore be zero locally at every mark. At a transverse point, the product of the real and marked dyadic local line problems has dimension $2D+5m$, by \cite[Lemma~3.12]{s:OAI}. The relative global presentation of \cite[Proposition~3.6]{s:OAI} has number of generators minus relations at least $-6m$. Every component of the completed chart consequently has dimension at least $2D-m$, and its components can be joined with intersections of dimension at least $2D-m-1$, by complete-local connectedness \cite[Lemma~52.14.1, Tag~0ECR]{s:Stacks}. The globally reducible locus has dimension at most \[ 2D-2m+u_F+3. \] Indeed, fixed-product diagonal characters contribute at most one parameter by Leopoldt for the abelian field $F$; moving the global invariant line and an upper coboundary contribute at most two; the strict extension class contributes at most $2D-3m+u_F$ by \eqref{s:eq:scalarStrictBound}; and moving the $m$ marked lines contributes at most $m$. This is a relative generator bound after freezing the diagonal characters. Adding the global invariant line by proper incidence and then forgetting it bounds the entire reducible image. By \eqref{s:eq:scalarDegreeChoices}, this bound is strictly less than $2D-m-1$. Therefore a maximal-dimensional intersection in a component chain is generically globally irreducible. Forgetting the global frame and marked lines loses at most $3+m$ dimensions. The remaining trace image has dimension at least $2D-2m-4$. Prescribing one Frobenius trace at each allowed odd place costs at most $u_F$ equations. The consequence of \eqref{s:eq:scalarDegreeChoices} displayed above leaves dimension greater than $D$. The dihedral-locus bound in \cite[Lemma~3.8]{s:OAI} and its curve-avoidance argument supply a non-virtually-solvable characteristic-two curve. The finite local-image conclusion is exactly \cite[Lemma~3.9]{s:OAI}, applied after its fixed preliminary local extensions; the determinant on the characteristic-two curve is one. The prescribed residual Frobenius traces thus make all allowed odd local images finite. Theorem~5.1 of \cite{s:OAI} then propagates potential pro-modularity across this intersection. Thus all components through a transverse residual point have the same potential pro-modularity status. The transverse residual fibre is geometrically irreducible. Fixing the global line, its extensions form the nonzero vectors of the strict scalar cohomology space. This space contains the strict space with no odd ramification, of dimension $2D-3m>0$ by the same duality calculation. The complementary marked lines form $\mathbb A^m$. Conjugation gives a surjection from the product of these two spaces with $\GL_2$, an irreducible parameter space. Since the representation scheme is noetherian, one irreducible component $Y_0$ contains this whole transverse residual fibre. Local chart propagation at the seed and target entry points, through $Y_0$, therefore connects their components. Completion is faithfully flat, so the completed-chart conclusion applies to components of the representation scheme. \smallskip \noindent\emph{Entering the transverse charts from the seed and target.} We include this step because connecting deformation components for one fixed residual extension would not suffice. Work first over $A$. The seed gives a pro-modular component in the fixed-determinant pseudo-base of dimension at least $1+2D_A$, by \cite[Proposition~4.4]{s:OAI}. An unrestricted fixed-determinant framed component through $r|_{G_A}$ has trace image of dimension at least $2D_A-2$, by the relative presentation over the real local rings and the loss of three frame parameters. Its generic representation is absolutely irreducible because it contains the given target. In each component pass to the special fibre, prescribe the residual Frobenius traces at the allowed odd places, and prescribe the constant scalar local pseudorepresentation at the selected dyadic place of $A$. The last condition costs at most the $c$ local generators fixed above. The total cost is at most $1+u_A+c$ equations. Even for the smaller target component, the remaining dimension is at least \[ 2D_A-2-(1+u_A+c)=2D_A-u_A-c-3>D_A \] by \eqref{s:eq:scalarDegreeChoices}. The bad-locus bound and curve avoidance therefore supply a non-virtually-solvable characteristic-two curve in each component. Lemma~3.9 of \cite{s:OAI} makes its odd local images finite, as above. At the selected dyadic place the curve representation has an invariant line after a finite coefficient extension. Choose a nonzero vector on that line and take its full $G_A$-orbit lattice over the curve's valuation ring. Compactness bounds it, and global irreducibility makes it full rank. The vector is primitive: otherwise all its translates lie in the maximal ideal times the orbit lattice, contradicting Nakayama's lemma. Its residual orbit spans the whole rank-two residual space. Consequently its residual line is not globally invariant. The residual representation is a nonzero additive extension. Injectivity of scalar extension classes under $A\subset F$ ensures that it remains nonzero over $F$. Transporting the local line to all marks gives transverse lines, because $F/A$ is Galois. Thus both curves enter the transverse residual fibre over $F$. The seed curve is potentially pro-modular by solvable base change. Propagation through the transverse charts, through the common component $Y_0$, makes the target curve potentially pro-modular. Applying \cite[Theorem~5.1]{s:OAI} back over $A$ to the original target curve makes its entire containing component potentially pro-modular. In particular, $r$ becomes pro-modular over one finite solvable totally real extension of $K$, completely split at $2$. Only finitely many component chains and propagation steps occur; their field extensions and tame levels can be replaced by one common Galois compositum and one common deeper tame level. This proves the scalar version of the pro-modularity input directly. \smallskip \noindent\emph{Classicality and descent.} At every dyadic place of the resulting field the local representation is still the original absolutely irreducible regular de Rham representation over $\mathbb Q_2$, since the auxiliary extensions split completely above $2$. Proposition~4.2 of \cite{s:OAI}, together with the all-prime local correspondence \cite[Theorems~1.1 and~1.3]{s:CDP}, gives a nonzero locally algebraic vector in its Hecke eigenmodule, including when the residual local block is scalar. Proposition~4.1 identifies this vector with a regular algebraic definite-quaternionic eigensystem. Global irreducibility excludes reduced-norm characters, so Jacquet--Langlands produces a Hilbert cusp form. Local-global compatibility and Chebotarev identify its Galois representation with the restriction of $r$. Finally descend through a tower of prime-degree cyclic extensions to $K$. At each step, Galois invariance of $r$ and strong multiplicity one make the upstairs cuspidal representation invariant under the cyclic Galois group. Cyclic automorphic descent \cite[Chapter~III, Theorems~4.2(d) and~5.1]{r:s:AC} yields a cuspidal representation downstairs. Its Galois representation differs from the given one by a character of the cyclic quotient, because their restrictions are absolutely irreducible and isomorphic upstairs; twisting corrects that character. The archimedean types remain regular algebraic because every real place splits in a totally real extension. Iterating gives modularity of $r$ over $K$. This is the locally irreducible part of the classicality and descent argument in \cite[Section~8.8]{s:OAI}; none of its ordinary dense-point arguments over $\mathbb Q$ is used here. \end{proof} \section{Dyadic lifts and compatible systems}\label{s:sec:lifts} We construct two local types which suffice at every dyadic place and then globalize them without enlarging the prescribed ramification support. Steinberg lifts enter only in the local multiplicity calculation; the global lifts will all be potentially Barsotti--Tate and potentially nonordinary. Let $k$ be finite of characteristic two. For a continuous $\bar s:G_{\Q_2}\to\GL_2(k)$ put $\psi_{\bar s}=\eps_2[\det\bar s]$, with brackets denoting Teichm\"uller lift. Choose $E/\Q_2$ with residue field $k$, enlarging both when necessary, and write $\OO$ for its ring of integers and $\varpi$ for a uniformizer. Let $\tau_0=\widetilde\omega_2\oplus\widetilde\omega_2^{2}$, where $\omega_2$ has niveau two, and let $\tau_*$ be the inertial type of the elliptic curve \[ y^2=x^3-2x, \] with Cremona label \href{https://www.lmfdb.org/EllipticCurve/Q/256/b/1}{\texttt{256b1}} (LMFDB label \texttt{256.b1})~\cite{h:LMFDB}. \begin{proposition}\label{s:prop:types} Every $\bar s$ has a potentially Barsotti--Tate lift of determinant $\psi_{\bar s}$ and labelled weights $\{0,1\}$, with type $\tau_0$ if it admits a crystalline lift of those weights, and $\tau_*$ otherwise. Both types force actual local irreducibility and potential nonordinarity. The inertia image of $\tau_*$ is $Q_8$. \end{proposition} \begin{proof} Every character $G_{\Q_2}\to k^\times$ is unramified: by local class field theory its inertial image is a quotient of the pro-two group $\Z_2^\times$, whereas $k^\times$ has odd order. If a crystalline weight-two lift exists, its determinant differs from $\psi_{\bar s}$ by an unramified character $\delta$ reducing to one. After enlarging coefficients, an unramified character $\nu$ with $\nu^2=\delta^{-1}$ also reduces to one. Twisting by $\nu$ therefore fixes the determinant without changing the residue, weights or inertial type. We first identify the residual representations for which no such crystalline lift exists. An irreducible $\bar s$ has trivial wild inertia, since the fixed space of this normal two-subgroup is nonzero and $G_{\Q_2}$-stable. After an unramified scalar twist its determinant is one; its two tame inertia characters are exchanged by squaring and have order three. It is the unique such representation, realized by $A[2]$ for the good-reduction curve $A:y^2+y=x^3$ used in the proof of Lemma~\ref{c:dyadiclocal}. Thus it has a crystalline weight-two lift. A split residue $\bar\alpha\oplus\bar\beta$ lifts to $\eps_2[\bar\alpha]\oplus[\bar\beta]$. For a nonsplit reducible residue, untwist its quotient and write its character ratio as $\bar\eta$. If $\bar\eta\ne1$, local Euler characteristic and the Bloch--Kato formula \cite[Corollary~3.8.4]{h:BK90} give \[ \dim_E H^1(G_{\Q_2},E(\eps_2[\bar\eta])) =\dim_E H_f^1(G_{\Q_2},E(\eps_2[\bar\eta]))=1. \] Moreover $H^2(G_{\Q_2},\OO(\eps_2[\bar\eta]))=0$ by local Tate duality, since $[\bar\eta]^{-1}-1$ takes a unit value. Reduction of $H^1$ is consequently surjective, and the actual residual extension lifts crystallinely. Hence failure of crystalline lifting is possible only for \[ \bar s=\bar\alpha\otimes \begin{pmatrix}1&c\\0&1\end{pmatrix},\qquad 0\ne c\in H^1(G_{\Q_2},k), \] with $\bar\alpha$ unramified. For this last representation, Kummer theory identifies $H^1(G_{\Q_2},\OO(\eps_2))$ with $\widehat{\Q_2^\times}\otimes_{\Z_2}\OO$, where the hat denotes pro-two completion. Its reduction onto $H^1(G_{\Q_2},k)$ is surjective. Choose a lift $\kappa$ of $c$ with nonzero valuation; adding a suitable $\varpi\OO$-multiple of the Kummer class of two achieves this without changing its reduction. The extension of $1$ by $\eps_2$ defined by $\kappa$ is semistable and noncrystalline: Kummer classes are semistable, and the crystalline subspace is the unit subspace. Twisting by $[\bar\alpha]$ gives determinant $\eps_2[\bar\alpha]^2=\psi_{\bar s}$ and preserves the actual residual extension. We now apply the full fixed-determinant Breuil--M\'ezard formula \cite[Conjectures~5.1.2--5.1.3 and Corollary~8.0.2]{s:Tung2}. The two Serre weights are $W_0=1$ and $W_1=\Sym^1$. Write $\mu_0(\bar s),\mu_1(\bar s)$ for their nonnegative multiplicities. For a compact type $\sigma$ with semisimplified reduction containing $W_i$ with multiplicity $a_i$, the Hilbert--Samuel multiplicity of the corresponding fixed-determinant weight-two ring modulo $\varpi$ is \[ a_0\mu_0(\bar s)+a_1\mu_1(\bar s). \] These deformation rings are taken with their reduced, $\OO$-torsion-free structure; positivity is equivalent to a nonempty characteristic-zero fibre. The crystalline trivial type has reduction $W_0$, so $\mu_0>0$ exactly when the crystalline lift exists, with determinant fixed as above. The compact type of $\tau_0$ is the inflation of the sign character of $\GL_2(\F_2)\simeq S_3$, whose reduction is also $W_0$. This proves the $\tau_0$ assertion. Otherwise $\mu_0=0$, whereas the semistable lift just constructed gives $\mu_1>0$: the compact type for unrestricted monodromy and trivial inertia is the finite Steinberg representation, whose reduction is $W_1$. The determinant compatibility in both comparisons follows from $\det\tau_0=1$ and the unramifiedness of $[\det\bar s]$. For $\tau_*$ we use one test representation. The two-torsion of the displayed curve has Kummer class two, which has odd valuation and is not finite flat \cite[Definition~A.3 and Lemma~A.4]{s:CEGS}; in particular it has no crystalline weight-two lift. Its three-division polynomial is $3X^4-12X^2-4$. Choose $u^2=-3$, $i^2=-1$, $s=iu$. Its roots are $\pm\alpha,\pm\beta$, with \[ \alpha^2=2+4/s,\qquad \beta^2=2-4/s, \qquad \alpha\beta=2u/3. \] They recover $s,u$, so their splitting field is $L=\Q_2(u,i,\alpha)$. The field $U=\Q_2(u)$ is unramified quadratic, and $T=U(i)$ is ramified quadratic. For $\pi=1-i$, one has $\alpha^2/\pi^2=i+2/u$ and $v_\pi(i+2/u-1)=1$. This unit is nonsquare even after unramified extension of $T$: a putative square root has residue one, while its square is one modulo $\pi^2$. Hence $L/T$ is ramified quadratic, with $e(L/\Q_2)=4$, $f(L/\Q_2)=2$. Projective inertia on the four roots is $V_4\subset A_4$. Its inverse image in $\SL_2(\F_3)$ is $Q_8$, with no proper subgroup surjecting onto $V_4$. The curve has potentially good reduction since $j=1728$; its finite inertia on $T_3$ reduces injectively modulo three. Compatibility identifies its faithful irreducible $Q_8$ inertia with $\tau_*$. The Tate module of this curve gives a nonempty type-$\tau_*$ ring, while its residue has $\mu_0=0$. The multiplicity formula therefore forces $a_1(\tau_*)>0$. For every $\bar s$ without a crystalline weight-two lift we already have $\mu_1(\bar s)>0$, so the same formula gives a nonempty type-$\tau_*$ fibre. An unramified scalar twist removes and restores $\bar\alpha$ without changing the compact type, giving the required determinant in general. The two characters of $\tau_0$ are exchanged by Frobenius, and $\tau_*$ is irreducible on inertia. Their Weil parameters are therefore irreducible, which forces Galois irreducibility. An ordinary line after finite Galois base change would be the unique line of labelled weight one and would descend, a contradiction. Thus neither type is potentially ordinary. \end{proof} \begin{lemma}\label{s:lem:potential2} Let $F$ be totally real and let $r:G_F\to\GL_2(\overline\Q_2)$ be finitely ramified, with nonsolvable residue, $\det r=\eps_2\eta$ for a finite totally even character $\eta$, and potentially Barsotti--Tate and potentially nonordinary at every dyadic place. Then $r$ is potentially modular. \end{lemma} \begin{proof} A finite character twist makes the residual determinant one. Apply \cite[Proposition~6.7]{g:Thorne2}, avoiding its kernel field, to make the residue modular without shrinking its image. A further solvable totally real extension makes every dyadic source Barsotti--Tate. Nonsolvability and residual modularity persist. Kisin's theorem \cite[Theorem~3.3.5]{h:Kisin2} applies: its condition that a completion equal $\Q_2$ concerns only potentially ordinary places, and there are none. Restore the finite twist. \end{proof} \Needspace{10\baselineskip} \begin{corollary}[Prescribed dyadic lifts in compatible systems] \label{s:cor:prescribed-dyadic-system} Let $K$ be real quadratic with two split, and let $\bar r:G_K\to\SL_2(k)$ be continuous with nonsolvable image. Fix a finite set $S$ containing the dyadic places and all residual ramification. At each odd place in $S$, choose a nonempty union of components of the generic fibre of the local lifting ring with determinant $\eps_2$. Then $\bar r$ lifts to the two-adic member of a potentially modular weight-two strongly compatible system of cyclotomic determinant, unramified outside $S$, on the chosen odd local components, and with dyadic types prescribed by Proposition~\ref{s:prop:types}. \end{corollary} \begin{proof} Lemma~\ref{c:awaytwo} makes the odd local lifting rings nonempty and permits an unramified choice wherever the residue is unramified. At the dyadic places use Proposition~\ref{s:prop:types}. Choose a finite extension of each completion killing its type. In the corresponding potentially semistable generic fibre of weights $\{0,1\}$, the type is locally constant, so the prescribed type selects a nonempty union of components. The associated Weil parameters are irreducible, so monodromy vanishes, and Proposition~\ref{s:prop:types} gives potential nonordinarity. The global determinant $\eps_2$ lifts the residual determinant and is totally odd. Since $K(\zeta_2)=K$, the nonsolvable image gives the required absolute irreducibility. The prescribed-lift theorem \cite[Theorem~4.2.1]{h:EGPZ} therefore supplies a characteristic-zero point unramified outside $S$ on these components. Lemma~\ref{s:lem:potential2} makes this actual lift potentially modular. The compatible-system construction in the proof of Lemma~\ref{shared:global} then gives the assertion, including the local conditions and ramification support. \end{proof} \section{A Serre criterion when two splits}\label{s:sec:chain} \begin{theorem}[The split-dyadic criterion]\label{struct:split-criterion} Let $K$ be real quadratic with two split. Use the lifting and propagation inputs of Sections~\ref{shared:systems} and~\ref{external:dyadic}, and suppose: \begin{enumerate}[label=(\roman*)] \item $V^+_{w_0}(K)=0$; \item the ordinary ideal class group $\Cl(K)$ has two-power order; \item every continuous absolutely irreducible representation $G_K\to\GL_2(\overline\F_2)$ of determinant one, unramified outside two, is modular. \end{enumerate} Then every continuous absolutely irreducible totally odd representation $G_K\to\GL_2(\overline\F_p)$, for every odd prime $p$ and arbitrary conductor, is modular. \end{theorem} Condition (iii) is a residual base case, separate from the scalar source theorem. The proof below has no analytic assumption beyond whatever is needed to establish that condition. Strong compatibility includes the coefficient prime, so a crystalline replacement clears that conductor from every companion. \begin{proof} \subsection{Initial lifts and the protective place} Use the global construction and compatible systems of Lemma~\ref{shared:global}, the modular witnesses of Lemma~\ref{shared:weightzero}, and the backward lifting statements in Lemma~\ref{shared:oddMLT}. For arbitrary regular dyadic sources at the split completions $\Q_2$, we use \cite[Theorem~8.0.3]{s:Tung2}. The initial characteristic-five exception is absent for this $K$. Indeed $K\ne\Q(\sqrt5)$, so $[K(\zeta_5):K]=4$, whereas $\PGL_2(\F_5)$ has abelianization of order two. Its projective kernel field therefore cannot contain $K(\zeta_5)$; this verifies the nonexceptional hypothesis of \cite[Definition~4.1.1 and Proposition~4.1.2]{h:EGPZ}. The subsequent protective images exclude all small exceptional images as well. The protective place in Proposition~\ref{shared:protect} can be chosen with the additional property that the order of two modulo its rational norm is odd. We give the simultaneous field conditions, since this property is needed when its own inertia ceases to protect the last congruence. Write $K=\Q(\sqrt d)$ with $d>1$ squarefree. Start with the initial weight-two compatible system. A reversible finite twist removes the odd-order part of its finite determinant, leaving $\det r_t=\eps_t\psi_2$, with $\psi_2$ of two-power order and with no new rational conductor prime. Choose a fresh odd reset prime $s>5$, above the needed prime bounds and split in $K$, by Chebotarev. Include it, two, three, five, every conductor prime and every prime dividing $d$ in a finite set $S$ before choosing the protective place. Choose $N$ and the projective residual field as in the proof of Proposition~\ref{shared:protect}, with $N$ large enough for the fixed weight bounds. Thus $N$ splits in $K$ and the coefficient field, the projective image is $\PGL_2(\F_N)$, and $\psi_2$ is square-valued modulo $N$. Let $K_S/K$ be the maximal elementary quadratic extension unramified outside $S$ and infinity. Choose a degree-one place $\lambda$ of rational norm $\ell$ with the standard good-dihedral Frobenius prescription and \begin{equation}\label{s:eq:protective} \ell\equiv-1\pmod N,\qquad \lambda\text{ split in }K_S,\qquad \ell\equiv9\pmod{16},\qquad 2\in(\F_\ell^\times)^8. \end{equation} The last two conditions make $\operatorname{ord}_\ell(2)$ odd. We verify compatibility of the simultaneous conditions. The Galois group of $B=\Q(\zeta_8,2^{1/8})$ has order sixteen. Writing its elements as $(a,b)$ for the actions on $2^{1/8}$ and $\zeta_8$, multiplication is $(a,b)(c,e)=(a+bc,be)$. The commutator of $(1,3),(0,7)$ is $(2,1)$, generating $\Gal(B/\Q(\zeta_8))$. Thus its maximal abelian subfield is $\Q(\zeta_8)$. Since $B$ is ramified only at two and $K$ is real quadratic with two split, $K\cap B=\Q$. The same statement holds with $\Q(\zeta_{16})$ in place of $B$. The compositum with the elementary extension $K_S/K$ has abelianization of exponent two: the same commutators survive with trivial $K_S$ coordinate. Hence $K(\zeta_{16})\not\subset BK_S$ and its intersection with $BK_S$ is $K(\zeta_8)$. Splitting in $BK_S$ is compatible with Frobenius nine on $\zeta_{16}$. This two-extension is unramified at $N$; the only nontrivial solvable quotient of the projective residual field is its quadratic cyclotomic quotient, ramified at $N$. Its intersection with the compositum of the preceding two-extension and $K(\zeta_N)$ is exactly that quotient. Projective complex conjugation and inversion on $\zeta_N$ agree on it, since $N\equiv1\pmod4$. Chebotarev therefore gives \eqref{s:eq:protective}, with $\ell$ split in $K$ and arbitrarily large. Install the order-$N$ induced type at $\lambda$ and apply the forward construction and split-$N$ backward lifting of Proposition~\ref{shared:protect}. Its image argument makes every residue in $S$ nonsolvable, with adequate cyclotomic restriction in odd characteristic. The additional conditions \eqref{s:eq:protective} do not alter that argument. \subsection{Dyadic preparation and odd conductor} Reduce the protected system modulo two. Its determinant is one, since $\psi_2$ has two-power order. At each odd ramified place use Lemma~\ref{c:awaytwo} to choose a component of determinant $\eps_2$; its type need not equal the old type. At $\lambda$ retain the order-$N$ induction: take inertia $\operatorname{diag}(\zeta_N,\zeta_N^{-1})$ and antidiagonal Frobenius with off-diagonal product $-\ell$, giving determinant $\ell$ and the same residual local representation. Corollary~\ref{s:cor:prescribed-dyadic-system} gives a potentially modular weight-two system of exact cyclotomic determinant, with dyadic types $\tau_0$ or $\tau_*$ and no new conductor prime. For the backward arrow apply Tung over $K$ itself, where the dyadic completions are $\Q_2$: the old regular source may have arbitrary dyadic conductor. Once the replacement is modular, the common nonsolvable residue is modular and the theorem applies. Retain the dyadic types $\tau_0,\tau_*$ in every subsequent odd-characteristic congruence. Remove all odd rational conductor primes except $s,\ell$. Before removing a nonsplit prime $t$, change coefficient characteristic to $s$ and choose a weight-two replacement, retaining the dyadic and protective types and every previously cleared place away from $s$. Weight-two local existence is \cite[Proposition~7.8.1]{g:Snowden}, followed by prescribed lifting. At $s$, the finite determinant correction reduces to one and has $s$-power order; its inverse square root fixes the determinant without changing the residue or weights \cite[Section~7.9]{g:Snowden}. Unramified local components preserve cleared places, and the old components preserve retained types. Since $s$ splits, Tung gives the backward implication for the actual old regular source. In characteristic $t$, replace the source by regular crystalline potentially diagonalizable lifts at all places above $t$. This clears its conductor in the new compatible system. For nonsplit $t$, the old source has weight two by the reset, and \cite[Theorem~2.3]{h:AKT}, with the witness from Lemma~\ref{shared:weightzero}, gives the backward implication. For split $t$, use Tung without a weight restriction on the old source. The forward construction applies in characteristic three regardless of its splitting in $K$. All conditions away from $t$ retain the dyadic and protective types. Finally clear $s$ by a regular crystalline replacement in characteristic $s$ and split-prime lifting. The weights are allowed to grow; no crystalline weight-two replacement has been asserted at a removal prime. \subsection{The last prime and the terminal source} Remove the protective place last, in characteristic $\ell$. Its own order-$N$ type can no longer protect that residue, so use the dyadic types. If one is $\tau_*$, its $Q_8$ inertia stays absolutely irreducible modulo odd $\ell$: the center acts as $-1$, while every one-dimensional character kills it. The local cyclotomic extension at two is unramified and preserves this inertia. If both are $\tau_0$, the two order-three characters remain distinct modulo $\ell>3$ and Frobenius interchanges them. By \eqref{s:eq:protective}, the cyclotomic residue degree is odd, so that interchange survives restriction to $G_{K(\zeta_\ell)}$. The global residual cyclotomic restriction is therefore absolutely irreducible in both cases. If the residue is solvable, its Artin lift and regular weight raising give a modular witness; Tung at the split prime $\ell$ already proves modularity of the old source, completing the backward chain. Otherwise $\ell$ is large enough for adequacy. Choose regular crystalline potentially diagonalizable lifts above $\ell$, preserving all dyadic types and cleared places. Prescribed lifting and potential modularity give the system with $\lambda$ removed. Tung gives the backward implication. The terminal system has cyclotomic determinant, conductor supported above two, and both dyadic supercuspidal types. Its actual $2$-adic member $r_2$ is locally absolutely irreducible, de Rham and regular, with no weight-two restriction. If its residue is irreducible and nonsolvable, hypothesis (iii) and Tung give modularity. If the residue is irreducible and solvable, apply \cite[Theorem~10.6]{s:FM}: $K/\Q$ is abelian, two splits, all places belong to $\Sigma_n$, and $\Sigma_o=\varnothing$, so its nonsplitness requirement is vacuous. If the residual semisimplification is reducible, both characters have odd order. At the dyadic completions $\Q_2$ they kill the pro-two unit group, and they also kill the real sign groups. They factor through the ordinary class group and are trivial by (ii). Hypothesis (i) and Theorem~\ref{s:thm:scalarFM} apply. Virtually solvable actual sources are alternatively covered by the CM-induction argument in its proof. In every case the actual terminal source is modular. Compatibility identifies its entire system with the automorphic system, and all backward implications return to the initial lift. Reducing proves Theorem~\ref{struct:split-criterion}. \end{proof} \section{Integral orders and finite-flat groups}\label{r:orders} For the quadratic applications write $F_d=\Q(\sqrt d)$. We first retain the full coefficient information in an integral order, then establish the finite-flat bounds used on its lattices. \subsection{Trace orders} We use simultaneous realizations at every dyadic coefficient place. This is essential when a local calculation bounds a global trace-field degree. Let $\mathcal R_1,\ldots,\mathcal R_m$ be distinct coefficient orbits of absolutely irreducible strongly compatible rank-two systems, with trace fields $D_i$ and cyclotomic determinants. Write $d_i=[D_i:\Q]$. Let $R_0$ be the closed $\Z_2$-algebra generated by their simultaneous traces and let $E_0$ be the actual order generated by their simultaneous group images and the trace scalars, in integral matrix realizations at all the coefficient places above two. \begin{lemma}\label{r:shared:trace-order} The rings $R_0,E_0$ are finite free $\Z_2$-modules, and \[ R_0[1/2]=\prod_i(D_i\otimes_\Q\Q_2),\qquad \operatorname{rank}_{\Z_2}R_0=\sum_i d_i,\qquad \operatorname{rank}_{\Z_2}E_0=4\sum_i d_i. \] The order $E_0$ is the closed $\Z_2$-span of the group image. Its trace and determinant laws take values in $R_0$. If every residual semisimplification is scalar, $E_0/2E_0$ is a local algebra with residue field $\F_2$. \end{lemma} \begin{proof} All the chosen matrices preserve integral lattices, so their span is contained in a finite free $\Z_2$-module. Cayley--Hamilton gives \[ T(g)1=g+\det(g)g^{-1}. \] Since $\det(g)\in\Z_2^\times$, the closed group span already contains every trace scalar and is an order. The determinant of a sum lies in $R_0$ by the polarization identity \[ \det(x+y)=\det(x)+\det(y)+T(x)T(y)-T(xy), \] first for linear combinations of group elements and then by continuity. Good Frobenius traces generate each $D_i$. A rational subalgebra of $\prod_iD_i$ surjecting onto every factor identifies two simple factors only if an embedding matches their good traces. Brauer--Nesbitt would then identify their coefficient orbits. The rational simultaneous trace algebra is therefore the displayed product. On each simple trace factor, absolute irreducibility makes the image algebra central simple of dimension four; it becomes $M_2$ after a coefficient extension. This proves both ranks. In characteristic two a scalar residual semisimplification with cyclotomic determinant is trivial. Each actual residual image is unipotent, and the principal congruence kernels are pro-two. The simultaneous image is therefore pro-two. Its finite image in $E_0/2E_0$ spans that algebra, which is consequently a quotient of a finite two-group algebra over $\F_2$. Such a group algebra is local with residue field $\F_2$. \end{proof} The next finite-flat assertion is restricted to a category in which the order-two model is unique. Let $F/\Q_2$ have absolute ramification index two, and allow a finite unramified extension of $F$. We call a finite flat group biconnected if its special fibre and the special fibre of its Cartier dual are connected. \begin{lemma}\label{r:middle} Among biconnected finite flat groups killed by two whose generic Jordan--H\"older factors over $\F_2$ are trivial, every morphism of generic fibres extends uniquely. Generic subobjects and quotients have biconnected finite-flat models, and their exact sequences extend to those models. The simple model has Breuil--Kisin Frobenius $u$. Every self-extension in this category has generic quadratic characters of conductor at most two. \end{lemma} \begin{proof} Schematic closure and finite-flat quotient preserve biconnectedness. Indeed a closed subgroup of a connected finite group over the perfect residue field has local coordinate algebra and is connected; connectedness of quotients follows from surjectivity. Cartier duality gives the other two assertions. For a group of order two, the height-one exponent is $j$ with $0\le j\le2$. Biconnectedness excludes $j=0,2$, leaving $j=1$. The unit factor is removed by change of generator when the generic character is trivial. This is the unique middle model, including its generic identification; see the connected classification in \cite[Theorem~4.1, Proposition~4.2 and Corollaries~4.3--4.4]{r:t:Kim}. A generic isomorphism between two objects extends as an isomorphism by induction on their common order. Close its graph in the product. For each projection, choose a generic subgroup of order two and close it in source and target. The restriction of the projection is an isomorphism by uniqueness of the middle model. Its induced map on the finite-flat quotients is an isomorphism by induction, hence the projection is an isomorphism. For an arbitrary generic morphism, apply this to the first projection of its graph closure. The second projection then extends the morphism. Flatness gives uniqueness. Taking schematic closures and finite-flat quotients gives the asserted exact sequences. Let $k_0$ be the residue field. Kim's exact classification \cite[Corollary~4.3 and Proposition~5.7.3]{r:t:Kim} represents a self-extension of the simple model by \[ \Phi=\begin{pmatrix}u&h(u)\\0&u\end{pmatrix},\qquad h\in k_0[[u]]. \] Changing the upper lift replaces $h$ by $h+u(\varphi(f)-f)$. The map $\varphi-1$ is bijective on $uk_0[[u]]$, so we may write $h=h_0+h_1u$. Rescaling the endpoint bases by $u^{-1}$ gives the \'{e}tale $\varphi$-module class $h_0/u+h_1$, whose positive upper break is at most one over the field of norms. To compare ramification, choose a uniformizer $\pi$ of $F$ and put $F_n=F(\pi^{1/2^n})$. The transition $F_n/F_{n-1}$ has different exponent $2e(F_{n-1})+1$ and upper break $4\cdot2^{n-1}$. Fontaine's bound gives upper break at most three for the generic quadratic character of our finite flat group \cite[Theorem~A]{c:FontaineBT}; his numbering is shifted by one from classical upper numbering. Herbrand's theorem therefore preserves this character's break at each transition. Wintenberger's comparison \cite[Proposition~3.3.2 and Corollary~3.3.6]{r:t:Wintenberger} identifies it with the field-of-norms break. Its conductor is consequently at most two. The argument is unchanged after finite unramified extension. \end{proof} \begin{lemma}\label{r:regular-bt} Let $F/\Q_2$ be finite, and let $V$ be a direct sum of crystalline rank-two representations, with finite coefficient fields, labelled weights $\{0,1\}$ and determinant $\chi_2$. Every stable $\Z_2$-lattice in $V$ is Barsotti--Tate. If all the constituents are nonordinary, its Barsotti--Tate group and its finite-flat subquotients are biconnected. This applies, in particular, to the left regular lattice of $E_0$. \end{lemma} \begin{proof} The crystalline lattice classification gives a Barsotti--Tate group for each constituent. Every stable lattice in their sum is realized by an isogenous group, by schematic closure of the finite generic isogeny kernel. Connectedness and dual connectedness are invariant under isogeny. The connected--\'{e}tale decomposition is stable under coefficient actions. A nonzero rational \'{e}tale factor in a rank-two constituent has coefficient rank one and gives an unramified quotient; the determinant and weights then give the ordinary filtration. Nonordinarity excludes this. Applying the same argument to $r^\vee(1)\simeq r$ excludes an \'{e}tale factor of the dual. Subgroups and quotients preserve biconnectedness by the first paragraph of the proof of Lemma~\ref{r:middle}. After extending coefficients, the rational regular representation of $E_0$ is a sum of columns of the original constituents, so the same argument applies to it. \end{proof} Here the trace and determinant are those of the original integral order. A subsequent nonfaithful matrix quotient need not preserve the information they carry. In particular a vanishing matrix in such a quotient does not imply vanishing of its original trace. \begin{lemma}\label{r:regular-kisin} Suppose every dyadic coefficient constituent is crystalline of labelled weights $\{0,1\}$ and determinant $\chi_2$, is nonordinary, and has scalar residual semisimplification. Let $m=2$ or $6$ and work at the dyadic completion of $F_m$. There the regular lattice \(E_0\) is Barsotti--Tate. Its Breuil--Kisin module is free of rank one as a right \(E_0[[u]]\)-module. If its Frobenius generator is \(\Phi(u)\), then \[ \mathcal D(\Phi(u))\in (u^2-m) R_0[[u]], \qquad T(\Phi(0))=a_2\in R_0 . \] The point field of \(E_0/2E_0\) has upper breaks at most three over \((F_m)_2\). \end{lemma} \begin{proof} The regular lattice is Barsotti--Tate and biconnected by Lemma~\ref{r:regular-bt}. Lemma~\ref{r:middle} gives exactness and full faithfulness for its torsion subquotients with trivial simple generic factors. We use the covariant Breuil--Kisin functor $\mathfrak M_{\mathrm{cov}}(G)=\mathfrak M_{\mathrm{contra}}(G^D)$. Cartier duality followed by Kim's anti-equivalence is covariant and exact \cite[Corollary~4.3 and Proposition~5.7.3]{r:t:Kim}, so it preserves the right action and the quotient used below. The commuting right action of \(E_0\) extends by Tate full faithfulness. Write \(J_0=\operatorname{rad}(E_0)\) and let \(\mathfrak M\) be the Breuil--Kisin module of the regular group. In the scheme killed by two take the sum of the right multiplication maps for a set of generators of \(J_0/2E_0\). Its finite-flat quotient has generic fibre $E_0/J_0=\F_2$ and is the unique middle model. Exactness of the covariant functor gives \(\mathfrak M/\mathfrak M J_0=\mathbf F_2[[u]]\) with Frobenius \(u\). A generator lifts by Nakayama to a surjection \(E_0[[u]]\to\mathfrak M\) of right modules. These have the same rank over \(\mathbf Z_2[[u]]\), so this is an isomorphism. Height one supplies \(\Psi\) with \(\Phi\Psi=\Psi\Phi=u^2-m\). In each integral matrix realization these two identities give a rank-two height-one matrix. Its rational \'{e}tale Frobenius module is the corresponding column constituent of the regular representation, by its commuting right matrix action after extending scalars. The connected finite-flat comparison thus identifies its rank-two generic realization with the given crystalline member. Kisin's determinant comparison \cite[Lemmas~2.3.2 and~2.3.4]{h:Kisin2} gives divisibility of its determinant by \(u^2-m\). Division by this distinguished monic polynomial commutes with each actual coefficient embedding, hence its remainder already vanishes in \(R_0[[u]]\). At \(u=0\) the covariant comparison gives crystalline Frobenius on \(V(-1)=V^\vee\) for the rank-two member \(V\). Its determinant is \(2\) and its trace is the arithmetic coefficient \(a_2\), since $V^\vee\simeq V(-1)$ has the arithmetic Frobenius normalization. This fixes the convention in both identities. Finally Fontaine's finite-flat ramification bound for the group killed by two gives classical upper break at most \(2e((F_m)_2)-1=3\). \end{proof} \begin{lemma}\label{r:fontaine-bound} Let $F/\Q_2$ be ramified quadratic and let $M/F$ be the point field of a finite flat group scheme killed by two. Every positive upper break is at most three, and the normalized relative different contribution is strictly less than two, with $v_2(2)=1$. In particular this applies to the reduction of any stable coefficient lattice of a crystalline rank-two source of weights $\{0,1\}$. \end{lemma} \begin{proof} Fontaine's bound~\cite[Theorem~A]{c:FontaineBT}, converted to classical upper numbering, is $2e(F)-1=3$. Integration of the upper filtration bounds the relative different exponent divided by the relative ramification index strictly below four. Dividing by $e(F/\Q_2)=2$ gives the normalized bound. For the last assertion, realize the stable crystalline lattice by a Barsotti--Tate group. Its coefficient-uniformizer torsion is a finite-flat subgroup of rational two-torsion. \end{proof} \subsection{Rank-one selfextensions at the endpoint exponents} \begin{lemma}\label{shared:endpoint-selfextensions} Let $F/\Q_p$ have absolute ramification index $e$ and residue degree $f$, and let $k$ be a finite coefficient field containing its residue field. Consider a rank-one Breuil--Kisin module with coefficient factors indexed by $i\in\Z/f\Z$, whose Frobenius maps are $b_i u^{r_i}$, where $b_i\in k^\times$ and $r_i\in\{0,e\}$. On these factors Frobenius is $k$-linear and sends $u$ to $u^p$. Every height-one selfextension, exact on the underlying free $k[[u]]$-modules, has Frobenius matrices \[ A_i=b_i u^{r_i} \begin{pmatrix}1&a_i\\0&1\end{pmatrix},\qquad a_i\in k, \] after changing the lifts of the quotient generators. The extension classes form a one-dimensional $k$-space. Whenever the exact finite-flat comparison identifies the given models with these modules, their generic selfextensions become unramified after dividing by the common character. \end{lemma} \begin{proof} Before changing lifts, write $A_i=\bigl(\begin{smallmatrix}b_i u^{r_i}&h_i\\0&b_i u^{r_i}\end{smallmatrix}\bigr)$. The height-one condition makes $u^eA_i^{-1}$ integral. Its upper-right entry gives $u^e\mid h_i$ when $r_i=e$; when $r_i=0$ no divisibility is needed. Thus in either case $h_i=b_i u^{r_i}f_i(u)$ with $f_i\in k[[u]]$. Changing the quotient lift by $\mu_i$ changes $f_i$ by $\varphi(\mu_{i-1})-\mu_i$. On $uk[[u]]$ this cyclic coboundary map is bijective: after one cycle its solution is obtained by inverting $1-\varphi^f$, using the convergent series $\sum_{n\ge0}\varphi^{fn}$. Hence all positive powers of $u$ can be removed. On constants the coboundaries are $(\mu_{i-1}-\mu_i)_i$, whose cokernel is $k$, measured by $\sum_i a_i$. This proves the dimension assertion. Unramified selfextensions of the common character form a one-dimensional space and have finite-flat models: tensor the rank-one model with an unramified two-dimensional unipotent representation and descend from a finite unramified extension. The exact comparison identifies this nonzero line with the whole extension space, proving the generic assertion. \end{proof} The restriction $r_i\in\{0,e\}$ is used in the divisibility of $h_i$. The middle-exponent calculations below are separate. \subsection{Ramification of nonordinary torsion}\label{r:t:shared:dyadic} We first prove a local statement independent of $K$. \begin{theorem}\label{r:break} Let $F/\Q_2$ be ramified quadratic, and let $r_F:G_F\to\GL_2(E)$ be nonordinary, crystalline of labelled weights $\{0,1\}$ and determinant $\chi_2$. For any stable lattice, the point field of its reduction has positive upper breaks at most one and normalized relative different contribution less than one, with $v_2(2)=1$. If the reduction is locally absolutely irreducible, its full image is $S_3$ and its wild inertia is trivial. \end{theorem} \begin{proof} As in Lemma~\ref{r:regular-bt}, the lattice is realized by a biconnected Barsotti--Tate group $G$. The same holds after isogeny, and $G[\varpi_E]$ and all its finite-flat subquotients are biconnected. After enlarging the residue coefficient field, suppose first that the reduction is reducible. Its diagonal characters are unramified, since $\OO_F^\times$ is pro-two. After an unramified extension it is $1+c(g)E_{12}$, where $c$ has values in a finite field $k$ of characteristic two. For any $\F_2$-linear functional $\lambda:k\to\F_2$, pull the quotient line back to $\F_2$ and quotient the invariant line by $\ker\lambda$. This gives a biconnected self-extension of the simple model with generic character $\lambda\circ c$. Lemma~\ref{r:middle} bounds its conductor by two. These characters separate the elementary abelian point-field group, so every positive upper break is at most one. If its ramification index is $e$, its different exponent is at most $2(e-1)$; division by $2e$ gives the asserted normalized bound. Unramified base change preserves these ramification bounds. If the reduction is locally absolutely irreducible, the local classification in Lemma~\ref{c:dyadiclocal}(ii) gives full image $S_3$ and trivial wild inertia. Tame index three contributes $1/3$ to the normalized relative different. \end{proof} \subsection{Inert odd coefficient primes} \begin{lemma}\label{r:t:shared:ordinary-inert} Let $(K,p)$ be one of $(F_2,5),(F_3,5),(F_3,7)$. Let $r$ be a member of a compatible system as above, crystalline of weight two at $p$ and of cyclotomic determinant. Suppose its residue is globally reducible with constituents $\psi,\omega_p\psi^{-1}$ and $\psi^2$ is unramified at every finite place away from $p$. Then $r$ is ordinary and modular. \end{lemma} \begin{proof} The completion $K_p$ is unramified quadratic. A residual character is prime to $p$ and therefore kills wild inertia. If $a$ is the tame exponent of $\psi$ modulo $p^2-1$, the rank-one finite-flat exponents and the unit relation for the totally even character $\psi^2$ give \[ \begin{array}{c|c|c|c|c} K&p&\operatorname{ord}(\epsilon\bmod p)&a\text{ allowed by units} &a\text{ finite flat}\\\hline F_2&5&12&6\mid a&0,1,5,6\\ F_3&5&3&3\mid a&0,1,5,6\\ F_3&7&8&4\mid a&0,1,7,8 \end{array} \] For the first order, $(1+\sqrt2)^3=2$ and $(1+\sqrt2)^6=-1$ modulo five. For the second, $\epsilon_3^2+\epsilon_3+1=0$ modulo five. Modulo seven, $\epsilon_3^4=-1$ and $\epsilon_3^8=1$. In each row only $a=0,p+1$ survive: these are the etale and multiplicative endpoints. The tame exponent list is the rank-one finite-flat classification over the unramified quadratic base \cite{r:t:Raynaud}. This implies actual ordinarity. Otherwise the Barsotti--Tate group of a stable coefficient lattice and its dual would be connected: a coefficient-stable etale factor would give the ordinary coefficient-rank-one filtration. Schematic closure and finite-flat quotient, after killing unramified character factors, produce a rank-one finite-flat subquotient with one of the two endpoint generic characters. At absolute ramification index one, finite-flat uniqueness \cite[Theorem~3.3.3 and Corollary~3.3.6]{r:t:Raynaud} identifies its model with the etale or multiplicative endpoint, contradicting biconnectedness. The unique ordinary filtration descends through the unramified extension. Apply \cite[Corollary~1.2.2]{h:DT} to $r^\vee$: $K$ is totally real abelian, $p$ is unramified, the actual source is ordinary, its determinant is inverse cyclotomic, and its residue is reducible. This proves modularity of $r$. \end{proof} \begin{lemma}\label{r:t:shared:inert-cyclotomic} Let $p\ge5$ be inert and unramified in a real quadratic field $K$, with $[K(\zeta_p):K]=p-1$. An absolutely irreducible residue of an actual weight-two crystalline cyclotomic $p$-adic source remains absolutely irreducible over $K(\zeta_p)$. \end{lemma} \begin{proof} Otherwise Clifford theory induces it from the unique quadratic subfield of $K(\zeta_p)/K$. The image is dihedral of order prime to $p$, and that quadratic field is ramified at $p$. Cyclic tame projective inertia therefore contains a reflection and has order two. This contradicts the finite-flat inertia possibilities. Over the unramified quadratic completion the rank-one exponents are $0,1,p,p+1$ modulo $p^2-1$. Cyclotomic determinant gives pairs $(0,p+1)$ or $(1,p)$, with projective orders $p-1$ or $p+1$. In the locally irreducible case the niveau-four exponents are \[ p+1,\quad p^2+p,\quad p^3+1,\quad p^3+p^2 \pmod{p^4-1}. \] They are paired by multiplication by $p^2$. Their projective inertia order is $(p^2+1)/2$: subtracting the paired exponent reduces the computation to $\gcd(p^2+1,p\pm1)=2$. None of the displayed orders is two for $p\ge5$. These are precisely the labelled finite-flat tame profiles \cite{r:t:Raynaud}, so the induced case is impossible. \end{proof} For either field at five, the cyclotomic degree is four; for $F_3$ at seven it is six. A small-field modular witness \cite[Theorem~2.5]{h:AKT}, or the projective witness specified below, can therefore be lifted to the actual source by \cite[Theorem~2.3]{h:AKT} using Lemma~\ref{r:t:shared:inert-cyclotomic}. The prime three is treated separately, since its order $p-1=2$ would not give this exclusion. \subsection{A common cubic coefficient-field screen} \begin{lemma}\label{r:cubic-screen} Let $D$ be a totally real number field of degree $d\le3$. If $d=3$, suppose $D=\Q(\alpha)$, where $\alpha$ is an algebraic integer, every conjugate of $\alpha$ lies in $[-\sqrt8,\sqrt8]$, and its minimal polynomial is congruent to $X^3$ modulo two. Then $D$ has a prime above three with residue field $\F_3$ or $\F_9$, or a prime above five with residue field $\F_5$. \end{lemma} \begin{proof} For $d\le2$, any prime above three has the required residue field. In degree three write the minimal polynomial as $f(z)=z^3+Az^2+Bz+C$. Its nonleading coefficients are even, and purity gives $|A|\le8$, $-8\le B\le24$, $|C|\le22$. The bound on $B$ follows by evaluating the pairwise-product sum at the vertices of the cube $[-\sqrt8,\sqrt8]^3$. Exact Sturm counting among these even coefficient triples leaves four irreducible polynomials and their conjugates $-f(-z)$: \[ \begin{array}{ccl} f(z)&p&\text{local certificate}\\\hline z^3-2z^2-4z+6&3&f(0)\equiv0,\quad f'(0)\not\equiv0\pmod3,\\ z^3-6z-2&3&f(z-1)=z^3-3z^2-3z+3\text{ is Eisenstein},\\ z^3-2z^2-2z+2&5&f(3)\equiv0,\quad f'(3)\not\equiv0\pmod5,\\ z^3-4z-2&5&f(1)\equiv0,\quad f'(1)\not\equiv0\pmod5. \end{array} \] The enumeration discards $C=0$ and reducible polynomials and counts roots on the indicated interval by Sturm's theorem; endpoint signs are evaluated exactly in $\Q(\sqrt8)$. Hensel's lemma supplies a degree-one completion in the first, third and fourth rows. The second has a totally ramified cubic completion at three and hence residue field $\F_3$. The same conclusions hold for $-f(-z)$. The exact calculation is supplied in~\cite[\path{common}]{code}. \end{proof} \section{Symmetry and quadratic descent}\label{struct:symmetry} The integral-order bounds can force either a small coefficient field or an isomorphism with a quadratic conjugate. We record the two rigidity mechanisms used below. The prime-to-characteristic cotangent argument is distinct from the dyadic order argument. \subsection{Finite-order actions on complete local rings} \begin{lemma}[Cotangent rigidity]\label{t:involution}\label{struct:cotangent} Let $\mathcal O$ be a complete noetherian local ring, and let $A$ be a complete noetherian local $\mathcal O$-algebra with the same residue field. Let a finite group $\Gamma$, with $|\Gamma|\in\mathcal O^\times$, act continuously by local $\mathcal O$-algebra automorphisms on $A$. If its action on $\mathfrak m_A/(\mathfrak m_A^2+\mathfrak m_{\mathcal O}A)$ is trivial, its action on $A$ is trivial. \end{lemma} \begin{proof} Average lifts $x_1,\ldots,x_n$ of a relative cotangent basis over $\Gamma$. The resulting invariant elements have the same cotangent classes. Nakayama and completeness give a surjection from $\mathcal O[[X_1,\ldots,X_n]]$ to $A$ sending its variables to these elements. The entire image is fixed. \end{proof} The invertibility hypothesis is essential: $T\mapsto-T$ is a nontrivial involution of $\Z_2[[T]]$ acting trivially on its relative cotangent space. In applications to pseudodeformations, the cotangent calculation must include products of opposite extensions, not only tangent vectors of representations. \begin{lemma}\label{shared:tensor-involution} Let $k$ be a field of characteristic different from two, and let $V=ks\oplus kd$ carry the involution $\sigma(s)=s$, $\sigma(d)=-d$. On $V^\vee\otimes_k V^\vee$ put \[ \iota(a\otimes b)=\sigma^\vee(b)\otimes\sigma^\vee(a), \qquad \ell_\pm=s^\vee\pm d^\vee. \] If an $\iota$-equivariant quotient kills both $\ell_+\otimes\ell_+$ and $\ell_-\otimes\ell_-$, then $\iota$ acts trivially on that quotient. \end{lemma} \begin{proof} The $(-1)$-eigenspace of $\iota$ is the line generated by $s^\vee\otimes d^\vee+d^\vee\otimes s^\vee$. It is killed because \[ \ell_+\otimes\ell_+-\ell_-\otimes\ell_- =2(s^\vee\otimes d^\vee+d^\vee\otimes s^\vee). \] Since two is invertible, the two eigenspaces split and the quotient has only the eigenvalue one. \end{proof} \subsection{Descent from a twisted conjugacy} \begin{lemma}[Quadratic descent]\label{r:quadratic-descent} Let $K/\Q$ be real quadratic, with nontrivial automorphism $\sigma$. Let $\mathcal R=(r_\lambda)$ be a totally odd, absolutely irreducible strongly compatible rank-two system over $K$, with cyclotomic determinant and regular labelled Hodge--Tate weights. Suppose that, for one coefficient place, \[ r_\lambda^\sigma\simeq r_\lambda\otimes\eta, \] where $\eta$ is the restriction of a quadratic character of $G_\Q$ (the trivial character is allowed). Then $\mathcal R$ is modular. \end{lemma} \begin{proof} The identity of good Frobenius traces transfers the displayed isomorphism to a companion $r$ in characteristic $p\ge5$, by compatibility and Chebotarev. Write $H=G_K$ and choose $t\in G_\Q\setminus H$. There is an intertwiner $A$ with \[ A r(h)A^{-1}=\eta(h)r(tht^{-1})\qquad(h\in H). \] Since $\eta^\sigma\eta=1$, Schur's lemma gives $A^2r(t^2)^{-1}$ scalar. Thus $[r]$ and $[A]$ define a continuous projective representation of $G_\Q$. After a finite coefficient extension, scale $A$ so that $\det A=\chi_p(t)$. Define a continuous choice of matrices by $L(h)=r(h)$ and $L(ht)=r(h)A$ for $h\in H$. Then $\det L(g)=\chi_p(g)$ for every $g$, so its scalar cocycle \[ c(g_1,g_2)=L(g_1)L(g_2)L(g_1g_2)^{-1} \] takes values in $\{\pm1\}$. It is locally constant and is trivial on $H\times H$. Tate's theorem in \cite[Theorem~6.3]{r:s:Ribet}, for the \emph{discrete trivial} $G_\Q$-module $\overline\Q^{\times}$, gives a continuous cochain $b:G_\Q\to\overline\Q^{\times}$ such that $c=\delta b$. Compactness makes the image of $b$ finite. Fix an embedding $\overline\Q\hookrightarrow\overline\Q_p$ and enlarge the coefficient field by these finitely many algebraic values. Then $\widetilde r=b^{-1}L$ is a continuous representation of $G_\Q$. Because $c|_{H\times H}=1$, the restriction $b|_H$ is a finite-order character. In fact $c^2=1$ implies that $b^2$ is a finite-order character of $G_\Q$. The representation $\widetilde r$ is absolutely irreducible, since its restriction to $H$ is a finite twist of $r$. The locally constant function $b$ factors through a finite Galois quotient. Outside the ramification of that quotient, $K/\Q$ and $r$, inertia lies in $H$ and acts trivially, so $\widetilde r$ is finitely ramified. Finite twists preserve de Rham representations and their weights; de Rhamness descends through finite local extensions. Thus $\widetilde r$ is regular de Rham at $p$. Every complex conjugation belongs to $H$, and the square of its value under $b|_H$ is one, so $\widetilde r$ is odd. Apply \cite[Theorem~1.0.4]{r:s:Pan}. Cyclic base change to $K$ is cuspidal because its Galois restriction is irreducible; untwisting by $b|_H$ proves modularity of $r$, and hence of the compatible system. \end{proof} \subsection{Coefficient orbits and the integral-order bound} \begin{proof}[Proof of Theorem~\ref{struct:packet}] The two operations commute and are involutions on isomorphism classes. Let $V=C_2^2$ be the resulting abstract group, let $m\in\{1,2,4\}$ be the number of coefficient orbits, and put $e=[D:\Q]$. Each operation preserves the trace-field degree, since it permutes the good primes and changes their traces only by rational signs. Lemma~\ref{r:shared:trace-order} gives $4me=\dim_{\F_2}(E_0/2E_0)<32$. The stabilizer $V_0$ of the coefficient orbit of $\mathcal R$ has order $4/m$. Its elements induce unique automorphisms of $D$, because the good traces generate $D$. These automorphisms form a homomorphism $V_0\to\Aut(D)$ whose kernel consists of actual isomorphisms of systems. The element $H$ is not in the kernel. Any nontrivial kernel therefore contains $G$ or $GH$ and gives one of the asserted twisted-conjugacy identities. Otherwise $V_0$ embeds in $\Aut(D)$. The fixed-field theorem gives $4/m\mid e$, hence $4\mid me$. Since $056.06306031$, a contradiction in both fields. Apply \cite[Theorem~6.1]{g:Thorne2} to the dual actual crystalline two-adic source and dualize back. The nonsolvable modular residue and checked real-place condition meet its hypotheses. It imposes no $\mathbf Q_2$-completion requirement and allows ordinary and nonordinary sources. Strong compatibility proves modularity of $\mathcal R$. \end{proof} \section{\texorpdfstring{Scalar and nonscalar sources over $\Q(\sqrt2)$}{Sources over Q(sqrt2)}}\label{r:t:sec:field-two} In this section $K=F_2$, $s=\sqrt2$, $\ell=7$ and $\epsilon=1+s$. A special type includes its exact quadratic scalar character. We first normalize the ramified scalar factors. \begin{lemma}\label{r:t:d2:quadratic-corrections} Put $\alpha_\pm=5\pm4s$. The quadratic characters attached to $K(\sqrt{\alpha_\pm})/K$ ramify at exactly one of the two seven-primes, are unramified at two and three, and their product is the character of $K(\sqrt{-7})/K$, split at two. \end{lemma} \begin{proof} We have $\alpha_+=\epsilon(3+s)$ and $\alpha_- =\alpha_+^\sigma$, with $\alpha_+\alpha_-=-7$. Both are units away from seven. The integral element $(1+\sqrt{\alpha_\pm})/2$ has polynomial $X^2-X-(1\pm s)$. Its reduction at the dyadic prime is $X^2+X+1$, so the quadratic extensions there are unramified. At three the defining radicands are units, hence the quadratic extensions are unramified. Their odd seven-adic valuations give exactly the asserted seven-ramification. Finally $-7$ is a square in $\mathbf Q_2$ because it is one modulo eight. \end{proof} \subsection{Nonscalar dyadic residues} \begin{lemma}\label{r:t:d2:lem:solvable-two-exclusion} Let $K=\mathbf Q(\sqrt2)$ and let $\bar r:G_K\to\mathrm{SL}_2(\overline{\mathbf F}_2)$ be finite flat at two, unramified away from two and seven, with inertia of order at most two at both seven-primes. If its image is solvable, then $\bar r^{\mathrm{ss}}=1\oplus1$. \end{lemma} \begin{proof} If the representation is reducible, its diagonal characters have odd order and are unramified at seven. Their inertia at two is trivial because the dyadic residue field is $\mathbf F_2$ and the wild inertia is pro-two. They are therefore everywhere unramified; narrow class number one makes them trivial. Otherwise its image is an odd dihedral group, and it is induced from an odd-order character $\theta$ of a quadratic extension $L/K$. The quadratic quotient is unramified outside two and seven. Since $K$ has class number one and unit group $\langle-1,\epsilon\rangle$, its squareclass is generated by $-1,\epsilon,s,3+s,3-s$, where $s^2=2$. Fontaine's finite-flat bound makes the quadratic conductor at two at most four. A quadratic extension of $K_2$ defined by an odd valuation radicand has conductor $2e(K_2)+1=5$. Thus the exponent of $s$ is zero, and there are precisely the fifteen nontrivial classes generated by \[ -1,\quad\epsilon,\quad3+s,\quad3-s. \] The character $\theta$ is unramified at seven. If $L/K$ ramifies there, its residual inertia is a reflection of order two and has trivial intersection with the odd rotation subgroup. If $L/K$ is unramified there, inertia lies in that odd subgroup and has order dividing two, so is trivial. Its wild inertia at two is also trivial. Consequently $\theta$ factors through the odd quotient of the ray group of $L$ modulo the primes above two. That quotient is trivial in all fifteen cases. Here is a complete finite certificate: use the bits of an integer $1\le m\le15$ to select the four displayed generators, form their product $A(y)$, and take \[ P_m(X)=\operatorname{Res}_y\big((X-y)^2-A(y),y^2-2\big). \] The accompanying program \texttt{verify\_d2\_solvable\_two.gp} computes the maximal order and certified class group of each quartic field, and its ray group modulo two. It applies \texttt{bnfcertify} in every row and asserts that every ray invariant is a power of two. The possible absolute discriminants and ray invariants are \[ \begin{array}{r|l} \operatorname{disc}(L)&\operatorname{Cl}_{(2)}(L)\\\hline 256&1\\ -1024&1\\ 7168&1\text{ or }C_4\times C_2\\ -448&1\\ -1792&1\\ 12544&1\\ 3136&C_2^2\\ -50176&C_2^2. \end{array} \] The modulus $(2)$ may contain higher dyadic powers; these add only two-primary ray factors, so its odd quotient equals that of the radical modulus required here. Infinite-place factors are also two-primary. Thus $\theta$ is trivial, contradicting the irreducibility of its induction. The exact calculation is supplied in~\cite[\path{ramified/d2}]{code}. \end{proof} \subsection{Ordinary scalar sources} \begin{proposition}[Ordinary scalar sources]\label{r:t:d2:ordinary-scalar} Let $\mathcal R$ be a pure, totally odd, absolutely irreducible, strongly compatible rank-two system over $K=\mathbf Q(\sqrt2)$, of cyclotomic determinant and labelled weights $\{0,1\}$, good outside seven and special at both seven-primes with unramified quadratic scalar characters. If some dyadic member is ordinary with scalar residual semisimplification, the actual system is modular. \end{proposition} \begin{proof} Write $s=\sqrt2$, $\epsilon=1+s$, $\pi=3+s$, $\pi'=3-s$ and $\delta=-7$. The supported quadratic classes with even dyadic valuation form \[ V=\langle[-1],[\epsilon],[\pi],[\pi']\rangle. \] Localization at $K_2$ has kernel $\langle[\delta]\rangle$. Indeed unit squares modulo $8\mathcal O_K$ have coordinate pairs $(1,0),(1,4),(3,2),(3,6)$ in the basis $1,s$. Testing the sixteen classes gives precisely $1,\delta$ in the square kernel. The accuracy exceeds twice the valuation of two, so Hensel's lemma makes the test exact. Moreover no nonzero class in $V$ is split at both seven-primes: vanishing valuations removes $\pi,\pi'$, and the Legendre-symbol vectors of $-1,\epsilon$ are $(-,-)$ and $(-,+)$. Let $r$ be an ordinary dyadic member over a finite extension $E/\mathbf Q_2$, with uniformizer $\varpi$ and residue field $k$. Choose a vector $v$ on its ordinary line and take the lattice generated by its global orbit. Its residue cannot be trivial, because a cyclic trivial module has dimension one. It is therefore a nonzero unipotent extension, whose invariant line is transverse to the reduction of the ordinary line. Both lines are locally preserved at two, forcing the actual local residue to be trivial. The localization kernel above makes the global residue $1+\chi_\delta E_{12}$. Since $\delta$ is negative at each real place, choose a real conjugation $c$ and the integral basis $((r(c)-1)v,v)$. Its reductions are independent, and it gives \[ r(c)=C:=\begin{pmatrix}-1&1\\0&1\end{pmatrix},\qquad \text{ordinary line }e_2,\qquad \bar r=1+\chi_\delta E_{12}. \] The normalized residue is defined over $\mathbf F_2$. We compare two actual sources with these normalizations and the same exact unramified ordinary quotient $\nu$ at two. Suppose $r_1\equiv r_2\pmod{\varpi^n}$, $n\ge1$, and write \[ r_2(g)=(1+\varpi^nX(g))r_1(g)\pmod{\varpi^{n+1}},\qquad X(g)=\begin{pmatrix}x(g)&y(g)\\z(g)&x(g)\end{pmatrix}. \] The determinant gives the trace-zero form. The adjoint action of $1+\chi_\delta E_{12}$ adds $\chi_\delta(g)z(h)(1+E_{12})$; hence $z$ is additive. Equality of the exact ordinary characters gives $x|_{G_{K_2}}=y|_{G_{K_2}}=0$, and the local class of $z$ is a unit Kummer class. Here is the required assertion for the divided difference. The two local ordinary extensions have identical characters $\nu,\chi_2\nu^{-1}$. The difference of their integral extension cocycles divided by $\varpi^n$ is integral, and its rational class remains in the $E$-linear crystalline extension space. The corresponding stable lattice is again ordinary Barsotti--Tate. Its reduction is an extension of the etale model by the multiplicative model, hence a unit Kummer class. This uses actual sources and does not assert exactness for arbitrary models. At a seven-place residual inertia has a unit upper entry, so its special line is uniquely determined. Congruent representations have congruent special lines: take the kernel of a rank-one inertia difference with a unit entry. If the exact special characters agree modulo $\varpi^{n+1}$, variation of that line by $t_ve_2$ gives \[ z|_{G_v}=0,\qquad x|_{G_v}=t_v\chi_\delta|_{G_v}. \] If they differ by an unramified quadratic sign, put $e=v_\varpi(2)$. The same homogeneous equations hold for $n0$ and a tame inertia element at $\mathfrak q$, and call their differences $A,C$. Their evaluations on $(\alpha,\beta)$ are $(1,0),(0,1)$, so $A,C$ generate $E$. The real matrix and the exact special identities give \[ A^2=C^2=0,\qquad AC+CA=X=T(AC). \] All words lie in the $R$-span of $1,A,C,AC$. At $\mathfrak q$, $-1$ is unramified nonsquare; change an arithmetic Frobenius lift by inertia so that its value on $\beta$ is zero. Its difference $B$ has the same first-layer values as $A$, hence $B-A\in J^2$. The special identity $BC=0$ gives $AC\in J^3$. Work over $R'=R/\mathfrak m_R^2$. The relative coefficient cotangent is generated by $X$, since the coefficient order is generated by the actual traces. All traces of $1,A,C$ vanish modulo two. The normal relations show that $T(J^3)=0$ over $R'$: the length-three words are $ACA=XA$ and $CAC=XC$, with trace zero, and the length-four trace $T(ACAC)=X^2$ vanishes in $R'$. Higher words reduce to the same relations; factors from $\mathfrak m_{R'}E$ contribute trace zero because all traces lie in $\mathfrak m_{R'}$. It follows that $X=T(AC)=0$ in $R'$. Nakayama gives $R=\mathbf F_2$, so $R_0$ has rank one and the trace field is $\mathbf Q$. Lemma~\ref{r:t:d2:small-coefficient} now proves modularity. \end{proof} Put $K=\mathbf Q(\sqrt2)$, $s=\sqrt2$, $\epsilon=1+s$, and $\pi=3+s$, $\pi'=3-s$. A special condition includes its exact quadratic twist, including the unramified character. \begin{proposition}\label{r:t:d2:prop:scalar-trace-degree} Let $\mathcal R$ be a pure, totally odd, absolutely irreducible, strongly compatible rank-two system over $K$ of determinant cyclotomic and labelled weights $\{0,1\}$. Suppose it is good outside seven and special at both seven-primes. If every dyadic coefficient member is nonordinary with scalar residual semisimplification, its global trace field is $D=\Q(a_2)$ of degree $d\le3$, and its dyadic trace order is $R_0=\Z_2[a_2]$ with $R_0/2R_0\simeq\F_2[t]/(t^d)$, where $t=\overline{a_2}$. \end{proposition} \begin{proof} Let $D$ be the global trace field, $d=[D:\mathbf Q]$, and apply the common simultaneous integral trace-order construction to all dyadic coefficient embeddings. Write its coefficient order as $R_0$ and its regular order as $E_0$, with trace $T$ and norm $\mathcal D$. Thus $\operatorname{rank}_{\mathbf Z_2}R_0=d$, and the regular lattice is an actual nonordinary Barsotti--Tate lattice. Write $E=E_0/2E_0$, $R=R_0/2R_0$, and $J=\operatorname{rad}E$. Lemma~\ref{r:middle} bounds the dyadic conductors of the quadratic characters in $(J/J^2)^\vee$ by two. The elementary quadratic quotient of the narrow ray group of modulus $\mathfrak p_2^2\pi\pi'\infty_1\infty_2$ has rank three. A basis is \[ \alpha=[-1],\qquad \beta=[5+4s]=[\epsilon\pi],\qquad \delta=[-7]. \] Indeed the full narrow ray group has invariants $(6,2,2)$; these classes are independent by their seven-valuations and real signs. Their relative discriminant norms are $4,7,49$, respectively. At two, $\alpha$ has break one, $\beta$ is the unramified quadratic class, and $\delta$ is trivial. The latter follows already from $-7\equiv1\pmod8$ in $\mathbf Q_2$. Choose the seven-place where $s\equiv3\pmod7$, a real conjugation at the embedding $s>0$, an arithmetic Frobenius at this seven-place, and a tame inertia generator there. Frobenius can be chosen to be trivial on the ramified quadratic class $[7]$. Their evaluation vectors on $(\alpha,\beta,\delta)$ are \[ (1,0,1),\qquad(1,1,1),\qquad(0,0,1). \] Let $A,B,C$ be their differences from one in $E$. These evaluation vectors are independent, so $A,B,C$ span $J/J^2$ and generate $E$. At the fixed special place the normalized characteristic-zero local differences $L(g)$ obey $L(g)L(h)=(\chi_2(g)-1)L(h)$. Their exact quadratic sign disappears modulo two, yielding zero products and trace zero for the corresponding local differences in $E$. Polarized Cayley--Hamilton therefore gives \[ A^2=B^2=C^2=BC=CB=0,\quad AB+BA=X=T(AB),\quad AC+CA=Y=T(AC). \] Every word is in the $R$-span of $1,A,B,C,AB,AC$. We calculate in $R'=R/\mathfrak m_R^2$, retaining the original trace and norm laws. The traces of the six normal words show that its cotangent space is generated by $X,Y$. The regular Kisin module is free of rank one on the right over $E_0[[u]]$; its Frobenius generator $\Phi(u)$ has norm divisible by $u^2-2$ and trace at $u=0$ equal to the arithmetic coefficient $a_2$. The ramified first-layer functional is $(1,0,0)$ on $(\alpha,\beta,\delta)$, so \[ \Phi=N+u(1+M)+O(u^2),\quad N=A+C+Z,\quad M=m_AA+m_BB+m_CC+Z',\quad Z,Z'\in J^2. \] The unramified functional takes value one on $\beta$, hence $m_B=1$. The norm of $Z$, its polar term with $A+C$, and all trace terms of length at least three vanish in $R'$ by the six-word relations. The constant coefficient of the norm gives $0=\mathcal D(N)=\mathcal D(A+C)=Y$. The linear coefficient of the norm gives \[ a_2=T(N)=T(NM)=m_BX+(m_A+m_C)Y=X\quad\text{in }R'. \] Nakayama proves \[ R=\mathbf F_2[t]/(t^d),\qquad t=\overline{a_2}, \qquad D=\mathbf Q(a_2). \] If $d\ge2$, $X$ is another generator of its maximal ideal and $Y=vX$ with $v\in\mathfrak m_R$. The degree assertion follows because $1,a_2,\ldots,a_2^{d-1}$ span $R_0$ by Nakayama. Suppose $d\ge4$. Put $Z_0=C+vB$. It is central and satisfies $Z_0^2=BZ_0=XZ_0=0$ and $T(Z_0)=T(AZ_0)=0$. Over $S=\mathbf F_2[X]/(X^3)$ use the abstract free rank-four algebra \[ Q=S\langle a,b\rangle/(a^2,b^2,ba+ab-X). \] There is a surjection $E\to Q$ taking $A,B,Z_0$ to $a,b,0$. To verify that this map is well-defined, pair any relation in the normal words $1,A,B,AB,Z_0,AZ_0$ with $1,A,B,AB$ by the original trace. For its first four coefficients $f_i$ the pairings are $Xf_3,Xf_2,Xf_1,Xf_0+X^2f_3$. Their vanishing puts every $f_i$ in $(X^{d-1})$, which vanishes after base change to $S$. The normal-word multiplication then proves the assertion. Compose this quotient with $a\mapsto E_{12}$ and $b\mapsto XE_{21}$ to obtain $\theta:G_K\to\mathrm{GL}_2(S)$. Its trace is the original trace reduced to $S$, and comparison of the two Cayley--Hamilton laws gives determinant one. The two visible linear characters are $\alpha+\beta$ and $\beta$: their values on $A,B,C$ are $(1,0,0)$ and $(0,1,0)$, since $v$ is in the radical. Put \[ F=K(i,\sqrt{5+4s}). \] On $G_F$ every matrix has the form $I+XM+X^2N$, where the lower-left entry of $M$ is zero and its diagonal entries agree. These $M$'s lie in a commutative algebra, so $\theta(G_F)$ is abelian of exponent at most four. If $r$ is the product of the chosen real and seven-Frobenius elements, $\operatorname{tr}\theta(r)=X$, whence \[ \theta(r)^4=I+X^2I\ne I. \] Since $r^2\in G_F$, the abelian image over $F$ has an element of order four. This contradicts a small ray calculation. The field $F$ has degree eight, class number one, absolute discriminant $3211264$, one dyadic prime $\mathfrak P$ with absolute $(e,f)=(4,2)$, and three seven-primes with $(e,f)=(2,2),(1,2),(1,2)$. The only ramified break of $F/K$ at two is one. The original regular point field has upper break at most three over $K_2$, so Herbrand's subgroup formula bounds the break over $F$ by five. At seven the special identities already show that the whole local image in $E$ has exponent two. Hence $\theta|_{G_F}$ is a quotient of the ray group $H$ of modulus $\mathfrak P^6\prod_{w\mid7}w$, with $4H$ and $2I_7$ killed, where $I_7$ is generated by the three local tame unit images. A defining polynomial, and the exact ray quotient, are \[ \begin{gathered} z^8-4z^7+8z^6-8z^5+7z^4-8z^3+8z^2-4z+1,\\ H\simeq C_{48}\times C_8\times C_2^3, \qquad H/(4H+2I_7)\simeq C_2^5. \end{gathered} \] Equivalently, let $H_0$ be the ray group for $\mathfrak P^6$ alone. The canonical map $H\to H_0$ has kernel $I_7$, and $H_0\simeq C_2^2$. Computing the Smith form of $4H+2\ker(H\to H_0)$ gives the displayed quotient; this is the coordinate-free calculation in \texttt{verify\_sp7\_ray\_quotient\_independent.gp}. The quotient has exponent two, contradicting the order-four image just constructed. Hence $d\le3$. The exact calculation is supplied in~\cite[\path{ramified/d2}]{code}. \end{proof} \begin{lemma}\label{r:t:d2:small-coefficient} A good/special-seven system as above is modular if its trace field has degree at most three and is generated by $a_2$, with $\mathbf Z_2[a_2]/2=\mathbf F_2[t]/(t^d)$ when $d=3$. \end{lemma} \begin{proof} Lemma~\ref{r:cubic-screen} gives a coefficient prime over three with residue field $\F_3$ or $\F_9$, or a coefficient prime over five with residue field $\F_5$. For an irreducible residue, finite-field Brauer descent realizes the representation over this trace residue field. At three a reducible residue has constituents $1,\omega_3$ and the actual source is ordinary and modular by Lemma~\ref{r:t:d2:three-source-characters}. An irreducible residue remains cyclotomically irreducible by that lemma. Its small-field witness \cite[Theorem~2.5]{h:AKT}, followed by \cite[Theorem~2.3]{h:AKT}, modularizes the actual weight-two source. At the selected prime over five the residue field is $\mathbf F_5$. In the reducible case the square of a constituent is unramified away from five, since every seven-type is good or quadratic-special. Lemma~\ref{r:t:shared:ordinary-inert} modularizes the actual source. In the irreducible case Lemma~\ref{r:t:shared:inert-cyclotomic}, the small-field witness and the actual weight-two AKT theorem apply. \end{proof} \begin{theorem}\label{r:t:d2:good-special-source} Assume GRH. Every pure totally odd absolutely irreducible weight-two cyclotomic compatible system over $K$, good outside seven, with good or quadratic-special type at both seven-primes, is modular. \end{theorem} \begin{proof} First remove any ramified quadratic special scalar factors using Lemma~\ref{r:t:d2:quadratic-corrections}; the corrections are unramified at two and three and at the other seven-prime. Hence all actual special scalar characters may be taken unramified. If some dyadic coefficient residue is nonsolvable, apply Proposition~\ref{r:t:shared:nonsolvable-source}, which covers both ordinary and nonordinary sources under the present GRH hypothesis. The irreducible solvable alternative is excluded by Lemma~\ref{r:t:d2:lem:solvable-two-exclusion}; hence we may assume every dyadic coefficient residue has scalar semisimplification. If some dyadic member is ordinary, the ordinary proposition or its one-good-place corollary proves the assertion. Otherwise all are nonordinary. With exactly one special place, Proposition~\ref{r:t:d2:prop:one-special-source} applies. With two special places, Proposition~\ref{r:t:d2:prop:scalar-trace-degree} and Lemma~\ref{r:t:d2:small-coefficient} apply. If both seven-places are good, the first-layer quadratic space of the regular nonordinary trace algebra is just $\langle[-1]\rangle$. Indeed its characters have dyadic conductor at most two by the middle-model calculation, and the narrow ray group of modulus $\mathfrak p_2^2\infty_1\infty_2$ has order two. The real difference $A=c-1$ therefore generates the algebra $E=E_0/2E_0$ and satisfies $A^2=0$. The original trace of both $1$ and $A$ is zero modulo two. All group traces consequently vanish, so the coefficient algebra $R=R_0/2R_0$ is $\mathbf F_2$. This makes $E$ spanned by $1,A$, contradicting $\dim_{\mathbf F_2}E=4[D:\mathbf Q]\ge4$. This treats the last case. Undo the finite quadratic twists. \end{proof} \paragraph{An unconditional observation.} The separate nonordinary exclusion remains available without GRH. If a nonordinary dyadic member in the preceding theorem had nonsolvable residue, its kernel field $M$ would have degree at least $120$. Theorem~\ref{r:break} gives $\rd(M)<4\sqrt{14}<14.967$, whereas \eqref{shared:eq:unconditional} with $b=7$ gives $\rd(M)>16.661$. This observation is not needed in the streamlined GRH proof above. \subsection{\texorpdfstring{The supported-seven problem over $\mathbf Q(\sqrt2)$}{The supported-seven problem}} Put $K=\mathbf Q(s)$ with $s^2=2$, $\epsilon=1+s$ and $\mathfrak l_\pm=(3\pm s)$. The prime three is inert and $K_3/\mathbf Q_3$ is unramified quadratic. \begin{lemma}\label{r:t:d2:three-source-characters} Let $\mathcal R$ be a pure, irreducible, totally odd weight-two strongly compatible system with cyclotomic determinant, good outside seven, whose seven-types are good, quadratic-special, or principal series of order three after the corrections above. If its three-adic residue is reducible, the actual system is modular. If that residue is irreducible, it remains irreducible on $G_{K(\zeta_3)}$. \end{lemma} \begin{proof} Twist away the scalar quadratic inertia using Lemma~\ref{r:t:d2:quadratic-corrections}. Modulo three the inertia at both seven-primes is unipotent. A reducible global residue therefore has characters unramified outside three. The prime-to-three narrow ray group modulo three is $C_2$: $\epsilon$ has order eight in $\mathbf F_9^\times$ and signs $(+,-)$, while $-1$ supplies the independent sign relation. Its characters are $1,\omega_3$. Thus the residual constituents are exactly $1,\omega_3$. Since $e(K_3)=1<3-1$, finite-flat uniqueness identifies the rank-one models with the etale and multiplicative models. A nonordinary Barsotti--Tate source would be biconnected, as would its finite-flat subquotients; these rank-one models contradict biconnectedness. The actual source is ordinary, with unramified quotient reducing to one. Skinner--Wiles \cite[\S4.5, Theorem~A]{r:t:SW99} applies: the ratio is locally nontrivial, totally odd, and its splitting field $K(\zeta_3)$ is abelian over $\mathbf Q$. The determinant is $\chi_3$ and the local quotient has weight zero. Undo the twists. For the second assertion, a cyclotomic-reducible irreducible residue would be induced from a prime-to-three character $\theta$ of $H=K(\zeta_3)$. Its restriction to seven-inertia is trivial since that inertia has order dividing three. The character is therefore unramified outside three and tame there. The absolute discriminant of $H$ is $8\cdot(-3)\cdot(-24)=576$, and its degree-four Minkowski bound is $36/\pi^2<4$. There is no prime of norm two or three: their norms are four and nine. Hence $H$ has class number one. Its unique prime above three has residue field $\mathbf F_9$, and $\epsilon$ generates the residue units. The tame ray group is trivial. This makes $\theta$ trivial, contradicting irreducibility. \end{proof} \begin{lemma}\label{r:t:d2:seven-menu} Every local determinant-$\omega_7$ residue over $\mathbf Q_7$ has a determinant-$\chi_7$ weight-two lift that is good, special, or principal series of order three after a scalar quadratic twist. \end{lemma} \begin{proof} The central-trivial Serre weights are $\operatorname{Sym}^r\det^b$, with $r=0,2,4,6$ and $r+2b=0\pmod6$. The two principal-series types with characters $\widetilde\omega_7^t,\widetilde\omega_7^{-t}$ have reductions \[ \begin{array}{c|cc} t&\text{first constituent}&\text{second constituent}\\\hline 1&(2,5)&(4,1)\\ 2&(4,4)&(2,2). \end{array} \] Together with the two scalar crystalline and two scalar semistable weights, these cover all eight weights. The fixed-determinant formula \cite[Theorem~1.2]{g:Tung} therefore gives a nonempty weight-two local ring in this menu, for the actual residual extension. A scalar quadratic twist interchanges the principal-series rows and changes order six to order three. The independent global characters in Lemma~\ref{r:t:d2:quadratic-corrections} implement these choices at the two split seven-primes without disturbing good reduction at two or three. \end{proof} \begin{lemma}\label{r:t:d2:seven-special-switch} Let $\bar r$ be a local characteristic-three representation at a seven-prime, of determinant one and with tame inertia unipotent. It admits an unramified or genuine special determinant-cyclotomic lift preserving the full residual representation. \end{lemma} \begin{proof} If inertia is trivial, lift the entire unramified Frobenius matrix, with fixed determinant seven; unequal Frobenius eigenvalues are allowed. Otherwise write tame inertia as $1+tE_{12}$, $t\ne0$. Because $7=1\pmod3$, Frobenius centralizes this matrix and has shape $aI+bE_{12}$, $a=\pm1$. Its lift \[ \Phi=\begin{pmatrix}7a&\widetilde b\\0&a\end{pmatrix}, \qquad U(x)=1+\widetilde t xE_{12} \] has determinant seven, satisfies $\Phi U(x)\Phi^{-1}=U(7x)$, and has nonzero monodromy. It retains both the residual inertia parameter and the Frobenius extension parameter. \end{proof} \begin{proposition}\label{r:t:d2:bounded-reduction} Assume GRH. Every irreducible determinant-$\omega_7$ residue supported at seven is modular. \end{proposition} \begin{proof} A solvable image has a modular Artin witness. For a nonsolvable image the cyclotomic restriction is irreducible. Use Lemma~\ref{r:t:d2:seven-menu}, prescribed globalization, actual avoiding potential modularity and Brauer descent to construct a weight-two compatible system with the stated bounded menu, good outside seven. Normalize the independent principal-series signs as above. Reduce at three. A reducible residue already modularizes the actual source by Lemma~\ref{r:t:d2:three-source-characters}. Otherwise that lemma gives cyclotomic irreducibility. There is no exceptional-five exclusion in characteristic three. Prescribe the actual good/special lifts of Lemma~\ref{r:t:d2:seven-special-switch}, retaining the old Barsotti--Tate component at three, and globalize by EGPZ. Actual potential modularity and compatibility give the new system. Purity ensures nonzero special monodromy wherever the special component was selected, since its two Frobenius eigenvalues have ratio seven. By Theorem~\ref{r:t:d2:good-special-source} this new system is modular. AKT applies backwards to the old weight-two three-adic source. Undoing the quadratic twists and reducing at seven proves the assertion. \end{proof} \section{\texorpdfstring{Integral-order rigidity over $\Q(\sqrt3)$}{Integral-order rigidity over Q(sqrt3)}}\label{r:t:sec:field-three} In this section $K=F_3$, $\ell=13$ and $\epsilon=2+\sqrt3$. Every special type retains its exact quadratic scalar character. No global correction of the two signs is required. \subsection{Nonscalar dyadic sources} \begin{proposition}\label{r:t:d3:solvable-source} Let a pure cyclotomic weight-two system over $K$ be good outside thirteen and quadratic-special at both thirteen-primes. If some dyadic residue is irreducible and solvable, its actual member is ordinary and modular. \end{proposition} \begin{proof} An irreducible solvable residue is necessarily odd dihedral. Write it as the induction of an odd-order character $\theta$ from a quadratic field $L/K$. Its quadratic quotient is unramified outside two and thirteen. The local finite-flat bound excludes odd dyadic valuation, whose quadratic conductor would be five. Since $K$ has class number one, its possible quadratic squareclasses are in \[ \langle-1,\epsilon,a,b\rangle,\qquad \epsilon=2+\sqrt3,\quad a=4+\sqrt3,\quad b=4-\sqrt3, \quad ab=13. \] If the source is nonordinary, Theorem~\ref{r:break} makes the inducing dyadic conductor at most two. The seven nontrivial possibilities are the masks $1,6,7,10,11,12,13$ in this displayed basis. Their ordinary ray groups modulo $(2)$ are, respectively, \[ 1,\ 1,\ C_8,\ 1,\ C_8,\ C_2,\ C_4\times C_2. \] The odd rotation character $\theta$ is unramified at thirteen because its inertia image there has order at most two, and at two it is tame. Its conductor divides the radical of $(2)$, hence divides $(2)$. Every displayed ray group is two-primary, so $\theta$ is trivial. This contradicts irreducibility. Therefore no solvable irreducible residue occurs in the nonordinary case. For an ordinary source use the conductor-at-most-four list, which includes all fifteen nontrivial classes above. Their ray groups modulo $(2)$ are two-primary except for mask fifteen, \[ L=K(\sqrt{-13\epsilon})=K(\sqrt{-26}), \qquad \mathrm{Cl}_{(2)}(L)=C_{12}\times C_2\times C_2. \] Here the equality of fields follows from $\epsilon$ having squareclass $2$ in $K$. This field is CM and is ramified at the dyadic place: its relative quadratic conductor there is four. In particular it does not split dyadically. Allen's ordinary residually dihedral theorem, including its ordinary residual-witness construction, applies to the actual source. This proves the second assertion. The program \texttt{d3\_quads13\_all.gp} computes these fifteen ray groups, certifying each class group with \texttt{bnfcertify}. The exact calculation is supplied in~\cite[\path{ramified/d3}]{code}. \end{proof} \subsection{The scalar envelope} \begin{lemma}\label{r:t:d3:scalar-envelope} Put $K=\mathbf Q(\sqrt3)$, $\epsilon=2+\sqrt3$, $\delta_1=\epsilon(4+\sqrt3)$ and $\delta_2=\epsilon(4-\sqrt3)$, and let \[ F_0=K(i,\sqrt{\delta_1},\sqrt{\delta_2}). \] This is a Galois extension of $K$ of degree eight. Its absolute discriminant is $2^{24}3^8 13^8$, and its class group is $C_2$. Let $F_1$ be its Hilbert class field. Then $F_1/K$ is Galois of degree sixteen. Write $\Gamma=\operatorname{Gal}(F_1/K)$. Its group has generators $a,b,c,t$ with \[ \begin{gathered} a^2=b^2=c^2=t^2=1,\qquad t\text{ central},\\ [a,b]=[a,c]=t,\qquad [b,c]=1. \end{gathered} \] Here $b$ and $c$ can be inertia generators at the two thirteen-primes, and the corresponding decomposition groups are $\langle b,t\rangle$ and $\langle c,t\rangle$. Thus the group is $D_8\times C_2$. Let $\mathfrak P_1,\mathfrak P_2$ be the dyadic primes of $F_0$. The ray group for the dyadic modulus \[ \mathfrak P_1^6\mathfrak P_2^6 \] is $C_4$. The class of every prime above thirteen is $3$ in a common cyclic generator. Its quotient imposing Frobenius order at most two at all these primes is precisely the Hilbert quotient $C_2$. \end{lemma} \begin{proof} We have $\delta_1\delta_2=13\epsilon^2$, and both $\delta_j$ are totally positive. Thus $F_0=K(i,\sqrt{\delta_1},\sqrt{13})$. The independent real sign and thirteen-adic valuation classes give $[F_0:K]=8$. Its dyadic completions have absolute $(e,f)=(4,2)$; its eight thirteen-primes have $(e,f)=(2,1)$. The discriminant and these local degrees give absolute dyadic different exponent $6$. Subtracting twice the different exponent $2$ of $K_2/\mathbf Q_2$ gives relative different exponent $2$; the local ramification index is two, so its unique positive upper break is $1$. Hence the local inverse Herbrand function has $\psi(3)=1+2(3-1)=5$. The exact arithmetic computation starts from the displayed radical description of $F_0$. It verifies degree sixteen and absolute discriminant $2^{24}3^8 13^8$, and certifies its class group $C_2$ by \texttt{bnfcertify}=1. Computing the ray group at the stated dyadic modulus gives $C_4$, with all eight thirteen-prime classes equal to $3$. Adding twice those classes to the relation lattice gives $C_2$, necessarily the Hilbert quotient. The program \texttt{verify\_d3\_classfield\_source.gp} constructs the Hilbert quadratic extension from the certified class field of $F_0$. Its absolute degree is $32$ and its discriminant is $2^{48}3^{16}13^{16}$. Exact Galois computations, restricting through an embedding of $X^2-3$, give an order-sixteen group with derived subgroup $\langle t\rangle$. The thirteen-adic inertia generators $b,c$ commute, and an involution $a$ outside $\langle b,c,t\rangle$ has both commutators equal to $t$. The decomposition groups are $\langle b,t\rangle$ and $\langle c,t\rangle$. These computations use no class group of $F_1$. The exact calculation is supplied in~\cite[\path{ramified/d3}]{code}. The group relation follows; putting $z=bc$ gives a central involution independent of $t$, with $\langle a,b\rangle\simeq D_8$ and $\Gamma=\langle a,b\rangle\times\langle z\rangle$. \end{proof} \begin{lemma}\label{r:t:d3:scalar-group-algebra} Let $\Gamma$ be the order-sixteen group of Lemma~\ref{r:t:d3:scalar-envelope}. In its binary group algebra the quotient by the two-sided ideal generated by \[ (1+t)(1+b),\qquad (1+t)(1+c) \] has dimension ten over $\mathbf F_2$. \end{lemma} \begin{proof} Put $z=bc$. The center is $\langle t,z\rangle\simeq C_2^2$, and $\mathbf F_2[\Gamma]$ is free of rank four over its central group algebra, with basis $1,a,b,ab$. Write $D=1+t$ and $Z=1+z$. Both are central and square to zero. The ideal in the statement is generated equivalently by $D(1+b)$ and $DZ$, since $c=bz$. The central ideal generated by $D$ has dimension eight. Quotienting it by $DZ$ first leaves dimension four, with basis $D,Da,Db,Dab$. The remaining relation and its product with $a$ identify $Db=D$ and $Dab=Da$, leaving dimension two. The quotient modulo $D$ has dimension eight. The required total dimension is therefore $8+2=10$. The multiplication-table computation in \texttt{verify\_d3\_group\_algebra\_independent.py} gives the same quotient directly. The exact calculation is supplied in~\cite[\path{ramified}]{code}. \end{proof} \subsection{The characteristic-five inner-twist endpoint} \begin{lemma}[Finite-field projective descent]\label{r:t:d3:finitefield-inner-descent} Let $k=\mathbf F_{25}$, let $\phi:x\mapsto x^5$, and let $\bar r:G\to\mathrm{GL}_2(k)$ be absolutely irreducible. Suppose $\eta:G\to\{\pm1\}$ is a quadratic character and \[ \bar r^\phi\simeq\bar r\otimes\eta. \] Then the projectivization of $\bar r$ is conjugate to a representation with values in $\mathrm{PGL}_2(\mathbf F_5)$. If $G=G_F$ for a totally real field, $\bar r$ is totally odd and $\eta$ is totally odd, the determinant squareclass of this descended projective representation is totally odd. \end{lemma} \begin{proof} Choose $A\in\mathrm{GL}_2(k)$ such that $\bar r^\phi=A\bar rA^{-1}\eta$. Applying $\phi$ twice shows that $A^\phi A$ commutes with the linear representation $\bar r$. Absolute irreducibility gives $A^\phi A=cI$. Applying $\phi$ and conjugating by $A$ gives $c^\phi=c$, so $c\in\mathbf F_5^\times$. The norm $k^\times\to\mathbf F_5^\times$ is surjective. Rescale $A$ so that $A^\phi A=I$. Matrix Hilbert 90 supplies $B\in\mathrm{GL}_2(k)$ with $A=B^\phi B^{-1}$. Hence \[ \bar r'=B^{-1}\bar rB,\qquad (\bar r')^\phi=\eta\bar r'. \] Every matrix class $[\bar r'(g)]$ is therefore fixed by $\phi$ and belongs to $\mathrm{PGL}_2(\mathbf F_5)$. More explicitly, when $\eta(g)=-1$, multiply $\bar r'(g)$ by a nonzero $u\in k$ satisfying $u^\phi=-u$; the resulting matrix has entries in $\mathbf F_5$. This proves descent without any hypothesis on the projective centralizer. For a real conjugation $c_v$, put $C=\bar r'(c_v)$. Total oddness and $c_v^2=1$ give $C^2=I$ and $\det C=-1$. If $\eta(c_v)=-1$, then $C^\phi=-C$. The preceding $\mathbf F_5$-representative $uC$ has determinant $-u^2$. Here $u^2\in\mathbf F_5^\times$ is a nonsquare: otherwise $u^2=a^2$ with $a\in\mathbf F_5$, forcing $u=\pm a\in\mathbf F_5$. Since $-1$ is a square in $\mathbf F_5$, $\det(uC)$ is a nonsquare. This holds at every real place. \end{proof} \begin{proposition}[The quadratic inner-twist source at five] \label{r:t:d3:quadratic-inner-twist-source} Put $K=\mathbf Q(\sqrt3)$ and $h=\chi_{-1}$, the character of $K(i)/K$. Let $\mathcal R$ be a pure, totally odd, absolutely irreducible, parallel-weight-two strongly compatible system over $K$, with cyclotomic determinant. Assume it is good outside thirteen and has quadratic-special type, with nonzero monodromy, at the two thirteen-primes. Let $D$ be its trace field. Suppose either $D=\mathbf Q$, or $D$ is quadratic and its nontrivial automorphism $\tau$ satisfies \[ \mathcal R^\tau\simeq\mathcal R\otimes h. \tag{\(*\)} \] Then $\mathcal R$ is modular. \end{proposition} \begin{proof} Choose a coefficient place above five and write $r$ for the actual five-adic source and $\bar r$ for its semisimple residue. The source is crystalline of labelled weights $\{0,1\}$ at five. Its residual trace field is contained in $\mathbf F_5$ or $\mathbf F_{25}$. If $\bar r$ is absolutely irreducible, finite-field Brauer descent realizes it over that field. If it is reducible, its two characters are individually defined over that field: coefficient Frobenius cannot interchange them because their values at a real conjugation are the distinct elements $1,-1$. If $\bar r$ is globally reducible, the square of a constituent is unramified away from five: good places give no inertia and each special thirteen-place contributes at most a common quadratic scalar and unipotent inertia. Lemma~\ref{r:t:shared:ordinary-inert} therefore proves actual-source modularity. Now suppose $\bar r$ is absolutely irreducible. If $D=\mathbf Q$, or five splits or ramifies in the quadratic field $D$, its residual field is $\mathbf F_5$. The small-field modularity theorem \cite[Theorem~2.5]{h:AKT} supplies a modular witness. In the remaining case five is inert in $D$ and $\tau$ induces the fifth-power automorphism of $\mathbf F_{25}$. Reducing $(*)$ gives \[ \bar r^\phi\simeq\bar r\otimes h. \] Lemma~\ref{r:t:d3:finitefield-inner-descent} descends its projectivization to $\mathrm{PGL}_2(\mathbf F_5)$, with totally odd determinant squareclass. Moreover $[K(\zeta_5):K]=4$, since the unique quadratic subfield of $\mathbf Q(\zeta_5)$ is $\mathbf Q(\sqrt5)$. The projective modularity theorem \cite[Theorem~1.2(3)]{h:AKT} applies. Its Lemma~2.1(2) returns a modular witness for the original linear representation $\bar r$, including its scalar character. Lemma~\ref{r:t:shared:inert-cyclotomic} proves the required absolute irreducibility over $K(\zeta_5)$. The actual weight-two automorphy lifting theorem \cite[Theorem~2.3]{h:AKT} now applies to $r$ (or to its dual to match the cohomological normalization): it is finitely ramified, de Rham of weights $\{0,1\}$ and has a modular, cyclotomically irreducible residue. This proves modularity of $\mathcal R$. \end{proof} \subsection{All scalar dyadic residues: a ten-dimensional group algebra} Put $K=\mathbf Q(\sqrt3)$ and let $h$ be the character of $K(i)/K$. It is unramified at every finite place, nontrivial unramified at two, and trivial at both thirteen-places. Indeed the absolute discriminants of $K(i)$ and $K$ are $144$ and $12$, and $-1$ is a square in $\mathbf Q_{13}$. A special condition includes its exact quadratic scalar character. \begin{proposition}\label{r:t:d3:scalar-source} Let $\mathcal R$ be a pure, totally odd, absolutely irreducible, strongly compatible rank-two system over $K$, with cyclotomic determinant and labelled weights $\{0,1\}$. Suppose it is good outside thirteen and special at both thirteen-places. If every dyadic coefficient member has scalar residual semisimplification, then the actual system is modular. No dyadic ordinarity assumption is needed. \end{proposition} \begin{proof} We first give a bound for a finite packet of such systems whose exact quadratic special characters agree at each thirteen-place. Apply Lemmas~\ref{r:shared:trace-order} and \ref{r:regular-bt} to the whole simultaneous coefficient tuple. Write $R_0$ for its trace order, $E_0$ for its actual matrix order, and $s$ for the sum of the degrees of the distinct coefficient orbits. Then \[ \operatorname{rank}_{\mathbf Z_2}R_0=s,\qquad \operatorname{rank}_{\mathbf Z_2}E_0=4s. \] Put $E=E_0/2E_0$. Its joint Galois image $G$ is a finite two-group and spans $E$, since every coefficient residue has scalar semisimplification. Let $L/K$ be its point field. The left regular lattice is actual Barsotti--Tate. The general Fontaine bound on its rational-two torsion gives upper break at most three over $K_2$ for the whole field $L$; no coefficient-uniformizer improvement is used. At a thirteen-place the integral special matrices have the form \[ \xi(\tau)\begin{pmatrix}1&v\\0&1\end{pmatrix},\qquad \xi(F)\begin{pmatrix}13&w\\0&1\end{pmatrix}, \quad \xi(\tau),\xi(F)\in\{\pm1\}. \] Normalize by the common rational signs. The inertia difference is in the actual order and squares to zero. The normalized Frobenius has polynomial $X^2-14X+13$ and hence squares to one modulo two. It commutes with inertia there. Its whole local image is therefore elementary abelian of exponent at most two. Moreover \begin{equation}\label{r:t:d3:exact-special-zero} (F-1)(\tau-1)=(\tau-1)(F-1)=0\quad\text{in }E. \end{equation} The integral normalized products are zero in one order and twelve times the inertia difference in the other. Exact common signs justify these normalizations in the actual joint order. Put $\epsilon=2+\sqrt3$, $\delta_1=\epsilon(4+\sqrt3)$, $\delta_2=\epsilon(4-\sqrt3)$, and $F_0=K(i,\sqrt{\delta_1},\sqrt{\delta_2})$. Lemma~\ref{r:t:d3:scalar-envelope} gives $[F_0:K]=8$, dyadic ramified break one, and ray group $C_4$ for all dyadic primes to exponent six. Its quotient killing twice every thirteen-prime class is $C_2$. The group $H=\operatorname{Gal}(LF_0/F_0)$ is a finite two-group. The compositum has upper dyadic break at most three over $K_2$, since $F_0/K$ has break one. Herbrand's formula $\psi_{F_0/K}(3)=1+2(3-1)=5$ bounds the conductors of characters of $H^{\rm ab}$ by six. At thirteen the local field of $L$ lies in the unique maximal elementary biquadratic extension of $\mathbf Q_{13}$. The field $F_0$ already has ramification index two, so $LF_0/F_0$ is unramified there and its Frobenii square to one. There is no other finite ramification. Thus $H^{\rm ab}$ is a quotient of the displayed $C_2$. By the finite-two-group Frattini theorem $H$ itself is cyclic of order at most two. The unique possible quadratic ray extension is the Hilbert class field $F_1/F_0$, since $\operatorname{Cl}(F_0)=C_2$. Consequently $L\subset F_1$. This does not assume $F_0\subset L$ or that $F_0$ contains all quadratic characters of $G$. Lemma~\ref{r:t:d3:scalar-envelope} identifies $\Gamma=\operatorname{Gal}(F_1/K)$, of order sixteen. Its thirteen-inertias are generated by $b,c$, and the corresponding decomposition groups are $\langle b,t\rangle,\langle c,t\rangle$ for one central involution $t$. Its Frobenius classes are $t$ modulo inertia. Equation~\eqref{r:t:d3:exact-special-zero} makes $E$ a quotient of \[ \mathbf F_2[\Gamma]\big/ \big((1+t)(1+b),(1+t)(1+c)\big). \] By Lemma~\ref{r:t:d3:scalar-group-algebra} this algebra has dimension ten. We have proved the packet bound \begin{equation}\label{r:t:d3:scalar-packet-bound}4s\le10.\end{equation} It bounds the actual matrix algebra and assumes no faithfulness of the scalar action $R_0/2\longrightarrow E_0/2$. Let $D$ be the global trace field. Apply the bound also to $\{\mathcal R,\mathcal R\otimes h\}$. The twist preserves every exact thirteen-character and is unramified at two. An actual $h$-selftwist would give induction from $K(i)$, contrary to nonzero monodromy at the split thirteen-places. Corollary~\ref{struct:two-orbit} therefore gives a rational trace field, or a quadratic trace field $D$ with nontrivial automorphism $\iota$ satisfying \begin{equation}\label{r:t:d3:quadratic-inner-twist} \mathcal R^\iota\simeq\mathcal R\otimes h. \end{equation} The hypotheses of Proposition~\ref{r:t:d3:quadratic-inner-twist-source} are now satisfied: either $D=\mathbf Q$, or $D$ is quadratic with the displayed inner twist. That proposition proves actual-source modularity in characteristic five and hence modularity of the system. \end{proof} \begin{theorem}\label{r:t:d3:two-special-source} Assume GRH. Every pure, totally odd, absolutely irreducible, cyclotomic-determinant weight-two strongly compatible system over $K$, good outside thirteen and quadratic-special at both thirteen-primes, is modular. The exact quadratic scalar signs are arbitrary. \end{theorem} \begin{proof} If a dyadic residue is nonsolvable, Proposition~\ref{r:t:shared:nonsolvable-source} applies. If it is irreducible and solvable, Proposition~\ref{r:t:d3:solvable-source} applies. Otherwise every dyadic coefficient residue is reducible. Its characters have odd order and are unramified at thirteen because inertia has order at most two. At two wild inertia is pro-two and the residue unit group is $\mathbf F_2^\times=1$, so those characters are unramified there too. They are unramified everywhere else and have trivial real-place action. Class number one makes them trivial. Thus every residue has scalar semisimplification, and Proposition~\ref{r:t:d3:scalar-source} proves modularity. \end{proof} \subsection{The supported residual base at thirteen} \begin{lemma}\label{r:t:d3:cuspidal13-menu} Every local residual representation of $G_{\mathbf Q_{13}}$ with cyclotomic determinant admits a weight-two lift of exact cyclotomic determinant whose type is quadratic special or determinant-one tame cuspidal with inertial order seven or fourteen. \end{lemma} \begin{proof} Write $W(r,b)=\operatorname{Sym}^r\mathbf F_{13}^2\otimes\det^b$. There are fourteen central-trivial weights: $r+2b=0$ modulo twelve, with $r\in\{0,2,4,6,8,10,12\}$ and $b$ modulo twelve. For the cuspidal type indexed by $t=1,\ldots,6$, with niveau-two character exponent $12t$, the compact-type reduction has constituents \[ W(2t-2,1-t)\quad\hbox{and}\quad W(12-2t,t). \] These twelve weights are distinct and exhaust those with $r<12$. The two remaining Steinberg weights occur in the scalar semistable compact types. The corresponding crystalline scalar compact types also occur in the cuspidal menu, so if a semistable component has only crystalline points one may choose a cuspidal lift instead. The full fixed-determinant formula \cite[Theorem~1.2]{g:Tung}, with its nonnegative multiplicities therefore gives the required nonempty local lifting ring for the actual residual extension. The inertial orders are fourteen for odd $t$ and seven for even $t$. No global quadratic correction of these signs is required. \end{proof} \begin{lemma}\label{r:t:d3:seven-residue} Let $\mathcal R$ be a determinant-cyclotomic weight-two strongly compatible system over $K=\mathbf Q(\sqrt3)$, good outside thirteen, with the preceding cuspidal or special menu at thirteen. If a seven-adic residue is globally reducible, the actual seven-adic source is modular. If it is irreducible, its cyclotomic restriction remains absolutely irreducible. \end{lemma} \begin{proof} Suppose first that the constituents are $\psi$ and $\omega_7\psi^{-1}$. At a thirteen-place both finite inertial eigencharacters reduce to the same quadratic sign; any surviving seventh-order inertia is unipotent. Consequently $\psi^2$ is unramified at thirteen. It is unramified at every other finite place away from seven as well. Lemma~\ref{r:t:shared:ordinary-inert} with $(K,p)=(F_3,7)$ proves actual ordinarity and modularity, without removing the scalar signs by a global twist. If the residue is irreducible, Lemma~\ref{r:t:shared:inert-cyclotomic} gives cyclotomic irreducibility. \end{proof} \begin{lemma}\label{r:t:d3:seven-special-switch} At either thirteen-place, the actual reduction in characteristic seven of any member of the cuspidal/special menu admits a genuine quadratic-special lift of cyclotomic determinant. This includes the case of trivial residual inertia. \end{lemma} \begin{proof} Retain the exact quadratic scalar inertial character and first remove it locally. Cuspidal Frobenius has trace zero and determinant thirteen, so its residual eigenvalues are $1,-1$, which are distinct modulo seven. Special Frobenius has the same pair, up to an unramified sign. The relation with inertia, since $13=-1$ modulo seven, places any nontrivial unipotent inertia in one offdiagonal orientation in this Frobenius eigenbasis. Choose $u=\pm1$ for that orientation and take \[ T(x)=1+t xE_{12},\qquad \Phi=\operatorname{diag}(13u,u), \] interchanging the basis vectors if needed. Choose $t\ne0$ with the required residual value; if inertia was trivial choose $t$ divisible by the coefficient uniformizer. These matrices have determinant one on inertia and thirteen on Frobenius and satisfy $\Phi T(x)\Phi^{-1}=T(13x)$. Restoring the quadratic character gives the asserted actual special lift, with nonzero monodromy. \end{proof} \begin{proposition}\label{r:t:d3:supported13-reduction} Assume GRH. Every irreducible cyclotomic-determinant residual representation unramified outside thirteen is modular. \end{proposition} \begin{proof} Solvable residual images have modular Artin witnesses. Otherwise the cyclotomic residual restriction in characteristic thirteen is irreducible. Use Lemma~\ref{r:t:d3:cuspidal13-menu} and prescribed global lifting to construct a weight-two source, then establish its actual avoiding potential modularity and compatible system. In characteristic seven, a reducible residue gives modularity by Lemma~\ref{r:t:d3:seven-residue}. For an irreducible residue that lemma verifies the cyclotomic hypothesis. Prescribe the genuine special lifts of Lemma~\ref{r:t:d3:seven-special-switch} at both thirteen-primes and retain the old Barsotti--Tate component above seven. Global lift existence in characteristic seven and actual potential modularity give the replacement system. Purity makes its special monodromy nonzero. Theorem~\ref{r:t:d3:two-special-source} modularizes this actual target. Allen--Khare--Thorne then applies backwards to the old actual weight-two seven-adic source. Compatibility and reduction at thirteen return the required original residual witness. \end{proof} \subsection{\texorpdfstring{The bounded source conditions for $d=6,11$}{The bounded source conditions for d=6,11}} For $K=F_6$ or $F_{11}$, a bounded system is pure, totally odd, absolutely irreducible, of cyclotomic determinant and labelled weights $\{0,1\}$, good outside five. Each five-adic type is quadratic-special, denoted $\SP$, or tame determinant-one cuspidal, denoted $\SC$, with inertial eigencharacters of order three or six. A special condition retains its exact quadratic character, including the unramified sign, and has nonzero monodromy. Its characters are $\alpha\chi_p,\alpha$ with $\alpha$ quadratic. At a coefficient-five place its monodromy-stable line gives the special filtration by weak admissibility. \section{\texorpdfstring{Trace relations over $\Q(\sqrt6)$}{Trace relations over Q(sqrt(6))}}\label{r:s:sec:d6} Put $K=\Q(s)$, $s^2=6$. \begin{theorem}\label{r:s:d6:thm:three} For a bounded source with types $\SC/\SC$ or $\SP/\SP$, a solvable residual image at a coefficient prime above three implies modularity. \end{theorem} \begin{theorem}\label{r:s:d6:thm:two} Assume GRH. A bounded source with types $\SP/\SP$ and nonsolvable residual image at a coefficient prime above two is modular. GRH is unnecessary if that dyadic member is nonordinary. \end{theorem} \subsection{Characters and cubic classes} Write \[ \epsilon=5+2s,\qquad a=1+s,\qquad b=1-s,\qquad \pi=3+s, \qquad\gamma=3+2s,\qquad H=K(\zeta_3). \] Then $N(\epsilon)=1$, $ab=-5$, $N(\pi)=3$ and $N(\gamma)=-15$. Let $\sigma$ be the nontrivial automorphism of $K/\Q$ and put $\omega=\bar\chi_3$. We identify quadratic characters with squareclasses when convenient. \begin{lemma}\label{r:s:d6:lem:characters} The class number of $K$ is one. The quadratic classes supported at three and five and unramified at two form the group generated by \[ [-3]=[-\epsilon],\quad [5],\quad[\gamma]. \] For a reducible $\bar r_3$ with determinant $\omega$ arising from a bounded source, its constituents are quadratic. Up to interchange and twisting by the character of $K(\sqrt5)/K$, their pair is either $\{1,\omega\}$ or $\{\psi_\gamma,\psi_\gamma\omega\}$. \end{lemma} \begin{proof} The Minkowski bound is $\sqrt{24}/2<2.45$, and the prime of norm two is generated by $2+s$, whose norm is $-2$. The unit $\epsilon$ is fundamental, by the least positive solution $(5,2)$ of $x^2-6y^2=1$; norm $-1$ is impossible modulo three. The squareclasses supported at three and five have basis $[-1],[\epsilon],[\pi],[a],[b]$. All five generators are dyadic units. Such a unit $u$ gives an unramified quadratic extension, possibly split, precisely when $u$ is a square modulo $4\OO_{K_2}$. A square of a unit $A+Bs$ has coordinates $(1,0)$ or $(3,2)$ modulo four, according as $B$ is even or odd. Enumerating the thirty-two products gives exactly eight allowable classes, generated as stated. The identity $3/\epsilon=(3-s)^2$ identifies $[-\epsilon]=[-3]$. Let a residual constituent be $\psi$. At five, its square is unramified because semisimplified residual inertia of the bounded types is scalar quadratic. At three, a finite characteristic-three character kills principal units and the residue field is $\F_3$, so its square is unramified there too. It is unramified elsewhere. The narrow class group has order two; its nontrivial character is $\omega$, which is odd at both real places. A square character is even, so $\psi^2=1$. The preceding eight classes give the two pairs. Finally, $K_3=\Q_3(\zeta_3)$ because $-2$ is a square in $\Q_3$. The first pair is locally unramified, while both characters of the second pair are ramified quadratic. \end{proof} Let $F=K_3$, normalize $v_F(s)=1$, and put $U_i=1+\mathfrak p_F^i$. \begin{lemma}\label{r:s:d6:lem:cubes} The global $S$-unit cubeclass space for $S=\{v:v\mid15\}$ has basis $[\epsilon],[\pi],[a],[b]$ and maps isomorphically to $F^\times/F^{\times3}$. The preimage of $U_2/U_4$ is \[ V=\langle q,d\rangle,\qquad q=[5],\qquad d=[5]+[\epsilon b]. \] We have $\sigma(q)=q$, $\sigma(d)=-d$. In this basis the two valuations at five are $(1,1)$ and $(1,-1)$, in some order. \end{lemma} \begin{proof} Since $\mu_3\subset F$, the class space has dimension four. Moreover $U_1=\mu_3U_2$, and the logarithm identifies cubing $U_2\to U_4$ with multiplication by three. Hence a unit is a cube exactly when it is congruent to $1$ or $-1$ modulo $9\OO_F$. The $\pi$ exponent is detected by valuation. Of the twenty-seven products $\epsilon^i a^j b^k$, $0\le i,j,k<3$, only the trivial one is $\pm1$ modulo nine. This finite calculation takes place in $(\Z/9\Z)[s]/(s^2-6)$ and proves injectivity, hence the asserted isomorphism. Modulo three, $\epsilon=-(1+s)$, $a=1+s$, $b=1-s$ and $s^2=0$. Membership in $\{\pm1\}U_2$ is therefore the equation $i+j-k=0$. This plane is generated by $[5]=[ab]$ and $w=[\epsilon b]$. Since $N(\epsilon b)=-5$, we have $\sigma(w)=q-w$. The assertions for $d=q+w$ and the valuations follow. \end{proof} \begin{lemma}\label{r:s:d6:lem:diagonal} The three-part of the ray class group modulo $\mathfrak p_3^3$, with or without the real places, is $C_3$. The two Frobenius classes at five coincide and generate it. The involution $\sigma$ acts trivially on its three-part. \end{lemma} \begin{proof} The unit group of $\OO_K/\mathfrak p_3^3$ has order eighteen. The images of $-1$ and $\epsilon$ generate a subgroup of order six, giving $C_3$. Here $\mathfrak p_3^3=(9,3s)$. The residues of $a$ and $b$ belong to neither the image of the units nor its trivial class: the unit residues are $\pm1,\pm(5+2s),\pm(4+2s)$ in this quotient. Thus both ray classes are nontrivial. The identity $a-\epsilon^2b=72+30s\in(9,3s)$ makes their classes equal. The involution $\sigma$ exchanges the two primes and fixes their common generating class, hence acts trivially on the three-part. Infinite places add only two-primary data. \end{proof} \begin{lemma}\label{r:s:d6:lem:cyclo} No absolutely irreducible characteristic-three residue of a bounded source becomes reducible over $H$. \end{lemma} \begin{proof} The field $H=\Q(\sqrt{-2},\sqrt{-3})$ has discriminant $576$ and class number one. Indeed its Minkowski bound is $36/\pi^2<4$. There is no prime of norm two. If $A^2=-2$ and $u^2-u+1=0$, the element $1+Au$ has relative norm $-1+A$ over $\Q(A)$, hence absolute norm three. Its conjugate gives the other prime of norm three. A cyclotomic-reducible irreducible residue would be $\Ind_H^K\xi$ with $\xi^c=\xi^{-1}$. Its image has order prime to three. At five, its inertia is scalar quadratic, so $\xi^2$ is unramified. The two primes of $H$ over three have residue field $\F_3$, so $\xi^2$ is unramified there as well. It is unramified elsewhere, hence trivial by the class number. Then $\xi=\xi^c$, contradicting irreducibility of the induction. \end{proof} \subsection{Ordinary sources and an unramified quotient} We prove the ordinary part of Theorem~\ref{r:s:d6:thm:three} without a local-distinctness assumption at three. Such an assumption would fail here because $\omega|_{G_F}=1$. The argument gives a stronger result in which the two five-adic types need not agree. \begin{proposition}\label{r:s:d6:prop:ordinary} Suppose that the residual constituents are unramified at three. Then the actual system is modular. After the quadratic normalization of Lemma~\ref{r:s:d6:lem:characters}, it is invariant under $\sigma$. Consequently a genuinely mixed $\SC/\SP$ source cannot have this residual branch, and two special conditions in this branch have the same exact character after normalization. \end{proposition} \begin{proof} By Lemma~\ref{r:s:d6:lem:characters}, a twist by $\psi_5$ makes the residual pseudorepresentation $1+\omega$. This quadratic character is invariant under $\sigma$ and unramified at three. Lemma~\ref{t:21:local:ordinary3} makes the crystalline source at $F$ ordinary. Write its unramified ordinary quotient as $\nu$; the subcharacter is $\chi_3\nu^{-1}$. Put $A=\Z_3[[U]]$ and let $\psi_U$ be the unramified character of $G_F$ sending arithmetic Frobenius to $1+U$. Begin with the universal finite-flat-at-three fixed-determinant Cayley--Hamilton pair for $1+\omega$, unramified outside three and five, provided by \cite[Theorems~2.5.3 and~2.5.5]{r:s:WWE}, and extend its coefficients to $A$. Impose, for every $g,h\in G_F$, the algebra relations \begin{equation}\label{r:s:d6:eq:ordinary} (g-\chi_3(g)\psi_U(g)^{-1})(h-\psi_U(h))=0. \end{equation} No local type is imposed at five. To construct the Cayley--Hamilton quotient, take the closed two-sided ideal $I$ generated by these relations, then the scalar ideal generated by the nonconstant coefficients of $D(1-xt)$, $x\in I$, and use \cite[Lemma~2.4.5]{r:s:WWE}. Thus no coefficient-$A$ version of the printed representability theorem is assumed. Denote the scalar ring by $R$, its maximal ideal by $\mathfrak m$, and its GMA by $\left(\begin{smallmatrix}R&B\\C&R\end{smallmatrix}\right)$. The actual source factors through it with $U=\nu(\Phi_3)-1$, since upper triangular multiplication proves \eqref{r:s:d6:eq:ordinary}. The offdiagonal cotangent duals embed into the finite-flat global extension spaces by \cite[Theorem~4.3.5]{r:s:WWE}; scalar extension and further quotients retain these inclusions at $3=U=0$. Locally such an extension is unipotent. Fontaine's bound gives classical upper break at most two, hence cubic conductor at most three. A nonunit Kummer class instead has conductor four: after replacing the radicand by an inverse and a cube, $X^3-u$ is Eisenstein with $v_F(u)=1$, and its different exponent is eight. Thus both global extension spaces lie in \[ V_3=\langle[\epsilon],[a],[b]\rangle. \] Their restriction to $G_F$ is injective by Lemma~\ref{r:s:d6:lem:cubes}. For larger coefficient fields the same conclusion follows by extracting $\F_3$ coordinate extensions by pullback and quotient. Let $J$ be the reducibility ideal and put \[ Q=\mathfrak m/(3R+UR+\mathfrak m^2),\qquad Q_{\mathrm{red}}=\mathfrak m/(3R+UR+\mathfrak m^2+J). \] For $g,h\in G_F$ both residual diagonal characters equal one. The upper-left entry of \eqref{r:s:d6:eq:ordinary}, reduced in $Q$, is $b_gc_h=0$; its other term is a product of elements of $\mathfrak m$. Evaluations on all of $G_F$ span the dual of each allowed global extension space, by local injectivity. Hence every product in $(B/\mathfrak mB)\otimes(C/\mathfrak mC)$ has zero image in $Q$. These products generate the image of $J$, so $Q\simeq Q_{\mathrm{red}}$. A diagonal tangent relative to $A$ has characters $1+\eta f$ and $\omega(1-\eta f)$, with $\eta^2=0$. Finite-flatness and Fontaine give conductor at most $\mathfrak p_3^3$ for $f$. It cannot ramify at five: wild inertia is pro-five and the tame unit quotient has order four. It is unramified elsewhere by construction. Therefore the dual of $Q_{\mathrm{red}}$ is a subspace of $\operatorname{Hom}(\operatorname{Cl}_{\mathfrak p_3^3}(K),\F_3)$. The involution $\sigma$ acts trivially on this space by Lemma~\ref{r:s:d6:lem:diagonal}. It consequently acts trivially on $Q$; the diagonal tangent need not itself vanish. The involution $\sigma$ preserves the support, finite-flat condition, determinant and ordinary relations. It fixes $U$, because the unramified Frobenius quotient of $G_F$ is unchanged by the automorphism of $F/\Q_3$. Thus it acts on $R$ over $A$. Its action on $Q$ is the identity. Lemma~\ref{t:involution}, applied over $A$, makes the involution $\sigma$ trivial on $R$. Thus $r_3^\sigma\simeq r_3$. Lemma~\ref{r:quadratic-descent}, with $\eta=1$, now proves modularity of the original system. Finally, $\sigma$ exchanges the two places above five, and invariance identifies their local Weil--Deligne representations. Their monodromy cannot be zero at one place and nonzero at the other. For two special types it also preserves the exact quadratic character. This proves the asserted exclusions. \end{proof} \subsection{Nonordinary reducible sources} The other pair in Lemma~\ref{r:s:d6:lem:characters} requires a different finite-flat extension space. \begin{lemma}\label{r:s:d6:lem:ramflat} For either ramified quadratic character $\theta$ of $G_F$, the finite-flat selfextensions of $\theta$ form exactly the Kummer plane $U_2/U_4$. \end{lemma} \begin{proof} The rank-one model has Breuil--Kisin exponent one. An extension has Frobenius matrix \[ \begin{pmatrix}cu&h(u)\\0&cu\end{pmatrix}. \] The height-one condition imposes no further restriction. A change of upper lift changes $h$ by $cu(f(u^3)-f(u))$. Positive powers are removed recursively, while constants in $f$ disappear. The extension space has dimension at most two, represented by $1,u$. Schematic closure and flat quotient give the rank-one endpoints; the exact finite-flat description used here is \cite[Theorem~1.1.3, Proposition~1.1.11 and Lemmas~1.2.4--1.2.5]{r:s:Kisin}. Coordinate pullback and quotient give the same bound for arbitrary coefficients. The unramified selfextension is finite flat and gives $U_3/U_4$. A generator is $1+6s$: substituting $\alpha=1+sz$ in $\alpha^3=1+6s$ gives \[ z^3+(s/2)z^2+(1/2)z-1=0, \] whose reduction $z^3-z-1$ is irreducible and separable over $\F_3$. For an independent class take \[ E:y^2=x^3-\frac{s}{8}x^2+x. \] Its discriminant is $-125/2$, and its reduction is supersingular. Its division polynomial factors as \[ \psi_3(x)=(x-1/s)(3x^3+6x+s). \] At $x=1/s$, the ordinate square is $s(5/12)^2$, so the stable subgroup has character $\theta_s$. Cardano's radicand for the other three lines is $2/(3s)$, whose product with $s^3$ is four. Hence its actual extension line is $[2]$. It lies in $U_2\setminus U_3$ after multiplying by the cube $-1$. An unramified quadratic twist gives the other ramified character; specifically $\gamma/(2s)=1+s/4$ is a local square. These two lines exhaust $U_2/U_4$. \end{proof} \begin{proposition}\label{r:s:d6:prop:ramified} A pure bounded source with common ramified quadratic residual constituents at three is modular. \end{proposition} \begin{proof} Normalize its residual pair to $\psi_\gamma+\psi_\gamma\omega$. By Lemmas~\ref{r:s:d6:lem:cubes} and~\ref{r:s:d6:lem:ramflat}, the two global offdiagonal extension spaces are contained in $V=\langle q,d\rangle$, with five-adic evaluations $\ell_1=(1,1)$ and $\ell_2=(1,-1)$. Use the finite-flat Cayley--Hamilton construction of Proposition~\ref{r:s:d6:prop:ordinary}, now over $\Z_3$ without an ordinary parameter, and impose the exact five-adic conditions by the same closed-ideal construction. After removing a fixed scalar quadratic inertia character, a cuspidal condition kills wild and prime-to-three tame inertia and imposes $T^2+T+1=0$, $\Phi^2+5=0$. A special condition of exact character $\alpha_v$ imposes $(g-\chi_3(g)\alpha_v(g))(h-\alpha_v(h))=0$ for all local $g,h$. The diagonal cotangent vanishes by Fontaine and Lemma~\ref{r:s:d6:lem:diagonal}. Thus products of the two offdiagonal cotangents surject onto $\mathfrak m/(3R+\mathfrak m^2)$. Since $N(\gamma)=-15$, the involution $\sigma$ followed by twisting by $\psi_5$ interchanges the residual constituents. For $\SC/\SC$ it restores the ordered order-three and order-six types. Its action on the product cotangent is factor interchange composed with $\sigma(q)=q$, $\sigma(d)=-d$. At either five-place, after removing its scalar quadratic character, the exact relation $(T-1)^2=-3T$ kills $\ell_v\otimes\ell_v$ in the relative cotangent. Lemma~\ref{shared:tensor-involution}, with $s=q$, therefore makes the combined involution trivial on the cotangent. For two special conditions, choose the place with $s\equiv1\pmod5$ first. There $\psi_\gamma$ has character $RU$, while at the other place it is trivial; here $R$ is the character of $\Q_5(\sqrt5)$ and $U$ the unramified quadratic. Write the exact special characters as $RU^e,U^f$. The allowed first extension orientation is ramified at the first place exactly when $e=0$, and at the second exactly when $f=1$; the other orientation is complementary. If $e\ne f$, one orientation is forbidden at both places. Independence of $\ell_1,\ell_2$ then makes that entire global extension space zero. Nakayama makes the source reducible, a contradiction. Hence $e=f$. The product of the exact characters is $R$, and the involution $\sigma$ followed by $\psi_5$ preserves the conditions. It interchanges the two permitted extension lines and fixes their tensor product. Its action on the whole cotangent is again trivial. Lemma~\ref{t:involution} makes the combined involution trivial on the complete ring. We obtain $r_3^\sigma\simeq r_3\otimes\psi_5$. Since $\psi_5$ is restricted from $G_\Q$, Lemma~\ref{r:quadratic-descent} proves modularity. \end{proof} \begin{proof}[Proof of Theorem~\ref{r:s:d6:thm:three}] If the residue is reducible, Propositions~\ref{r:s:d6:prop:ordinary} and~\ref{r:s:d6:prop:ramified} apply. If it is irreducible, Lemma~\ref{r:s:d6:lem:cyclo} gives irreducibility on $G_H$. The Fong--Swan and solvable Artin argument supplies a regular cuspidal residual witness; Lemma~\ref{shared:weightzero} gives one in the weight-zero convention. Then \cite[Theorem~2.3]{h:AKT} applies to the actual weight-two three-adic source. Strong compatibility gives modularity of the system. \end{proof} \subsubsection{The mixed reducible case} \begin{proposition}\label{r:s:d6:prop:mixed-three} A mixed bounded source cannot have a globally reducible three-adic residue. \end{proposition} \begin{proof} The unramified constituent pair is excluded by Proposition~\ref{r:s:d6:prop:ordinary}, since the involution $\sigma$ exchanges the two monodromy operators. For the other pair normalize the special place to $\mathfrak q_-=(1-s)$ and the cuspidal place to $\mathfrak q_+=(1+s)$, with residual pair $\psi_\gamma+\psi_\gamma\omega$. The global finite-flat plane is $V=\langle q,d\rangle$, with evaluation functionals $\ell_+=(1,1),\ell_-=(1,-1)$. We need the relation $q^\vee\otimes q^\vee=0$ in the relative pseudocotangent. We prove it in its finite quotient generalized matrix algebra, retaining the coefficient-valued offdiagonal product. Here are the exactness and coefficient arguments for this relation. Every finite-flat group killed by three whose generic factors are the same ramified quadratic $\theta$ has a unique model with its generic identification. The rank-one Kisin exponent is one, and its unramified constant is determined by $\theta$. Closing a generic graph and a common generic $\theta$-line, then inducting on the finite-flat quotient, proves both full faithfulness and exactness of generic subquotients. This argument is confined to this filtered category; it does not assert full faithfulness on all groups at ramification index $p-1$. Let $R$ be the common finite-flat pseudoring, with generalized matrix algebra $(R\ B;C\ R)$. Put $R'=R/(3R+\mathfrak m^2)$, $B'=B/\mathfrak mB$, $C'=C/\mathfrak mC$, and let $E'$ be the resulting generalized matrix algebra. The products of $B',C'$ lie in the diagonal $\mathfrak m_{R'}$, and its radical $I$ satisfies $I^3=0$. The regular left Galois module $E'$ is finite flat by the stable condition of~\cite[Definition 2.5.2, Theorem 2.5.5]{r:s:WWE}. Every finite left $E'$-module, being a quotient of a sum of regular modules, has the same finite-flat property and local factors $\theta$. For such a module $V$ let $\mathfrak M(V)$ be the covariant Kisin module, obtained by applying the contravariant classification to the Cartier dual of its finite-flat model. The graph argument above extends every short exact sequence of these generic modules to a short exact sequence of finite-flat groups. Kim's exact classification~\cite[Corollary~4.3 and Proposition~5.7.3(2)]{r:t:Kim} therefore gives an exact sequence of the underlying Kisin modules. These are finite free $\F_3[[u]]$-modules, so the sequence splits as modules and remains exact modulo $u$. The presentation argument for an exact linear functor now gives $\mathfrak M(V)/u=P_0\otimes_{E'}V$ for a projective right $E'$-module $P_0$. Each of the two simple modules has a rank-one model, so each indecomposable projective occurs in $P_0$ once. Thus $P_0\simeq E'$. Lifting a right generator and using Nakayama and equality of ranks gives $\mathfrak M(E')\simeq E'[[u]]$ as right modules. Normalize the simple Frobenius maps to the same $cu$ and write the regular Frobenius as $cu+X(u)$, with $X(u)\in I[[u]]$. Its height-one inverse is \[ u^2(cu+X(u))^{-1} =u/c-X(u)/c^2+X(u)^2/(c^3u). \] Integrality forces $X(0)^2=0$. The diagonal radical is square zero and annihilates the offdiagonal modules, so this is exactly the vanishing of the product of the two constant offdiagonal coordinates in $R'$. On a rank-two extension those coordinates are $h(0)$ in Lemma~\ref{r:s:d6:lem:ramflat}. The constant functional vanishes on the unramified line $d$, represented by $5\epsilon(1-s)\equiv1+3s\pmod9$, and is nonzero on $q$. This identifies the asserted relation with $q^\vee\otimes q^\vee$ in the global product cotangent. In the mixed problem the special relation kills one extension orientation at $\mathfrak q_-$, while the cuspidal relation kills $\ell_+\otimes\ell_+$. Together with the displayed product relation these kill the product cotangent. The diagonal quotient has the one $C_3$ ray direction of Lemma~\ref{r:s:d6:lem:diagonal}, killed by the exact cuspidal Frobenius trace zero. Thus the relative pseudocotangent is zero, and every characteristic-zero point factors through $\Z_3$. Compare such a point with the CM curve \[ E:y^2=x^3+\gamma^3(7+3s)^2/4. \] It is the $\gamma$-twist of $y^2+(7+3s)y=x^3$, and has exactly the same residual pair. Its minimal conductor is $\mathfrak q_+^2$: it is good at two, three and $\mathfrak q_-$, and has type II$^*$ at $\mathfrak q_+$. Its division points with $x=0$ give the character $\psi_\gamma$; its trace at $\mathfrak q_-$ is zero, also forced by its CM and the inertness of five in $\Q(\sqrt{-3})$. These assertions are checked by the exact Weierstrass and minimal conductor calculation supplied with the proof. The unknown source has rational three-adic traces by the vanishing cotangent. Its trace algebra is split: nonzero special monodromy provides a nonzero nilpotent, impossible in a division algebra. Realize it over $\Q_3$. For it and $E$ choose the idempotents of a Frobenius at $\mathfrak q_-$; its two eigenvalues are distinct modulo three. Their generalized matrix orders have reducibility ideal $(3)$. Indeed the normalized cuspidal inertia has diagonal entries $1+x,1+y$ with $x,y\in3\Z_3$, trace $-1$, determinant one, and offdiagonal product $xy-3$, of valuation one. Choose an Eichler lattice whose primitive upper residual extension is the orientation forbidden at the special place. For both sources that class spans $\ker\ell_-$. Scale it and fix the two idempotents so the two residual upper cochains agree exactly, say $\beta$. Both $\beta(q^\vee)$ and $\beta(\ell_+)$ are nonzero. Let $f,g$ be the two pseudopoints modulo nine of the common finite-flat problem with the cuspidal condition at $\mathfrak q_+$ and no imposed condition at $\mathfrak q_-$. Their difference $\delta=(f-g)/3$ is a derivation on the relative pseudocotangent. Writing $c_f,c_g$ for their divided lower residual functionals gives $\delta(BC)=\beta(B)(c_f-c_g)(C)$. The finite-flat $q^\vee$ product relation and the equality of the two cuspidal inertia traces give \[ (c_f-c_g)(q^\vee)=(c_f-c_g)(\ell_+)=0 . \] These functionals are independent on $V$, so the derivation kills every offdiagonal product. Its diagonal quotient is zero as above. The two trace functions are equal modulo nine. At $\mathfrak q_-$ this asserts $\pm6=0$ modulo nine, a contradiction. \end{proof} \subsection{Nonsolvable characteristic-two sources} We give the complete ramification and explicit-formula argument for Theorem~\ref{r:s:d6:thm:two}. The quadratic extension $H/K$ is unramified at all finite places, unramified quadratic at two and at both places over five. \begin{lemma}\label{r:s:d6:lem:dyadic} Let a finite-flat representation $G_{K_2}\to\SL_2(\F_q)$ have point field $L$. Its normalized dyadic different contribution is at most $2-2/q$. \end{lemma} \begin{proof} Fontaine's classical upper-break bound is $2(1+1/(2-1))-1=3$. If wild inertia is nontrivial, its unique invariant line is preserved by the decomposition group. The diagonal characters are characters of $G_{K_2}$; their odd-order restrictions to units factor through $\F_2^\times$ and are trivial. Thus inertia is elementary abelian of order at most $q$. The upper-filtration formula, divided by the absolute ramification index two of the base, gives \[ d_2\le\tfrac12\{1-1/q+3(1-1/q)\}=2-2/q. \] If wild inertia is trivial, a nontrivial irreducible tame inertia has order three, since Frobenius squares and exchanges its determinant-one eigencharacters. Its contribution is $1/3$, which is smaller. \end{proof} At a special five-place the full residual decomposition image is contained in $C_2^2$. Adjoin $H=K(\zeta_3)$, which is unramified at every finite place and unramified quadratic at two and both five-places. For $M/K$ with nonsolvable image $\SL_2(\F_q)$, simplicity gives $M\cap H=K$, and \[ [MH:\Q]=4q(q^2-1),\quad \rd(MH)\le\sqrt{24}\,2^{2-2/q}\sqrt5,\quad 2N_{25}/[MH:\Q]\ge\tfrac12 . \] If the actual dyadic member is nonordinary, Theorem~\ref{r:break} improves the upper bound to $\rd(MH)<2\sqrt{120}<21.910$. For $q\ge8$, \eqref{shared:eq:unconditional} with $b=10$, $n\ge2016$ and just the first norm-$25$ term gives $\rd(MH)>23.29$, a contradiction. A totally real kernel field of degree at least $120$ has $\rd(M)>41.69$ by the same formula with $b=7$, contradicting the same upper bound. These two comparisons require no GRH. For an arbitrary dyadic member, assume GRH and apply the explicit formula of Section~\ref{shared:explicit}, with $B_{b,100}$, retaining four nonnegative prime-power terms at norm $25$. For the totally real comparison retain $A_{b,4}$ and no prime terms. Outward interval evaluation at 256 bits gives \[ \begin{array}{lrrrr} \text{case}&n\ge&b&\text{lower }\rd&\text{upper }\rd\\\hline q=8&2016&7.36&41.44651216&36.84623482\\ q\ge16&16320&9.4&49.15326236&43.81780461\\ \text{totally real kernel}&120&4.8&56.06306031&43.81780461 \end{array} \] Each lower entry is strict and each upper entry is a strict outward upper bound. At fixed $b$, the right side of the explicit formula increases with $n$, since $C_b>0$. The common dyadic factor four bounds the unbounded ranges. These comparisons exclude every $q\ge8$. The exact calculation is supplied in~\cite[\path{ramified/d6_d10_d11/shared}]{code}. For $q=4$, \cite[Theorem~2.5]{h:AKT} supplies a regular cuspidal residual witness. If all real conjugations acted trivially, the kernel field would be totally real of degree at least $120$, contradicting the preceding unconditional comparison for a nonordinary member, or the last table row under GRH. Thus some residual real conjugation is nontrivial. All hypotheses of \cite[Theorem~6.1]{g:Thorne2} now hold for the actual crystalline two-adic source, including nonsolvability and this real-place condition. Its modularity proves Theorem~\ref{r:s:d6:thm:two}. \subsection{Two special conditions at two} \begin{proposition}\label{r:s:d6:prop:sptwo} For a source of types $\SP/\SP$, the semisimplified residue $\bar r_2$ is either nonsolvable or $1\oplus1$. In particular an absolutely irreducible $\bar r_2$ is nonsolvable and, under GRH, the source is modular. \end{proposition} \begin{proof} An absolutely irreducible solvable subgroup of $\GL_2(\overline{\F}_2)$ with determinant one is odd dihedral, by Dickson's classification. Indeed a normal unipotent two-subgroup would give a nonzero invariant subspace fixed by the whole group, while the irreducible subfield groups $\SL_2(\F_{2^r})$, $r\ge2$, are nonsolvable. Thus the residue is $\Ind_{G_L}^{G_K}\xi$ for a quadratic $L/K$ and a nontrivial odd-order character $\xi$. Special inertia at five reduces to the identity or a group of order two: its scalar quadratic factor becomes trivial in characteristic two. Its intersection with the odd rotation subgroup is trivial. Consequently $\xi$ is unramified at all primes over five, even when $L/K$ is ramified there. It is unramified elsewhere away from two. At two it is tame, so its conductor divides the radical of $(2)$ in $L$, hence divides $(2)$. Fontaine's upper-break bound three gives dyadic conductor at most four for $L/K$. An odd-valuation quadratic Kummer class has conductor $2e+1=5$, since $e(K_2/\Q_2)=2$. Therefore the possible nontrivial radicands are precisely the fifteen classes in $\langle-1,\epsilon,a,b\rangle$. In mask order for this displayed basis their ordinary ray groups modulo $(2)$ are \[ C_4,1,C_2,C_4,C_4,C_4,C_4,C_4,C_4,C_4,C_4, C_8\times C_4,C_2,C_4\times C_2,C_2. \] These are exact degree-four ray computations; certification of each class group returns one. All groups are two-primary, so none admits the odd-order character $\xi$, a contradiction. The exact calculation is supplied in~\cite[\path{ramified}]{code}. If the residue is reducible, both constituent characters have odd order. They are unramified at five by the special calculation, and at two since the residue units are $\F_2^\times$ and wild inertia is pro-two. They are unramified at every other finite place and at the real places. The class number one statement makes both trivial. The final modularity assertion follows from Theorem~\ref{r:s:d6:thm:two}. \end{proof} \subsubsection{Ordinary scalar sources} \begin{proposition}\label{r:s:d6:prop:ordinary-scalar-two} A bounded source with two special places and scalar residual semisimplification at two is modular if some two-adic member is ordinary. \end{proposition} \begin{proof} Let $r$ be that member. Put $H_0=\chi_{-3}|_{G_K}$ and $\psi_5=\chi_5|_{G_K}$. The exact two special characters $\theta_1,\theta_2$ determine a unique $\eta\in\{1,H_0,\psi_5,H_0\psi_5\}$ whose restriction at both five-places is $\theta_1\theta_2$. Both $r$ and $r^\sigma\otimes\eta$ have the same exact special characters, ordinary crystalline condition, determinant and support. Choose a vector $v$ on the ordinary line and the lattice generated by its global orbit. The vector is primitive and cyclic. Hence the actual residue is a nonzero unipotent extension, and the reduction of its ordinary line is transverse to its unique global invariant line. Its restriction at two is trivial. The kernel of localization of the supported quadratic classes at two is generated by $[-15]$: in the unit basis $-1,\epsilon,1+s,1-s$ this follows by taking squares modulo $8\OO_{K_2}$. The squares have coordinates $(1,0),(1,4),(7,2),(7,6)$, and only the identity and $[-15]$ occur. Thus the actual residue is $1+\chi_{-15}E_{12}$. For a real involution $c$, the basis $((r(c)-1)v,v)$ is integral and gives \[ r(c)=\begin{pmatrix}-1&1\\0&1\end{pmatrix}, \qquad \text{ordinary mark } e_2 . \] It puts both aligned representations in the same marked problem. Kisin's marked ordinary functor, Proposition~2.4.4 and Corollary~2.4.5 of~\cite{h:Kisin2}, is represented at arbitrary ramification, including two. Pull it back to the global matrix functor with the displayed real matrix and the ordinary section $e_2$. At five impose the exact character line equations. The line there is unique since a residual unipotent element has unit upper entry, so these too are represented conditions. We compute the homogeneous tangent in this gauge. Write its cocycle as $X(g)=\left(\begin{smallmatrix}x(g)&y(g)\\z(g)&x(g)\end{smallmatrix}\right)$ relative to $1+\chi_{-15}(g)E_{12}$. The adjoint action adds $\chi_{-15}(g)z(g)(1+E_{12})$, so $z$ is additive. The ordinary finite-flat condition gives $y|_{G_{K_2}}=0$, $x|_{G_{K_2}}$ unramified, and $z|_{G_{K_2}}$ a unit Kummer class. At a special place allow the line to move to $e_1+t_ve_2$. The exact characters give $z|_{G_v}=0$ and $x|_{G_v}=t_v\chi_{-15}|_{G_v}$. The supported squareclass table then gives $z\in k\chi_{-1}$. The real involution relation forces $z(c)=0$, hence $z=0$. Now $x,y$ are additive, and the dyadic conditions give $y=B\chi_{-15}$ and $x=A\chi_{-15}+D\psi_5$. At five $\chi_{-15}=R+U$ and $\psi_5=R$, where $R,U$ are the independent ramified and unramified quadratic characters; thus $D=0$. The fixed real matrix forces $A=B=0$. The represented ring has zero relative tangent and is a quotient of $W(k)$. Its actual characteristic-zero point makes it $W(k)$, so the aligned pair is identical. We have $r^\sigma=r\otimes\eta$, with $\eta$ restricted from $G_\Q$. Lemma~\ref{r:quadratic-descent} proves modularity. \end{proof} \subsection{The scalar two-adic source with two special places} \label{r:s:d6:sec:quartic-ray-scalar} Let \(K=\mathbf Q(\sqrt6)\). A special place in this section has nonzero monodromy and its exact scalar twist is quadratic, including its unramified part. Let \(a_v\) denote the good-place trace of a pure parallel weight-two system, with arithmetic Frobenius normalization. \begin{theorem}\label{r:s:d6:thm:quartic-ray-scalar} Let \(\mathcal R\) be a totally odd, absolutely irreducible, pure, strongly compatible rank-two system over \(K\), with compatibility including monodromy. Suppose it has cyclotomic determinant, labelled weights \(\{0,1\}\), is good outside five and is special at both primes above five. Suppose every member at a coefficient place above two is nonordinary and has scalar residual semisimplification. Then the global trace field has degree at most three. The actual system is modular. \end{theorem} \subsubsection{The integral trace order and its regular group} Let $D$ be the global trace field and $d=[D:\Q]$. Apply Lemma~\ref{r:shared:trace-order} to the tuple at all dyadic coefficient places, and use its notation $R_0,E_0,T,\mathcal D$. Then $R_0$ has $\Z_2$-rank $d$, while $E_0/2E_0$ is local with residue field $\F_2$. \subsubsection{The two special generators and the dyadic trace} Put \(E=E_0/2E_0,\ R=R_0/2R_0\), and let \(J\) be the radical of \(E\). A functional on $J/J^2$ gives a middle/middle extension by quotienting the regular module. In the calculation of Theorem~\ref{r:break}, its Kisin matrix is $\left(\begin{smallmatrix}u&h\\0&u\end{smallmatrix}\right)$, and its Artin--Schreier class has the unique form $n/u+m$, with $n,m\in\F_2$. The first coordinate is ramified of break one when nonzero; the second is the unramified character. At five these quadratic characters are tame. They belong to the quadratic dual of the narrow ray group modulo the square of the dyadic prime and both five-primes. Its invariants are \(4,2,2\). Three independent characters, in squareclass notation, are \[ \alpha=[-1],\qquad \beta=[\epsilon]=[2],\qquad \gamma=[-15],\qquad \epsilon=5+2\sqrt6. \] Their relative discriminant norms are \(4,4,25\); independence also follows from the independent quadratic generators \(i,\sqrt2,\sqrt5\) over \(K\). Thus they span every allowed first-layer character. At a fixed five-place, \(-1\) is square, \(2\) is an unramified nonsquare and \(-15\) gives a ramified character. Choose a Frobenius lift trivial on \(\gamma\), and a tame inertia generator. Write \(A,B,C\) for the differences from one of a real involution, this Frobenius and this inertia. Their evaluations on \((\alpha,\beta,\gamma)\) are respectively \((1,0,1),(0,1,0),(0,0,1)\). Since the group spans \(E\), the first-layer functional is determined by the resulting character; these three elements therefore span \(J/J^2\). Nilpotence of \(J\) gives algebra generation of \(E\). At any one five-place normalize by its fixed rational quadratic special sign. In a basis of the special line, for local \(g\) the normalized difference \(L(g)\) has top row \((\chi_2(g)-1,*)\) and bottom row zero. Thus \(L(g)L(h)=(\chi_2(g)-1)L(h)\) in the original joint order. The signs disappear modulo two and all local differences have zero products and trace zero there. This uses equality of the exact quadratic characters across coefficient conjugates. For the real element \((1+A)^2=1,\ T(A)=0\) modulo two. Polarized Cayley--Hamilton now gives in \(E\) \begin{equation}\label{r:s:d6:eq:quartic-special-rel} \begin{gathered} A^2=B^2=C^2=BC=CB=0,\\ AB+BA=X=T(AB),\qquad AC+CA=Y=T(AC). \end{gathered} \end{equation} In particular all words reduce to the \(R\)-span of \(1,A,B,C,AB,AC\). \begin{lemma}\label{r:s:d6:lem:quartic-trace-generator} The ring \(R\) is \(\mathbf F_2[t]/(t^d)\), where \(t\) is the image of \(a_2\). When \(d\ge2\), \(X\) is another generator of its maximal ideal and \(Y/X\) can be chosen to be a unit. In characteristic zero \(D=\mathbf Q(a_2)\). \end{lemma} \begin{proof} First work in the relative cotangent quotient \(R'=R/\mathfrak m_R^2\), retaining the full algebra over it and its trace and determinant laws. The normal form above shows that the relative cotangent is generated by \(X,Y\): all traces reduce to their \(R'\)-span and the trace order is generated by the actual traces. All traces are in \(\mathfrak m_{R'}\). The ramified functional \(n\) in the preceding middle-extension calculation on \((\alpha,\beta,\gamma)\) is \((1,1,0)\). Indeed \(\alpha\) is dyadically ramified, \(\gamma\) splits there, and \(\alpha+\beta+\gamma=[5]\) is the unramified quadratic character. It follows in \(J/J^2\) that the constant term of \(\Phi\) is \(A+B+C\). In the \(u\)-coefficient of \(\Phi\) the corresponding unramified functional takes value one on \([5]\). These statements are unaffected by a first-order basis change, whose correction is \(u(\varphi-1)z\) with zero constant and linear coefficients. Write \[ \begin{gathered} \Phi=N+u(1+M)+O(u^2),\qquad N=A+B+C+Z,\\ M=m_A A+m_B B+m_C C+Z',\qquad Z,Z'\in J^2 . \end{gathered} \] Evaluation of \([5]\) on \(A,B,C\) gives \(0,1,1\), hence \(m_B+m_C=1\). More explicitly, take the functional on the full allowed character space obtained by restricting to the dyadic tower and reading the constant Artin--Schreier coefficient. The local ramification break bound ensures that only the terms of degrees \(-1,0\) can occur. Its restriction to the actual middle extensions gives the coefficients of \(M\), so one may use this lift of the functional even when the actual first layer is smaller. We give the determinant calculation in \(R'\) itself. By the six-word normal form, \(J^2\subset\mathfrak m_{R'}E+R'AB+R'AC\). A term from \(\mathfrak m_{R'}E\) contributes neither a norm nor a polar term, since \(\mathfrak m_{R'}^2=0\) and all traces are in \(\mathfrak m_{R'}\). The norms of \(AB,AC\) are zero by multiplicativity, and their mutual polar term is \(XY-T(ABAC)=0\). Their trace pairings with \(A+B+C\) are zero by~\eqref{r:s:d6:eq:quartic-special-rel}. Thus \(\mathcal D(N)=\mathcal D(A+B+C)=X+Y\). The constant term of the determinant in Lemma~\ref{r:regular-kisin} gives \(X=Y\). The same normal form gives \(T(J^3)=0\): each word of length three reduces to a multiple of a trace-zero generator, and longer products contribute either such terms or quadratic products of \(X,Y\). The coefficient of \(u\) in the norm is \(T(N)T(1+M)-T(N(1+M))\). It vanishes by the determinant divisibility, and the product of the two traces vanishes in \(R'\). Hence \[ T(N)=T(NM)=(m_B+m_C)X=X\qquad\text{in }R'. \] Lemma~\ref{r:regular-kisin} identifies \(T(N)\) with \(a_2\). Nakayama now makes \(R\) a quotient of \(\mathbf F_2[t]\), with radical generated by \(t\). Its dimension is the rank \(d\) of \(R_0\), giving the asserted relation \(t^d=0\). The same calculation gives relative generators \(X\) and \(Y\) whenever \(d\ge2\), with identical initial coefficient. Finally \(1,a_2,\ldots,a_2^{d-1}\) span \(R_0\) over \(\mathbf Z_2\) by Nakayama. The rational degree of \(a_2\in D\) is therefore at least \(d\), proving \(D=\mathbf Q(a_2)\). \end{proof} \subsubsection{An abelian restriction of the regular image} Suppose \(d\ge4\) and use \(X\) as the parameter of \(R\). Choose \(v\in R^\times\) with \(Y=vX\), and put \(Z=C+vB\). By~\eqref{r:s:d6:eq:quartic-special-rel}, \(Z\) is central, \(Z^2=BZ=XZ=0\), and \(T(Z)=T(AZ)=0\). Here \(XC=(AB+BA)C=BAC=B(Y+CA)=YB\). Consequently \(E\) is spanned over \(R\) by \(1,A,B,AB,Z,AZ\). Let \(S=\mathbf F_2[X]/(X^3)\) and \(Q=S\langle a,b\rangle/(a^2,b^2,ba+ab-X)\), as in the proof of Proposition~\ref{r:t:d2:prop:scalar-trace-degree}. The same trace-pairing argument gives a surjection \(E\to Q\), taking \(A,B,Z\) to \(a,b,0\). Indeed, pairing a relation in \(1,A,B,AB,Z,AZ\) with \(1,A,B,AB\) puts its first four coefficients in \((X^{d-1})\), which vanishes in \(S\). The free normal basis \(1,a,b,ab\) then makes the map multiplicative. This does not require faithfulness of the scalar action on \(E\). Compose with \(a\mapsto E_{12},\ b\mapsto XE_{21}\) to obtain \(\theta:G_K\to\operatorname{GL}_2(S)\). On formal coefficients its matrix is \[ \begin{pmatrix} f_0+Xf_3 & f_1\\ Xf_2 & f_0\end{pmatrix}. \] Its trace is the reduction of \(T\). Comparing the two Cayley--Hamilton identities for a group element gives \(\det\theta(g)=1\). The two linear characters \(f_1(0),f_2(0)\) are visible on actual regular quotients: send \(a\), respectively \(b\), to the generator of \(\mathbf F_2[z]/(z^2)\), and send the other generator and \(X\) to zero. Their values on \(A,B,C\) are \((1,0,0),(0,1,1)\). The calculated first-layer space gives \(f_1(0)=\alpha,\ f_2(0)=[5]\). Put \(F=K(i,\sqrt5)\). Each matrix over \(G_F\) is \(I+XM+X^2N\); the lower left entry of \(M\) is zero by \(f_2(0)=0\), and its diagonal entries agree by the determinant. All these \(M\)'s lie in the commutative algebra generated by \(I,E_{12}\). Multiplication modulo \(X^3\) thus makes \(\theta(G_F)\) abelian of exponent at most four. It has a specified nonzero double. Let \(r\) be the product of the chosen real and special Frobenius elements. Its image has trace \(X\), so \[ \theta(r)^2=I+X\theta(r),\qquad \theta(r)^4=I+X^2I . \] The element \(r^2\) lies in \(G_F\), since \(F/K\) is elementary abelian, and therefore gives an element of order four there. \subsubsection{The four Frobenius classes above five} Fix a five-prime \(w\) of \(F\). Choose a Frobenius \(h\) over \(F_w\) and a tame inertia element \(k\) in the underlying \(K\)-decomposition group with \([5](k)=1\) (additive notation). The special identities give \[ T(h-1)=T(k-1)=0,\qquad (h-1)(k-1)=(k-1)^2=0\quad\text{in }E , \] where the trace equalities are in \(R\). In particular the formal \(ab\)-coefficient of each difference vanishes in \(Q\), by the original trace pairing \(Xf_3=0\) in \(R\). Write \(h-1=P+Ua+Vb,\ k-1=P'+U'a+V'b\) in \(Q\). The \(b\)- and \(ab\)-coefficients of their zero product give \[ PV'+VP'=UV'+VU'=0 . \] Now \(V'\) is a unit and \(V\in XS\), by the values of \([5]\). Thus \(h-1=(V/V')(k-1)\) in the \emph{abstract} algebra \(Q\). The coefficient comparison here requires its free normal basis. The matrix of \(k\) is \(I\) modulo \(X\), since \(-1\) is square at five. If \(\theta(k)-I=XM_0+O(X^2)\), its square gives \(M_0^2=0\). We have proved at all four primes that \begin{equation}\label{r:s:d6:eq:quartic-frobenius-nil} \theta(h)=I+X^2 N_w,\qquad N_w^2=0 . \end{equation} There is one dyadic prime \(\mathfrak P\) of \(F\), with absolute \((e,f)=(4,2)\). Over \(K\), the quadratic \(K(i)\) has dyadic conductor two: its absolute discriminant is the product of \(24,-4,-24\), and division by \(24^2\) gives relative discriminant norm four. The factor \(K(\sqrt5)\) is dyadically unramified. Thus the ramified break of \(F/K\) is one. By the finite-flat bound and Herbrand's subgroup formula the upper break over \(F\) is at most \(1+2(3-1)=5\). At five the elementary local image over \(K\) has exponent two and its possible quadratic tame inertia is killed by \(K(\sqrt5)\). Hence the abelian representation over \(F\) is unramified at five and all its five-Frobenius elements have order dividing two. The field is totally imaginary. Write \(H\) for the ray class group of modulus \(\mathfrak P^6\) of \(F\). Class field theory makes \(\theta|_{G_F}\) a quotient of \[ H/(4H+\langle 2[w]:w\mid5\rangle). \] The following coordinates are from the ordinary ray group, with the ordered primes as in the certificate. They concern abelian extensions of the degree-eight field \(F\). \[ \begin{array}{c|c} \text{group or prime}&\text{invariants or class}\\ \hline H & (8,4,2,2)\\ H/(4H+\langle 2[w]\rangle) & (4,2,2,2)\\ \hline w_1 &(0,1,0,0)\\ w_2 &(0,1,0,1)\\ w_3 &(2,1,0,1)\\ w_4 &(2,1,0,0) \end{array} \] For reproducibility one defining polynomial for \(F\) is \[ y^8-4y^7+2y^6+8y^5+13y^4-44y^3+164y^2-140y+145 . \] The certificate computes a certified class group and units, obtains the ray cyclic generators and the ideal logarithms \(L\) of the four primes, and takes the Smith form of the relation lattice generated by \(\operatorname{diag}(8,4,2,2),\,4I,\,2L\). Thus \([w_1]-[w_4]\) is the unique nonzero double of the quotient. It maps to \(I+X^2I\), by the square of \(\theta(r^2)\). By \eqref{r:s:d6:eq:quartic-frobenius-nil} this gives \(N_{w_1}+N_{w_4}=I\) for square-zero matrices, impossible since \((I+N_{w_1})^2=I\). This proves \(d\le3\). The exact calculation is supplied in~\cite[\path{ramified/d6_d10_d11/d6}]{code}. \subsubsection{Passage to an odd coefficient prime} The trace-field bound and Lemma~\ref{r:cubic-screen} give a coefficient prime above three with residue field $\F_3$ or $\F_9$, or above five with residue field $\F_5$. Choose the indicated coefficient place. For an absolutely irreducible residue, the traces in its finite residue field give a realization over that field by finite-field descent (the Brauer group of a finite field is zero). Allen--Khare--Thorne, \emph{Modularity of \(\mathrm{PGL}_2(\mathbf F_p)\)-representations over totally real fields}, Theorems~1.1 and 2.3, give residual automorphy and lift the actual weight-two member when its restriction to the cyclotomic subgroup is absolutely irreducible. In the nonsolvable case this restriction is automatic: reducibility on a normal cyclic quotient would force an induced dihedral representation, with solvable image. At three all solvable cases, including globally reducible residues, are covered by the pure special characteristic-three source theorem. At five a solvable irreducible residue either remains irreducible on the cyclotomic subgroup or is induced from \(K(\sqrt5)\). Such induction at a five-place is incompatible with the special ordinary line: a tame generator outside the inducing subgroup has scalar square and projective order two, while the ratio of the ordinary inertial characters has order four. This gives cyclotomic irreducibility in this case as well. When the five-adic residual semisimplification is reducible, the dual source has determinant $\chi_5^{-1}$ and at both five-places has an invariant line of finite character with quotient of character that finite character times $\chi_5^{-1}$. Thus it is ordinary of weight $(0,0)$ in the convention of~\cite[\S1.6]{h:DT}. Corollary~1.2.2 there applies since $K$ is totally real abelian and five is unramified in $K$. This proves modularity without a calculation of the constituent ratio. Strong compatibility transfers modularity to the entire system, proving Theorem~\ref{r:s:d6:thm:quartic-ray-scalar}. \subsection{Local lifts and a switch in characteristic three} \begin{lemma}[Bounded five-adic lifts]\label{r:s:d6:local:menu} Every continuous representation $\bar r:G_{\Q_5}\to\GL_2(\overline\F_5)$ of determinant $\bar\chi_5$ has a lift of determinant $\chi_5$, labelled weights $\{0,1\}$ and type $\SC$ or $\SP$. \end{lemma} \begin{proof} The weights of $\GL_2(\F_5)$ with trivial scalar central character are \[ 1,\quad\det^2,\quad\Sym^2\det,\quad\Sym^2\det^3, \quad\Sym^4,\quad\Sym^4\det^2. \] The two cuspidal compact types reduce with constituents \[ \overline\sigma(\tau_6)^{\rm ss}=1\oplus\Sym^2\det, \qquad \overline\sigma(\tau_3)^{\rm ss}=\det^2\oplus\Sym^2\det^3. \] Here $\tau_3=\widetilde\omega_2^8\oplus\widetilde\omega_2^{16}$ and $\tau_6=\widetilde\omega_2^4\oplus\widetilde\omega_2^{20}$, for a niveau-two character $\omega_2$ of order 24. The identities follow by evaluating the cuspidal Brauer characters on split and nonsplit regular elements. These finite character identities determine the semisimplified reductions. The full fixed-determinant Breuil--M\'ezard formula~\cite[Theorem 1.2]{g:Tung} has nonnegative multiplicities; some multiplicity is positive because a potentially Barsotti--Tate lift exists. The determinant confines it to the six displayed weights. The first four therefore give the cuspidal alternatives. The last two give the semistable compact type for a scalar quadratic inertial type. If the resulting point has nonzero monodromy, it is $\SP$. If all such points were crystalline, the corresponding crystalline compact type, $1$ or $\det^2$, would have positive multiplicity, and the cuspidal alternative would again exist. This is a statement about the actual residual extension, with any finite coefficient field. The determinant is fixed in the deformation rings; equivalently an unramified determinant error reducing to one is corrected by its square root. \end{proof} \begin{lemma}[The actual-residual switch]\label{r:s:d6:local:switch} Let $\bar s$ be the reduction of a determinant-$\chi_3$ representation of $G_{\Q_5}$ of type $\SC$ or $\SP$, with any stable lattice. After finite coefficient extension, $\bar s$ has lifts of both types, with determinant exactly $\chi_3$. The scalar quadratic inertia character determines whether the cuspidal lift has type $\tau_3$ or $\tau_6$. \end{lemma} \begin{proof}Apply Lemma~\ref{t:triadic-switch} with $\ell=5$. Its hypotheses include the actual residual lattice, both nonsplit orientations and the split case. Restoring the scalar quadratic character gives $\tau_3$ or $\tau_6$ and preserves the fixed determinant.\end{proof} \subsection{The supported residual base} \begin{proposition}\label{r:s:d6:prop:bounded} Assume GRH. Every bounded system over $K=\Q(\sqrt6)$ is modular. Every absolutely irreducible determinant-$\omega_5$ residual representation unramified outside five is modular. \end{proposition} \begin{proof} Consider first a bounded system of two special types. The quartic-ray calculation excludes an irreducible solvable two-residue. A nonsolvable two-residue gives modularity by the bounded special source theorem under GRH. In the remaining case its semisimplification is $1\oplus1$. If some two-adic member is ordinary, the marked ordinary rigidity Proposition~\ref{r:s:d6:prop:ordinary-scalar-two}, followed by projective descent, gives modularity. Otherwise Theorem~\ref{r:s:d6:thm:quartic-ray-scalar} applies to all dyadic coefficient places together and proves modularity. Thus every two-special bounded system is modular. Now consider any bounded system at three. A reducible residue is handled by Theorem~\ref{r:s:d6:thm:three} for a pure menu and is excluded by Proposition~\ref{r:s:d6:prop:mixed-three} for a mixed menu. An irreducible solvable residue has a regular cuspidal Artin witness; Lemma~\ref{r:s:d6:lem:cyclo} and \cite[Theorem~2.3]{h:AKT} then modularize the actual weight-two source. In the nonsolvable case the residue stays absolutely irreducible over $K(\zeta_3)$. At both five-places use the genuine special lift of Lemma~\ref{r:s:d6:local:switch}, which retains the entire actual residue and its quadratic scalar character. Keep the old crystalline weight-two component at the three-place. Emerton--Gee--Pan--Zhu (EGPZ), Theorem~4.2.1, gives a global point on the selected components. Lemma~\ref{shared:global} gives a pure compatible system. On the special components the Frobenius eigenvalue ratio is five; zero monodromy at a pure point would contradict purity. Thus both local monodromies are nonzero and the new system is modular by the previous paragraph. AKT, Theorem~2.3, applies backwards to the old weight-two three-adic member. This proves modularity of every bounded system. Let $\bar r$ be absolutely irreducible of determinant $\bar\chi_5$, unramified outside five. Solvable image has a regular cuspidal Artin witness after weight raising. Otherwise cyclotomic restriction is absolutely irreducible. The EGPZ exception cannot occur: $[K(\zeta_5):K]=4$, whereas the abelianization of $\mathrm{PGL}_2(\mathbf F_5)\simeq S_5$ has order two. The fixed determinant bounded local lift lemma at each completion $K_w=\mathbf Q_5$ supplies a weight-two lift of one of the indicated types for the entire local representation. EGPZ produces the global lift, with unramified conditions away from five. Lemma~\ref{shared:global} and purity yield a bounded system. Its modularity, followed by reduction, proves the claimed residual base at five. \end{proof} \section{\texorpdfstring{Dihedral sources over $\Q(\sqrt{10})$}{Dihedral sources over Q(sqrt(10))}}\label{r:s:sec:d10} Put $K=\Q(s)$, $s^2=10$. \begin{theorem}\label{r:s:d10:thm:residual} Assume GRH. Every continuous irreducible representation $G_K\to\GL_2(\overline\F_3)$ of determinant $\bar\chi_3$ unramified outside three is modular. \end{theorem} The two places $v_+,v_-$ above three are distinguished by $s\equiv 1,-1$, respectively. Both completions are $\Q_3$, and the unique dyadic completion $F=K_2$ has ramification index two and residue field $\F_2$. In proving Theorem~\ref{r:s:d10:thm:residual}, we prescribe wild binary dihedral inertia at $v_+$ and either a special or a quartic supercuspidal type at $v_-$. A discriminant bound reduces the nonsolvable characteristic-two case to an $A_5$ lifting theorem. The soluble case requires the modularity of an actual lift, since the relevant CM inducing field splits at the dyadic prime. We prove the special case by bounding the global trace field across all of its dyadic embeddings. Theorem~\ref{r:thm:ramified-fields} then follows by a conductor-removal argument with protected residual images. \subsection{The local conditions at three} \begin{lemma}\label{r:s:d10:lem:semistable} Let $\br:G_{\Q_3}\to\GL_2(\overline\F_3)$ have determinant $\om$. It has a lift of weights $\{0,1\}$ and determinant $\chi_3$ which is either a quadratic twist of a semistable representation with nonzero monodromy, or potentially crystalline of tame supercuspidal type \[ \tau_4=\widetilde\omega_2^{\,2}\oplus\widetilde\omega_2^{\,6}, \] where $\omega_2$ is a fundamental character of niveau two. For either choice in a determinant-cyclotomic compatible system, the characteristic-two residual local kernel field is contained in \begin{equation}\label{r:s:d10:eq:quadratic-envelope} E'=\Q_3(\sqrt3,\sqrt{-1}),\qquad e(E'/\Q_3)=2,\quad f(E'/\Q_3)=2,\quad d(E'/\Q_3)=\tfrac12. \end{equation} \end{lemma} \begin{proof} Apply Lemma~\ref{c:semistable}; its two local choices have the same determinant, weights and characteristic-two kernel envelope. \end{proof} At $v_+$ use the affine-generic construction of Lemma~\ref{c:type}, with its notation $K_0,J,\vartheta,\sigma_+$ and $\tau_+$. The sign determining the characteristic-two quadratic subfield is the same as in that lemma. \begin{proposition}\label{r:s:d10:prop:type} The representation $\sigma_+=\Ind_J^{K_0}\vartheta$ is the eight-dimensional type of a supercuspidal inertial type $\tau_+$ with determinant one. Every $\br:G_{\Q_3}\to\GL_2(\overline\F_3)$ with determinant $\om$ has a potentially crystalline lift of weights $\{0,1\}$, determinant $\chi_3$, and inertial type $\tau_+$. Its inertia remains irreducible after reduction in every characteristic greater than three. For a determinant-cyclotomic compatible system of this type, its residual characteristic-two kernel field at this place is the unique totally ramified $S_3$ extension $E_+/\Q_3$ with quadratic subfield $\Q_3(\sqrt3)$ and normalized different $7/6$. \end{proposition} \begin{proof} The construction is that of Lemma~\ref{c:type}, and Proposition~\ref{c:wildlift} gives its lifts with the specified determinant and weights. Lemma~\ref{c:type} also gives the irreducibility of inertia in characteristic greater than three and the unique characteristic-two kernel field $E_+$, with its quadratic subfield and normalized different. \end{proof} \begin{proposition}\label{r:s:d10:prop:system} Let $\brho$ satisfy the hypotheses of Theorem~\ref{r:s:d10:thm:residual}, and suppose its image is nonsolvable. It lifts to a totally odd, potentially modular, strongly compatible system $\cR=(r_\lambda)$ with determinant $\chi$ and weights $\{0,1\}$. The system is unramified away from three, has type $\tau_+$ at $v_+$, and has one of the two local conditions of Lemma~\ref{r:s:d10:lem:semistable} at $v_-$. \end{proposition} \begin{proof} Proposition~\ref{r:s:d10:prop:type} and Lemma~\ref{r:s:d10:lem:semistable} supply the local lifts at $v_+$ and $v_-$, respectively. At the second place prescribe the tame supercuspidal type or a component through the special lift supplied by the lemma; no global quadratic character is required. In the latter case, the closure of the special component fixes Weil Frobenius eigenvalue ratio three, so the purity observation in Section~\ref{shared:pure-special} retains nonzero monodromy. Nonsolvability implies that the restriction to $G_{K(\zeta_3)}$ is irreducible, since an irreducible representation becoming reducible over a quadratic extension has dihedral projective image. Apply Lemma~\ref{shared:global} to these nonempty weight-two components, with determinant $\chi_3$ and unramified conditions away from three. It supplies an actual potentially modular lift and its strongly compatible system, including local compatibility at the coefficient places. In particular its two-adic member is crystalline of weights $\{0,1\}$ at two. \end{proof} Choose a place above two in the coefficient field and write $r_2$ for this member and $\br_2$ for its reduction. Proposition~\ref{r:s:d10:prop:type} makes $\br_2$ irreducible, with determinant one. Let $M$ be its kernel field. Its completion at $v_+$ is $E_+$, while Lemma~\ref{r:s:d10:lem:semistable} puts its completion at $v_-$ inside $E'$. \subsection{The controlled characteristic-two image} \begin{lemma}\label{r:s:d10:lem:q10-cubic} The splitting field $A/K$ of $X^3-3X+1+s$ has Galois group $S_3$. It is unramified away from three, and its dyadic completion is the unramified cubic extension of $K_2$. Its completions at $v_+$ and $v_-$ are $E_+$ and $\mathbb Q_3(\sqrt3)$, respectively. \end{lemma} \begin{proof} The cubic reduces to $X^3+X+1$ at the unique dyadic prime. This polynomial is irreducible and separable over $\mathbb F_2$, proving both global irreducibility and the dyadic assertion. Its discriminant is \[ -27(7+2s)=-6\left(3+\frac32s\right)^2, \qquad N_{K/\mathbb Q}\bigl(-27(7+2s)\bigr)=3^8. \] It follows that $\operatorname{Gal}(A/K)=S_3$, with quadratic subfield $K(\sqrt{-6})$, and that no finite prime away from three ramifies. At $v_+$ put $d=s-1$. Since $d(s+1)=9$, we have $v_3(d)=2$. Writing $X=1+y$ gives $y^3+3y^2+d=0$, whose roots have valuation $2/3$. The element $z=y^2/3$ satisfies \[ z^3-3z^2-\frac{2d}{3}z-\frac{d^2}{27}=0. \] This polynomial is Eisenstein. In the resulting cubic extension the three derivative terms have valuations $5,4,3$, respectively, so its different exponent is three. The lower ramification groups of the Galois closure consequently have orders $6,3,1$, giving normalized different $(5+2)/6=7/6$. Its quadratic subfield is $\mathbb Q_3(\sqrt{-6})=\mathbb Q_3(\sqrt3)$, so the closure is $E_+$. At $v_-$, the constant term $1+s$ has valuation two. The Newton polygon has slopes $-1$ and $-1/2$, of lengths one and two. There is one linear factor and one ramified quadratic factor. Its splitting field is the quadratic field defined by the discriminant, namely $\mathbb Q_3(\sqrt3)$. \end{proof} \begin{proposition}\label{r:s:d10:prop:q10-ray-auxiliary} There is a solvable Galois extension $N/K$ of degree $768$, unramified outside two and three, whose dyadic completions have no upper break greater than three. Its completion at $v_+$ is $E_+$, and its completion at $v_-$ is contained in $E'$. \end{proposition} \begin{proof} We use an abelian ray extension of the field $A$ in Lemma~\ref{r:s:d10:lem:q10-cubic}. An absolute equation for $A$ is \[ \begin{split} P_A(X)={}&X^{12}-2X^{11}+5X^{10}+19X^8+2X^7+31X^6\\ &\hspace{20mm}+2X^5+19X^4+5X^2-2X+1. \end{split} \] It is obtained by taking the compositum of $X^6-6X^4+2X^3+9X^2-6X-9$ with $X^2+6$; if $u$ is the root of the degree-six polynomial, the embedding of $K$ is given explicitly by $s=-u^3+3u-1$. The class number is four. There are two primes $\mathfrak P_1,\mathfrak P_2$ above two, each of absolute ramification index two and residue degree three. Above three there is one prime of absolute ramification index six and residue degree one, lying over $v_+$, and three primes of ramification index two and residue degree one, lying over $v_-$. The field has discriminant $241864704000000$ and signature $(0,6)$. Let $R$ be the ray class group of modulus $\mathfrak P_1^4\mathfrak P_2^4$. Quotient $R$ by the class of the prime over $v_+$ and by twice the class of each prime over $v_-$. Exact ideal and unit arithmetic gives \[ R\simeq C_4^2\times C_2^5, \qquad R/\langle[\mathfrak q_+],2[\mathfrak q_-]\rangle \simeq C_4\times C_2^5. \] Here the bracket in the denominator includes all three primes over $v_-$. These are calculations in degree twelve. The accompanying PARI/GP certificate computes maximal orders and ray class groups, checks the quotient order $128$ by Hermite and Smith normal forms, and returns $\texttt{bnfcertify}=1$ for the class-group calculation. The exact conductor of the quotient has exponent four at both dyadic primes. The exact calculation is supplied in~\cite[\path{ramified/d6_d10_d11/d10}]{code}. Let $N/A$ be the associated ray extension. The modulus and the relations are preserved by $\operatorname{Gal}(A/K)$, so $N/K$ is Galois. Its group is an extension of an abelian group by $S_3$, hence is solvable, and $[N:K]=128\cdot6=768$. At $v_+$ the ray relation makes $N/A$ split, so its completion remains $E_+$. At each prime over $v_-$ it is unramified of degree at most two, giving a subfield of $E'$. At two, conductor exponent at most four gives upper break at most three over $A$. Since $A/K$ is unramified there, the same upper bound holds over $K_2$. \end{proof} \begin{theorem}\label{r:s:d10:thm:q10-controlled-images} Assume GRH for Dedekind zeta functions. There is no finite Galois extension $M/K$, unramified outside two and three, satisfying all the following conditions: \begin{enumerate} \item $\operatorname{Gal}(M/K)\simeq\operatorname{SL}_2(\mathbb F_q)$ for $q=2^r\ge8$; \item its dyadic completions have no upper break greater than three; \item its completion at $v_+$ is $E_+$ and its completion at $v_-$ is contained in $E'$. \end{enumerate} \end{theorem} \begin{proof} The group $\operatorname{SL}_2(\mathbb F_q)$ is nonabelian simple in characteristic two. Thus $M\cap N=K$ for the solvable Galois extension in Proposition~\ref{r:s:d10:prop:q10-ray-auxiliary}. For $L=MN$ we obtain \[ [L:\mathbb Q]=2\cdot768\,q(q^2-1)\ge774144. \] Upper numbering is compatible with quotients, so the compositum still has no dyadic upper break greater than three. Since $K_2$ has absolute ramification index two, its normalized relative different is strictly less than two. The completions at three remain in the stated extensions. The global discriminant formula therefore gives \[ \operatorname{rd}(L)<4\sqrt{40}\,3^{5/6} =63.1962098567\ldots. \] Apply the GRH explicit formula of Section~\ref{shared:explicit} with the retained coefficients $c_3=1/6$ and $c_9=1/4$. The $v_+$ primes have norm three. If the other completion has residue degree two, its norm-nine primes give the second coefficient. If it has residue degree one, retaining only the even powers of its norm-three Euler terms gives at least the same contribution. At $b=13.3$ and degree $774144$, interval arithmetic gives $\operatorname{rd}(L)>64.5683408714$. The lower expression increases with the degree. This contradicts the preceding upper bound and proves the theorem. The exact calculation is supplied in~\cite[\path{ramified/d6_d10_d11/shared}]{code}. \end{proof} \begin{corollary}\label{r:s:d10:cor:q10-nonsolvable-lift} Assume GRH. Let $r:G_K\longrightarrow\operatorname{GL}_2(\overline{\mathbb Q}_2)$ be a continuous representation of determinant $\chi_2$, unramified outside finitely many primes and Barsotti--Tate at two. Suppose that its residual representation is irreducible and nonsolvable, is unramified outside two and three, and has the local extensions specified in Theorem~\ref{r:s:d10:thm:q10-controlled-images}. Then $r$ is modular. \end{corollary} \begin{proof} Dickson's classification and Theorem~\ref{r:s:d10:thm:q10-controlled-images} leave only residual image $\operatorname{SL}_2(\mathbb F_4)\simeq A_5$. The characteristic-two $A_5$ modularity theorem~\cite[Theorem~2.5]{h:AKT} supplies residual modularity. If its kernel field $M$ were totally real, it would have degree $120$ and root discriminant less than $63.19621$. The same explicit formula, now retaining the real-place term and only the selected norm-three primes, gives $\operatorname{rd}(M)>66.10839$ at $b=5$. Thus some real complex conjugation has nontrivial residual image. All the hypotheses of \cite[Theorem~6.1]{g:Thorne2} are satisfied, so $r$ is modular. \end{proof} \begin{lemma}\label{r:s:d10:lem:q10-character-descent} Every finite characteristic-three character $\psi$ of $G_K$, unramified outside three, has square equal to $1$ or to the quadratic character $\nu$ of $K(\sqrt2)/K$. In particular $\bar\chi_3\psi^{-2}$ is totally odd and extends to $G_{\mathbb Q}$. \end{lemma} \begin{proof} The Minkowski bound is $\sqrt{40}/2=\sqrt{10}<3.17$. The dyadic prime $\mathfrak p_2$ is nonprincipal: an element of norm $\pm2$ would give $x^2\equiv\pm2\pmod5$. Its square is $(2)$. The element $2+s$ has norm $-6$, so each prime above three represents the same ideal class as $\mathfrak p_2$; the two triadic classes are inverse and have order two. Thus $h_K=2$. The unit $3+s$ has norm $-1$, so the narrow class number is also two. The principal units at three are pro-three, and each residue field is $\mathbb F_3$. Consequently $\psi^2$ is unramified at every finite place. The narrow class-group calculation makes it $1$ or $\nu$. The biquadratic field $K(\sqrt2)=\mathbb Q(\sqrt2,\sqrt5)$ has discriminant $8\cdot5\cdot40=40^2$; hence it is the quadratic Hilbert class field of $K$. Its character is the restriction of the quadratic character of $\mathbb Q(\sqrt2)/\mathbb Q$, which is totally even. Multiplication by $\bar\chi_3$ proves the last assertion. \end{proof} \subsection{The residual field at two} \label{r:s:d10:sec:dihedral} Let $K=\mathbf Q(s)$, $s^2=10$, and label the places $v_+$ and $v_-$ above three by $s\equiv 1$ and $s\equiv-1$, respectively. Let the two-adic member of the chosen weight-two system have the binary dihedral type at $v_+$ and the special or quartic type at $v_-$. Its determinant is cyclotomic and it is crystalline of labelled weights $\{0,1\}$ at the dyadic place. Throughout this section the residual point field is the field cut out by the actual representation on a stable lattice. The local group at $v_+$ has standard image $S_3$ in characteristic two. In particular the residual representation is absolutely irreducible. \subsubsection{The possible inducing quadratics} For a nonsolvable residual image, Proposition~\ref{r:s:d10:prop:q10-ray-auxiliary} and Theorem~\ref{r:s:d10:thm:q10-controlled-images} leave only $\mathrm{SL}_2(\mathbf F_4)$; Corollary~\ref{r:s:d10:cor:q10-nonsolvable-lift} proves modularity of the actual source in that case. These statements apply to either choice of the mild type, including special monodromy. In the solvable case Dickson's classification gives induction from an odd-order character $\theta$ over a quadratic field $H/K$. The character defining $H$ is unramified outside two and three, has conductor exponent at most four at two by the finite-flat bound, and cuts out $\mathbf Q_3(\sqrt3)$ at $v_+$. The squareclasses unramified away from this support have basis \[ -1,\qquad \epsilon=3+s,\qquad 2,\qquad 3,\qquad 2+s . \] Lemma~\ref{r:s:d10:lem:q10-character-descent} gives the class group of order two, generated by the dyadic class, and identifies both triadic classes with it. The fundamental unit is $\epsilon=3+s$. The valuations of $2,3,2+s$ generate the admissible index-two lattice on these three primes, proving the displayed squareclass basis. For a representative $a$ of each of the $31$ nonzero classes, test $3a$ for squareness at $v_+$ and evaluate the dyadic valuation of the relative discriminant of $X^2-a$. The complete list at conductor at most four is \[ \begin{array}{c|rrrr} a & 3&3\epsilon&-6&-6\epsilon\\ \hline \text{dyadic conductor}&2&4&0&4 . \end{array} \] Only $L=K(\sqrt{-6})$ is CM. If the source is ordinary, Allen's residually dihedral theorem~\cite[Lemma~5.1.2 and Theorem~5.2.1]{g:Allen} applies for each of the three non-CM choices, with its ordinary witness construction and without a dyadic nonsplitting hypothesis for these choices. If the source is nonordinary, its residual dyadic point field has upper breaks at most one, by Theorem~\ref{r:break}. Both conductor-four quadratics are therefore impossible. \subsubsection{The odd character and the cubic} It remains to consider $H=K(\sqrt d)$ for $d=3,-6$. Every dyadic completion of either field has residue field $\mathbf F_2$. Its local units are pro-two, and hence $\theta$ is unramified there. At $v_+$ the relative quadratic different exponent is one and the residue degree is one. The conductor formula for induction gives $3=1+a_{H}(\theta)$, so its conductor exponent is two. At the prime above $v_-$ the full local image is a two-group for either mild type; thus $\theta$ kills the entire decomposition group there. At real places an odd-order character is trivial. Write $P_+,P_-$ for the two primes of $H$ above $v_+,v_-$. The table records the ordinary ray group for $P_+^2$, the class of $P_-$ in the indicated generator, and the quotient by $[P_-]$ and $1+\sigma$, where $\sigma$ generates $\mathrm{Gal}(H/K)$: \[ \begin{array}{c|c|c|c|c} d & (e,f)\text{ above }2 & \mathrm{Cl}_{P_+^2}(H)&[P_-] &\text{quotient}\\ \hline 3 &(4,1)& C_2&1&0\\ -6 &(2,1),(2,1)& C_{12}&9&C_3 . \end{array} \] Here $1+\sigma$ imposes inversion on the rotation character, since the determinant is trivial modulo two. These computations can be performed in the quartic model $z=s+\sqrt d$ with polynomial $z^4-2(10+d)z^2+(10-d)^2$. The expression $s=(z+(10-d)/z)/2$ identifies the two triadic primes; $z\mapsto 2s-z$ identifies $\sigma$. Compute the ray group and its automorphism by \texttt{bnrinit} and \texttt{bnrgaloismatrix}, adjoin the column \texttt{bnrisprincipal}$(P_-)$ and the columns of $1+\sigma$ to the diagonal relation matrix, and take its integer Smith form. The class groups of $H$ are $C_2$ and $C_4$, respectively, and \texttt{bnfcertify} returns one in both fields. For $L$ either triadic prime has class $1$ in the chosen generator of $\mathrm{Cl}(L)=C_4$. The exact calculation is supplied in~\cite[\path{ramified/d6_d10_d11/d10}]{code}. There is consequently no such residual representation induced from $K(\sqrt3)$. The two nontrivial characters of the cubic quotient over $L$ cut out a single field $A$. Indeed the splitting field of $X^3-3X+1+s$ has the requisite local data by Lemma~\ref{r:s:d10:lem:q10-cubic}, so it realizes this cubic quotient. In particular, once a conjugate with one of the modular images has been disposed of, every remaining two-adic residual representation has this same kernel and is the standard representation over $\mathbf F_2$. \subsubsection{The two normal ray modules} \label{r:s:d10:subsec:dten-normal-ray} \begin{proposition}\label{r:s:d10:prop:normal-ray} Let $V_n$ be the binary quotient of the ray group of $A$ of dyadic depth $n$ in which the positive triadic prime splits. For the standard $S_3$-module $W_0$ we have $\mathrm{Hom}_{S_3}(V_2,W_0)=0$. At depth four this space has dimension two, and projection onto either nontrivial $C_3$-eigencharacter after scalar extension to $\mathbf F_4$ is injective on local units at either dyadic prime. \end{proposition} \begin{proof} Write $W_0$ for the two-dimensional standard module over $\mathbf F_2$ for $\mathrm{Gal}(A/K)=S_3$. Let $\mathfrak P_1,\mathfrak P_2$ lie above two and $\mathfrak q_+$ above $v_+$. Set \[ D_n=\mathrm{Cl}_{\mathfrak P_1^n\mathfrak P_2^n}(A),\qquad V_n=D_n/(2D_n+\langle[\mathfrak q_+]\rangle). \] There is no infinite modulus. The normal first-error module in the basis of the wild order-three element is $W_0$ after extension of residue scalars. A class unramified at the mild prime, split at $\mathfrak q_+$, and of dyadic conductor at most $n$ therefore belongs to $\mathrm{Hom}_{S_3}(V_n,W_0)$ after extending the coefficients. To see the identification on the global cohomology class, restrict to $G_A$. The kernel and cokernel terms in inflation--restriction vanish since $W_0$ restricts to the regular module for a Sylow two-subgroup of $S_3$ and is projective. The restricted class is an equivariant additive character. Use the maximal order and certified class group of $A$ from Proposition~\ref{r:s:d10:prop:q10-ray-auxiliary}. Computing the binary ray quotient and its nontrivial $C_3$ part gives \[ \begin{array}{c|c|c|c} n&D_n&\dim V_n&\dim( V_n)_{C_3\ne1}\\ \hline 1&C_4&0&0\\ 2&C_4\times C_2&1&0\\ 3&C_4^2&1&0\\ 4&C_4^2\times C_2^5&6&4 . \end{array} \] In particular $\mathrm{Hom}_{S_3}(V_2,W_0)=0$. The exact matrices and local unit columns in Appendix~\ref{r:s:app:normal-ray} give a two-dimensional space of equivariant maps. For $\alpha^2+\alpha+1=0$, the determinants of the two local evaluation matrices for either $C_3$-eigencharacter are $1+\alpha$ and $\alpha$, up to Galois conjugation. Both are nonzero. Vanishing of either eigenprojection on local units at either dyadic prime therefore kills the global depth-four class. These statements concern the normal error in characteristic two; the tame inertia in the actual joint trace order is checked below. \end{proof} \begin{lemma}\label{r:s:d10:lem:normal-comparison} Let $R$ be a complete local order in a finite product of finite extensions of $\Q_2$, finite free over the integers of an unramified extension containing $\zeta_3$. Let $r:G_K\to\GL_2(R)$ be an integral realization of a tuple of the controlled weight-two systems of Proposition~\ref{r:s:d10:prop:system}. Suppose its residue is the standard representation with kernel field $A$. At the dyadic place, suppose either all constituents are nonordinary, or all are ordinary with the same residual quotient character. Let $J$ be a proper closed ideal of $R$, and assume that the image of inertia above $v_-$ in $G_A$ is scalar modulo $J$. Then the image of $r$ modulo $J$ is monomial, with component character $\eta=\chi_{-6}$, in the fixed eigenbasis of the order-three element at $v_+$. \end{lemma} \begin{proof} The order-three eigenprojectors are integral. In their basis the normal space to the diagonal torus is the standard $S_3$-module $W_0$, extended to the residue field; it has no invariants under the fixed order-three element $u$, and $H^i(S_3,W_0)=0$ for $i>0$. Write $\mathfrak m$ for the maximal ideal of $R/J$. Assume monomiality modulo $\mathfrak m^n$ and work modulo $\mathfrak m^{n+1}$, with square-zero ideal $I=\mathfrak m^n/\mathfrak m^{n+1}$. Erasing the entries forbidden by $\eta(g)$ in each matrix gives a monomial representation, since products of two erased entries vanish. Its first error is a cocycle $\kappa$ with values in $W_0\otimes I$, and $\kappa(u)=0$. At $v_+$ the actual source normalizes the $u$-eigenspaces, so the error vanishes on the full decomposition group. At $v_-$ the hypothesis on inertia makes $\kappa|_{G_A}$ unramified. Thus, once its dyadic conductor is bounded, inflation--restriction places the class in the normal ray module of Proposition~\ref{r:s:d10:prop:normal-ray}. Here the conductor bounds apply to actual finite-flat subquotients. At the dyadic place $L/K$ is split, so the local component character is trivial. Put $S=(R/J)/\mathfrak m^{n+1}$. The module $S^2$ is a finite $2$-primary quotient of the stable $\Z_2$-lattice underlying $R^2$; schematic closure and quotient in its Barsotti--Tate group realize it as a finite-flat group. The submodule $Se_2+Ie_1$ and its submodule $\mathfrak m S e_2$ are stable. Their quotient, followed by a residue-linear functional on $I$, extracts one offdiagonal extension; interchanging the indices extracts the other. Over the unramified cubic completion of $A$, its prime-field coordinates are quadratic extensions. In a nonordinary tuple they have conductor at most two by the middle-model calculation in the proof of Theorem~\ref{r:break}. The depth-two vanishing in Proposition~\ref{r:s:d10:prop:normal-ray} therefore kills the global class. For an ordinary tuple, the two dyadic residual Frobenius eigenvalues are distinct. The common ordinary line is consequently the same Frobenius projector in every constituent, and extends over $R$. Straightening it at first order makes one normal orientation a local coboundary, which vanishes on $G_A$. Fontaine's bound puts both quadratic coordinates at conductor at most four. The injectivity of the corresponding local eigenprojection in Proposition~\ref{r:s:d10:prop:normal-ray} again kills the global class. The unramified local class is zero because Frobenius minus one is invertible on the nontrivial unramified cubic character. Thus $\kappa$ is a coboundary. Its value at $u$ forces the matrix defining that coboundary into $W_0^u\otimes I=0$, so the error itself vanishes in the fixed basis. Induction and completeness prove the claim. \end{proof} \subsection{The quartic mild type} \begin{proposition}\label{r:s:d10:prop:quarticsource} Suppose that the controlled two-adic source has the residual field $A$ and the tame quartic supercuspidal type at $v_-$. Then it is induced from a geometric character of $G_L$, where $L=K(\sqrt{-6})$, and is modular. \end{proposition} \begin{proof} Choose a coefficient ring containing $\zeta_3$ and the integral eigenbasis of the actual order-three wild inertia at $v_+$. At $v_-$ the quartic tame generator has scalar square. Its projective quadratic character is $\eta=\chi_{-6}$, so inertia inside $G_A$ has scalar image. Apply Lemma~\ref{r:s:d10:lem:normal-comparison} with $J=0$ to this single source, using its actual ordinary or nonordinary condition at two. The image is monomial in characteristic zero. The component character is $\eta$, so $r=\Ind_{G_L}^{G_K}\theta$ for a character $\theta$. It is de Rham and finitely ramified, as a summand of the restriction of $r$. By global class field theory it is attached to an algebraic Hecke character~\cite[Chapter~III]{r:s:Serre}. Its conjugate is distinct, by the absolute irreducibility of $r$. Automorphic induction from the CM field $L$ is cuspidal and gives $r$, with its regular weights~\cite[Chapter~3]{r:s:AC}. \end{proof} \subsection{The coefficient field of the special source}\label{r:s:d10:sec:d10-trace-blocks} Let $\mathcal R$ be a weight-two strongly compatible system with determinant cyclotomic, good outside the two places above three. Suppose that its type at $v_+$ is the binary supercuspidal type of Proposition~\ref{r:s:d10:prop:type} and its type at $v_-$ is special, with nonzero monodromy and a fixed quadratic scalar character. Let $B$ be the number field generated by its traces at the good places. We consider the case in which all its residual representations at coefficient places above two are the standard representation of $\Gal(A/K)$. Put $k=\F_4$, let $\mathcal W=W(k)$, and fix a primitive cube root of unity $\zeta\in\mathcal W$. \begin{proposition}\label{r:s:d10:prop:d10-trace-degree} Let $d_0$ be the number of nonordinary embeddings of $B$ in $\overline{\Q}_2$. Then $d_0\le2$. For each of the two possible residual unramified quotient characters at the dyadic place, there is at most one ordinary embedding. Every coefficient completion associated to a nonordinary embedding is ramified over $\Q_2$. In particular $[B:\Q]\le4$. \end{proposition} We first describe the integral representation to which we shall apply the finite-flat conditions. Partition the embeddings of $B$ into a nonordinary set and two ordinary sets, according to the eigenvalue $\bar\zeta$ or $\bar\zeta^{-1}$ of residual Frobenius on the unramified quotient. Each set is stable under the absolute Galois group of $\mathcal W[1/2]$. For any nonempty set of this partition, project $B\otimes_{\Q}\mathcal W[1/2]$ to the product of field factors corresponding to the set. Let $R$ be the closed $\mathcal W$-subalgebra of this product generated by the traces of the associated tuple of representations. Its fraction algebra is the entire product, since already the traces at the good places generate $B$ over $\Q$. Thus $R$ is finite free over $\mathcal W$ of rank the cardinality of the set. It is local with residue field $k$. Indeed all constituent reductions give the same residual trace, and every maximal ideal of this integral order lifts to a prime of its normalization. The joint representation has a model on $R^2$. To see this, choose four group elements whose residual matrices form a $k$-basis of $M_2(k)$, using absolute irreducibility. Their trace-pairing Gram determinant is a unit. Pairing any group matrix with these four elements expresses it uniquely in their joint span, with coefficients in $R$; the same argument applies to their products. Their span is thus a free rank-four algebra with reduction $M_2(k)$. Lift a primitive matrix idempotent to this complete algebra and take the corresponding left ideal. This ideal is free of rank two over $R$, and the action on it identifies the algebra with $M_2(R)$ by reduction and Nakayama. Denote the resulting representation by $r$. Its rational constituents are those of the original tuple. In particular the stable $\Z_2$-lattice underlying $R^2$ in their sum is realized by an isogenous Barsotti--Tate group over $K_2$, by the lattice classification \cite[Theorem~4.1]{r:t:Kim}. All its generic subquotients killed by two acquire finite-flat models by schematic closure and quotient. For the nonordinary set their special fibres and their Cartier duals are connected, by the isogeny argument in the proof of Theorem~\ref{r:break}. \subsubsection{Induction modulo rational two} Write $T=R/2R$ with maximal ideal $\mathfrak n$. Choose a wild inertia element $u$ at $v_+$ whose image generates the actual order-three subgroup and use its two integral eigenprojectors to put $r(u)=\diag(\zeta,\zeta^{-1})$. The standard residual representation in this basis is monomial with component character $\eta=\chi_{-6}$. Here $\chi_a$ denotes the quadratic character of $K(\sqrt a)/K$. The offdiagonal matrices with residual conjugation action form the standard module $W\otimes_{\F_2}k$ of Proposition~\ref{r:s:d10:prop:normal-ray}. In particular $W^u=0$ and $H^i(S_3,W)=0$ for $i>0$, since $W$ is projective. \begin{lemma}\label{r:s:d10:lem:d10-monomial-mod-two} In the fixed $u$-eigenbasis the image of $r$ over $T$ is monomial with component character $\eta$. \end{lemma} \begin{proof} A tame generator $U$ at $v_-$ has the two characteristic-zero eigenvalues $\epsilon,\epsilon$, with one fixed quadratic sign $\epsilon=\pm1$ across the tuple. Hence $U^2=2\epsilon U-1$ in the joint order and $U^2=1$ modulo rational two. The inertia in $G_A$ is cut out by the ramified quadratic subfield, so it acts trivially modulo two; higher inertia already acts trivially in the special source. Lemma~\ref{r:s:d10:lem:normal-comparison}, with $J=2R$, proves the assertion for the nonordinary block and for each ordinary orientation. \end{proof} \subsubsection{The residual scalar character} \begin{lemma}\label{r:s:d10:lem:d10-scalar-cotangent} For the nonordinary set $\dim_k\mathfrak n/\mathfrak n^2\le1$. If this space is nonzero, its visible first-order scalar character is a nonzero multiple of $\chi_{-2}$ modulo $k\eta$. For either ordinary set $\mathfrak n=0$. \end{lemma} \begin{proof} A $k$-linear functional on $\mathfrak n/\mathfrak n^2$ gives a specialization to $k[\delta]/(\delta^2)$. Lemma~\ref{r:s:d10:lem:d10-monomial-mod-two} makes the specialized representation monomial. Compare it with the constant residual matrices. Their determinant agrees, since the cyclotomic determinant is one in characteristic two. The diagonal infinitesimal error therefore has equal entries, even on an antidiagonal coset after multiplying by the inverse residual matrix. The error is $\delta c(g)$ times the identity for an additive character $c:G_K\to k$. Changing basis by an infinitesimal diagonal matrix can add a multiple of $\eta$; conjugation of such a matrix by an antidiagonal element interchanges its diagonal entries. Accordingly the class of $c$ modulo $k\eta$ depends linearly on the functional. This map is injective: if the character is a multiple of $\eta$, the representation is conjugate to its constant reduction, so every trace has zero infinitesimal part and the trace generators force the functional to vanish. At $v_+$ choose an antidiagonal inertia element $t$ with $t^2=-1$. Multiply an arithmetic Frobenius lift by $t$ if necessary so that it centralizes $u$. It takes $t$ to $-u^i t$ for some $i$, since it acts by cubing on the tame quotient of order four. Multiplying the lift by a power of $u$ makes its diagonal ratio exactly $-1$. Its eigenvalues are then $\lambda,-\lambda$ with $\lambda^2=-3$. The trace after multiplication by $u$ is $\lambda(\zeta-\zeta^{-1})=\pm3$. It is rational and common to the tuple by compatibility of the local Weil representations at $v_+$. Thus $\lambda$ is one fixed value $\pm(\zeta-\zeta^{-1})$ in $\mathcal W$. The scalar error vanishes on this Frobenius and on $u$, so $c|_{D_{v_+}}$ is a multiple of $\eta$. At the dyadic place the nonordinary scalar errors have conductor at most two, by extracting the middle self-extension on the unramified cubic base as in the proof of Lemma~\ref{r:s:d10:lem:normal-comparison}. Unramified base change preserves positive breaks. In an ordinary set the exact unramified quotient shows instead that $c$ is unramified at two. Use the supported squareclass basis $\langle-1,3+s,2,3,2+s\rangle$ of Section~\ref{r:s:d10:sec:dihedral}. The condition at $v_+$ and dyadic conductor at most two leave \[ \begin{array}{c|rrrr} a&1&-2&3&-6\\ \hline \text{dyadic conductor exponent}&0&2&2&0 . \end{array} \] For an explicit verification form the $32$ products $a$ of the displayed basis, keep those with $a$ or $-6a$ a square at $v_+$, and compute the dyadic exponent of the relative discriminant of $x^2-a$ (zero for the trivial class). Equivalently this is the calculation in \texttt{scalar\_characters.gp}. Scalar extension from $\F_2$ gives the same kernel conditions over $k$. Modulo $\eta$ only $\chi_{-2}$ survives, and it is ramified at two. Hence the nonordinary cotangent dimension is at most one and an ordinary set has zero cotangent dimension. Nakayama's lemma gives $\mathfrak n=0$ in the latter case. The exact calculation is supplied in~\cite[\path{ramified}]{code}. \end{proof} \subsubsection{The local height condition} \begin{proof}[Proof of Proposition~\ref{r:s:d10:prop:d10-trace-degree}] Lemma~\ref{r:s:d10:lem:d10-scalar-cotangent} shows that each nonempty ordinary set has cardinality $\dim_k T=1$. For the nonordinary set we may assume the cotangent dimension is one. Then $T=k[z]/(z^h)$, where $h=\dim_k T$, and its nonzero tangent is represented by $\chi_{-2}$ modulo $k\eta$ after scaling $z$. Restrict to the unramified cubic extension $F/K_2$ killing the residual dyadic characters. Since $L/K$ splits at $2$, Lemma~\ref{r:s:d10:lem:d10-monomial-mod-two} splits the local representation as two rank-one $T$-characters. The restriction of $\eta$ is trivial, and the first-order error of each character is ramified over $F$, since $\chi_{-2}$ has conductor two there. Take one of the rank-one summands as a finite-flat subquotient of the regular lattice modulo two. It and its dual have connected special fibres. By Lemma~\ref{r:middle}, the generic coefficient action, the images of powers of $z$, and the resulting exact filtrations extend to the finite-flat model. Its Kisin module is free of rank one over $T[[u]]$ on every residue embedding factor after enlarging the finite constant field to accommodate the residue field of $F$. Indeed the $z$-filtration is exact, with rank-one successive quotients on each factor. The quotient by $z$ has rank one over the residue power-series ring, so lifting its generator gives a surjection by Nakayama. The ranks over that power-series ring are equal, giving the assertion. Contravariance interchanges the subobject and quotient filtrations and gives the same lengths. Frobenius permutes the residue embedding factors. The Frobenius multiplier on each factor reduces modulo $z$ to $u$ times a unit. Weierstrass preparation writes it as a unit times $u-\lambda_i$ with $\lambda_i\in(z)$. Height one implies $u^2\in(u-\lambda_i)$, since the Eisenstein polynomial of a uniformizer reduces modulo two to $u^2$. Evaluating at $\lambda_i$ gives $\lambda_i^2=0$. If all $\lambda_i$ vanished modulo $z^2$, the first-order multipliers would each be $u$ times a power-series unit. After the middle exponent is removed, the corresponding Artin--Schreier class has no pole on any residue factor and is unramified. This is the middle self-extension calculation in the proof of Theorem~\ref{r:break}, including its comparison of breaks with the field of norms. It contradicts the ramification just proved. Thus one $\lambda_i$ is a unit times $z$, and $z^2=0$. This gives $h\le2$. Finally suppose that the completion of $B$ belonging to a nonordinary embedding is unramified over $\Q_2$. The trace-order construction above, applied to a single completion before and after unramified coefficient extension, realizes its representation on an integral rank-two lattice there. The residual local representation splits into two distinct unramified lines after unramified coefficient extension. Over the unramified cubic base, Lemma~\ref{r:middle} splits its finite-flat model as the direct sum of middle models. Therefore Frobenius at $u=0$ on the integral Kisin module vanishes modulo two. This can be checked after unramified base extension by faithful flatness. Over the original base with residue field $\F_2$, the integral module has coefficient rank two: the underlying finite-free module with unramified coefficient action is free over the coefficient power-series ring, by reduction and Nakayama. Since two is a uniformizer, every entry of its Frobenius-at-zero matrix belongs to $2\OO_E$, where $E$ is the enlarged unramified coefficient field. Its determinant would have valuation at least two. On the other hand the determinant of crystalline Frobenius has valuation one because the determinant character is cyclotomic \cite[Proposition~4.2]{r:t:Kim}. This contradiction proves the ramification assertion. Summing the three bounds proves the proposition. \end{proof} \subsection{The small coefficient prime} Let $B$ be the minimal global trace field of the system with a special condition at $v_-$. We are in the case in which the residual representation at every embedding $B\hookrightarrow\overline\Q_2$ is the standard representation of the same $S_3$ extension $A/K$. Modularity at any one coefficient prime suffices for modularity of the entire system. \begin{proposition}\label{r:s:d10:prop:fielddegree} If all the dyadic embeddings are ordinary, or all are nonordinary, then $[B:\Q]\le2$. Otherwise $B$ is quartic, ramified at two, and is generated by the Frobenius trace $a_2$ at the dyadic prime. Its minimal polynomial is congruent to $X^2(X^2+X+1)$ modulo two. \end{proposition} \begin{proof} The trace order in each set bounds the number of embeddings in that set. Residue Frobenius interchanges the ordinary sets, so if they occur each has exactly one embedding. A nonordinary completion must ramify and its block has at most two embeddings. In the mixed case it has exactly two. At an ordinary embedding one crystalline Frobenius root is the unramified quotient and the other is divisible by the maximal ideal; $a_2$ thus reduces to the two distinct primitive cube roots in the two orientations. At each nonordinary embedding it is in the maximal ideal by the Newton slope criterion. Strong compatibility supplies $a_2\in B$ by taking a coefficient prime away from two. Each ordinary root occurs once, whereas restricting the embeddings to a proper subfield would repeat every root $[B:\Q(a_2)]$ times. This proves the proposition. \end{proof} \begin{lemma}\label{r:s:d10:lem:screen} Let $B$ be a mixed quartic field as in Proposition~\ref{r:s:d10:prop:fielddegree}. Then at least one of the following holds: \begin{enumerate}[label=(\roman*)] \item Every prime of $B$ above three has residue degree at most two. \item There is a degree-one prime of $B$ above five or seven. \item Up to replacing $a_2$ by $-a_2$, its minimal polynomial is \[ X^4-X^3-7X^2+8X+2 \quad\text{or}\quad X^4-X^3-5X^2+2X+2. \] \end{enumerate} \end{lemma} \begin{proof} Every conjugate of the algebraic integer $a_2$ lies in $[-\sqrt8,\sqrt8]$. Write its minimal polynomial as $P=X^4+aX^3+bX^2+cX+d$. By Proposition~\ref{r:s:d10:prop:fielddegree}, $a,b$ are odd and $c,d$ are even. The elementary symmetric functions and Cauchy's inequality give \[ |a|\le11,\quad |b|\le48,\quad |c|\le90,\quad 0<|d|\le64,\quad 0\le a^2-2b\le32,\quad 8b\le3a^2. \] Replacing $a_2$ by $-a_2$, we take $a>0$. Exact rational Sturm sequences on $[-283/100,283/100]$, with the integral coefficient bounds above, leave the twelve irreducible polynomials in the following table. For a row marked $(p,r,e)$ we have $P(r)\equiv0$ and $P'(r)\equiv e\ne0\pmod p$. \begin{center} \begin{tabular}{rrrrl} \toprule $a$&$b$&$c$&$d$&Local check\\ \midrule 1&-9&-6&14&$(5,2,2)$\\ 1&-7&-10&-2&one prime above $3$, of residue degree $2$\\ 1&-7&-8&2&$(3,2,2)$; first exceptional polynomial\\ 1&-7&-6&2&$(7,2,3)$\\ 1&-7&-6&4&$(7,1,1)$\\ 1&-7&-4&4&$(5,1,4)$\\ 1&-7&-4&8&$(5,4,4)$\\ 1&-7&-4&10&$(5,0,1)$\\ 1&-5&-4&2&$(5,1,3)$\\ 1&-5&-2&2&$(3,1,1)$; second exceptional polynomial\\ 1&-5&-2&4&odd field discriminant\\ 3&-3&-12&-6&Eisenstein at $3$\\ \bottomrule \end{tabular} \end{center} The accompanying certificate uses only integer and rational arithmetic: it tests irreducibility by Gauss's lemma and counts real roots by Sturm's theorem. Thus this step requires no maximal-order computation. A simple root modulo five or seven gives an embedding $B\hookrightarrow\Q_p$ by Hensel's lemma, proving~(ii). For the second row, set $Q=X^2-X-1$. Then \[ P=Q^2+3(X^3-2X^2-4X-1). \] The polynomial $Q$ is irreducible modulo three, and the remainder of the second factor modulo $Q$ is $-4X-2$. Dedekind's index criterion shows that three does not divide $[\mathcal O_B:\mathbf Z[a_2]]$; hence three has one prime in $B$, of ramification index two and residue degree two. The final row is Eisenstein at three and has one prime of residue degree one. Both rows give~(i). For the penultimate row, put $\alpha=a_2$ and $\beta=(\alpha^3+\alpha^2+\alpha)/2$. Direct reduction modulo $P$ gives \[ \beta^4-29\beta^2+10\beta+19=0. \] Thus $\beta$ is integral. Since $\operatorname{disc}P=8900$, the integral lattice spanned by $1,\alpha,\alpha^2,\beta$ has discriminant $2225$. The field discriminant is therefore odd, contrary to Proposition~\ref{r:s:d10:prop:fielddegree}. The two remaining rows are exactly the polynomials in~(iii), after replacing $X$ by $-X$. \end{proof} \begin{lemma}\label{r:s:d10:lem:threecyclo} If an irreducible residual representation of determinant $\bar\chi_3$, unramified outside three, is restricted to $G_{K(\zeta_3)}$, it remains irreducible. \end{lemma} \begin{proof} Otherwise the representation is induced from a character $\theta$ of $H=K(\zeta_3)$ with $\theta^\sigma=\theta^{-1}$. Such a character has order prime to three. Its ray conductor divides the product of the two primes above three, both of residue degree one and ramification index two. An absolute equation of $H$ is $X^4+10X^2+100$. Its class group is cyclic of order four, certified by \texttt{bnfcertify}; its ray group for this modulus is also $C_4$. If $S$ is the action of the nonidentity automorphism over $K$, the Smith invariant of the lattice generated by $4$ and $1+S$ is two. Equivalently this ray group modulo $1+\sigma$ is $C_2$. Thus $\theta^2=1$ and $\theta=\theta^\sigma$, contradicting irreducibility. These data are reproduced by \texttt{cyclotomic\_three\_ray.gp}. \end{proof} \begin{proof}[Proof of Theorem~\ref{r:s:d10:thm:residual}] Soluble residues have solvable odd Artin lifts, which give modularity by Hecke induction or Langlands--Tunnell and regular weight raising. Assume therefore that the prescribed three-residue is nonsolvable. The bounded lift, the discriminant estimate and the soluble-image classification reduce the proof to the special source with the fixed $S_3$ residue, the quartic source having already been treated. Apply Proposition~\ref{r:s:d10:prop:fielddegree}. The exact calculation is supplied in~\cite[\path{ramified}]{code}. If the degree of $B$ is at most two, or every coefficient prime above three has degree at most two, choose the originally prescribed prime. Its residue is realizable over $\F_3$ or $\F_9$ by its traces, absolute irreducibility and the vanishing of the Brauer group of a finite field. The small-image theorem of Allen--Khare--Thorne~\cite[Theorem~2.5]{h:AKT} gives the required cuspidal residual witness. At a degree-one coefficient prime of characteristic $p=5$ or $7$, the binary inertia at $v_+$ remains irreducible modulo $p$; the order-three element has distinct eigenvalues and a tame element interchanges its eigenlines. The local cyclotomic extension at $v_+$ is unramified, so the residual representation is absolutely irreducible even on $G_{K(\zeta_p)}$. Theorems~2.5 and~2.3 of~\cite{h:AKT} give respectively modularity of its residue and modularity of the actual de Rham member with weights $\{0,1\}$. This includes the ramified prime five. Strong compatibility gives modularity at the original prime. The last two polynomials in Lemma~\ref{r:s:d10:lem:screen} have a coefficient place above three with residue field $\F_3$. If its residue were reducible, write the characters as $\psi$ and $\bar\chi_3\psi^{-1}$. Lemma~\ref{r:s:d10:lem:q10-character-descent} gives $\psi^2=1$ or $\nu$, the Hilbert class character, which is nontrivial unramified at both triadic places. On $G_{K_{v_-}}$ a quadratic Steinberg twist has semisimplification $\alpha\oplus\alpha\chi_3$ with $\alpha$ quadratic, so $\psi^2$ is locally trivial. It follows that $\psi^2=1$. At the good dyadic place the residual trace would be $\psi(\Frob_2)(1+2)=0$. For the selected coefficient prime $\lambda\mid3$, this gives $a_2\in\lambda$. Hence $3\mid N_{B/\Q}(a_2)$. Both polynomials have constant term $2$, so $N_{B/\Q}(a_2)=2$, a contradiction. The residue is therefore irreducible. Lemma~\ref{r:s:d10:lem:threecyclo} and \cite[Theorems~2.5 and~2.3]{h:AKT} make the actual member modular. This proves the residual theorem. \end{proof} \section{\texorpdfstring{Quadratic symmetry over $\Q(\sqrt{11})$}{Quadratic symmetry over Q(sqrt(11))}}\label{r:s:sec:d11} Put $K=\Q(s)$, $s^2=11$. \subsection{Arithmetic and modular witnesses} We will use the elements \[ \epsilon=10+3s,\qquad \pi=3+s,\qquad \alpha=4+s,\qquad \alpha'=4-s. \] They satisfy $N(\epsilon)=1$, $N(\pi)=-2$ and $N(\alpha)=N(\alpha')=5$. \begin{lemma}\label{r:s:d11:lem:arithmetic} The class number of $K$ is one, and its units are generated by $-1$ and $\epsilon$. Its finite ray class groups of moduli $(3)$ and $(5)$ are both cyclic of order four. With both real places added to the modulus, the groups are $C_8\times C_2$ at three and $C_4\times C_2$ at five. \end{lemma} \begin{proof} Minkowski's bound is $\sqrt{44}/2<3.32$. Three is inert, and the dyadic prime is generated by $\pi$, so the class number is one. The least nontrivial positive solution of $x^2-11y^2=1$ is $(10,3)$; checking $y=1,2$ proves that $\epsilon$ is fundamental. The negative Pell equation has no solution because $-1$ is not a square modulo eleven. Modulo three the residue group is $\F_9^\times$ and $\epsilon=1$. The image of the units is $\{\pm1\}$, giving a cyclic quotient of order four. Adding the real signs gives \[ (C_8\times C_2^2)/\langle(-1,-1,-1)\rangle \simeq C_8\times C_2. \] At the two five-places, ordered by $s=1,-1\bmod5$, $\epsilon$ has residues $(3,2)$, of order four, and its square is the image $(-1,-1)$ of $-1$. The product map $(u,v)\mapsto uv$ identifies the finite ray quotient with $\F_5^\times$. Adding the signs leaves one independent sign factor and gives $C_4\times C_2$. \end{proof} \subsection{Characteristic-three sources} \begin{lemma}\label{r:s:d11:lem:ordinary3} Let $F/\Q_3$ be unramified quadratic, and let $r$ be crystalline of labelled weights $\{0,1\}$ with determinant $\chi_3$. If its residual constituents have inertia characters $1,\omega_3$, then $r$ is ordinary. \end{lemma} \begin{proof} Since $e(F)=1<3-1$, the rank-one finite-flat models of these characters are uniquely etale and multiplicative. Choose a Barsotti--Tate group with coefficient action realizing a stable lattice. If it has a nonzero etale rational quotient, the coefficient rank is one, and the determinant gives the ordinary filtration. Otherwise the group and its Cartier dual are connected, using $r^\vee(1)\simeq r$ and invariance of connectedness under isogeny. The closure of a residual invariant line is etale or multiplicative. The former contradicts connectedness, and the latter gives an etale quotient of the connected dual. Thus the source is ordinary. \end{proof} \begin{proposition}\label{r:s:d11:prop:reducible3} If $\bar r_3^{\ssm}=\psi\oplus\omega_3\psi^{-1}$ for a bounded system $\mathcal R$, then $\psi^2=1$ and the actual source $r_3$ is ordinary and modular. \end{proposition} \begin{proof} At a five-place the semisimplified residual inertia of our two types is scalar quadratic: the cubic part of the cuspidal inertia disappears modulo three, as does the unipotent part of the special condition. Hence $\psi^2$ is unramified there. A characteristic-three finite character kills principal units at three, so $\psi^2$ factors through the finite ray group of modulus $(3)$ from Lemma~\ref{r:s:d11:lem:arithmetic}. Fontaine--Laffaille theory \cite{r:s:FL} over the unramified quadratic $K_3$ gives the possible inertia exponents $0,1,3,4\bmod8$ of a rank-one constituent in weight two. The determinant has exponent four. Thus the reducible pairs are $\{0,4\}$ and $\{1,3\}$. In the second case $\psi^2$ is a faithful quartic ray character. Both five-primes generate that ray group: their positive generators $\alpha,\alpha'$ reduce to $1+s,1-s$ in $\F_9$, and \[ (1+s)^2=-s,\qquad (1+s)^4=-1,\qquad 1-s=(1+s)^3. \] It follows that $\psi^2(\Frob_v)$ is a primitive fourth root at either five-place. The quotient of the two residual local characters would therefore have Frobenius value of order four. For both of our five-adic types that quotient is instead $-1$ in characteristic three: the special eigenvalues have ratio five, while the cuspidal parameter has trace zero and determinant five. Scalar quadratic factors do not change the quotient. This excludes $\{1,3\}$. For the pair $\{0,4\}$, $\psi^2$ is unramified also at three and is even at both real places. The ordinary class number is one, so $\psi^2=1$. Lemma~\ref{r:s:d11:lem:ordinary3} makes the actual source ordinary, with unramified quotient. Its residual constituent ratio is $\omega_3$, which is totally odd, extends to $G_\Q$, and is nontrivial at $K_3$. The field $K(\zeta_3)$ is abelian over $\Q$. All the hypotheses of Skinner--Wiles \cite[Theorem~A]{r:t:SW99} are satisfied, so $r_3$ is modular. \end{proof} \begin{proposition}\label{r:s:d11:prop:cyclo3} Let $r:G_K\to\GL_2(\overline\Q_3)$ be continuous, irreducible, totally odd and finitely ramified, with determinant $\chi_3$, and crystalline of weight two at three. If $\bar r$ is absolutely irreducible but becomes reducible on $G_{K(\zeta_3)}$, then $r$ is ordinary at three and modular. \end{proposition} \begin{proof} Put $H=K(\zeta_3)$. Clifford theory gives $\bar r=\Ind_{G_H}^{G_K}\xi$, where $\xi$ has finite order prime to three. Wild inertia at three is therefore trivial. The extension $H_3/K_3$ is ramified quadratic. A tame generator $t$ outside $G_{H_3}$ acts antidiagonally in an inducing basis, and \[ \det\bar r(t)=-\xi(t^2)=\omega_3(t)=-1. \] Thus $\bar r(t)^2=1$. A lift of Frobenius can be chosen in $G_{H_3}$, and its diagonal matrix commutes with $\bar r(t)$ because the residue cardinality is nine. Its two eigenvalues are consequently equal, and the determinant makes that common value $\pm1$. The local pair is exactly $\beta,\beta\omega_3$ with $\beta$ unramified quadratic. Lemma~\ref{r:s:d11:lem:ordinary3} applies to the actual source. Induce the Teichmuller lift of $\xi$ to obtain a finite-image, irreducible, totally odd Artin representation. Automorphic induction supplies its weight-one Hilbert theta series. At three its Euler factor has degree one and unramified root $\pm1$, so the form is ordinary and has the correct residual ordinary quotient. Wiles's Hilbert ordinary family theorem \cite[Theorem~1.4.1]{r:s:Wiles} gives a regular ordinary specialization with the same residual quotient. This application is also explicit in \cite[pp.~1133--1134]{r:s:Ellenberg}. We have obtained the quotient-compatible modular lift required by \cite[Theorem~5.1]{r:s:SW01}. The actual source is ordinary and distinguished at three, its determinant is cyclotomic, and its remaining geometric and parity hypotheses are the stated ones. That theorem permits reducibility on the cyclotomic subgroup and proves modularity of $r$. \end{proof} \subsection{Nonsolvable and dihedral dyadic residues} In this section the system is special at both five-primes. Let $M$ be the field of a semisimple dyadic residue. At five its full decomposition image is elementary abelian of exponent at most two: removing its quadratic sign leaves an upper unipotent group, generated by tame inertia and Frobenius. In particular every local field is contained in the maximal elementary quadratic extension of $\Q_5$. \begin{lemma}\label{r:s:d11:lem:aux} Put $F=K(i,\sqrt5)$ and let $H$ be its Hilbert class field. Then $[H:\Q]=16$, its dyadic primes have absolute $(e,f)=(2,4)$ and its four five-primes have $(e,f)=(2,2)$. Its fourth-power dyadic ray, with all five-prime classes killed, has group $C_4^2\times C_2^4$. \end{lemma} \begin{proof} The defining polynomials for $F,H$ are respectively \[ \begin{split} &y^8-4y^7+24y^6-58y^5+141y^4-190y^3+186y^2-100y+20,\\ &y^{16}-9y^{14}+39y^{12}-111y^{10}+214y^8-189y^6+69y^4-6y^2+1 . \end{split} \] The class groups are $C_2,1$; the second field is the class field of the first and has its squared discriminant. Over $H$ the full ray invariants are $12,4,4,2,2,2,2,2$. Adding the four reciprocity columns of the five-primes to the diagonal relation matrix gives the displayed quotient by Smith reduction. The certificate \texttt{sp\_auxiliary.gp} checks the polynomial identification, both class groups with \texttt{bnfcertify}, the prime decompositions and the Smith factors. The exact calculation is supplied in~\cite[\path{ramified/d6_d10_d11/d11}]{code}. \end{proof} \begin{proposition}\label{r:s:d11:prop:nonsolv} Assume GRH. A two-special bounded system with a nonsolvable dyadic residue is modular. \end{proposition} \begin{proof} Its image is $\SL_2(q)$, $q=2^f\ge4$, by Dickson and determinant one. Let $B$ be the ray field of Lemma~\ref{r:s:d11:lem:aux}, of absolute degree 4096 and Galois over $K$ by invariance of the modulus and quotient. The dyadic breaks of $BM/K$ are at most three: this follows from Fontaine's bound for the actual Barsotti--Tate lattice, and from the fourth-power ray over the unramified dyadic extension $H/K$. The two-adic contribution to the logarithmic relative different, in valuation $v_2(2)=1$, is at most $4/2=2$. At five the field $H$ contains the ramified field $\Q_5(\sqrt5)$ and its unramified quadratic extension. This is already the full elementary quadratic envelope. Consequently for $L=BM$ and also $L=BMM^\sigma$, \[ \rd(L)\le4\sqrt{44\cdot5}<59.329588,\qquad 2N_{25}(L)/[L:\Q]=1/2. \] Here $N_Q$ counts primes of norm $Q$. Apply Section~\ref{shared:explicit} with $B_{b,100}$ and the retained prime powers specified below. For the totally real comparison retain $A_{b,4}$. Directed 256-bit evaluation gives \[ \begin{array}{lrrrc} &n\ge &b&\text{powers of 25}&\rd(L)>\\ \hline q\ge16&4096\cdot4080&16.4&9&59.981834\\ q=8,\ M\ne M^\sigma&4096\cdot504^2&20.5&10&62.360509\\ \text{totally real field}&120&4.8&0&56.063060 \end{array} \] Simplicity gives disjointness from the solvable auxiliary. Distinct simple conjugate fields are disjoint over $K$, and their product still has no nontrivial solvable quotient. The first two lines exclude their respective cases. The exact calculation is supplied in~\cite[\path{ramified/d6_d10_d11/shared}]{code}. When $q=8$ and $M=M^\sigma$, conjugation by $\Gal(M/\Q)$ induces an inner automorphism of $\SL_2(8)$ since its outer group has order three. The group is centerless, so conjugation realizes an extension of the actual residue to $G_\Q$. Rational Serre modularity \cite{h:KWI,h:KWII} and quadratic base change give a regular cuspidal witness over $K$; cuspidality follows by irreducibility of the restriction. For $q=4$ use \cite[Theorem~2.5]{h:AKT}. A real conjugation is nontrivial in either case: otherwise $M$ would be totally real of degree at least 120. If wild inertia at two is nontrivial, it fixes a unique line, the diagonal characters are unramified since the base residue field is $\F_2$, and inertia has order at most $q$. Integrating the upper bound three gives exponent at most $2-2/q$; the tame case is smaller. Thus $\rd(M)\le\sqrt{44\cdot5}\,2^{7/4}<49.890038$, contrary to the last table entry. Lemma~\ref{shared:weightzero} and \cite[Theorem~6.1]{g:Thorne2}, applied to the dual in the opposite weight convention, give modularity of the actual dyadic source. \end{proof} \begin{proposition}\label{r:s:d11:prop:dihedral} A two-special system with irreducible solvable two-residue is modular. Every reducible two-residue has trivial semisimplification. \end{proposition} \begin{proof} An irreducible solvable subgroup in determinant one is odd dihedral. Let $L/K$ be its inducing quadratic field. The odd rotation character is unramified away from two, trivial at every five-decomposition group over $L$, and at two is tame. Its conductor divides the product of the dyadic primes of $L$. The quadratic squareclasses are supported on $-1,\epsilon,\pi,\alpha,\alpha'$ by class number one. Odd dyadic valuation gives conductor five by an Eisenstein square-root polynomial and its derivative, contradicting Fontaine's upper break three. Among the fifteen nontrivial classes in $\langle-1,\epsilon,\alpha,\alpha'\rangle$, the odd part of the ray modulo all dyadic primes with all five-primes split is trivial except at $-5\epsilon$, where it is $C_3$. The certificate \texttt{sp\_dihedral\_rays.gp} verifies these quartic ray quotients. Since $\pi^2=2\epsilon$ the survivor is $K(\sqrt{-10})$. Its dyadic relative conductor is four. Theorem~\ref{r:break} excludes a nonordinary actual source. The ordinary source has a nonsplit dyadic prime in its imaginary inducing field and the specified near-ordinary line; Allen's ordinary dihedral theorem \cite[Theorem~5.2.1]{g:Allen} applies. For a reducible residue the odd-order constituents are unramified at two and five and everywhere else, so class number one makes both trivial. The exact calculation is supplied in~\cite[\path{ramified/d6_d10_d11/d11}]{code}. \end{proof} \subsection{Ordinary scalar systems} \begin{proposition}\label{r:s:d11:prop:ordinaryscalar} A two-special system with a scalar semisimple two-residue and an ordinary member above two is modular. \end{proposition} \begin{proof} In that member choose a vector $v$ of the ordinary line and the lattice generated by the global orbit of $v$. Its reduction is cyclic, with unique global invariant line transverse to $v$. The local decomposition image at two preserves both lines, so it is trivial. The supported quadratic characters splitting locally at two form the single line $\chi=\chi_{-5}$. Indeed the sixteen classes $\langle-1,\epsilon,\alpha,\alpha'\rangle$ contain precisely $1,5,-1,-5$ unramified at two and precisely $1,-5$ split there; a class of odd dyadic valuation cannot split. These local computations are recorded in \texttt{ordinary\_squareclasses.gp}. Every coordinate of the additive residual extension is a multiple of $\chi$. For a real involution $c$, the basis $((r(c)-1)v,v)$ is therefore integral and normalizes the residue, the real matrix and the ordinary mark to \[ 1+\chi E_{12},\qquad \begin{pmatrix}-1&1\\0&1\end{pmatrix},\qquad e_2 . \] Use the marked ordinary crystalline functor of \cite[Proposition~2.4.4, Corollary~2.4.5]{h:Kisin2}, pulled back to the matrix problem with this fixed real matrix, support and determinant. At five fix the quadratic characters on inertia, and allow the unramified special signs to vary. The special line is represented without an extra free parameter: for a tame generator normalized by the given sign, its difference from one has unit upper-right entry, so the coordinate of a fixed line in the $e_1$ chart is determined by its first row. Stability and the character equations on that line and its quotient, including square one of the latter, are then closed equations. The exact calculation is supplied in~\cite[\path{ramified}]{code}. Write a homogeneous tangent as $(1+tX(g))\bar r(g)$, where $X(g)=\left(\begin{smallmatrix}x(g)&y(g)\\z(g)&x(g)\end{smallmatrix}\right)$. Conjugation adds $\chi z(1+E_{12})$, so $z$ is additive. At two, $y=0$, $x$ is unramified and $z$ is a unit Kummer class by the finite-flat ordinary extension condition. At both five-primes the special line has the form $e_1+t t_w e_2$, whence $z=0$ on the whole decomposition group. The two valuations exclude $\alpha,\alpha'$; $\epsilon$ is a nonsquare at both places. Thus $z$ is a multiple of $\chi_{-1}$ and the fixed real matrix kills it. The remaining $x,y$ are additive. The split-at-two calculation gives $y\in k\chi_{-5}$, and $y(c)=0$ kills it. The unramified-at-two calculation and $x(c)=0$ leave $x\in k\chi_5$. This last direction is detected by the marked unramified quotient $\nu$ on dyadic Frobenius, since $\chi_5$ is locally nontrivial unramified. Cotangent Nakayama makes the complete ring a quotient of $W(k)[[\nu(\Frob_2)-1]]$. Choose $\eta=1$ or $\chi_{-5}$ aligning the inertial characters of $r$ and $r^\sigma\otimes\eta$ at both five-primes. The twist is dyadically trivial. The two normalized points agree on the sole generator, since the local involution fixes unramified Frobenius. They agree as marked representations. Lemma~\ref{r:quadratic-descent} applies. \end{proof} \subsection{The integral order of nonordinary scalar systems} Take the tuple at all dyadic coefficient places of the bounded systems under consideration. Every member is nonordinary with trivial residual semisimplification, and the systems have common quadratic inertial signs at each special place. Use the orders $R_0,E_0$ of Lemma~\ref{r:shared:trace-order}. For $m$ distinct coefficient orbits of common trace-field degree $d$, \[ \operatorname{rank}_{\Z_2}R_0=md,\qquad \operatorname{rank}_{\Z_2}E_0=4md. \] Write $E=E_0/2E_0$; its radical has residue field $\F_2$. \begin{lemma}\label{r:s:d11:lem:regular} The left regular lattice of $E_0$ is Barsotti--Tate at two and biconnected. For the simple factors of its torsion with trivial generic character, the middle order-two model is unique. If $J$ is the radical of $E$, the differences $A,B,C$ from one of a real involution and of the two normalized tame five-inertias generate $E$ as an $\F_2$-algebra, and all three have square zero. \end{lemma} \begin{proof} Apply Lemmas~\ref{r:regular-bt} and~\ref{r:middle} to the regular lattice and its trivial-simple torsion subquotients. These lemmas also hold after the unramified base changes used below. A radical functional gives a middle/middle extension by quotienting the regular model. At the dyadic place its quadratic conductor is at most two by the height-one Artin--Schreier calculation in the proof of Theorem~\ref{r:break}. These characters belong to the narrow ray modulo $\pi^2(5)$, with invariants $4,2,2$. Its quadratic dual is spanned by \[ [-1],\quad [\epsilon\alpha],\quad [\epsilon\alpha'] . \] Their dyadic discriminant exponents are respectively $0,2,2$. The latter two ramify at distinct five-primes and split at the opposite prime. Real conjugation and the two tame inertias evaluate in the three coordinate directions. Hence $A,B,C$ span $J/J^2$. Nilpotence gives algebra generation. The real involution squares to one and each normalized tame generator is unipotent with square-zero difference, proving the lemma. The arithmetic is certified in \texttt{scalar\_regular\_rays.gp}. The exact calculation is supplied in~\cite[\path{ramified}]{code}. \end{proof} \subsection{A quaternion obstruction over the Hilbert class field} The next result is used both for the original reduction and a central quotient of it. Suppose $E'$ is such an actual regular finite-flat quotient, with elementary abelian dyadic-characteristic decomposition image at both five-primes of exponent two. Let $M'$ be its point field, and $P=\Gal(M'H/H)$. All five-primes split in $M'H/H$ because $H$ contains the maximal elementary quadratic completion at five. Its dyadic primes have absolute ramification index two; Fontaine's bound gives upper breaks at most three. \begin{lemma}\label{r:s:d11:lem:quat} Assume GRH. The group $P$ is elementary abelian of rank at most two. Its span in $E'$ has dimension at most four and at most two radical generators. \end{lemma} \begin{proof} Denote the dyadic primes of $H$ by $v_1,v_2$. The ray groups with all five-prime classes killed at dyadic conductor exponents $(4,0)$ and $(0,4)$ are trivial; the corresponding $(2,2)$ group is $C_2^2$. If either inertia image were proper in the finite two-group $P$, it would lie in an index-two subgroup. The associated quadratic extension would contradict one of the first two ray calculations. Thus inertia at either dyadic place is all of $P$. Let $C_0$ be the span of $P$ in $E'$ and $J_0$ its radical. Its regular representation is the actual submodule $C_0\subset E'$. At either dyadic completion it therefore has a biconnected finite-flat model. A functional on $J_0/J_0^2$ gives a middle/middle extension, whose conductor is at most two by the calculation in the proof of Theorem~\ref{r:break}. The $(2,2)$ ray calculation gives $\dim J_0/J_0^2\le2$. At each dyadic place the ramified Artin--Schreier coordinates of independent such functionals are independent over $\F_2$: a nonzero linear combination with zero ramified coordinate would be unramified, whereas inertia is all of $P$. The right action of $C_0$ extends to its finite-flat model by Lemma~\ref{r:middle}. The quotient by the right radical is the unique middle model. Exactness, Nakayama's lemma and equality of ranks give a Kisin module free of rank one over $(k\otimes_{\F_2}C_0)[[u]]$, where $k=\F_{16}$. Normalize its Frobenius to $u+V(u)$, with $V(u)\in(k\otimes J_0)[[u]]$. Height one makes $u^2(u+V)^{-1}$ integral. Modulo $J_0^3$ this expression is $u-V+V^2/u$, so $V(0)^2=0$. With one radical generator its nonzero ramified coefficient gives $J_0^2=J_0^3$, hence $J_0^2=0$. With two generators write $V(0)=\alpha X+\beta Y\pmod{J_0^2}$. Here $\alpha,\beta\in\F_{16}$ are $\F_2$-independent. If $\beta/\alpha\notin\F_4$, then $1,\beta/\alpha,(\beta/\alpha)^2$ are independent; otherwise their unique relation is $1+t+t^2=0$. Consequently the classes of $X^2,XY+YX,Y^2$ in $J_0^2/J_0^3$ are equal to a common element $z$, which is zero in the first case. If $z=0$, the associated graded algebra is a quotient of $\F_2[X,Y]/(X^2,Y^2)$, so $J_0^3=0$ and $C_0$ is itself a quotient of this commutative algebra. If $z\ne0$, choose two group elements whose augmentation differences form a cotangent basis. In the quotient modulo $J_0^3$ their squares and commutator are $1+z$. The kernel over the two-dimensional additive quotient is central elementary abelian. Quotienting it by a complement to $\langle1+z\rangle$ gives a quaternion quotient of $P$. Its three quadratic subfields are the subfields of the unique split-five conductor-$(2,2)$ biquadratic extension of $H$. For the selected quadratic subfield $H'/H$, the certificate \texttt{scalar\_quaternion.gp} gives dyadic break one at both places and the split-five ray group $C_4\times C_2^5$ at conductor $(6,6)$. Its quotient on which the involution of $H'/H$ acts by inversion is $C_2^6$; the latter assertion is the Smith reduction obtained by adding the columns of $S+I$ to the ray relation matrix. A quaternion quotient would supply an order-four character of this quotient: Herbrand's formula bounds its break by $1+2(3-1)=5$, and conjugation inverts the rotation character. This is impossible. The class and unit calculation in degree thirty-two uses GRH. The exact calculation is supplied in~\cite[\path{ramified/d6_d10_d11/d11}]{code}. We have proved that $C_0$ is a quotient of $\F_2[X,Y]/(X^2,Y^2)$, so $P$ is elementary abelian and $|P|\le8$. To exclude order eight, suppose it occurs. Then $C_0$ is exactly this four-dimensional algebra and $P$ is its group of augmentation-one units. At either dyadic place write the regular Frobenius as \[ u\bigl(1+fX+gY+hXY\bigr), \qquad f,g,h\in u^{-1}k[[u]]. \] After removing the simple factor $u$, a fundamental solution $1+xX+yY+wXY$ satisfies \[ x^2-x=f,\qquad y^2-y=g,\qquad (w+xy)^2-(w+xy)=h+fg. \] Multiplication by $1+aX+bY+cXY\in P$ translates $(x,y,w+xy)$ by $(a,b,c+ab)$. These three additive coordinates therefore form a basis of the quadratic characters of $P$. The first two right sides have poles of order at most one, and the third has order at most two. A double pole is equivalent to a simple pole modulo Frobenius minus identity because $k$ is perfect. Every character of $P$, being an $\F_2$-linear combination of these three, thus has field-of-norms upper break at most one. Independently, the actual finite-flat point field gives classical upper break at most three for every such character by Fontaine's bound. The first break of the uniformizer tower is four and all later transition breaks are larger. Herbrand base change preserves these original breaks, and the ramification comparison in the proof of Theorem~\ref{r:break} identifies them with the field-of-norms breaks. Every character consequently has conductor at most two over each $H_{v_i}$. Thus $P$ is a quotient of the split-five ray group of conductor $(2,2)$, which is $C_2^2$, a contradiction. \end{proof} \subsection{Quadratic conjugation and the scalar theorem} \begin{lemma}\label{r:s:d11:lem:finite} In the setting of Lemma~\ref{r:s:d11:lem:quat}, if $E'$ is generated by the three real and tame elements above, then $\dim_{\F_2}E'\le24$. \end{lemma} \begin{proof} The group $\Gal(H/K)$ is dihedral of order eight: it has the biquadratic quotient $\Gal(F/K)$, a dyadic Frobenius of order four, and involutions given by real conjugation and five-inertia with independent images in that quotient. The two five-inertias belong to the same reflection class. Set $\Gamma=\Gal(M'H/K)$. Choose real and tame involutions $a,b,c$ so that $b,c$ have the same image in $\Gal(H/K)$, and put $r=ab$, $c=jb$ with $j\in P$. Write $R,B$ for conjugation by $r,b$ on $P$. Thus $brb=r^{-1}$ and $Bj=j$. The image of the two-group $\Gamma$ in $\mathrm{Aut}(P)$ is either trivial or generated by one transvection on a two-dimensional space. In particular $R^2=1$. At the five-place belonging to $b$, a Frobenius lift, after multiplication by inertia if necessary, has the form $f=p r^2$, $p\in P$. The five-decomposition image in the compositum $M'H$ is elementary abelian, so $f^2=1$. Since $R^2=1$ and $P$ has exponent two, $f^2=r^4$, and therefore $r^4=1$. Thus $\langle r,b\rangle$ maps isomorphically to the dihedral quotient of order eight. The subgroup $\Gamma_0=\langle a,b,c\rangle$ is $P_0\rtimes\langle r,b\rangle$, where $P_0=\langle j,Rj\rangle$; indeed $B$ fixes $j$ and commutes with $R$. Its group span maps onto $E'$, even when $\Gamma_0$ is proper in $\Gamma$. If $\dim P_0\le1$, this gives $\dim E'\le|\Gamma_0|\le16$. Otherwise $P_0=P$ has rank two, $R=U$ is a transvection, and $j$ does not belong to its fixed line. Since $B\in\{1,U\}$ fixes $j$, we have $B=1$. Now $\Gamma=\Gamma_0=P\rtimes D_8$ has order thirty-two. Let $q=R(j)j$ in multiplicative notation. The elements $q$ and $r^2$ are independent central involutions. The Cayley--Hamilton identity with determinant one makes $r+r^{-1}$ central in $E'$. Commuting it with $j$ gives \[ (1+q)(1+r^2)=0\qquad\text{in }E'. \] Indeed $rj=qjr$ and $r^{-1}j=qjr^{-1}$, and the other factors in this commutator are units. The group algebra $\F_2[\Gamma]$ is free of rank eight over the central subalgebra $\F_2[\langle q,r^2\rangle]$. In the latter, the ideal generated by $(1+q)(1+r^2)$ has dimension one. It therefore generates an eight-dimensional ideal of $\F_2[\Gamma]$, and $\dim E'\le32-8=24$. \end{proof} \begin{theorem}\label{r:s:d11:thm:scalar} Assume GRH. A two-special bounded system all of whose dyadic members are nonordinary and residually scalar is modular. \end{theorem} \begin{proof} Let $h=\chi_{-1}$ and choose $\eta=1$ or $\chi_{-5}$ aligning the two exact quadratic \emph{inertia} signs of $r$ and $r^\sigma\otimes\eta$. Both characters are invariant under $\sigma$; $h$ is unramified at two, trivial at five, and $\eta$ is trivial at two. The commuting operations $r\mapsto r\otimes h$ and $g:r\mapsto r^\sigma\otimes\eta$ give a packet of at most four coefficient orbits. Denote their number by $m$ and the common degree of their trace fields by $d$. Each member retains the biconnected dyadic hypothesis. Use their joint reduction $E$. If their full special signs agree at both places, the five-group already has exponent two. Otherwise at each place, with Frobenius lifts corresponding under $\sigma$, the signs on the $r,rh$ packet differ from those on the $g(r),g(r)h$ packet. The discrepancy is the same at both places, since the involution $\sigma$ exchanges their characters, $h$ splits at five, and $\eta$ is invariant under $\sigma$. A difference of rational signs separates the two coefficient packets. Write $S$ for the tuple of $+1$ on the first packet and $-1$ on the second. At each five-place $T(F)=\pm6S$, so $t=2S\in R_0$. For normalized tame differences $B,C$ at the two places, the exact special matrices give $BF=\pm SB$ or $CF=\pm SC$. In $E$ this yields $tB=tC=0$ and $t^2=0$. The equation $F^2-T(F)F+5=0$ makes the Frobenius an involution in $E'=E/tE$. Tame inertia there is an involution commuting with Frobenius by the fifth-power relation. If all signs agree, take $E'=E$ instead. In both cases the five decomposition groups have exponent two. The stable central quotient inherits the finite-flat regular model at two, by closure and quotient of the actual right radical maps; its regular group span, or its restricted group span over $H$, is precisely the torsion representation used in Lemma~\ref{r:s:d11:lem:quat}. Lemma~\ref{r:s:d11:lem:regular} gives generation of $E$ by $A,B,C$ independently of the Frobenius signs, and therefore generation of $E'$. Put $R=R_0/2R_0$ and let $\mathfrak m$ be its maximal ideal. The polarized identities identify the anticommutators of $A,B,C$ with the scalars $X=T(AB)$, $Y=T(AC)$, $Z=T(BC)$. Ordering words with these relations and the three zero squares leaves eight square-free monomials. Their traces show that $X,Y,Z$ and $W=T(ABC)$ generate $\mathfrak m$: the actual trace order is generated by traces of group elements, and each such element belongs to the algebra generated by $A,B,C$ over $\F_2$. Applying the original $R$-valued trace to $tAB,tAC,tBC,tABC$ gives $tX=tY=tZ=tW=0$ in $R$, independently of faithfulness of its reduced scalar action. Thus $t\mathfrak m=0$. Since $t$ is central and kills $B,C$, the ideal $tE$ is spanned over $\F_2$ by $t,tA$. Hence Lemma~\ref{r:s:d11:lem:finite} gives \[ 4md=\dim E\le 24+2=26. \] An actual selftwist by $h$ is impossible: it would give induction from $K(i)$, while this quadratic extension splits at five and the source has nonzero monodromy there. The bound $\dim E\le26<32$ and Theorem~\ref{struct:packet} now give either an actual twisted-conjugacy identity involving $\sigma$, in which case Lemma~\ref{r:quadratic-descent} proves modularity, or a trace field of degree at most two or biquadratic over $\Q$. Every prime above three then has residue degree at most two. In the small-residue-field cases use the triadic source results above for a solvable residue. A nonsolvable residue stays absolutely irreducible over $K(\zeta_3)$ by Clifford theory. Its trace field realizes it over $\F_3$ or $\F_9$ by Brauer--Nesbitt and the vanishing of the Brauer group of a finite field. Theorems~2.5 and~2.3 of \cite{h:AKT}, with Lemma~\ref{shared:weightzero}, give modularity of the actual weight-two member and hence of the system. \end{proof} \subsection{The supported residual base at five} \begin{proposition}\label{r:s:d11:prop:bounded} Assume GRH. Every bounded system is modular. Every absolutely irreducible $\bar r:G_K\to\GL_2(\overline\F_5)$ of determinant $\omega_5$, unramified outside five, is modular. \end{proposition} \begin{proof} For the two-special menu, Propositions~\ref{r:s:d11:prop:nonsolv}, \ref{r:s:d11:prop:dihedral}, \ref{r:s:d11:prop:ordinaryscalar} and Theorem~\ref{r:s:d11:thm:scalar} exhaust the dyadic reductions. Consider now any bounded system in characteristic three. The preceding triadic results handle reducible and cyclotomic-dihedral residues. For an irreducible solvable residue not of the latter form, solvable Artin modularity and weight raising supply a weight-zero witness, and \cite[Theorem~2.3]{h:AKT} applies. Otherwise the residue remains absolutely irreducible on $K(\zeta_3)$. At both five-places prescribe the genuine special lifts of Lemma~\ref{r:s:d6:local:switch}, and retain the given crystalline weight-two component at the unique triadic place. The prescribed-component theorem \cite[Theorem~4.2.1]{h:EGPZ} and the weight-two construction in Lemma~\ref{shared:global} give a pure compatible replacement. A point on the special component has Frobenius eigenvalue ratio five; with zero monodromy this would contradict purity. Thus the replacement is special at both places and modular. A common modular residual witness lifts the actual old weight-two triadic member by \cite[Theorem~2.3]{h:AKT}. Finally, an irreducible solvable supported five-residue has a regular cuspidal witness by solvable Artin modularity. In the nonsolvable case its restriction to $K(\zeta_5)$ is irreducible. The EGPZ exceptional case cannot occur: $[K(\zeta_5):K]=4$, whereas the abelianization of $\PGL_2(\F_5)$ has order two. Use Lemma~\ref{r:s:d6:local:menu} at the split five-places to fix bounded weight-two lifts of the actual local representations. EGPZ gives a global lift, unramified off five, and Lemma~\ref{shared:global} gives a bounded strongly compatible system. Its modularity proves the supported residual base by reduction. \end{proof} \section{\texorpdfstring{Exceptional lifts and sources over $\Q(\sqrt5)$}{Exceptional lifts and sources over Q(sqrt(5))}}\label{t:5:section} Throughout this section put $K=\Q(\sqrt5)$. The prime eleven supplies both a bounded residual base and the wild reset used in the common conductor reduction. The initial characteristic-five exception is constructed before any actual-source modularity theorem is applied. \subsection{Arithmetic and terminal characters}\label{t:5:sec:preliminaries} Write \[ \phi=(1+\sqrt5)/2,\quad \pi=3+\phi,\quad\pi'=4-\phi, \qquad \mathfrak p=(\pi),\quad\mathfrak p'=(\pi'). \] Then $\pi\pi'=11$, $N(\phi)=-1$, and $N(\pi)=11$. Minkowski's bound $\sqrt5/2<2$ gives class number one; the norm-minus-one unit gives narrow class number one. The units are $\{\pm\phi^n:n\in\mathbf Z\}$. Two and three are inert, five ramifies, and eleven splits. The completion $K_2$ is unramified quadratic over $\mathbf Q_2$. \begin{proposition}\label{t:5:prop:characters} Let $\bar r:G_K\to\GL_2(\overline\F_{11})$ be unramified outside eleven, with determinant $\bar\chi_{11}$ and reducible semisimplification. The ratio of its two constituent characters is the restriction of \[ \bar\chi_{11}^{\,e},\qquad e\in\{1,3,5,7,9\}, \] up to inversion. In particular it extends to $G_\Q$ and is nontrivial at both eleven-adic places. \end{proposition} \begin{proof} Write $\bar r^{\ssm}=\psi\oplus\bar\chi_{11}\psi^{-1}$. A finite characteristic-eleven character has order prime to eleven. Wild inertia at eleven is therefore killed. At the two primes above eleven, $\phi$ has residues $8=2^3$ and $4=2^2$ in $\F_{11}^\times$, where $2$ is a generator. The signs of $\phi$ at the two real places are $(+,-)$. Write $a,b\in\Z/10\Z$ for the two tame character exponents and $u,v\in\Z/2\Z$ for the signs. The relations imposed by the units $-1$ and $\phi$ are \[ a+b+u+v\equiv0\pmod2,\qquad 3a+2b+5v\equiv0\pmod{10}. \] They give $a\equiv b\pmod5$, $v\equiv a\pmod2$ and $u\equiv b\pmod2$. There are exactly twenty characters. Squaring makes the two tame exponents equal and kills the signs. Comparison with a power of $\bar\chi_{11}^2$ leaves a character unramified at all finite and infinite places, hence trivial. Thus $\psi^2$ is an even power of $\bar\chi_{11}$, and $\bar\chi_{11}\psi^{-2}$ is an odd power. \end{proof} Put \[ \alpha=\phi\pi=1+4\phi,\qquad \alpha'=5-4\phi, \qquad \alpha\alpha'=-11. \] The quadratic extensions $K(\sqrt\alpha)$ and $K(\sqrt{\alpha'})$ ramify at exactly one of the two primes above eleven, respectively, and are unramified at two. Indeed $(1+\sqrt\alpha)/2$ satisfies $X^2-X-\phi=0$; its reduction at two is irreducible over $\F_4$ because $\operatorname{Tr}_{\F_4/\F_2}(\phi)=1$. The same holds for $\alpha'$. These two characters permit independent quadratic corrections at eleven without destroying good reduction at two. Their product cuts out \[ H=K(\sqrt{-11}), \] which is split at two: $K_2$ contains $\sqrt5$, and $-11/5$ is a square in $\Q_2$ since it is $1$ modulo eight. We use Lemma~\ref{shared:global} for prescribed lifts and compatible systems; the exceptional initial characteristic-five point is constructed separately in Section~\ref{t:5:exceptional-start}. The notation $F_b,J_b,B_b,C_b$ and the real-place correction are as in Section~\ref{shared:explicit}. We use the explicit formula \eqref{shared:eq:grh}, including its finite-tail bound and retained prime densities. The unconditional comparison below uses the positive kernel $F_b(x)/\cosh(x/2)$. \subsection{Starting an exceptional characteristic-five representation}\label{t:5:exceptional-start} The ring-theoretic theorem of Khare--Thorne gives the finiteness needed below before a characteristic-zero global point has been found. We use its published nilpotent-kernel statement, rather than its finite-length approximation corollary. \begin{corollary}[A classical quotient of the prepared KT ring] \label{t:5:cor:KT-classical-finite} In the exact setup of \cite[\S7]{h:KT}, assume the fixed set $V_2$ of $D^{\mathrm{St}(\mathrm{uni})}$ places is nonempty. The quotient $R_{\mathcal S}^{\mathrm{cl}}$ of the global deformation ring at the classical weight-two augmentation of its ordinary Iwasawa algebra $\Lambda$ is finite over $\mathcal O$. \end{corollary} \begin{proof} In the notation of \cite[\S7]{h:KT}, let $\mathbb T$ be the localized ordinary Hecke algebra at infinite ordinary level. The paragraph preceding their Theorem~7.2 makes $\mathbb T$ finite over $\Lambda$. Theorem~7.2 in the final version of~\cite{h:KT} gives a surjection \[ R_{\mathcal S}\longrightarrow\mathbb T \] with nilpotent kernel $I$, since $V_2\ne\varnothing$. We use the final-version numbering of that reference. The complete deformation ring $R_{\mathcal S}$ is Noetherian. Thus each $I^j/I^{j+1}$ is a finite module over $R_{\mathcal S}/I=\mathbb T$, and hence finite over $\Lambda$. The finite filtration by powers of $I$ shows that $R_{\mathcal S}$ is finite over $\Lambda$. Tensoring along the classical augmentation $\Lambda\to\mathcal O$ gives the claimed finiteness. This argument includes the entire deformation ring, including possible characteristic-five components; it uses no assumed characteristic-zero point or flatness of $R_{\mathcal S}$. \end{proof} \begin{proposition}[An initial exceptional lift]\label{t:5:prop:initial-exceptional-lift} Let $K=\mathbf Q(\sqrt5)$, and let $\bar r:G_K\longrightarrow\mathrm{GL}_2(\overline{\mathbf F}_5)$ be continuous, absolutely irreducible and totally odd. Assume that its projective image is $\mathrm{PGL}_2(\mathbf F_5)$ and that its projective kernel field contains $K(\zeta_5)$. Put \[ \nu=[\det\bar r\,\bar\chi_5^{-1}], \] where brackets denote the Teichmuller lift. There is a continuous lift $r:G_K\longrightarrow\mathrm{GL}_2(\overline{\mathbf Q}_5)$ with determinant $\nu\chi_5$, ramified at finitely many places and de Rham of parallel weight two. It is totally odd. The lift is potentially modular over totally real extensions with avoidance of any prescribed finite extension, and belongs to a strongly compatible system. \end{proposition} \begin{proof} \emph{Fixed local rings over $K$.} Let $S$ contain the places above $5$ and the ramification of $\bar r$. At each $v\in S$, choose a definite-type local lift by \cite[Propositions~7.7.1 and~7.8.1]{g:Snowden}, weight two when $v\mid5$. The determinant correction of \cite[\S7.9]{g:Snowden} twists by the square root of a finite character reducing to $1$; hence it does not change the residual representation or Hodge--Tate weights and gives determinant $\nu\chi_5$. Choose a corresponding nonzero, reduced, $\mathcal O$-flat local fixed-type quotient $R_v^\dagger$ from \cite[Propositions~7.3.1 and~7.4.1]{g:Snowden}. Its framed dimension is $4$ for $v\nmid5$, and $4+[K_v:\mathbf Q_5]$ for $v\mid5$. At $5$ the type may be $A$, $B$ or $C$; we do not require a potentially Barsotti--Tate lift when only type $C$ was chosen. Fix finite Galois local extensions killing these fixed inertial types, the local residual representations, and their finite determinant factors. After these extensions, the local characteristic-zero points are unramified or in the Steinberg closure away from $5$; at $5$ they are ordinary semistable for types $A,C$, or nonordinary Barsotti--Tate for type $B$. A single finite extension works for every point of each chosen local ring: the finite inertial type is fixed, while its possible monodromy is retained. Let $K_\nu$ be the field cut out by $\nu$. It is totally real since $\nu$ is totally even. \emph{Preparation and the modular residual witness.} Real approximation and Krasner's lemma, followed by taking a totally real Galois closure, supply a totally real finite extension containing $K_\nu$, for which every completion at a controlled place contains the fixed local extension. This Galois closure is taken before constructing any modular witness. Apply the residual potential-modularity theorem \cite[Theorem~5.1.1]{g:Snowden} over this field. This theorem is for any odd residual representation and has no exceptional-image exclusion. It gives a weight-two modular residual witness after a further totally real extension. Let $R$ now denote the set of places away from $5$ at which the restrictions of the original chosen local rings have Steinberg type. Let $\sigma$ denote the places above $5$ lying over a local type $A$ or $C$. Apply \cite[Lemma~9.2]{h:KT} to the known modular witness, this $R$, and this $\sigma$. After a soluble totally real extension it supplies a weight-two witness $\pi$ that is ordinary and Iwahori at $\sigma$, nonordinary and spherical at the other places above $5$, Steinberg at the pullback of $R$, and unramified at every other finite place. Thus incidental bad primes of the first potential-modularity witness have disappeared. The residual representation is trivial at the places above $5$ and $R$. The already imposed original local conditions remain of the stated type under all these restrictions. Finally make a totally real quadratic extension split at the current places above $5$ and $R$, and base change $\pi$. Denote the resulting field by $F$. Then $[F:K]$ and $\#R_F$ are even. Its central character is everywhere unramified. In the dual normalization its ratio with the required Teichmuller character $\psi=1$ is a finite unramified character of $5$-power order; twisting by the inverse square root makes the central character exactly $1$, without changing residues or local types. The residual representation over $F$ is unramified away from $5$ and trivial at $5$ and $R_F$. \emph{A fresh fixed unipotent prime.} Choose the auxiliary prime only after $F$ has been fixed. Let $M/K$ be the projective kernel field and let $F^{\mathrm g}/K$ be the Galois closure of $F/K$. Since $F^{\mathrm g}$ is totally real, $M\cap F^{\mathrm g}=K$: the only proper nontrivial normal subextension of the $S_5$-extension $M/K$ is its imaginary quadratic subfield $K(\zeta_5)$, and $M$ itself is not totally real. Let $L/K$ be the full linear residual kernel field, put $G=\Gal(L/K)$, let $Z$ be its scalar subgroup, and set $N=\Gal(L/(L\cap F^{\mathrm g}))$. Projective disjointness gives $ZN=G$, so $G/N$ is a quotient of the prime-to-five group $Z$. Every element of order five in $G$ therefore lies in $N$ and is an actual nontrivial unipotent matrix. It is trivial on $K(\zeta_5)$. Chebotarev therefore gives a prime $q\notin S$, split completely in $F^{\mathrm g}$, with actual residual Frobenius a nontrivial unipotent of order $5$. Then $\mathrm Nq\equiv1\pmod5$ and $\nu(q)=1$. Impose $D^{\mathrm{St}(\mathrm{uni})}$ at $q$ on the original $K$-deformation problem, as well as at every place above $q$ on its $F$-restriction. This is a nonzero formally smooth local problem of framed dimension $4$ by \cite[Proposition~6.6]{h:KT}. Write $Q_F$ for the places above $q$. Its cardinality is $[F:K]$, hence even and positive. Take $V_1=\varnothing$ and $V_2=Q_F$ in \cite[\S7]{h:KT}. No congruence modulo $25$ is needed for this fixed local problem. Choose the good-subgroup auxiliary place $a$ using \cite[Lemma~7.1]{h:KT}, away from the places above $5$, $R_F$ and $Q_F$. Apply \cite[Lemma~5.2 and the proof of Theorem~9.1]{h:KT} to $\pi$ at the even set $Q_F$. The ordinary/nonordinary variant of \cite[Lemma~4.13]{t:5:ThorneDihedral} preserves the local conditions at $5$ and $R_F$, makes every place of $Q_F$ Steinberg, and leaves the other finite places unramified. Its degeneracy maps and pairings commute with the Hecke operators at $5$. The place $a$ is used in this argument and adds no ramification to the deformation problem. An unramified five-power square-root twist, if necessary, restores central character one. This is the witness required in \cite[\S7]{h:KT}, with $V_1=\varnothing$ and $V_2=Q_F$. \emph{Restriction at the level of local rings.} Let $R_{K,q}^{\dagger}$ be the global fixed-determinant deformation ring over $K$, with the original local quotients $R_v^\dagger$, the new $D^{\mathrm{St}(\mathrm{uni})}$ condition at $q$, and unramifiedness outside $S\cup\{q\}$. Let $R_F^{\mathrm{cl}}$ be the classical weight-two quotient of the prepared KT ring just constructed. There is a restriction homomorphism \[ R_F^{\mathrm{cl}}\longrightarrow R_{K,q}^{\dagger}. \] Here it is essential to verify the type conditions before passing to the global ring. On each original reduced, $\mathcal O$-flat local ring $R_v^\dagger$, restriction kills the defining ideal of the corresponding prepared local condition: this holds on all finite characteristic-zero points by the choice of the fixed local extensions, and these points are dense in the reduced flat local generic fiber. Thus it holds on the whole local ring, as in \cite[Lemma~6.4.1]{g:Snowden}. The ordinary local ring is the reduced, $\mathcal O$-torsion-free image quotient of the framed ring obtained from the flag scheme; it forgets the flag \cite[\S5.4.1 and Proposition~5.13]{t:5:ThorneDihedral}. Its classical characteristic-zero points require an ordinary line with trivial inertial character, without marking a particular line. After preparation and dualization each type $A$ or $C$ point has such a line. The same density argument therefore gives the ordinary quotient-ring map on the original local ring. Type $B$ maps to the nonordinary Barsotti--Tate problem. Away from $5$, finite-inertia type maps to the unramified problem and Steinberg type to its Steinberg closure. The determinant is identical, and the unramified sign at residual-trivial Steinberg places is forced by the residue. At the new $q$ the local fields are unchanged because $q$ splits in $F$, so the local restriction map preserves $D^{\mathrm{St}(\mathrm{uni})}$ directly. The resulting local ring homomorphisms remain valid after tensoring with any global ring, including one with $5$-torsion. No density of global characteristic-zero points has been assumed. The projective image over $F$ remains $S_5$, so the residual restriction is absolutely irreducible. Restriction is finite by \cite[Lemma~6.3.1]{g:Snowden}. Passing to the local quotients preserves finiteness, since their defining ideals have already been shown to vanish. Corollary~\ref{t:5:cor:KT-classical-finite} therefore makes $R_{K,q}^{\dagger}$ finite over $\mathcal O$. \emph{A characteristic-zero point.} Put $n=\#(S\cup\{q\})$. The completed tensor product of the local rings has dimension $1+[K:\Q]+3n$. The framed presentation in the proof of \cite[Proposition~6.2.1]{g:Snowden} uses $r+n-1$ variables and $r+[K:\Q]$ relations, so its global quotient has dimension at least $4n$. Forgetting the frames has relative dimension $4n-1$; hence $\dim R_{K,q}^{\dagger}\ge1$. The invariant vanishing used in that calculation follows from absolute irreducibility over $K(\zeta_5)$ and Schur's lemma; the exceptional-image exclusion is not used. A finite $\mathcal O$-algebra of positive dimension has nonzero generic fibre. A closed generic point gives the required weight-two lift, and its fixed determinant makes it totally odd. \emph{Actual potential modularity and the compatible system.} For any prescribed finite extension $B/K$, apply the avoiding version of \cite[Theorem~5.1.1]{g:Snowden} to $\bar r$, including the residual and cyclotomic kernel fields in the avoidance data. It gives a totally real Galois field $K'/K$, disjoint from the prescribed avoidance field, and a weight-two modular residual witness over $K'$. The actual lift $r|_{G_{K'}}$ is finitely ramified and weight two, and its residual restriction to $G_{K'(\zeta_5)}$ is absolutely irreducible. Apply \cite[Theorem~1.1]{h:KT}, after dualizing, to obtain the automorphy of this actual lift. The proof of \cite[Proposition~9.2.1 and Remark~9.2.2]{g:Snowden}, using solvable descent and Brauer induction, now constructs the strongly compatible system. In this use the proposition's old exceptional exclusion is replaced in its proof by the actual avoiding potential modularity just proved; the remaining argument uses only that input and absolute irreducibility, not an adequacy hypothesis. In particular this is an application of that proof with its modularity input supplied by KT, not an invocation of the original proposition outside its stated hypotheses. \end{proof} \begin{remark}[Scope] This construction permits the extra prime $q$ in the initial lift. It asserts no minimal conductor or prescribed minimal weight theorem, and does not assume residual modularity over $K$. The auxiliary prime is imposed on the original deformation problem as well as the prepared one: restricting an arbitrary unramified deformation cannot create a new Steinberg eigenvalue relation. Its use of Theorem~7.2 is a finiteness argument with a fixed auxiliary prime; the varying-prime approximate level lowering in KT~\S9 is not needed. \end{remark} \subsection{A one-prime source obstruction}\label{t:5:sec:oneprime} \begin{proposition}\label{t:5:prop:oneprime-residual} Let $\bar r:G_K\to\mathrm{SL}_2(k)$ be finite flat at two, unramified outside $2\mathfrak p$, with inertia of order at most two at $\mathfrak p$, where $k$ is finite of characteristic two. Then $\bar r^{\mathrm{ss}}=1\oplus1$. This assertion is unconditional. \end{proposition} \begin{proof} Fontaine's finite-flat bound over the unramified dyadic base gives classical upper breaks at most one and normalized different $d_2<2$ \cite[\S2.1, Theorem~1, and \S1.2(i)]{c:FontaineBT}. The latter reference converts his upper numbering to the classical one. At eleven the ramification is tame of index at most two at just one of the two places. A nonsolvable image is $\mathrm{SL}_2(\mathbf F_q)$, $q\ge4$, so its kernel field has degree at least $120$ and \[ \operatorname{rd}(M)<4\sqrt5\,11^{1/4}<16.28896. \] With $b=7$ and $[M:\mathbf Q]\ge120$, \eqref{shared:eq:unconditional} gives $\rd(M)>16.661$, excluding the nonsolvable case. The exact calculation is supplied in~\cite[\path{d5/new}]{code}. An irreducible solvable image is odd dihedral, induced from an odd-order character $\theta$ of a quadratic field $L/K$. Its rotation character is unramified at eleven and tame at two. The quadratic quotient has dyadic conductor at most two; an odd dyadic valuation would have conductor three. Thus $L$ belongs to the seven nontrivial classes in $\langle-1,\phi,\pi\rangle$. The following table gives the fields up to rational conjugation. \[ \begin{array}{c|r|c|c} a&|D_{K(\sqrt a)}|&r_2&\text{Minkowski bound}\\\hline -1&400&2&3.1\\ \phi&400&1&2.4\\ \pi&4400&0&6.3\\ -\pi&4400&2&10.2\\ \phi\pi&275&1&2\\ -\phi\pi&4400&1&8 \end{array} \] The bound in the fifth row is strict. Integral bases are $1,\phi,t,\phi t$ with $t^2=a$, except in that row, where $t^2-t=\phi$. The relative discriminants at odd primes follow from Eisenstein; at two maximality follows by testing unit squares modulo four. The first, second and fifth rows have no prime ideal below their Minkowski bounds, hence class number one. In the $\pi$ row the possible generators have norms four or five; the dyadic class has square $(2)$ and $N(t-1)=5$ makes the norm-five classes principal. In the $-\pi$ row the identity $N(t-\phi)=20$ makes the norm-five classes equal to the dyadic class, and three has residue degree four. In the last row only the norm-four prime occurs; its square is principal. Thus all these class groups have two-power order. If $L/K$ is ramified at two, its involution acts trivially on the residue field. Anti-invariance makes the odd tame inertia character trivial. It is then everywhere unramified, hence trivial by the preceding class groups. The only nontrivial class in this list unramified at two is $\alpha=\phi\pi$. For $L=K(\sqrt\alpha)$ the integral unit $t=(1+\sqrt\alpha)/2$ reduces to a root of $X^4+X+1$ and has order fifteen in $\mathbf F_{16}^{\times}$. It kills the entire odd tame ray group at two. This also makes $\theta$ trivial, a contradiction. Finally, in the reducible case the odd-order constituent characters are unramified at eleven. At two their inertia factors through $\mathbf F_4^\times$, generated by the image of $\phi$. Class number one therefore makes both constituents trivial. \end{proof} Put $R=\OO_K[1/\pi]$ and $F=K_2$. \begin{lemma}\label{t:5:lem:ext} We have $\Ext_R^1(\mu_2,\Z/2\Z)=0$. \end{lemma} \begin{proof} Over the completion at two, an extension of $\mu_2$ by $\Z/2\Z$ splits by the connected--etale sequence. Its middle term is therefore killed by two locally, hence globally, by flatness. Its generic extension class lies in $H^1(K,\F_2)=K^\times/K^{\times2}$, is unramified away from the inverted primes and two, and is a square in $F$. For $R$, the global Kummer space has basis $-1,\phi,2,\pi$. Its localization in $F^\times/F^{\times2}$ is injective. An odd power of $2$ is excluded by valuation. Taking the norm to $\Q_2$ then excludes nonzero exponents of $\phi$ and $\pi$: the three possible nontrivial norm classes are $-1,11,-11$, all nonsquare in $\Q_2$. Finally $-1$ is nonsquare in $F$. Thus the generic extension is split. A generically split extension is split over $R$. Of its two generic splittings, one agrees with the local splitting at two; away from two both group schemes are etale and any generic splitting extends. These splittings glue. Equivalently this is the finite-flat gluing sequence of \cite[Proposition~2.3 and Corollary~2.4]{t:5:Schoof03}. \end{proof} \begin{lemma}\label{t:5:lem:ordinarygroup} Let $G$ be a $2$-divisible group over $R$. Suppose that every simple finite flat subquotient of every $G[2^n]$ has trivial generic Galois action. Then $G$ has a multiplicative $2$-divisible subgroup with etale quotient. \end{lemma} \begin{proof} A simple generic factor has order two. Schematic closure realizes a generic composition series as a finite flat composition series. At the unique dyadic prime, the prime-order classification over an unramified ring gives either the etale or the multiplicative model. Away from two the model is etale and its generic action is trivial. Since the generic group of order two has no nontrivial automorphism, gluing gives exactly $\Z/2\Z$ or $\mu_2$ over $R$. By Lemma~\ref{t:5:lem:ext}, adjacent factors in the order $\Z/2\Z$ below $\mu_2$ can be interchanged: the relevant extension splits. Repeating this operation orders the filtration with all multiplicative factors below all etale factors. Extensions of multiplicative groups are multiplicative, by Cartier duality, and extensions of etale groups are etale. Thus each $G[2^n]$ has a multiplicative subgroup with etale quotient. This subgroup is unique. Its localization at two is the connected part, since its quotient is etale; its global subgroup is the flat closure of that generic subgroup. Consequently it is functorial in $G[2^n]$ and compatible with the transition and multiplication maps. The local connected part is multiplicative at every level, so these subgroups form the multiplicative $2$-divisible part, with etale quotient. This is also the filtration argument used in finite-flat approaches to small-conductor abelian varieties \cite{t:5:Schoof03,t:5:Schoof11}. \end{proof} \begin{theorem}\label{t:5:thm:oneprime} Let $E_0/\Q_2$ be finite and let $r:G_K\to\GL_2(E_0)$ be continuous. Suppose that $\det r=\chi_2$, that $r$ is Barsotti--Tate at two, unramified outside $2\mathfrak p$, and semistable at $\mathfrak p$ with unipotent inertia. Then $r$ fits into an exact sequence \[ 0\longrightarrow E_0(1)\longrightarrow r\longrightarrow E_0\longrightarrow0. \] In particular there is no absolutely irreducible such representation. The same assertion holds with $\mathfrak p'$ in place of $\mathfrak p$. No GRH assumption is needed. \end{theorem} \begin{proof}[Proof of Theorem~\ref{t:5:thm:oneprime}] Choose a global stable $\OO_{E_0}$-lattice. The local Barsotti--Tate condition realizes such a lattice after an isogeny: taking the closure of a finite generic subgroup realizes any commensurable stable lattice. The integral coefficient action extends by full faithfulness for Tate modules of $2$-divisible groups. Glue this local group to the etale group defined by the lattice away from two. Levelwise finite-flat gluing \cite[Proposition~2.3]{t:5:Schoof03} gives a $2$-divisible group over $R$ with $\OO_{E_0}$ action. The reduction modulo the maximal ideal of $\OO_{E_0}$ satisfies the one-prime hypotheses of Proposition~\ref{t:5:prop:oneprime-residual}. At eleven, unipotent inertia factors through the procyclic tame pro-two quotient. Its reduction is cyclic of exponent two, hence has order at most two. Its semisimplification is trivial. Filtering by the coefficient uniformizer and then over $\F_2$ shows that every simple generic factor of every $G[2^n]$ is trivial. Lemma~\ref{t:5:lem:ordinarygroup} supplies a functorial multiplicative subgroup. The height of $G$ is $2[E_0:\Q_2]$, and its dimension is $[E_0:\Q_2]$, because its labelled Hodge--Tate pairs are $\{0,1\}$. The multiplicative part has height $[E_0:\Q_2]$. Functoriality makes its rational Tate module $E_0$-stable, hence an $E_0$-line. The quotient character is unramified at two, and also at eleven because inertia in $r$ is unipotent. It is unramified at every other finite prime by hypothesis. Class field theory and narrow class number one make this character trivial. The determinant identifies the subcharacter with $\chi_2$. \end{proof} \subsection{Two cuspidal places above eleven}\label{t:5:sec:cuspidal} We use one finite-flat ordinarity observation repeatedly. A nonordinary rank-two Barsotti--Tate source of cyclotomic determinant is biconnected up to isogeny. Its coefficient-uniformizer torsion and finite-flat subquotients are then biconnected. Over an unramified dyadic ring, a prime-order model with trivial generic action is $\Z/2\Z$ or $\mu_2$. For odd $p$ and absolute ramification index $e56.06306031$, a contradiction. Apply \cite[Theorem~6.1]{g:Thorne2} to the dual of the actual crystalline two-adic source, in his Hodge--Tate convention: its residual image is nonsolvable, its residue is modular, and the required real-place condition holds. This proves modularity of the source and hence of $\mathcal R$. \end{proof} \subsection{The supported residual base at eleven}\label{t:5:sec:supported} \paragraph{Fixed-determinant positivity.} We use the following observation for both eleven-adic menus. Let $\bar r:G_{\Q_{11}}\to\GL_2(\overline\F_{11})$ have determinant $\omega_{11}$. By \cite[Propositions~2.6--2.7]{c:BIP} it has a regular crystalline lift of determinant $\chi_{11}$. Write its weights as $h_1e(n+1/(p-1))$; the convention in the cited statement is $G^{(u)}=G^{u-1}$, where the latter is classical upper numbering \cite[pp.~515--516, Theorem~A]{c:FontaineBT}. For $p=3$, $n=1$, and $e=2$, the classical upper break is at most $2$. The abelian cubic character $f$ therefore has conductor at most $\mathfrak p_3^3$. Its tame quadratic factor does not change positive ramification groups. At a prime over $5$, a cyclic cubic character is unramified because $3\nmid5-1$. We need only the ray group of conductor $\mathfrak p_3^3$ and its Frobenius classes at five. These have the following elementary calculation. The group $(\OO_K/\mathfrak p_3^3)^\times$ has order $18$. The unit $\epsilon=1+\gamma$ has exact order $3$ modulo $\gamma^3$, so the image of $\langle-1,\epsilon\rangle$ has order $6$. Class number one then gives ray group $C_3$. The class of $\mathfrak q=(a)$ is nontrivial: $-a=1-\gamma$ differs from $1,\epsilon,\epsilon^2$ with valuations $1,1,2$, respectively, so it does not belong to the unit image modulo $\gamma^3$. Its conjugate also has nontrivial class because conjugation acts as an automorphism of the ray group. Infinite places contribute only $2$-primary data and do not affect this cubic quotient. In the SP case, \eqref{t:21:red3:SP-algebra-condition} fixes the two Frobenius characters, hence forces $f(\Phi)=0$. In the SC case, $\Phi^2=-5$ gives the same conclusion, since the residual Frobenius eigenvalues are $\pm1$ and $2$ is invertible. Therefore $f=0$, proving \eqref{t:21:red3:diagonal-zero}. The ideal $J$ is generated by products from $B$ and $C$, and $\mathfrak mJ\subseteq\mathfrak m^2$; this proves \eqref{t:21:red3:pseudotangent-surjection}. \end{proof} \subsubsection{The SP/SP action of quadratic conjugation} \label{t:21:red3:SP-geography} \begin{lemma}[The two SP signs agree] \label{t:21:red3:SP-signs} An irreducible actual source in the situation of Theorem~\ref{t:21:red3:source-theorem} has equal normalized unramified Steinberg signs at the two primes over $5$. For these signs, quadratic conjugation acts trivially on the relative cotangent space of $R$. \end{lemma} \begin{proof} Suppose first that both signs are $-1$. The invariant monodromy character is $\chi_3\alpha_v$ and reduces to the Frobenius eigenvalue $-\alpha_v$. Using \eqref{t:21:red3:frobenius-signs}, the permitted global extension spaces in \eqref{t:21:red3:offdiagonal-extensions} are contained in \begin{equation} \ker\ell_{\bar{\mathfrak q}}=\langle b\rangle, \qquad \ker\ell_{\mathfrak q}=\langle\sigma(b)\rangle, \label{t:21:red3:SP-lines} \end{equation} respectively. If both signs are $+1$, the roles of the lines are exchanged. If the signs are opposite, one orientation must be unramified at both $5$-places. Its global extension space is then contained in \[ \ker\ell_{\mathfrak q}\cap\ker\ell_{\bar{\mathfrak q}}=0. \] Nakayama gives $B=0$ or $C=0$, so $J=0$ and every point of $R$ is globally reducible. This contradicts the actual irreducible source. Equivalently, Ribet's lattice lemma would require nonsplit residual extensions in both orientations, while one of them has no permitted nonzero extension. The signs therefore agree. Quadratic conjugation preserves the resulting local conditions, interchanges the two residual characters, and interchanges the two lines \eqref{t:21:red3:SP-lines}. Its action on their product is $+1$. One can check the sign directly from $b=d-s$ and $\sigma(b)=s-b=-(d+s)$. For a basis-independent verification, choose a lift $h\in G_\Q$ of $\sigma$ which fixes $\zeta_3$; it exists since $K\cap\Q(\zeta_3)=\Q$. The coefficient action on $\omega$ is then trivial, and $\chi_1(h^2)=\chi_2(h^2)$. Squared transport is inner by a residual scalar, so the two exchange factors multiply to $+1$. The surjection \eqref{t:21:red3:pseudotangent-surjection} now shows that quadratic conjugation fixes the entire relative cotangent space. \end{proof} \subsubsection{The SC/SC action of quadratic conjugation} \label{t:21:red3:SC-geography} \begin{lemma}[The SC anti-invariant product vanishes] \label{t:21:red3:SC-anti} For the normalized SC/SC menu, quadratic conjugation acts trivially on the relative cotangent space of $R$. \end{lemma} \begin{proof} Both extension spaces in \eqref{t:21:red3:offdiagonal-extensions} are subspaces of $V\otimes_{\F_3}k$. Hence the tensor surjecting onto the relative cotangent space is a quotient of $V^\vee\otimes V^\vee$. Quadratic conjugation swaps the factors and acts on each by $s\mapsto s$, $d\mapsto-d$. The scalar in this statement is fixed by the same choice of $h$ as in the preceding proof. The local equation \eqref{t:21:red3:SC-square-zero} kills the image of $\ell_v\otimes\ell_v$ in the cotangent space. Indeed, in the universal generalized matrix algebra write \[ T-1=\begin{pmatrix}x&b_T\\c_T&y\end{pmatrix}, \qquad x,y\in\mathfrak m. \] Its upper-left squared entry is $x^2+b_Tc_T$. The exact relation over $R$ is $(T-1)^2=-3T$, so $b_Tc_T\in\mathfrak m^2+3R$; therefore $b_Tc_T$ is zero in $\mathfrak m/(3R+\mathfrak m^2)$. The classes of $b_T$ and $c_T$ are evaluation on local inertia, which are the functionals in \eqref{t:21:red3:local-evaluations}, up to a harmless nonzero change of local generator. The evaluations in \eqref{t:21:red3:local-evaluations} are $s^\vee-d^\vee$ and $s^\vee+d^\vee$. Thus Lemma~\ref{shared:tensor-involution} makes the action on the relative cotangent trivial. The local calculation above concerns products of opposite offdiagonal extensions, not only traces of representation tangents. \end{proof} \subsubsection{From the cotangent to the entire pseudoring} \label{t:21:red3:nakayama} Quadratic conjugation defines a $W(k)$-linear involution of $R$: its square is induced by an inner automorphism of $G_K$, which does not change a pseudorepresentation. Lemmas~\ref{t:21:red3:SP-signs} and~\ref{t:21:red3:SC-anti}, followed by Lemma~\ref{t:involution}, show that this involution is the identity. Consequently every normalized actual source satisfies \begin{equation} \operatorname{tr}r_3(g) =\operatorname{tr}r_3(hgh^{-1})\qquad(g\in G_K). \label{t:21:red3:integral-trace-invariance} \end{equation} No assertion that $R$ is finite over $W(k)$, that its whole cotangent space is zero, or that its traces are rational is needed. The essential fact is that an involution of order prime to the residue characteristic cannot survive when its anti-invariant cotangent is zero. \subsubsection{Whole-system descent and the rational theorem} \label{t:21:red3:descent} \begin{proof}[Completion of the proof of Theorem~\ref{t:21:red3:source-theorem}] Strong compatibility transports \eqref{t:21:red3:integral-trace-invariance}, first at every good Frobenius, to all companions of the normalized system. Chebotarev identifies the semisimplifications of the system and its quadratic conjugate. In particular its absolutely irreducible $5$-adic member is invariant under $\sigma$. Lemma~\ref{r:quadratic-descent} makes the normalized system modular. Undoing the finite quadratic normalization proves the claim. \end{proof} \begin{remark}\label{t:21:red3:certificates} The seed has $a_v=-1$ at both places over five and $a_{\mathfrak p_3}=0$. The displayed identities determine its local Kummer class and the local-cube kernel explicitly. The source proof uses these finite-flat and pseudodeformation calculations and no dyadic modularity assumption. \end{remark} The exact calculation is supplied in~\cite[\path{d21}]{code}. \subsection{Completion of the bounded source theorem} \begin{theorem}\label{t:21:source:allmenus} Assume GRH. Let $\mathcal R$ be a potentially modular compatible system over $K$, of determinant cyclotomic and labelled weights $\{0,1\}$, with conductor supported at five. Suppose each five-place has one of the bounded types $\SC$ or $\SP$. Then $\mathcal R$ is modular. \end{theorem} \begin{proof} The mixed case is Proposition~\ref{t:21:input:mixed}. A pure special source with reducible five-adic residue is covered directly by Proposition~\ref{t:21:source:ordinary5}. For the other pure sources, consider the three-adic member, which is crystalline at three. If its residue remains irreducible on $G_H$, Proposition~\ref{t:21:source:switchable} applies. If the residue is irreducible but its restriction is reducible, the dihedral source theorems of the preceding sections apply: induction from $H$ for the cuspidal pair, and fixed-character rigidity followed by descent under quadratic conjugation for the special pair. Otherwise Lemma~\ref{t:21:arith:characters3} gives the complete constituent list. The two unramified-at-three possibilities are modular by Proposition~\ref{t:21:source:ordinary3}. In the remaining two, twist by $\psi_5$ when necessary to remove the common quadratic inertia at five. This twist is unramified at three and preserves the determinant and crystalline condition. The nonordinary reducible source theorem proved above gives invariance under quadratic conjugation and modularity. These alternatives are exhaustive for every finite coefficient field and every residual lattice. \end{proof} \begin{theorem}[Unrestricted residual base at five]\label{t:21:base:five} Assume GRH. Every irreducible representation \[ \bar r:G_K\longrightarrow\GL_2(\overline\F_5),\qquad \det\bar r=\bar\chi_5, \] unramified outside five is modular. No local finite-flatness, tameness, coefficient-field or conductor-exponent restriction is imposed. \end{theorem} \begin{proof} Solvable projective image is handled by an odd Artin lift and solvable Artin automorphy, with weight raising to obtain a regular residual witness. In the nonsolvable case the restriction to $G_{K(\zeta_5)}$ is irreducible. The exceptional projective-$\PGL_2(\F_5)$ condition cannot occur: $[K(\zeta_5):K]=4$, whereas the abelianization of $\PGL_2(\F_5)$ has order two. Lemma~\ref{t:21:local:menu} provides actual weight-two local lifts at both five-places with determinant $\chi_5$. Prescribe their components and no ramification elsewhere. Theorem 4.2.1 of~\cite{h:EGPZ} gives a global lift. Snowden's potential-modularity and compatible-system construction gives a potentially modular weight-two system with the selected menu. At a selected special component, the purity observation in Section~\ref{shared:systems} retains nonzero monodromy. The system is crystalline at two and three by all-place compatibility. Theorem \ref{t:21:source:allmenus} makes the actual system modular, and its five-adic reduction is the original $\bar r$. \end{proof} \subsection{The arbitrary-conductor transfer} We verify the terminal character condition and apply Theorem~\ref{shared:terminal-criterion} with reset prime five. Its terminal source is allowed arbitrary regular weights. \subsubsection{Characters supported at five} \begin{lemma}\label{t:21:chain:characters5} Every finite characteristic-five character of $G_K$ unramified outside five extends to $G_\Q$. If a representation of determinant $\bar\chi_5$ and support at five is reducible, its constituent quotient is exactly $\bar\chi_5$ or $\bar\chi_5^{-1}$. \end{lemma} \begin{proof} Such a character kills principal units at five. Since $h_K=1$, it is determined by two exponents $m,n$ modulo four at the two residue groups and two real signs. The positive unit $\epsilon$ has residues $2,3=2^{-1}$ at the two five-places, so the unit relation gives $m=n$. The relation from $-1$ then says that the real signs agree. The ray-character data are invariant under quadratic conjugation, hence so is the character. An invariant character extends across the index-two group after adjoining a square root of its value on the square of a lift of the nontrivial automorphism. The target $\overline\F_5^\times$ is divisible, and the extension still has finite image. The equality of the tame exponents shows that $\psi^2$ has at both five-places either trivial inertia or the inertia of $\bar\chi_5^2$. It is trivial at the real places. Dividing by $1$ or $\bar\chi_5^2$ therefore gives an everywhere unramified character trivial at infinity, which is trivial since $h_K=1$. Thus $\psi^2\in\{1,\bar\chi_5^2\}$, and the quotient $\bar\chi_5\psi^{-2}$ is $\bar\chi_5^{\pm1}$. \end{proof} \begin{remark} The generic mixed-case theorem of \cite[Theorem~A]{t:21:OrdinaryMixed} excludes the local ratio $\bar\chi_5^{\pm1}$. It therefore does not replace the residually reducible endpoint used here. At three its complete-splitting hypothesis also fails. The terminal argument uses the locally irreducible source theorems of Zhang, not the generic mixed theorem. \end{remark} \begin{proof}[Removal of arbitrary conductor over $\Q(\sqrt{21})$] A solvable residual image is modular by an odd solvable Artin lift and regular weight raising. Otherwise its cyclotomic restriction is absolutely irreducible. The initial weight-two lift and Lemma~\ref{shared:global} give a potentially modular compatible system with cyclotomic determinant times a finite totally even character. In characteristic five the exception is excluded by $[K(\zeta_5):K]=4$, as in Theorem~\ref{t:21:base:five}. The nonsolvable residue makes the system nondihedral. Apply Theorem~\ref{shared:terminal-criterion} with $\ell=5$. Two is inert and five splits. Theorem~\ref{t:21:base:five} is the supported irreducible residual assertion, and Lemma~\ref{t:21:chain:characters5} gives constituent quotient $\bar\chi_5^{\pm1}$, which is totally odd and extends to $G_\Q$. The common criterion therefore proves modularity of the original residual representation. \end{proof} \Needspace{10\baselineskip} \part[Residual bases and applications]{Residual bases and field applications}\label{part:applications} \section{A common discriminant comparison}\label{shared:explicit} For $b>0$ put $a=\pi/(2b)$ and define the even function \[ F_b(x)=\begin{cases} (1-|x|/(2b))\cos(a|x|)+\pi^{-1}\sin(a|x|),&|x|\le2b,\\ 0,&|x|>2b. \end{cases} \] This is the normalized autocorrelation of $\cos(at)$ on $[-b,b]$, extended by zero. Thus $F_b(0)=1$, $0\le F_b\le1$, and its Fourier transform is nonnegative. Its Laplace transform is \begin{equation}\label{shared:eq:J} J_b(s)=\int_0^{2b}F_b(x)e^{-sx}\,dx =\frac{s}{s^2+a^2} +\frac{a^2(1+e^{-2bs})}{b(s^2+a^2)^2}. \end{equation} Put \[ B_b=\int_0^\infty\frac{1-F_b(x)}{2\sinh(x/2)}\,dx, \qquad C_b=2\{J_b(-1/2)+J_b(1/2)\}. \] Expansion of $(2\sinh(x/2))^{-1}$ gives $B_b=\sum_{j\ge0}((j+1/2)^{-1}-J_b(j+1/2))$. Every summand lies between zero and $a^2(j+1/2)^{-3}$. For $N\ge1$ this gives \begin{equation}\label{shared:eq:B} B_b\le B_{b,N}:= \sum_{j=0}^{N-1}\left((j+1/2)^{-1}-J_b(j+1/2)\right) +\frac{a^2}{2(N-1/2)^2}. \end{equation} The midpoint inequality for $x^{-3}$ also gives the convenient one-term bound $B_b\le2-J_b(1/2)+a^2/2$. The alternating four-term sum \[ A_{b,4}=J_b(1/2)-J_b(3/2)+J_b(5/2)-J_b(7/2) \] is a lower bound for $\int_0^\infty F_b(x)/(2\cosh(x/2))\,dx$. \begin{proposition}\label{shared:poitou} Assume GRH for the Dedekind zeta function of $L$. Let $n=[L:\mathbf Q]$, let $r_1$ be its number of real embeddings, and let $N_Q$ count its prime ideals of norm $Q$. If $B_b^*$ is either upper bound for $B_b$ just given, then \begin{equation}\label{shared:eq:grh} \log\operatorname{rd}(L)\ge\gamma+\log(8\pi)-B_b^* -\frac{C_b}{n}+\frac{r_1}{n}A_{b,4} +\sum_Q\frac{2N_Q}{n}\log Q \sum_{m\ge1}Q^{-m/2}F_b(m\log Q). \end{equation} Any subset of the prime terms may be retained. At fixed $b$, signature ratio and retained splitting densities, this lower bound increases with $n$. \end{proposition} \begin{proof} Apply Poitou's explicit formula \cite[equation~(7) and Proposition~4]{g:Poitou} with $F_b$. Under GRH its zero terms have the required sign. Replace $B_b$ by its upper bound and the real-place integral by its displayed lower bound. All omitted prime terms are nonnegative, and $C_b>0$. The inner sums are finite because $F_b$ has compact support. \end{proof} For comparisons without GRH, use $G_b(x)=F_b(x)/\cosh(x/2)$, and write $B(G_b),C(G_b)$ for the preceding integrals with $F_b$ replaced by $G_b$. Its zero terms in the explicit formula are nonnegative throughout the critical strip; see \cite[equation~(2.3)]{s:Odlyzko} and the strip positivity calculation in Section~\ref{s:sec:residual}. The bound $1-F_b(x)\le\pi^2x^2/(8b^2)$ gives \[ B(G_b)\le\log2+\frac{7\pi^2\zeta(3)}{16b^2},\qquad C(G_b)=\frac{32b}{\pi^2},\qquad \int_0^\infty\frac{F_b(x)}{2\cosh^2(x/2)}\,dx \ge1-\frac{\pi^4}{24b^2}. \] Consequently, without any hypothesis on the zeros, \begin{align}\label{shared:eq:unconditional} \log\rd(L)&\ge\log(4\pi)+\gamma -\frac{7\pi^2\zeta(3)}{16b^2}-\frac{32b}{\pi^2n} +\frac{r_1}{n}\left(1-\frac{\pi^4}{24b^2}\right)\notag\\ &\quad+\sum_Q\frac{2N_Q}{n}\log Q \sum_{m\ge1}\frac{2F_b(m\log Q)}{Q^m+1}. \end{align} As before, one may discard prime terms or the real-place term. The numerical uses below take $\zeta(3)<1.203$ and, when convenient, retain just the first prime-power term, using the same quadratic lower bound for $F_b$; their certificates are in \cite[\path{common/verify_unconditional_refinements.py}]{code}. The comparisons below specify the degree, signature, retained Euler densities and local discriminant upper bound. Their strict numerical inequalities include the tail in~\eqref{shared:eq:B}; no truncated infinite sum is used without a remainder bound. \section{Local constructions for a split terminal prime} \label{c:local} We record the local characteristic-three calculations used in the three supported residual bases below. The universal existence statement is Theorem~\ref{shared:wild-reset}; here we also need the exact characteristic-two reduction of its type over $\Q_3$. \begin{lemma}\label{c:characters} Let $K$ be a real quadratic field of ordinary class number one in which three splits. Every finite character $G_K\to\overline\F_3^\times$ unramified outside three has square one. \end{lemma} \begin{proof} Such a character has order prime to three and kills the principal units at the two places above three. The ray-class sequence, with both real places included, makes its group a quotient of $(\F_3^\times)^2\times\{\pm1\}^2$, which has exponent two. \end{proof} For a finite Galois extension $L/\Q_3$, write $d(L)=v_3(\Delta(L/\Q_3))/[L:\Q_3]$, with $v_3(3)=1$. Unramified base extension does not change this contribution. \begin{lemma}\label{c:localdifferent} Let $\br:G_{\Q_3}\to\GL_2(\overline\F_3)$ have determinant $\om$, and let $L$ be its linear kernel field. If wild inertia is nontrivial, its fixed line makes $\br$ triangular, with diagonal quotient $\om\alpha$ for an unramified character $\alpha$. Write $Q$ for the wild inertia order. \begin{enumerate}[label=(\roman*)] \item If $\alpha\ne1$, then $d(L)\le3/2-1/Q<3/2$. \item If $\alpha=1$, then $Q\in\{3,9\}$ and $d(L)\le13/6-1/Q$. Its projective kernel field is totally ramified of degree $2Q$. \item In the second case with $Q=9$, the projective kernel field is \begin{equation}\label{c:eq:E0} E_0=\Q_3(\zeta_3,\sqrt[3]3,\sqrt[3]4),\qquad [E_0:\Q_3]=18, \qquad d(E_0)=37/18. \end{equation} \end{enumerate} In the tame case $d(L)<1$. In all cases $d(L)\le37/18$; the same upper bounds hold for projective subfields. \end{lemma} \begin{proof} A nontrivial normal $3$-subgroup has a unique fixed line, which the decomposition group preserves. Rank-one characters over $\Q_3$ have inertia of order dividing two. The diagonal inertia characters are therefore $1,\om$ in some order, and their quotient is $\om\alpha$ with $\alpha$ unramified. The kernel $P$ of the two diagonal characters in the finite image consists of translations. It equals wild inertia. Indeed, diagonalize a tame inertia involution, whose conjugation on translations is inversion. The commutator with any translation belongs to inertia and is twice that translation. Since $2$ is invertible in characteristic three, every translation lies in wild inertia. In particular $P$ is elementary abelian of order $Q$. Put $T=L^P$. It contains $\mu_3$ by the determinant condition and has ramification index two. Thus $T=U(\zeta_3)$ for an unramified extension $U/\Q_3$ of degree prime to three. Let $k$ be its residue field and $\varpi=\zeta_3-1$. Local reciprocity gives an equivariant surjection $T^\times/(T^\times)^3\to P$. Write $U_j=1+\varpi^j\OO_T$. The logarithm gives $U_2^3=U_4$, and \[ (1+a\varpi)^3\equiv1+(\bar a^3-\bar a)\varpi^3\pmod{\varpi^4}, \] since $\varpi^2=-3(1+\varpi)$. Thus the possible unit grades modulo cubes are $1,2,3$, and the last is $k/\{a^3-a:a\in k\}$, of dimension one over $\F_3$. The tame involution acts on grade $j$ by $(-1)^j$, and on the valuation quotient trivially. Only grades one and three can map to $P$. A last nonzero grade $j$ gives conductor $j+1$; these are the usual local reciprocity and conductor conventions of \cite{c:SerreLocal,c:MT}. A Frobenius fixing $\zeta_3$ acts trivially on the top quotient. If $\alpha\ne1$, its action on $P$ is multiplication by a nontrivial scalar; none of the associated $\F_3$-eigenvalues is one. The top grade cannot contribute, so every nontrivial character of $P$ has conductor at most two. The tower and conductor--discriminant formulas give \begin{equation}\label{c:eq:localtower} d(L)=\frac12+\frac1{2Q}\sum_{\xi\ne1}a_T(\xi) \le\frac32-\frac1Q. \end{equation} If $\alpha=1$, divide by the lower diagonal character. The extension class belongs to \[ H^1(\Q_3,\overline\F_3(\om)) =\bigl(\Q_3^\times/(\Q_3^\times)^3\bigr)\otimes_{\F_3}\overline\F_3. \] The rational Kummer space has basis the classes of $3$ and $4$, coming from valuation and principal units. The additive image therefore has rank at most two. At least $Q/3$ characters kill the image of the one-dimensional top unit grade. Their nontrivial members have conductor at most two; the others have conductor at most four. Hence \[ \sum_{\xi\ne1}a_T(\xi)\le2(Q/3-1)+4(2Q/3)=10Q/3-2. \] Equation~\eqref{c:eq:localtower} gives $d(L)\le13/6-1/Q\le37/18$. In the projective representation, a Frobenius fixing $\zeta_3$ is a translation. Its commutator with the tame inertia involution again shows that it belongs to inertia. Projective decomposition therefore equals inertia, of order $2Q$. When $Q=9$, the two Kummer coordinates are independent, so their common kernel cuts out precisely $E_0$. Over $\Q_3(\zeta_3)$ its eight nontrivial characters have conductors two, twice, and four, six times. Thus \[ d(E_0)=\frac12+\frac{2\cdot2+6\cdot4}{18}=\frac{37}{18}. \] The tame formula is $1-1/e<1$, and normalized differents cannot increase on passage to a subfield. \end{proof} The bound $Q\le9$ here is confined to an exactly cyclotomic diagonal quotient. For an arbitrary determinant, the bound we use is Moon--Taguchi's inequality \begin{equation}\label{c:eq:MT} d(L)\le\frac{13}{6}-\frac1Q \end{equation} when the wild ramification index is $Q>1$; tame extensions have $d(L)<1$~\cite[Theorem~3 and the following Remark~(1)]{c:MT}. \subsection{\texorpdfstring{A wild type over $\Q_3$}{A wild type over Q3}} Specialize the construction of Lemma~\ref{a:struct:lem:type} to $F=\Q_3$, with uniformizer $+3$ and additive character $\psi(x)=\zeta_3^x$. Retain its notation $K_0,K_1,J$ and write $\vartheta$ for the affine generic character. The sign of this uniformizer fixes the quadratic subfield in the following lemma. \begin{lemma}\label{c:type} The representation $\sigma_+:=\Ind_J^{K_0}\vartheta$ is the irreducible eight-dimensional type of a supercuspidal inertial type $\tau_+$ with determinant one. Its reduction satisfies \begin{equation}\label{c:eq:weights} \overline\sigma_+^{\ssm} =1\oplus\det\oplus\Sym^2\F_3^2\oplus(\det\otimes\Sym^2\F_3^2). \end{equation} The inertia representation $\tau_+$ is irreducible after reduction in every characteristic greater than three. For a determinant-cyclotomic compatible system of this type, its characteristic-two residual local kernel field is the unique totally ramified $S_3$ extension $E_+/\Q_3$ with quadratic subfield $\Q_3(\sqrt3)$ and normalized different $7/6$. \end{lemma} \begin{proof} The construction of Theorem~\ref{shared:wild-reset} at $\ell=3$ identifies $\sigma_+$ as the irreducible compact type of dimension eight, with determinant-one inertial parameter $\tau_+$. In the induced integral lattice, $\vartheta$ reduces to one, so $\overline\sigma_+=\Ind_{ZU}^{\GL_2(\F_3)}1$. If $B$ is the upper triangular Borel, then $B/ZU$ has order two, with characters $1,\det$. Induction in stages gives \[ \overline\sigma_+=\Ind_B^{\GL_2(\F_3)}1 \oplus\det\otimes\Ind_B^{\GL_2(\F_3)}1. \] The first summand is the permutation module on $\mathbf P^1(\F_3)$. Its constant line splits, since four is invertible in characteristic three, and its augmentation module is the Steinberg representation $\Sym^2\F_3^2$. For example, evaluating a quadratic at the four projective points identifies it with that augmentation module. This proves \eqref{c:eq:weights}; its four constituents are exactly the Serre weights with trivial scalar central character. By \cite[Section~3.4 and Theorem~3.18]{c:AL}, the parameter is induced from a character of a ramified quadratic extension of $\Q_3$, of order three on the first principal-unit quotient, with conjugate the inverse there. Its inertia contains, in a suitable basis, \[ u=\begin{pmatrix}\zeta_3&0\\0&\zeta_3^{-1}\end{pmatrix}, \qquad v=\begin{pmatrix}0&1\\-1&0\end{pmatrix}. \] The off-diagonal matrix has this form because inertia has determinant one. These matrices generate a group of order twelve. In characteristic $l>3$, the two $u$-eigenlines remain distinct and $v$ exchanges them. Thus reduction is irreducible; semisimplicity for this group of order prime to $l$ makes the conclusion independent of the stable lattice. We check the quadratic field in the local parameter. In the parametrization of~\cite[Sections~3.1, 3.4 and Theorem~3.18]{c:AL}, the affine-generic character $b+c$ uses uniformizer $+3$. The inducing quadratic extension is consequently $\Q_3(\sqrt3)$. Reduction modulo two gives the irreducible group $S_3$. Its normalizer in $\SL_2(\overline\F_2)$ is itself: all automorphisms of $S_3$ are inner, and its centralizer consists of determinant-one scalars, hence is trivial. Thus the entire local image equals its inertia image and the residual field has $e=6$, $f=1$. The wild character has conductor two over the ramified quadratic field. The two-dimensional Artin conductor is three and remains three after reduction modulo two: the wild inertia has odd order and the inertia invariants remain zero. The lower groups of the residual field have orders $6,3,1$, giving normalized different $(5+2)/6=7/6$. Finally, over $\Q_3(\sqrt3)$ the conductor-two anti-invariant character space has dimension one over $\F_3$. Its unique nonzero line gives a unique $S_3$ extension. This proves the stated uniqueness; the splitting field of $x^3-3x-3$ is one realization. \end{proof} \begin{proposition}\label{c:wildlift} Every continuous $\br:G_{\Q_3}\to\GL_2(\overline\F_3)$ with determinant $\bchi_3$ has a potentially crystalline lift of determinant $\chi_3$, Hodge--Tate weights $\{0,1\}$, and inertial type $\tau_+$. \end{proposition} \begin{proof} Apply Theorem~\ref{shared:wild-reset} with $F=\Q_3$ and the affine-generic character fixed above. \end{proof} \begin{lemma}\label{c:semistable} Let $\br:G_{\Q_3}\to\GL_2(\overline\F_3)$ have determinant $\om$. It has a lift of weights $\{0,1\}$ and determinant $\chi_3$ which is either a quadratic twist of a semistable representation with nonzero monodromy, or potentially crystalline of tame supercuspidal type \[ \tau_4=\widetilde\omega_2^{\,2}\oplus\widetilde\omega_2^{\,6}, \] where $\omega_2$ is a fundamental character of niveau two. For either choice in a determinant-cyclotomic compatible system, the characteristic-two residual local kernel field is contained in \begin{equation}\label{c:eq:quadratic-envelope} E'=\Q_3(\sqrt3,\sqrt{-1}),\qquad e(E'/\Q_3)=2,\quad f(E'/\Q_3)=2,\quad d(E'/\Q_3)=\tfrac12. \end{equation} \end{lemma} \begin{proof} We first obtain a quadratic twist of a semistable lift. If $\br$ is reducible, a ramified quadratic twist, when necessary, puts its invariant character in the form $\om\lambda$ and its quotient in the form $\lambda^{-1}$, with $\lambda$ unramified. Let $\eta$ be the unramified Teichm\"uller lift of $\lambda$. If $\lambda^2\ne1$, local duality gives $H^2(G_{\Q_3},\OO(\chi_3\eta^2))=0$. Thus the actual residual extension lifts to an extension of $\eta^{-1}$ by $\chi_3\eta$; the one-dimensional characteristic-zero extension space is crystalline. If $\lambda^2=1$, take $\eta^2=1$. Kummer theory lifts both the unit and valuation directions of $\Q_3^\times/(\Q_3^\times)^3$; the former is crystalline and the latter is semistable with nonzero monodromy. If $\br$ is irreducible, its wild inertia is trivial and its tame exponents are $\{1,3\}$ or $\{5,7\}$ modulo eight. A ramified quadratic twist interchanges the pairs. The first lifts by induction from a crystalline character with labelled weights $(1,0)$ over the unramified quadratic extension. An unramified scalar correction fixes the determinant. Twisting back gives the claimed semistable lift; compare \cite[proof of Proposition~7.8.1]{g:Snowden}. Retain this lift if its monodromy is nonzero. Otherwise its inertial type is scalar quadratic, say $\kappa\oplus\kappa$. Its potentially crystalline compact type is $\kappa\circ\det$, with reduction $1$ or $\det$. The fixed-determinant formula of \cite[Theorem~1.2]{g:Tung}, throughout with determinant $\chi_3$, gives $\mu_1(\br)>0$ or $\mu_{\det}(\br)>0$. The cuspidal type for $\tau_4$ has dimension two and trivial central character. It is the two-dimensional representation of $\PGL_2(\F_3)\simeq S_4$ obtained from the standard representation of $S_3$. The integral augmentation lattice in $\Z_3^3$ reduces to a module with a constant line and sign quotient. Sign inflates to $\det$, so its semisimplification is $1\oplus\det$. Breuil--M\'ezard for Hodge type $(1,0)$ and $\tau_4$ has multiplicity $\mu_1(\br)+\mu_{\det}(\br)>0$. Its three-torsion-free deformation ring gives the required potentially crystalline lift, with determinant exactly $\chi_3$. For the semistable choice, the characteristic-two local member is a quadratic twist of a special representation with diagonal characters $\chi_2\nu$ and $\nu$, where $\nu$ is unramified quadratic. Both reduce to one, as does the quadratic twist. The full residual image is therefore an elementary abelian two-group. For the tame choice, inertia and arithmetic Frobenius matrices $T,\Phi$ satisfy \[ T^2=-I,\qquad \Phi T\Phi^{-1}=T^{-1},\qquad \Phi^2=-3I. \] The last equality follows from trace zero and determinant three. Modulo two these matrices become commuting involutions, again giving an elementary abelian two-group for every stable lattice. The field $E'$ is the maximal elementary quadratic extension of $\Q_3$, so it contains both residual kernel fields. It is the compositum of the ramified and unramified quadratic extensions; its tame ramification index two gives the stated different. \end{proof} Over a real quadratic field with a single dyadic place, Lemma~\ref{shared:global} applies to every finitely ramified, totally odd lift of determinant $\chi_2$ with nonsolvable residue that is crystalline nearly ordinary of regular labelled weights, or crystalline of weight two and potentially nonordinary. It gives potential modularity and a strongly compatible system, with the ordinary ordering of the actual source. \section{\texorpdfstring{The characteristic-three base over $\Q(\sqrt7)$}{The characteristic-three base over Q(sqrt(7))}} \label{c:7:section} Throughout this section $K=\Q(\sqrt7)$, $C=K(\zeta_3)$, and $\mathfrak p_1=(2+\sqrt7)$, $\mathfrak p_2=(2-\sqrt7)$. \begin{theorem}\label{c:7:thm:base} Assume GRH. Every continuous irreducible representation $\brho:G_K\to\GL_2(\overline\F_3)$ with determinant $\bchi_3$, unramified outside $3$, is modular. \end{theorem} \begin{lemma}\label{c:7:lem:characters} The ordinary class number of $K$ is one and its narrow class number is two. The extension $K(i)/K$ is the unique nontrivial quadratic extension unramified at every finite place, and is inert at both places above $3$. Every finite character $G_K\to\overline\F_3^\times$ unramified outside $3$ has square one. \end{lemma} \begin{proof} The Minkowski bound is $\sqrt7<3$. The prime above $2$ is principal, generated by $3+\sqrt7$ of norm two, so the ordinary class number is one. The equation $x^2-7y^2=-1$ has no solution modulo seven. The least positive solution of $x^2-7y^2=1$ with $y>0$ is $(8,3)$, so $\epsilon=8+3\sqrt7$ generates the positive units and the narrow class number is two. The discriminant of the biquadratic field $K(i)$ is $28(-4)(-7)=28^2$. Thus its relative discriminant over $K$ is trivial. Narrow class field theory gives uniqueness, and its completion at either place above three is unramified quadratic because $-1$ is nonsquare in $\F_3$. The character assertion follows from Lemma~\ref{c:characters}. \end{proof} \begin{lemma}\label{c:7:lem:bounds} Assume GRH. Retain only a proved number $N_3$ of primes of norm three, and put $c=2N_3/[L:\Q]$. The following lower bounds hold when the degree and $c$ are at least the indicated values: \[ \begin{array}{c|r|c|c} b&[L:\Q]\text{ at least}&c&\rd(L)\text{ greater than}\\\hline 21/2&60000&0&37.33078\\ 15/2&2880&0&31.93013\\ 9&19656&2/9&48.00416\\ 23&20\,000\,000\,000&1/9&50.67997. \end{array} \] In particular, every totally imaginary $K$-Galois field unramified outside three whose completions at both places above three are $E_0$ satisfies \begin{equation}\label{c:7:eq:cutoff} [L:\Q]<20\,000\,000\,000. \end{equation} \end{lemma} \begin{proof} Use the explicit formula of Section~\ref{shared:explicit}, discarding the real-place term and retaining only the specified primes of norm three. Its one-term bound for $B_b$ gives \begin{equation}\label{c:7:eq:positive} \log\rd(L)\ge\gamma+\log(8\pi)-B_b^*-\frac{C_b}{n} +c\log3\sum_{m\ge1}3^{-m/2}F_b(m\log3). \end{equation} The prime sum is finite. Outward evaluation gives the four rows; the bound increases with $n$ and $c$. For the last assertion, the local degree is eighteen and residue degree one at both primes. Thus $N_3=n/18$, $c=1/9$, and the discriminant tower gives \begin{equation}\label{c:7:eq:U} \rd(L)=\sqrt{28}\,3^{37/18}<50.62072. \end{equation} The last row contradicts this equality for degree at least twenty billion. \end{proof} \begin{lemma}\label{c:7:lem:residualshape} Assume GRH. A representation as in Theorem~\ref{c:7:thm:base} is modular unless its projective group is \[ \PSL_2(\F_{3^{2e}})\quad(e\ge2), \qquad\text{or}\qquad \PGL_2(\F_{3^{2e+1}})\quad(e\ge1). \] In every remaining case its projective completion at each place above three is exactly $E_0$. In the projective general linear case the unique quadratic subfield over $K$ is $C$. \end{lemma} \begin{proof} Let $L$ and $M$ be the linear and projective kernel fields, with groups $G$ and $H$ over $K$, and let $Z$ be the scalar subgroup of $G$. Both fields are totally imaginary, since complex conjugation has eigenvalues $1,-1$. Solvable images are modular. Dickson's classification leaves $A_5$, $\PSL_2(\F_q)$ and $\PGL_2(\F_q)$ with $q=3^f\ge9$. We first consider $q\ge27$. If $H$ is perfect, then $|Z|\ge4$: for $|Z|\le2$ the determinant would kill $Z$ and factor through the perfect group $H$, contradicting oddness; scalar order is prime to three. For $H=\PGL_2(\F_q)$, $Z$ is nontrivial. Otherwise its perfect subgroup $\PSL_2(\F_q)$ would embed in $\SL_2(\overline\F_3)$, which has only one nonidentity involution. Therefore $[L:\Q]\ge78624$ in either case. If one place above three is tame or has nontrivial unramified factor $\alpha$, Lemma~\ref{c:localdifferent} gives \[ \rd(L)<\sqrt{28}\,3^{(3/2+37/18)/2} =\sqrt{28}\,3^{16/9}<37.30737, \] contrary to the first row of Lemma~\ref{c:7:lem:bounds}. Both diagonal quotients are therefore cyclotomic with nontrivial wild inertia. If one wild rank is one, the projective field satisfies \[ \rd(M)\le\sqrt{28}\,3^{35/18}<44.80380. \] Its degree is at least $2|\PSL_2(\F_{27})|=19656$, and its local degrees are at most $6$ and $18$, both totally ramified. Thus $c=1/e_1+1/e_2\ge2/9$, contradicting the third row. Both wild ranks are two, so both local fields are $E_0$. For odd $f$, the tame inertia involution has Borel torus coordinate $-1$, nonsquare in $\F_q$, and cannot belong to $\PSL_2(\F_q)$. For even $f$, a full $\PGL_2(\F_q)$ image would have a nontrivial squareclass quotient trivial on both local decomposition groups: these equal inertia, and $-1$ is a square. Its quadratic field would be unramified at every finite place and split at three, contrary to Lemma~\ref{c:7:lem:characters}. This gives the asserted parity restrictions. In the remaining projective general linear case, the squareclass quotient agrees with $\om$ on both decomposition groups at three. Their quotient is unramified at all finite places and split at three, so is trivial by the same lemma. The quadratic field is therefore $C$. It remains to consider full $\PGL_2(\F_9)$. If its quadratic squareclass character ramifies at a place above three, projective wild inertia there must vanish: otherwise the decomposition group is in a Borel, and the tame Frobenius relation makes its inertia torus coordinates $\pm1$, both squares in $\F_9$. Linear wild inertia then vanishes as well, since a scalar in characteristic three has order prime to three. We obtain \[ \rd(L)<\sqrt{28}\,3^{55/36}<28.34747. \] If the squareclass character is unramified everywhere finite, it is the character of $K(i)$, nontrivial on Frobenius at both places above three. Neither local quotient can then be exactly cyclotomic with wild inertia, because its projective decomposition equals inertia. Hence $\rd(L)<\sqrt{28}\,3^{3/2}<27.49546$. But $[L:\Q]\ge2880$, contradicting the second row of Lemma~\ref{c:7:lem:bounds}. The remaining small nonsolvable groups are $A_5$ and $\PSL_2(\F_9)$. Their embeddings in $\PGL_2(\F_9)$ have trivial squareclass character, since they are perfect. Theorem~1.2(5) and Lemma~2.1 of \cite{h:AKT} give modularity of the projective representation and of its given linear lift. \end{proof} \subsection{A classical form of level eighty-four}\label{c:7:sec:classical} Let $\eta=\chi_{28}$ be the primitive quadratic Dirichlet character attached to $K/\Q$. \begin{lemma}\label{c:7:lem:form} There is a normalized newform $f\in S_2(\Gamma_0(84),\eta)$ with coefficient field $F=\Q(\alpha)$ such that \begin{equation}\label{c:7:eq:coeffs} \alpha^4+14\alpha^2+32=0,\qquad a_5(f)=\alpha,\qquad a_{11}(f)=\frac{\alpha^3+10\alpha}{4}. \end{equation} Moreover, $a_3(f)=1$. The prime three is inert in $F$. Complex conjugation sends $\alpha$ to $-\alpha$, and its fixed field is $\Q(\sqrt{17})$, with $\sqrt{17}=\alpha^2+7$. \end{lemma} \begin{proof} Take the normalized newform with LMFDB label \href{https://www.lmfdb.org/ModularForm/GL2/Q/holomorphic/84/2/b/b/}{\texttt{84.2.b.b}}~\cite{h:LMFDB}. Its nebentypus is $\chi_{84}(55,\cdot)$, induced by $\eta$. The tabulated coefficient field is $\Q(\nu)$, where $\nu^4-\nu^3-2\nu+4=0$, and the coefficients are \[ a_3(f)=1,\qquad a_5(f)=\nu^3+\nu-2, \qquad a_{11}(f)=\nu-\nu^2. \] Put $\alpha=\nu^3+\nu-2$. Reduction modulo the polynomial of $\nu$ gives \[ \alpha^4+14\alpha^2+32=0, \qquad 4(\nu-\nu^2)=\alpha^3+10\alpha, \] which proves \eqref{c:7:eq:coeffs}. Modulo three, $P(X)=X^4+14X^2+32$ becomes $X^4-X^2-1$. If $y^2=y+1$ in $\F_9$, then $y^4=-1$, so $y$ has order eight and is not a square in $\F_9$. Thus $P$ is irreducible modulo three and over $\Q$. It follows that $F=\Q(\nu)=\Q(\alpha)$. The polynomial discriminant is $2^{13}17^2$, prime to three, so three is inert in the full ring of integers of $F$. Finally $\alpha^2=-7\pm\sqrt{17}$ is negative at both real embeddings. Hence $F$ is CM, complex conjugation sends $\alpha$ to $-\alpha$, and its fixed field is $\Q(\sqrt{17})$. \end{proof} \begin{proposition}\label{c:7:prop:linearaux} There exists a finite Galois extension $B/K$ with \[ \Gal(B/K)\simeq D:=\{g\in\GL_2(\F_9):\det g\in\{1,-1\}\},\qquad [B:K]=1440. \] It is unramified outside three, and its two local completions at three are contained in $E_0$. Its maximal solvable quotient is $C_2$, with fixed field $C$. \end{proposition} \begin{proof} Let $\lambda$ be the unique prime of $F$ above three. Restrict the representation attached to $f$ to $G_K$, and let $\bar r$ be its semisimplified reduction at $\lambda$. Its determinant is $\om$, since the rational determinant is $\eta\chi_3$, and its initial coefficient field is $\F_{81}$. Write $t=\bar\alpha$ and $x=t^2$, so $x^2=x+1$. The primes $5$ and $11$ are inert in $K$, and the rational residual determinants at both are one. Squaring their Frobenii gives elements of $G_K$ with determinant one and traces \[ s_5=x+1,\qquad s_{11}=1-x. \] Here $\bar a_{11}^{\,2}=-x$, from \eqref{c:7:eq:coeffs}. Their projective invariants are \begin{equation}\label{c:7:eq:orders} s_5^2=2,\qquad s_{11}^2=-x-1,\qquad (-x-1)^2=-1. \end{equation} If $z$ is squared trace divided by determinant, the eigenvalue ratio $u$ satisfies $u+u^{-1}=z-2$. The first equality gives projective order four. In characteristic three, $z^2+1=0$ is equivalent to $u^4+u^3+u^2+u+1=0$, so the second element has projective order five. Both have distinct eigenvalues. All traces on $G_K$ lie in $\F_9$. For a good rational prime $q$, the Petersson adjoint identity gives $\overline{a_q(f)}=\eta(q)a_q(f)$. At a split prime of $K$ the trace is $a_q(f)$, and at an inert prime it is $a_q(f)^2-2\eta(q)q$. Both belong to the real field $F^+=\Q(\sqrt{17})$, so their reductions lie in its residue field $\F_9$. Chebotarev gives the assertion for every element of $G_K$. This also proves that $\bar r$ is irreducible. If its semisimplification were $\psi_1\oplus\psi_2$, trace invariance under the ninth-power coefficient automorphism would either fix both characters or interchange them. In the first case their quotient has order dividing eight; in the second its ninth power is its inverse, so its order divides ten. Neither alternative permits both projective orders in \eqref{c:7:eq:orders}. Absolute irreducibility, together with the traces and determinant in $\F_9$ and the triviality of its Brauer group, gives a model of $\bar r$ over $\F_9$. Both $1$ and $-1$ are squares in $\F_9$, so its projective image lies in $\PSL_2(\F_9)$. By Dickson's classification, a proper subgroup cannot contain elements of both orders four and five: the Borel, subfield, $A_4$ and $S_4$ alternatives have no element of order five, the dihedral alternatives have orders dividing eight or ten, and $A_5$ has no element of order four. Hence its projective image is all of $\PSL_2(\F_9)\simeq A_6$. The scalar subgroup of the linear image is contained in $\mu_4$, since the determinant has order two. It cannot have order at most two: then the determinant would factor through the perfect projective group, contradicting total oddness. Thus it has order four. The linear image has order $4\cdot360=1440$ and equals $D$. Moreover $D=\mu_4\SL_2(\F_9)$, with the scalar factor central and $\SL_2(\F_9)$ perfect. Therefore $D'=\SL_2(\F_9)$ and its maximal solvable quotient is its determinant quotient, which cuts out $C$. We check the local kernels, not merely the projective conductor. At $2$ and $7$, local--global compatibility for the newform gives conductor exponents respectively two and one \cite[equation~(1)]{g:Skinner}. These equal the exponents of its determinant character $\eta$. For a two-dimensional Weil--Deligne representation with ramified determinant, equality $a(V,N)=a(\det V)$ forces its inertia to be $1\oplus\det V$ and $N=0$. Indeed, if $V^I=0$, then \[ a(V,N)\ge2+\operatorname{Swan}(V) \ge2+\operatorname{Swan}(\det V)>a(\det V). \] Thus $V^I$ is a line. The other inertia character is the ramified determinant, and a commuting nilpotent operator is zero. Consequently the inertia at $2$ and $7$ is $1\oplus\eta$, which becomes trivial over $K$. At every other prime away from three the representation is already unramified. At three, $\eta|_{G_{\Q_3}}=1$. Since $3\Vert84$ and $a_3(f)=1$, the local representation is semistable of Steinberg type with unit eigenvalue one. Local--global compatibility and Lemma~\ref{c:monodromy} give the trivial unramified ordinary quotient at both places above three. Its determinant is $\chi_3$, so the reduction is an actual extension of $1$ by $\bchi_3$. Its extension class lies in $H^1(\Q_3,\overline\F_3(\bchi_3))$, the rational cube-class space tensored with $\overline\F_3$. That space is generated by $3$ and $4$, so its full linear local kernel field is contained in $E_0$. There is no additional unramified scalar factor. Taking $B=K^{\ker\bar r}$ proves the proposition. \end{proof} \subsection{Auxiliary composita and the residual theorem}\label{c:7:sec:baseproof} \begin{proposition}\label{c:7:prop:auxfield} There exists a totally imaginary $K$-Galois extension $\mathcal F$ of degree $38880$ over $K$, unramified outside three, with completion $E_0$ at both places above three. Its degree has prime factors only $2,3,5$, and it contains $C$. \end{proposition} \begin{proof} Put $\pi=2+\sqrt7$, $\pi'=2-\sqrt7$ and $\epsilon=8+3\sqrt7$, and consider \[ E=K(\zeta_3,\sqrt[3]\pi,\sqrt[3]{\pi'},\sqrt[3]\epsilon). \] The three cube classes are independent over $K$: valuations at $\mathfrak p_1,\mathfrak p_2$ first give independence of the two prime generators, and the fundamental unit is not a cube. They remain independent over $C$, since taking norms across a degree-two extension is injective on cube classes from $K$. Thus $[E:C]=27$ and $[E:K]=54$, with solvable group $(C_3)^3\rtimes C_2$, where the involution acts by inversion. The radicands are units outside three, so their Kummer extensions are unramified there. At the two three-adic embeddings, $\sqrt7$ reduces to $4,5$ modulo nine. At each place one of $\pi,\pi'$ has valuation one and the other is a unit congruent to seven modulo nine, hence is not a cube. The first two radicands therefore span the rational local cube-class space, giving completion $E_0$ at both places. The fundamental unit enlarges the global degree without enlarging either completion. Let $B$ be the field of Proposition~\ref{c:7:prop:linearaux}. Its maximal solvable quotient is $C_2$, whereas $E/K$ is solvable and contains $C$. Therefore $B\cap E=C$. Set $\mathcal F=BE$. We have \begin{equation}\label{c:7:eq:auxdegree} [\mathcal F:K]=\frac{1440\cdot54}{2}=38880=2^5\cdot3^5\cdot5. \end{equation} Its local completions are still $E_0$, because those of $B$ are contained in $E_0$ and those of $E$ equal $E_0$. All relative extensions away from three are unramified, and $C\subset\mathcal F$ makes it totally imaginary. \end{proof} Whenever two $K$-Galois fields have completion $E_0$ at the same place, their local compositum is again $E_0$. This is equality of subfields of $\overline\Q_3$: the field in \eqref{c:eq:E0} is normal and is the full Kummer extension of the rational cube-class space. We shall repeatedly apply the degree bound \eqref{c:7:eq:cutoff} after taking such composita. \begin{proof}[Proof of Theorem~\ref{c:7:thm:base}] By Lemma~\ref{c:7:lem:residualshape}, it suffices to treat the two displayed families there, with projective field $M$ locally equal to $E_0$. Let $H=\Gal(M/K)$ and use the auxiliary field $\mathcal F$ of Proposition~\ref{c:7:prop:auxfield}. Suppose $H=\PSL_2(\F_{3^{2e}})$, $e\ge2$. Its simplicity makes $M\cap\mathcal F$ either $K$ or $M$. The latter is impossible because $|H|\ge|\PSL_2(\F_{81})|=265680>38880$. Hence \[ [M\mathcal F:\Q]\ge2\cdot265680\cdot38880 =20\,659\,276\,800, \] contradicting \eqref{c:7:eq:cutoff}. Suppose $H=\PGL_2(\F_{3^{2e+1}})$ with $e\ge2$. Its only proper nontrivial normal quotient is $C_2$, and its order exceeds $38880$. Since both fields contain the unique quadratic subfield $C$, we have $M\cap\mathcal F=C$. Therefore \[ [M\mathcal F:\Q]\ge|\PGL_2(\F_{243})|\cdot38880 =557\,876\,056\,320, \] again impossible. It remains to consider $H=\PGL_2(\F_{27})$, of order $h=19656$. Let $\sigma$ be the nontrivial automorphism of $K/\Q$. If $M\ne M^\sigma$, the two fields intersect exactly in $C$: their only proper nontrivial common quotient is quadratic, and $C$ is invariant under rational conjugation. For $P=MM^\sigma$ we have \[ \Gal(P/K)\simeq H\times_{C_2}H, \qquad [P:K]=h^2/2. \] Its derived subgroup is $\PSL_2(\F_{27})^2$, which is perfect. Any quotient that does not kill this subgroup has a composition factor of order divisible by $7$ and $13$. Such a quotient cannot be a quotient of $\Gal(\mathcal F/K)$, whose order has prime factors only $2,3,5$. Thus $P\cap\mathcal F=C$, and \[ [P\mathcal F:\Q]=h^2\cdot19440 =7\,510\,806\,051\,840, \] contradicting \eqref{c:7:eq:cutoff}. Consequently $M=M^\sigma$ and $M/\Q$ is Galois. We now prove modularity rather than nonexistence. Write $\Gamma=\Gal(M/\Q)$. Conjugation induces a map $\Gamma\to\Aut(H)$. Its outer image is a quotient of $\Gamma/H=C_2$, whereas $\Out(\PGL_2(\F_{27}))\simeq C_3$ consists of field automorphisms \cite[Section~13, Lemma~52]{c:Gow}. The outer action is trivial. As $H$ is centreless, the conjugation map lands in $\operatorname{Inn}(H)\simeq H$. Composing it with $G_\Q\to\Gamma$ gives a two-dimensional projective representation of $G_\Q$ extending the given projective representation over $K$. Its complex conjugations are nontrivial projective involutions. Tate's lifting theorem gives a finite linear lift in characteristic three, necessarily odd and irreducible \cite[Section~2.1]{h:AKT}. Serre modularity over $\Q$ \cite{h:KWI,h:KWII} and cyclic base change give residual modularity over $K$. The base change is cuspidal: otherwise its Galois representation restricted to $K$, and hence its residual semisimplification, would be reducible. Finally, the given linear representation differs from this restricted lift by a finite character, which preserves modularity. This proves the residual theorem. \end{proof} \section[The residual base over \texorpdfstring{$\Q(\sqrt{13})$}{Q(sqrt(13))}]{\texorpdfstring{The characteristic-three base over $\Q(\sqrt{13})$}{The characteristic-three base over Q(sqrt(13))}} \label{c:13:section} In this section $K=\Q(\sqrt{13})$. Both completions above three are $\Q_3$, while two is inert. \begin{lemma}\label{c:13:lem:characters} The narrow class number of $K$ is one. Every finite character $G_K\to\overline\F_3^{\times}$ unramified outside $3$ has square one. \end{lemma} \begin{proof} The Minkowski bound $\sqrt{13}/2<2$ gives ordinary class number one. The unit $\varepsilon=(3+\sqrt{13})/2$ has norm $-1$; together with $-1$, it realizes all four real signatures. Thus the narrow class number is also one. The character assertion follows from Lemma~\ref{c:characters}. \end{proof} \begin{lemma}\label{c:13:lem:bounds} Under GRH, a totally imaginary number field of degree $n$ has root discriminant greater than $39.01$ if $n\ge531360$, greater than $37.63$ if $n\ge78624$, and greater than $28.02$ if $n\ge1440$. \end{lemma} \begin{proof} Use the explicit formula of Section~\ref{shared:explicit}, with no Euler terms or real-place term retained, using $B_b\le B_b^*$. The parameters give \[ \begin{array}{c|r|c} b&n&\rd(L)>\\\hline 45/4&531360&39.01373\\ 21/2&78624&37.63959\\ 8&1440&28.02230. \end{array} \] These are outward interval evaluations of the displayed finite expressions. Each bound increases with $n$, since $C_b>0$. The exact calculation is supplied in~\cite[\path{classical}]{code}. \end{proof} \begin{proposition}\label{c:13:prop:base} Assume GRH. Every continuous irreducible totally odd representation $\brho:G_K\to\GL_2(\overline\F_3)$ unramified outside $3$ is modular. \end{proposition} \begin{proof} Let $L/K$ be the linear kernel field, let $G=\Gal(L/K)$, and let $Z$ and $H=G/Z$ be its scalar subgroup and projective image. The field $L$ is totally imaginary. The scalar subgroup has order prime to three. Since three splits in $K$, the discriminant tower and \eqref{c:eq:MT} give \[ \rd(L)\le\sqrt{13}\,3^{13/6}<38.971. \] The solvable case is already modular. Otherwise Dickson's classification gives $H\simeq A_5$, $\PSL_2(\F_{3^f})$ or $\PGL_2(\F_{3^f})$, with $f\ge2$. If $3^f\ge81$, then $[L:\Q]\ge2|\PSL_2(\F_{81})|=531360$, contradicting Lemma~\ref{c:13:lem:bounds}. Suppose $3^f=27$. The wild inertia order at either prime above three is at most $27$, so \[ \rd(L)\le\sqrt{13}\,3^{13/6-1/27}<37.417. \] If $H=\PSL_2(\F_{27})$, then $|Z|\ge4$. Indeed, if $|Z|\le2$, determinant kills $Z$ and factors through the perfect group $H$, contradicting oddness; order three is impossible. If $H=\PGL_2(\F_{27})$, then $|Z|\ge2$: otherwise its perfect subgroup $\PSL_2(\F_{27})$ embeds in $\SL_2(\overline\F_3)$, which has only one nonidentity involution, namely $-I$. In either case $[L:\Q]\ge78624$, again contradicting Lemma~\ref{c:13:lem:bounds}. Suppose now that $H=\PGL_2(\F_9)$. Its nontrivial determinant-squareclass character is unramified outside three, so narrow class number one forces it to ramify at some $\mathfrak p_i$. At a place with nontrivial projective wild inertia, that inertia has a unique fixed point, defined over $\F_9$, and the local projective group lies in the corresponding Borel. Its torus quotient is abelian, so the tame Frobenius relation gives $a^3=a$ for every inertia torus coordinate. Thus $a=\pm1$, both squares in $\F_9^\times$. The squareclass character is unramified there. At the place where it ramifies, projective wild inertia, and hence linear wild inertia, is trivial. Consequently \[ \rd(L)<\sqrt{13}\,3^{(1+13/6)/2} =\sqrt{13}\,3^{19/12}<20.532. \] But $[L:\Q]\ge2|\PGL_2(\F_9)|=1440$, and \eqref{shared:eq:unconditional} with $b=16$ gives $\rd(L)>21.15$. Thus this image is excluded without GRH. The remaining nonsolvable groups are $A_5$ and $\PSL_2(\F_9)$. They embed in $\PGL_2(\F_9)$ with trivial determinant-squareclass character, since they are perfect. The projective representation is modular by \cite[Theorem~1.2(5)]{h:AKT}, and its given linear lift is modular by \cite[Lemma~2.1]{h:AKT}. \end{proof} \section[The residual base over \texorpdfstring{$\Q(\sqrt{37})$}{Q(sqrt(37))}]{\texorpdfstring{The characteristic-three base over $\Q(\sqrt{37})$}{The characteristic-three base over Q(sqrt(37))}} \label{c:37:section} Put $K=\Q(\sqrt{37})$. We use one wild type and one bounded quadratic or quartic type at the two three-primes. The compatible system thereby has a finite-flat characteristic-two member whose local fields fit inside a fixed solvable auxiliary extension. \begin{theorem}\label{c:37:thm:residual} Assume GRH. Every continuous irreducible representation $\brho:G_K\to\GL_2(\overline\F_3)$ with determinant $\om$, unramified outside three, is modular. \end{theorem} Write \[ w=\frac{1+\sqrt{37}}2,\qquad \epsilon=5+2w,\qquad \pi=3+w,\qquad \pi'=4-w. \] Thus $w^2-w-9=0$, $N_{K/\Q}(\epsilon)=-1$ and $\pi\pi'=3$. Put $\mathfrak p=(\pi)$ and $\mathfrak p'=(\pi')$. Both completions above three are $\Q_3$, and the unique dyadic completion is the unramified quadratic extension of $\Q_2$. \begin{lemma}\label{c:37:lem:arithmetic} The ordinary and narrow class numbers of $K$ are one. Every finite character $G_K\to\overline\F_3^{\times}$ unramified outside three has square one. \end{lemma} \begin{proof} The Minkowski bound is $\sqrt{37}/2<3.05$. There is no ideal of norm two, and the two ideals of norm three are principal. Thus the ordinary class number is one. The unit $\epsilon$ has norm $-1$, which also gives narrow class number one. The character assertion follows from Lemma~\ref{c:characters}. \end{proof} \begin{proposition}\label{c:37:prop:system} Let $\brho$ satisfy the hypotheses of Theorem~\ref{c:37:thm:residual}, and suppose its image is nonsolvable. It lifts to a totally odd, potentially modular, strongly compatible system $\cR=(r_\lambda)$ with determinant $\chi$ and weights $\{0,1\}$. The system is unramified away from three, has type $\tau_+$ at $\mathfrak p$, and has one of the two local conditions of Lemma~\ref{c:semistable} at $\mathfrak p'$. \end{proposition} \begin{proof} Proposition~\ref{c:wildlift} and Lemma~\ref{c:semistable} supply the local lifts at $\mathfrak p$ and $\mathfrak p'$, respectively. At the second place prescribe the tame supercuspidal type or a component through the special lift supplied by the lemma; no global quadratic character is required. In the latter case, the closure of the special component fixes Weil Frobenius eigenvalue ratio three, so the purity observation in Section~\ref{shared:pure-special} retains nonzero monodromy. Nonsolvability implies that the restriction to $G_{K(\zeta_3)}$ is irreducible, since an irreducible representation becoming reducible over a quadratic extension has dihedral projective image. Lemma~\ref{shared:global}(i) now globalizes these weight-two components with determinant $\chi_3$, no additional ramification, and compatibility at every finite place. In particular its characteristic-two member is crystalline of weights $\{0,1\}$ at two. \end{proof} Choose a place above two in the coefficient field and write $r_2$ for this member and $\br_2$ for its reduction. Lemma~\ref{c:type} makes $\br_2$ irreducible, with determinant one. Let $M$ be its kernel field. Its completion at $\mathfrak p$ is $E_+$, while Lemma~\ref{c:semistable} puts its completion at $\mathfrak p'$ inside $E'$. \subsection{Ramification at two}\label{c:37:sec:two} For a finite Galois extension $L/F$ of dyadic local fields, $d(L/F)$ denotes the valuation of its different, normalized by $v_F(2)=1$. Thus $d(L/F)$ is the local contribution to the exponent of two in the root discriminant. We use Serre's upper numbering for ramification groups. \begin{lemma}\label{c:37:lem:upperbreak} Let $F/\Q_2$ be unramified and let $J$ be a finite flat commutative group scheme over $\OO_F$ killed by two. The field generated by $J(\overline F)$ has no positive upper ramification break greater than one, and its normalized different is strictly less than two. \end{lemma} \begin{proof} Fontaine's Theorem~A and its corollary~\cite[pp.~515--516]{c:FontaineBT} apply to finite flat group schemes of any rank. His notation is $G^{(u)}=G^{u-1}$, where the right side uses Serre's upper numbering. For a group scheme killed by $p^n$ over a field of absolute ramification index $e$, his bound is \[ G^{(u)}\text{ acts trivially if }u>e\left(n+\frac1{p-1}\right). \] Here $e=n=1$ and $p=2$. Thus $G^v$ acts trivially for $v>1$. The different bound is the cited corollary. \end{proof} \begin{corollary}\label{c:37:cor:composite} A compositum of finite Galois extensions of $F$ having no upper break greater than one has the same property, and its normalized different is strictly less than two. \end{corollary} \begin{proof} Upper-numbering ramification is compatible with quotients. If all quotient fields have trivial groups above one, so does their compositum. The integral formula for the different then gives $d<2$; more precisely, for tame inertia order $t$ and wild inertia order $Q$, one has $d\le2-(t+1)/(tQ)$. \end{proof} We also need to recognize ordinary lifts in the dihedral case. \begin{lemma}\label{c:37:lem:ordinary} Let $F/\Q_2$ be unramified and let $r:G_F\to\GL_2(E)$ be Barsotti--Tate with determinant $\chi_2$. If its residual image is a two-group, then $r$ is ordinary. \end{lemma} \begin{proof} A two-group has only trivial irreducible representations in characteristic two, so both residual constituents are unramified. Apply Lemma~\ref{t:21:local:ordinary3} with $p=2$ and $e(F/\Q_2)=1$. \end{proof} \subsection{An explicit solvable auxiliary field}\label{c:37:sec:auxfields} Put \[ Q(X)=X^4-2X^3+(3-2w)X^2+(4+2w)X-w-2, \qquad C=K(\sqrt\pi,\sqrt{\pi'},\sqrt{-\epsilon}), \] and let $H$ be the splitting field of $Q$ over $K$. We use $N=CH$. \begin{proposition}\label{c:37:prop:auxdegree} The extension $N/K$ is solvable Galois of degree $96$, is unramified outside two and three, and has no dyadic upper break greater than one. Its completions at $\mathfrak p$ and $\mathfrak p'$ are $E_+$ and $E'$, respectively. \end{proposition} \begin{proof} The cubic resolvent and discriminant of $Q$ are \[ R(T)=T^3+(2w-3)T^2+4-4w,\qquad \operatorname{disc}(Q)=\operatorname{disc}(R)=144\delta, \quad \delta=99-28w=-\epsilon^{-3}\pi^3. \] The quartic is irreducible modulo the prime above eleven with $w\equiv7$: its reduction is $X^4-2X^3+7X+2$. This polynomial is relatively prime to $X^{121}-X$ and divides $X^{14641}-X$ over $\F_{11}$, so is irreducible. Hence $Q$ is irreducible over $K$. At $\mathfrak p$ we have $w\equiv18\pmod{27}$. The shifted polynomial is \[ Q(Y-1)=Y^4-6Y^3+(15-2w)Y^2+(6w-12)Y-5w. \] Its Newton polygon has segments of slopes $-1$ and $-1/3$, of lengths one and three. It therefore has one root in $\Q_3$ and an irreducible totally ramified cubic factor. The cubic roots have valuation $1/3$, whereas the linear root has valuation one. The resultant of these factors has valuation one. Since $v_{\mathfrak p}(\operatorname{disc}(Q))=5$, the cubic factor has discriminant exponent three. Its root is a uniformizer, so this is also the field discriminant exponent. Its discriminant square class is that of $-\epsilon\pi$, namely $3$. Thus its splitting field is totally ramified $S_3$, with normalized different $7/6$ and quadratic subfield $\Q_3(\sqrt3)$. By Lemma~\ref{c:type} it is $E_+$. The quartic's linear factor shows that its full local splitting field is the same field. The global quartic group is transitive and contains this point-stabilizer $S_3$, so it is $S_4$. Its unique quadratic subfield is $D=K(\sqrt{-\epsilon\pi})$. Valuations at $\mathfrak p,\mathfrak p'$ and the mixed real signs of $-\epsilon$ give $[C:K]=8$. The intersection $H\cap C$ is an elementary abelian two-extension and hence is $D$, the maximal such quotient field of $S_4$. Therefore $[N:K]=24\cdot8/2=96$. At $\mathfrak p'$ the reduction of $Q$ is $X^2(X-1)^2$. Hensel factorization gives two quadratic factors with unit resultant. Their discriminant valuations are each at least one and sum to $v_{\mathfrak p'}(\operatorname{disc}(Q))=2$, so each is one. Both define ramified quadratic extensions; the local splitting field lies in $E'$. The local square classes of $\pi,\pi',-\epsilon$ are respectively $3,1,1$ at $\mathfrak p$, and $1,3,-1$ at $\mathfrak p'$. Thus adjoining $C$ leaves $E_+$ unchanged and makes the other completion exactly $E'$. At two put $h=X^2+X+1-w$. Direct calculation gives \[ Q=h^2+2g,\qquad g=-2X^3+(2w+1)X-6. \] Over the residue field $\F_4$, the polynomial $\bar h$ is irreducible, since $\operatorname{Tr}_{\F_4/\F_2}(1-\bar w)=1$, and $\bar g=X$ is relatively prime to it. Dedekind's index criterion therefore shows that the quartic order is maximal at two. Its local field has ramification index two and residue degree two, and its relative discriminant exponent is four. These statements also follow by passing to the unramified quadratic extension defined by $h$ and using the Eisenstein factor of $Q$ near a root of $h$. The reduction of $R$ is $T^2(T+1)$, so it has a local linear factor and its local splitting group has order at most two. The local splitting group of $Q$ is consequently contained in a subgroup of order eight of $S_4$, and is a two-group. Thus its lower ramification groups satisfy $G_0=G_1$. In the permutation action on the four roots, inertia has two orbits, since the quartic field has residue degree two. The terms for $G_0$ and $G_1$ in the Artin conductor are therefore two each. They exhaust its discriminant exponent four, so faithfulness of the permutation action gives $G_2=1$. The largest upper break is at most one. Finally, the radicands defining $C$ are dyadic units. Every corresponding quadratic character has conductor at most two and upper break at most one. Corollary~\ref{c:37:cor:composite} gives the same bound for $N$. The displayed relative discriminant and the radicands also show that there is no ramification away from two and three. \end{proof} \subsection{The discriminant comparisons}\label{c:37:sec:discriminants} Let $L/K$ be finite Galois and unramified outside two and three. If its dyadic completion has no upper break greater than one, its dyadic different is less than two. If its completion at $\mathfrak p$ is $E_+$ and its completion at $\mathfrak p'$ is contained in $E'$, the discriminant tower gives \begin{equation}\label{c:37:eq:upper-nonzero} \rd(L)<4\sqrt{37}\,3^{5/6}<60.78017. \end{equation} Each three-adic exponent is multiplied by one half, since $[K:\Q]=2$. \begin{proposition}\label{c:37:prop:analytic} Assume GRH for the Dedekind zeta function of $L$, and put $n=[L:\Q]$. \begin{enumerate}[label=(\roman*)] \item If its completions above $\mathfrak p,\mathfrak p'$ are $E_+,E'$ and $n\ge96768$, then $\rd(L)>60.80125$. \item If $L$ is totally real, its completion above $\mathfrak p$ is $E_+$, and $n=120$, then $\rd(L)>66.10839$. \end{enumerate} \end{proposition} \begin{proof} Use the explicit formula of Section~\ref{shared:explicit} with its one-term bound $B_b\le B_b^*$ and the finite Euler sums. In (i), there are $n/12$ primes of norm three and $n/8$ of norm nine. Thus $c_3=1/6$, $c_9=1/4$, and we discard the real-place term. In (ii), retain only $c_3=1/6$, use $r_1/n=1$, and bound the real-place integral below by its first four alternating terms. The parameters and directed lower bounds are \[ \begin{array}{c|c|r|c|c} &b&n&\text{retained prime norms}&\rd(L)>\\\hline \mathrm{(i)}&56/5&96768&3,9&60.8012536506\\ \mathrm{(ii)}&5&120&3&66.1083983139. \end{array} \] The first expression increases with $n$, since $C_b>0$. \end{proof} \begin{proposition}\label{c:37:prop:largeimage} Assume GRH. The image of $\br_2$ constructed in Proposition~\ref{c:37:prop:system} is either solvable or conjugate to $\SL_2(\F_4)\simeq A_5$. \end{proposition} \begin{proof} Dickson's classification makes a nonsolvable finite subgroup of $\SL_2(\overline\F_2)$ conjugate to $\SL_2(\F_q)$, for $q=2^r$, $r\ge2$. Suppose $q\ge8$. This group is nonabelian simple, so its kernel field $M$ is linearly disjoint over $K$ from the solvable extension $N$. The compositum $L=MN$ satisfies \[ [L:\Q]=2\cdot96\cdot q(q^2-1)\ge96768. \] By Lemma~\ref{c:37:lem:upperbreak}, Proposition~\ref{c:37:prop:auxdegree} and Corollary~\ref{c:37:cor:composite}, its dyadic different is less than two. Its completions at three are exactly $E_+$ and $E'$. Equation~\eqref{c:37:eq:upper-nonzero} contradicts Proposition~\ref{c:37:prop:analytic}(i). \end{proof} \subsection{Modularity of the remaining images}\label{c:37:sec:smallimages} The sign of the type $\tau_+$ gives a direct treatment of the dihedral case. We make the required quadratic calculation explicit. \begin{lemma}\label{c:37:lem:quadratic} Let $D/K$ be quadratic, unramified at two and at every finite place away from three. If it ramifies at $\mathfrak p$, its completion there is $\Q_3(\sqrt{-3})$. \end{lemma} \begin{proof} Class number one shows that $D=K(\sqrt\alpha)$ with $\alpha$ a product of $-1,\epsilon,\pi,\pi'$ modulo squares. At the unramified dyadic place, a unit quadratic class is unramified precisely when it is a square modulo four. Calculation in $(\Z/4)[w]/(w^2-w-9)$ leaves exactly the four classes \[ 1,\qquad \epsilon\pi,\qquad -\pi\pi'=-3, \qquad -\epsilon\pi'. \] At $\mathfrak p$ we have $w=18\pmod{27}$, hence $\pi/3=1\pmod3$ and $\epsilon=-1\pmod3$. The two ramified classes in this list, $\epsilon\pi$ and $-3$, therefore both have local square class $-3$. \end{proof} \begin{proposition}\label{c:37:prop:modular2} Assume GRH. The characteristic-two member $r_2$ of Proposition~\ref{c:37:prop:system} is modular. \end{proposition} \begin{proof} Suppose first that the residual image is solvable. Since $\br_2$ is irreducible and has determinant one, its image is an odd dihedral group $D_{2m}$. Let $D/K$ be its quadratic inducing extension. Its local quadratic field at $\mathfrak p$ is $\Q_3(\sqrt3)$, by Lemma~\ref{c:type}. Lemma~\ref{c:37:lem:quadratic} shows that $D/K$ ramifies at the unique dyadic place. The dyadic wild inertia image is a nontrivial normal two-subgroup of the local decomposition image in $D_{2m}$. It is generated by a reflection. The normalizer of a reflection in an odd dihedral group is its own subgroup, so the entire local residual image is $C_2$. Lemma~\ref{c:37:lem:ordinary} makes $r_2$ ordinary. Allen's nearly ordinary residually dihedral theorem~\cite[Theorem~5.2.1 and Lemma~5.1.2]{g:Allen} now gives modularity. The extra condition when the inducing quadratic field is CM is satisfied because the dyadic place ramifies in $D$; in particular it does not split. Allen's dihedral argument supplies the ordinary modular witness required in that theorem. Suppose now that the residual image is nonsolvable. Proposition~\ref{c:37:prop:largeimage} makes it $A_5=\SL_2(\F_4)$. Its residual modularity follows from~\cite[Theorem~2.5]{h:AKT}. We check the real-place hypothesis in Thorne's characteristic-two lifting theorem. If the kernel field $M$ were totally real, then $[M:\Q]=120$, while its finite ramification gives \[ \rd(M)<4\sqrt{37}\,3^{5/6}<60.78017. \] This contradicts Proposition~\ref{c:37:prop:analytic}(ii). Hence $\br_2(c_v)\ne1$ for at least one real place $v$ of $K$. The representation $r_2$ is Barsotti--Tate at two, its residual image is nonsolvable and modular, and this real-place condition holds. Thorne's Theorem~6.1~\cite{g:Thorne2} applies directly. It permits both ordinary and nonordinary local representations and imposes no condition $K_2=\Q_2$. \end{proof} \begin{proof}[Proof of Theorem~\ref{c:37:thm:residual}] Solvable projective images are modular by automorphic induction and Langlands--Tunnell, with the characteristic-three lifting and twisting argument used in~\cite[proof of Theorem~2.5]{h:AKT}. We may therefore suppose that the image is nonsolvable. Proposition~\ref{c:37:prop:system} constructs a compatible system lifting $\brho$. Proposition~\ref{c:37:prop:modular2} makes its characteristic-two member modular. Strong compatibility, Chebotarev and strong multiplicity one then identify the system with the system of that Hilbert modular form. Its characteristic-three residual representation is $\brho$, which is consequently modular. \end{proof} \section{The split-three field applications}\label{c:mainchain} \label{c:chain} For the rest of this section let $K=\Q(\sqrt d)$ with $d\in\{7,13,37\}$, and denote its two places above three by $\mathfrak p_1,\mathfrak p_2$. \begin{theorem}\label{c:main} Assume GRH for Dedekind zeta functions. For $K=\Q(\sqrt d)$, $d\in\{7,13,37\}$, every continuous, absolutely irreducible, totally odd representation $G_K\to\GL_2(\overline\F_\ell)$, $\ell>2$, is modular. \end{theorem} \begin{proof}[Proof of Theorem~\ref{c:main}] For $\ell=3$, the supported irreducible residual assertion of Theorem~\ref{shared:terminal-criterion} is Theorem~\ref{c:7:thm:base}, Proposition~\ref{c:13:prop:base}, or Theorem~\ref{c:37:thm:residual}, according as $d=7,13,37$. Lemma~\ref{c:characters} makes both constituents of a reducible supported determinant-$\bar\chi_3$ residue quadratic, so their quotient is $\bar\chi_3$. It is totally odd and extends to $G_\Q$. Solvable residual images are already modular. Otherwise the initial weight-two lift and Lemma~\ref{shared:global} give a nondihedral potentially modular compatible system. Two does not split and three splits in all three fields, so Theorem~\ref{shared:terminal-criterion} proves its modularity. Reduce at the original coefficient place. GRH is used only in the supported residual assertions. \end{proof} \section[Ramified-dyadic field applications]{The ramified-dyadic field applications}\label{r:completion} \begin{proof}[Proof of Theorem~\ref{r:thm:ramified-fields}] For $d=2,3$, Proposition~\ref{r:t:d2:bounded-reduction} and Proposition~\ref{r:t:d3:supported13-reduction} give the supported irreducible residual assertions, and Lemma~\ref{r:t:lem:cc-ray} gives the constituent quotient. For $d=6,10,11$, the supported irreducible residual assertions required by Proposition~\ref{r:conductor-reduction} are Proposition~\ref{r:s:d6:prop:bounded}, Theorem~\ref{r:s:d10:thm:residual} and Proposition~\ref{r:s:d11:prop:bounded}. It remains to verify the constituent quotient in its second hypothesis. For $d=6$, write the constituents as $\psi$ and $\omega_5\psi^{-1}$. The unit $5+2\sqrt6$ has residues $2,3$ at the two five-places, so the two tame exponents of $\psi$ agree. The character $\psi^2$ is even and, after correction by $1$ or $\omega_5^2$, is everywhere unramified. Since the class number is one, $\psi^2$ is $1$ or $\omega_5^2$. The quotient is therefore $\omega_5$ or $\omega_5^{-1}$. For $d=10$, Lemma~\ref{r:s:d10:lem:q10-character-descent} gives the quotient $\omega_3$ or $\omega_3\nu$, where $\nu$ is the Hilbert class character. The lemma also gives its extension to $G_\Q$. These characters are totally odd. For $d=11$, Lemma~\ref{r:s:d11:lem:arithmetic} gives the finite ray group $C_4$ and the narrow ray group $C_4\times C_2$ at five. The even square $\psi^2$ factors through the finite ray group and has order at most two. Its unique nontrivial quadratic character is $\omega_5^2$, so the quotient is again $\omega_5^{\pm1}$. Both characters are totally odd and extend to $G_\Q$. All hypotheses of Proposition~\ref{r:conductor-reduction} hold. \end{proof} \section{The split-dyadic theorem}\label{s:sec:main} Put $\mathcal D=\{17,33,41,57,65,73,89\}$. All seven fundamental discriminants are equal to $d$, and two splits in $K=\Q(\sqrt d)$. The dyadic lifting and source theorems used below are recalled in Section~\ref{external:dyadic}. \begin{theorem}\label{s:thm:main} Assume GRH for Dedekind zeta functions. Let $K=\Q(\sqrt d)$, $d\in\mathcal D$. For every odd prime $p$, every continuous absolutely irreducible totally odd representation $\bar r:G_K\to\GL_2(\overline\F_p)$ is modular. There is no restriction on its conductor. \end{theorem} \begin{corollary}\label{s:cor:no-grh} Theorem~\ref{s:thm:main} holds without GRH for $d=17$. It also holds without GRH for $d=33$ if one assumes that every Dedekind zeta function has no zero in $\operatorname{Re}s>31/32$, away from its pole, as in \cite[Theorem~1.1]{s:CT}. \end{corollary} The general criterion is Theorem~\ref{struct:split-criterion}. The next two sections verify its arithmetic and residual hypotheses. The source theorems used for the terminal representation are already proved in the general part. \section{Arithmetic of the split-dyadic fields}\label{s:sec:arithmetic} \label{s:sec:field-arithmetic} Write $w_0,w_1$ for the dyadic places of $K$. We record the arithmetic needed in the scalar-residual field construction. \begin{lemma}\label{s:lem:kummer-basis} Let $S_2$ consist of the dyadic and real places of $K$. Every finite-order character $G_K\to\overline{\mathbf F}_2^\times$ unramified outside two is trivial. Moreover \[ H^1(G_{K,S_2},\mathbf F_2) =\langle[-1],[\epsilon],[2],[\alpha]\rangle_{\mathbf F_2} \] has dimension four, with the choices in Table~\ref{s:tab:units}. \end{lemma} \begin{table}[ht] \centering\small \begin{tabular}{@{}rllrrrr@{}} \toprule $d$&$\epsilon$&$\alpha$&$N\epsilon$&$N\alpha$&$h_K$&$h_K^+$\\ \midrule 17&$4+\sqrt{17}$&$(5+\sqrt{17})/2$&$-1$&2&1&1\\ 33&$23+4\sqrt{33}$&$(5+\sqrt{33})/2$&1&$-2$&1&2\\ 41&$32+5\sqrt{41}$&$(7+\sqrt{41})/2$&$-1$&2&1&1\\ 57&$151+20\sqrt{57}$&$(7+\sqrt{57})/2$&1&$-2$&1&2\\ 65&$8+\sqrt{65}$&$(9+\sqrt{65})/2$&$-1$&4&2&2\\ 73&$1068+125\sqrt{73}$&$(9+\sqrt{73})/2$&$-1$&2&1&1\\ 89&$500+53\sqrt{89}$&$(217+23\sqrt{89})/2$&$-1$&2&1&1\\ \bottomrule \end{tabular} \caption{Units, dyadic generators, and ordinary and narrow class numbers.} \label{s:tab:units} \end{table} \begin{proof} Minkowski's bound is $\sqrt d/2$. Except at $57$ and $73$, every prime ideal required by this bound is dyadic. At $41$, $65$ and $89$, three is inert, so its prime ideal has norm nine and lies outside the bound. The primes above three in the two remaining cases are principal, since \[ N(15+2\sqrt{57})=-3,\qquad N(17+2\sqrt{73})=-3. \] For $d\ne65$, the displayed $\alpha$ has norm $\pm2$ and generates a dyadic prime. At $65$, its norm is four and its trace is odd, so it generates the square of a dyadic prime. That prime is not principal: an element of norm $\pm2$ would yield $x^2-65y^2=\pm8$, impossible modulo five. Thus $h_{65}=2$ and the class group is generated by a dyadic prime. This proves the ordinary class numbers and shows that the $S_2$-ideal class group is trivial in every case. The unit squareclass group has dimension two. If $N\epsilon=-1$, the mixed signs of $\epsilon$ show that its class and $[-1]$ are independent. For $d=33,57$ the unit is totally positive and is not a square: a square root would have integral trace whose square is, respectively, $44$ or $48$, or $300$ or $304$. Thus $[-1]$ and $[\epsilon]$ are again independent and form a basis. No fundamental-unit calculation is needed. The norm-minus-one units give the narrow class numbers in the other five cases; for $33,57$, norm minus one is impossible modulo three, giving the remaining entries in the table. Kummer theory identifies the group in the lemma with $\mathcal O_{K,S_2}^\times/(\mathcal O_{K,S_2}^\times)^2$. For $d\ne65$, the valuation vectors of $2$ and $\alpha$ form a basis of the dyadic divisor lattice. For $65$, that lattice consists of pairs whose sum is even, and its basis is supplied by the vectors $(1,1)$ and $(2,0)$ of $2$ and $\alpha$. Together with $-1$ and the displayed unit squareclass, this proves the asserted basis. Finally, a characteristic-two finite character has odd-order image. At either dyadic completion $\mathbf Q_2$, it kills the pro-two unit group and is unramified. It also kills real sign groups. Thus it factors through the ordinary class group, whose order is a power of two. The character is trivial. \end{proof} For $i=0,1$ write $V_i(K)=V^+_{w_i}(K)$ for the space in \eqref{struct:selmer-space}. We verify its vanishing using the full quadratic Selmer group, including the ideal-class contribution. \begin{lemma}[Base scalar Selmer vanishing]\label{s:lemma:baseSelmer} For every $d\in\mathcal D$ and either dyadic choice, $V_i(K)=0$. \end{lemma} \begin{proof} Let $H^+$ be the subspace of $H^1(G_{K,S_2},\F_2)$ trivial at both real places. The displayed $\epsilon$ and $\alpha$ are positive in the embedding $\sqrt d>0$. For $d=17,41,65,73,89$, their norms have signs $-1,+1$; for $d=33,57$, the signs are $+1,-1$. The two real conditions therefore give \[ H^+=\langle[2],[\beta]\rangle,\qquad \beta=\begin{cases} \alpha,&d=17,41,65,73,89,\\ \epsilon,&d=33,57. \end{cases} \] Index the dyadic places by $w=(1+\sqrt d)/2\equiv0,1\pmod2$. Their local square classes are as follows: \begin{center} \begin{tabular}{@{}rcc@{}} \toprule $d$&$[\beta]_{w_0}$&$[\beta]_{w_1}$\\\midrule 17&$-2$&$-1$\\ 33&$-5$&$-5$\\ 41&$5$&$10$\\ 57&$-5$&$-5$\\ 65&$5$&$5$\\ 73&$-10$&$-5$\\ 89&$-5$&$-10$\\\bottomrule \end{tabular} \end{center} These entries follow by Hensel lifting $w^2-w+(1-d)/4$ and reading the valuation modulo two and the odd unit modulo eight. Each entry is outside the subgroup $\{1,[2]\}\subset\Q_2^\times/(\Q_2^\times)^2$. Thus restriction of $H^+$ to either dyadic completion is injective. Its kernel at $w_{1-i}$ is precisely $V_i(K)$. \end{proof} For $d=17,41,73,89$, Table~\ref{s:tab:units} gives narrow class number one. The totally positive element $\alpha$ of norm two has trace $5,7,9,217$, respectively. At the other dyadic place its unit root satisfies $u+2u^{-1}=\operatorname{Tr}(\alpha)$ and is congruent to $7,5,3,3$ modulo eight. Thus these four fields satisfy condition~\textup{(S)} of \cite[Lemma~9.1]{s:FM}. \section{Residual representations unramified outside two}\label{s:sec:residual} For a finite Galois extension $E/\Q_2$, let $d(E)$ be its normalized different exponent, with $v_2(2)=1$. Put $E_2=\Q_2(\sqrt{-1},\sqrt2,\sqrt5)$. \begin{lemma}\label{s:lem:two-local} Let $r:G_{\Q_2}\to\SL_2(\F_q)$, $q=2^a$, have finite image and kernel field $E$. Then $d(E)\le2$. More precisely, either $E\subset E_2$, or wild inertia is nontrivial of order $Q\le q$, all positive upper breaks are at most one and the local abelianization has odd order, or $E$ is unramified or tame with ramification index three. In the wild case outside $E_2$, $d(E)\le2(1-1/Q)$ and $E\cap\Q_2(\sqrt{-1},\sqrt5)=\Q_2$. In the tame ramified case the image is $S_3$ and its quadratic subfield is unramified. \end{lemma} \begin{proof} If wild inertia is trivial, its tame generator has eigencharacters $\theta,\theta^{-1}$, and Frobenius squares them. The normalizer acts by identity or inversion, so the nontrivial tame order is three; Frobenius is the Weyl involution and the image is $S_3$. Its different exponent is $2/3$. Otherwise the wild subgroup fixes a unique line, stable under the whole decomposition group. The diagonal characters are $\theta,\theta^{-1}$ and are unramified, since $\Z_2^\times$ is pro-two. If $\theta=1$, the image is elementary abelian unipotent and $E\subset E_2$. The seven quadratic characters of $E_2$ have conductors $0,2,2,3,3,3,3$, giving $d(E_2)=2$. If $\theta\ne1$, Frobenius acts on the translation subgroup $P$ by the nontrivial scalar $\theta(\mathrm{Frob}_2)^2$. The operator obtained by subtracting one is invertible, so $[P,\mathrm{Frob}_2]=P$, $P$ is wild inertia, and $E^P/\Q_2$ is unramified of odd degree. Its Kummer module has no Frobenius invariants. The equivariant valuation map to $\Z/2\Z$ is zero, so all its squareclasses are units and all its quadratic characters have conductor at most two. The different bound follows from conductor--discriminant. The commutator identity gives odd local abelianization and hence disjointness from the stated biquadratic field. \end{proof} \begin{lemma}\label{s:lem:compositum} Put $H=K(\sqrt{-1},\sqrt\epsilon)$. It is biquadratic over $K$, Galois over $\Q$ and unramified at odd places, and its dyadic completions lie in $\Q_2(\sqrt{-1},\sqrt5)$. If $L/K$ has group $\SL_2(\F_q)$, $q\ge8$, and is unramified outside two, put $M=LH$. Let $k$ count the dyadic places whose local kernel fields lie in $E_2$. Then \begin{equation}\label{s:eq:upper} [M:\Q]=8q(q^2-1),\qquad \rd(M)\le\sqrt d\,2^{2-1/q+k/(2q)}. \end{equation} Each of these $k$ places supplies in the explicit formula the contribution of at least $[M:\Q]/16$ primes of norm four. \end{lemma} \begin{proof} The classes $[-1],[\epsilon]$ are independent by their real signs and nonsquareness. Conjugation sends $\sqrt\epsilon$ to $1/\sqrt\epsilon$ or $\sqrt{-1}/\sqrt\epsilon$, so $H/\Q$ is Galois. Its defining polynomials have discriminants supported at two, and every unit squareclass over $\Q_2$ is generated by $[-1],[5]$. Simplicity gives $L\cap H=K$. At a dyadic place with $E\subset E_2$, the compositum still lies in $E_2$ and has different exponent at most two, ramification index at most four and residue degree at most two. Otherwise, in the wild case of Lemma~\ref{s:lem:two-local}, adjoining the ramified quadratic part of $H$ enlarges its elementary wild group from $P$ to at most $P\times C_2$. All characters still have conductor at most two. Its different exponent is at most $2(1-1/(2Q))\le2-1/q$. In the tame case the compositum has ramification index at most six, with only a possible quadratic positive interval of length one. Its different exponent is at most $(1-1/6)+(1-1/2)=4/3<2-1/q$. Averaging the two dyadic exponents proves \eqref{s:eq:upper}. At a counted place, residue degree two gives at least $[M:\Q]/16$ norm-four primes. Residue degree one gives at least $[M:\Q]/8$ norm-two primes; keeping their even Euler powers supplies the same lower contribution. \end{proof} Use the functions $F_b,J_b,B_b,C_b$ of Section~\ref{shared:explicit}. Put $a=\pi/(2b)$ and let \begin{align*} \log\mathcal L(n,b,k) &=\gamma+\log(8\pi) -\sum_{j=0}^3\left\{\frac1{j+1/2}-J_b(j+1/2)\right\}\\ &\quad-\frac{2a^2}{49}-\frac{C_b}{n} +\frac{k\log4}{8}\sum_{m\ge1}4^{-m/2}F_b(m\log4). \end{align*} The tail estimate in that theorem makes this a lower bound for the root discriminant, under GRH, when the indicated norm-four Euler contributions are present. It increases with $n$ at fixed $b,k$. We shall also write $B(G),C(G)$ for the corresponding integrals with an even test function $G$. \begin{theorem}\label{s:thm:residual} Assume GRH. For $d\in\mathcal D$, every absolutely irreducible characteristic-two representation of $G_K$ unramified outside two is modular. \end{theorem} \begin{proof} Twist by the inverse square root of the odd-order determinant to make the image determinant one. Linear and projective kernels then coincide. For a solvable irreducible image use \cite[Proposition~10.3]{s:FM}, since $K/\Q$ is abelian and two splits completely. The image $\SL_2(\F_4)$ is covered by \cite[Theorem~1.2(1)]{h:AKT} and its projective-to-linear modularity statement. Otherwise Dickson's classification gives $\SL_2(\F_q)$, $q\ge8$. For $M$ in Lemma~\ref{s:lem:compositum}, the following strict comparisons hold with $n=8q(q^2-1)$ and $d\le89$: \begin{center}\small \begin{tabular}{@{}cccrr@{}}\toprule $q$&$k$&$b$&$\mathcal L(n,b,k)>$&upper bound $<$\\\midrule 8&1&7.8&38.07399&36.13605\\ 16&0&9.8&36.64088&36.13605\\ $q\ge32$&0&12&38.91320&37.73593\\\bottomrule \end{tabular} \end{center} The last row drops Euler terms and uses $n\ge261888$ and $\rd(M)\le4\sqrt{89}$. In the first two rows, raising $k$ by one increases the logarithmic lower bound by $\frac{\log4}{8}\sum_m4^{-m/2}F_b(m\log4)$, exceeding $\log2/(2q)$ already in the first summand since $F_b(\log4)>1/2$. Thus the table excludes every case except $q=8,k=0$. The exact calculation is supplied in~\cite[\path{split_dyadic}]{code}. Let $\sigma$ be the nontrivial automorphism of $K/\Q$. If $L^\sigma\ne L$, simplicity and Goursat's lemma give $[LL^\sigma:\Q]=2\cdot504^2=508032$. Each local constituent has upper breaks at most one, since $k=0$. The same is true of the compositum by compatibility of upper numbering with quotients, so $\rd(LL^\sigma)\le4\sqrt{89}$. The last table row, by degree monotonicity, contradicts this. Hence $L/\Q$ is Galois. Since $\operatorname{Out}(\SL_2(\F_8))=C_3$, its quadratic conjugation action is inner. The group is centerless, so correcting a lift of $\sigma$ gives a centralizing involution and $\Gal(L/\Q)=\SL_2(\F_8)\times C_2$. Projection extends the residual representation to $G_\Q$. Serre's theorem \cite{h:KWI,h:KWII,h:Kisin2} makes that extension modular: in characteristic two oddness is the determinant condition $-1=1$, and no nontrivial complex-conjugation action is required. Cyclic base change to $K$ is cuspidal, since the restriction is absolutely irreducible. Undo the determinant twist. \end{proof} \subsection{Comparisons without GRH} For an even test function $G$ the unconditional explicit formula has, in addition to \eqref{shared:eq:grh}, the zero terms associated to $\Phi_G(s)=\int_{\mathbb R}G(x)e^{(s-1/2)x}\,dx$; see \cite[equation~(2.3)]{s:Odlyzko}. If the zeros lie in $|\operatorname{Re}s-1/2|\le a$, these terms may be discarded for $G=F_b/\cosh(ax)$. Indeed, for $|h|55.42$, whereas $\rd(L)\le4\sqrt{89}<37.736$. This does not treat the totally imaginary kernels. For $d=17$, the kernel field has degree at least $1008$ and $\rd(L)\le4\sqrt{17}<16.50$. Formula~\eqref{shared:eq:unconditional} with $b=10$, dropping prime and real terms, gives $\rd(L)>20.57$. This proves the unconditional residual theorem for $17$. For $d=33$, put $b=24$. If $k\ge1$, retain just the first norm-four term from Lemma~\ref{s:lem:compositum}. At degree at least $4032$, \eqref{shared:eq:unconditional} gives $\rd(M)>23.31$, whereas \eqref{s:eq:upper} gives $\rd(M)\le4\sqrt{33}<22.979$. If $q=8$ and $k=0$, the same formula without prime terms gives $\rd(M)>21.75>21.07115$, contradicting \eqref{s:eq:upper}. For $k=0$ use the original kernel field $L$: Lemma~\ref{s:lem:two-local} gives $[L:\Q]=2q(q^2-1)$ and $\rd(L)\le\sqrt{33}\,2^{2-2/q}$. The resulting unconditional comparisons are \[ \begin{array}{crrr} q&[L:\Q]&\text{lower }\rd(L)&\text{upper }\rd(L)\\\hline 16&8160&>21.97&<21.07115\\ 32&65472&>22.15&<22.00405 \end{array} \] Only $k=0$, $q\ge64$ remains without the additional zero-free input. Assume now the additional zero-free statement in Corollary~\ref{s:cor:no-grh}. The functional equation puts the zeros of every relevant zeta function in the strip with $a=15/32$. For $G=F_b/\cosh(15x/32)$ one has \[ B(G)<\frac{13}{20}+\frac{3\pi^2}{4b^2}, \qquad C(G)\le4J_b(-1/32). \] Here the first bound follows from $B(G)\le B_\infty(a)+\pi^2Q(a)/(8b^2)$, where \[ B_\infty(a)=\int_0^\infty \frac{1-\operatorname{sech}(ax)}{2\sinh(x/2)}\,dx<13/20, \qquad Q(a)=\int_0^\infty \frac{x^2}{2\sinh(x/2)\cosh(ax)}\,dx<6. \] For completeness these constants have finite rational bounds: expand $B_\infty(a)=\sum_{j\ge0}(-1)^jS_j$, with $S_j=\sum_{m\ge0}2a^2/[s(s+a)(s+2a)]$ and $s=m+1/2+2ja$. The decreasing alternating sums through $j=80,81$, truncating each inner sum at $500$ with error at most $a^2/(499.5+2ja)^2$, give $0.6491123.06000>4\sqrt{33}. \] This proves the residual assertion needed for the second part of Corollary~\ref{s:cor:no-grh}. All numerical inequalities in this section follow from the displayed finite expressions and tail bounds by directed interval arithmetic. The exact calculation is supplied in~\cite[\path{split_dyadic}]{code}. \begin{proof}[Proof of Theorem~\ref{s:thm:main} and Corollary~\ref{s:cor:no-grh}] Lemma~\ref{s:lem:kummer-basis} gives the two-power ordinary class number, and Lemma~\ref{s:lemma:baseSelmer} verifies \eqref{struct:SV}. Theorem~\ref{s:thm:residual} supplies the irreducible residual base. Apply Theorem~\ref{struct:split-criterion}. For $d=17$ use the unconditional residual comparison above; for $d=33$ use its stated zero-free variant. The scalar and conductor arguments introduce no additional analytic hypothesis. \end{proof} \section{Proof of the main theorem} \begin{proof}[Proof of Theorem~\ref{main:grh}] Theorem~\ref{c:main} proves the cases $d=7,13,37$, and Theorem~\ref{r:thm:ramified-fields} proves $d=2,3,6,10,11$. The source arguments and conductor reductions in Sections~\ref{t:5:section} and~\ref{t:21:section} give $d=5,21$. The remaining seven fields are covered by Theorem~\ref{s:thm:main}. Every congruence returns modularity to the original residual representation, without a restriction on its conductor. \end{proof} \clearpage \appendix \section{Depth-four normal ray calculation}\label{r:s:app:normal-ray} We give the matrices used in Proposition~\ref{r:s:d10:prop:normal-ray}. The fields $K,A$, the standard module $W_0$ and the binary ray coordinates are those in its proof. For the depth-four calculation write $S=\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)$ and $U=\left(\begin{smallmatrix}0&1\\1&1\end{smallmatrix}\right)$ on $W_0$. In the seven binary ray coordinates the class killed by $[\mathfrak q_+]$ is $v=(1,0,0,1,1,0,1)^t$. The source matrices for the actions of $S,U$ in those coordinates are \[ C_2=\begin{pmatrix} 1&0&0&0&0&0&0\\0&1&0&0&0&0&0\\0&1&0&0&0&1&0\\ 0&1&1&1&0&1&0\\0&0&0&0&1&0&0\\0&1&1&0&0&0&0\\ 0&0&1&1&1&0&1 \end{pmatrix},\qquad C_3=\begin{pmatrix} 1&0&0&0&0&0&0\\0&1&0&0&0&0&0\\1&0&1&0&0&1&1\\ 1&1&1&0&0&1&0\\0&0&0&0&1&0&0\\0&0&0&1&0&0&1\\ 0&0&1&1&1&0&1 \end{pmatrix}. \] The solution space to $Xv=0$, $XC_2=SX$, $XC_3=UX$ has basis \[ X_1=\begin{pmatrix}1&1&1&1&0&1&0\\1&0&0&1&0&0&0\end{pmatrix}, \quad X_2=\begin{pmatrix}1&0&1&1&1&1&1\\1&1&1&0&0&0&1\end{pmatrix}. \] Chinese remainder lifts of local units, chosen to be one at the other dyadic prime, have binary ray logarithm columns \[ g=(0,1,1,0,0,0,0)^t,\quad h=(0,0,0,1,0,0,0)^t,\quad j=(0,0,0,1,0,1,1)^t . \] The first prime supplies $g,h$ and the second supplies $g,j$. For $\alpha^2+\alpha+1=0$, evaluation of the two equivariant maps followed by the left $U$-eigenrow $(1,\alpha)$ gives, respectively, \[ \begin{pmatrix}0&1+\alpha\\1&1\end{pmatrix},\qquad \begin{pmatrix}0&\alpha\\1&1+\alpha\end{pmatrix}. \] Both determinants are nonzero, and the conjugate eigenrow has the same property. Hence vanishing on local units in either one of the two $C_3$ eigencharacters, at either prime, kills the global depth-four class. These conclusions concern the characteristic-two normal error; the order of tame inertia after reduction in the joint trace order is checked separately. The matrices and local logarithms are reproduced in~\cite[\path{ramified/d6_d10_d11/d10}]{code}. \section[Auxiliary fields over \texorpdfstring{$\Q(\sqrt{21})$}{Q(sqrt(21))}]{Auxiliary fields over the field of discriminant twenty-one}\label{t:21c:app:cert} The unit assertions involve only a number field of degree twelve. One defining polynomial for $A$ is \begin{align*} P(X)={}&X^{12}-3X^{11}+8X^{10}-9X^9+9X^8+6X^7+X^6\\ &+6X^5+9X^4-9X^3+8X^2-3X+1. \end{align*} The field defined by $P$ has discriminant $21^6 5^8$ and signature $(0,6)$. The certificate reconstructs the embeddings of $u$ and $z$, checks their equations and the norm-one relations, and tests the sixteen products at the two dyadic primes. Integrality and unit status follow directly from their monic minimal polynomials with constant coefficient $\pm1$. It uses only \texttt{nfinit}, exact field arithmetic, Galois conjugates and local-power tests; no computation of a full unit group or class group is used in the proof. The exact calculation is supplied in~\cite[\path{d21}]{code}. The arithmetic and interval certificates are: \begin{itemize} \item \path{check_small_unit_formulas.gp}: the two-orbit unit identities and local squareclass kernels in Lemma~\ref{t:21c:lem:units}; \item \path{verify_split5_local.py}: the two compact-type Brauer-character identities, calculated in $\Z[\zeta_{24}]$; \item \path{certify_K21_SC_SC.py}: real-ball evaluations of \eqref{shared:eq:grh}, with each comparison asserted only when its lower and upper balls are disjoint; \item \path{K21_SC_SC_arb_certificate.json}: the full-precision interval enclosures for Table~\ref{t:21c:tab:bounds}. \end{itemize} The smallest gap, in the distinct-conjugate row, is greater than $0.00974417509$. The analytic calculation certifies the displayed inequalities; the group and local hypotheses are proved in the preceding sections. The exact calculation is supplied in~\cite[\path{d21/inherited_python}, \path{d21/split5}]{code}. \section[Normal Selmer calculations over \texorpdfstring{$\Q(\sqrt{21})$}{Q(sqrt(21))}]{Normal Selmer calculations over the field of discriminant twenty-one}\label{t:21s:app:certificates} The following scripts in~\cite[\path{d21}]{code} print certificates for the normal Selmer calculations. All class and unit calculations have absolute degree at most twelve. The accompanying code includes output records, the certificate \path{s3_mixed_full_sunit_portable.gp}, and the diagnostic-checking runner \path{run_gp_checked.py}. The certificate uses relative discriminants and Hilbert symbols to test the stated invariant subspace conditions. \begin{center}\small \begin{tabular}{p{0.53\textwidth}p{0.39\textwidth}} \toprule file & certified assertion\\ \midrule \path{k21_cm_split_s3_classification.gp} & $H$ class number one; ray groups and four cubic quotients\\ \path{k21_cm_split_s3_galois.gp} & quadratic involution and residual fields\\ \path{s3_mixed_full_sunit_portable.gp} & full mixed normal space of dimension six; two necessary cuts intersect in zero\\ \path{anti_C3_conductors_A1.gp} & $A_1$ has class group $C_3$; all nonzero normal classes have dyadic conductor six somewhere\\ \bottomrule \end{tabular} \end{center} The displayed masks in Section~\ref{t:21s:sec:selmer} refer to the chosen unit and $S$-unit bases. A different PARI version or valid basis may change these coordinates. Invariantly, the nontrivial $C_3$-part of the $S_5$-unit squareclass space has dimension six over $\F_2$; its two cuts have dimensions two and four, and their intersection is zero. For $A_1$, only the zero normal class meets both dyadic conductor bounds. Reproduction should record the actual basis and masks and check these invariant assertions. Quadratic discriminant exponents from relative discriminants are the local quadratic conductors. Successful output records include \texttt{bnfcertify=1}, the full normal rank and every retained mask; a zero process exit status alone does not certify success in PARI/GP. The exact calculation is supplied in~\cite[\path{d21}]{code}. The nonsolvable branch uses the discriminant comparisons above and the \path{certify_K21_SC_SC.py} and \path{K21_SC_SC_arb_certificate.json}. It is used only for the nonsolvable branch of the mixed theorem; Theorem~\ref{t:21s:thm:A1} has no analytic hypothesis. The exact calculation is supplied in~\cite[\path{d21/split5}]{code}. \Needspace{12\baselineskip} \section*{AI disclosure} During the preparation of this article, the authors used ChatGPT 6-Astra and Sol 6.1 as an assistant in mathematical research and manuscript preparation. Its use included exploring proof strategies, checking mathematical arguments, investigating possible simplifications, assisting with computations and verification scripts, checking references, and drafting and revising portions of the exposition. The authors take full responsibility for the mathematical content, proofs, computations, and conclusions of the article. \clearpage \begin{thebibliography}{99} \small\raggedright\setlength{\itemsep}{3pt} \bibitem{c:AL} M.~Adrian and B.~Liu, \href{https://doi.org/10.1016/j.jnt.2015.08.002}{\emph{Some results on simple supercuspidal representations of $\GL_n(F)$}}, J. Number Theory \textbf{160} (2016), 117--147. 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