% main.tex -- Forced 3-adic anomalies of S3 cubic fields. Draft 2026-10-24. % Claim ledger convention (source comments only): every mathematical statement carries one of % [Proved] complete proof given here, from the standard results cited % [Numerical] computation; GRH-conditional unit groups unless said otherwise % [Heuristic] model or expectation, not a theorem % [Lit] quoted from the literature % Labels are removed from the prose; the honesty is kept in the wording ("we prove", "we observe", "a natural model"). % SINGLE-FILE VERSION generated by make_flat.py from main_src.tex: edit main_src.tex, not this file. \documentclass[11pt]{amsart} \usepackage[margin=1.15in]{geometry} \usepackage{amsmath,amssymb,amsthm,mathtools} \usepackage{booktabs} \usepackage{tikz-cd} \usepackage{pgfplots} \pgfplotsset{compat=1.17} \usepackage[hidelinks]{hyperref} \newtheorem{theorem}{Theorem} \renewcommand{\thetheorem}{\Alph{theorem}} \newtheorem{thm}{Theorem}[section] \newtheorem{prop}[thm]{Proposition} \newtheorem{lemma}[thm]{Lemma} \newtheorem{cor}[thm]{Corollary} \theoremstyle{definition} \newtheorem{definition}[thm]{Definition} \newtheorem{remark}[thm]{Remark} \newtheorem{heur}[thm]{Conjecture} \newcommand{\Q}{\mathbb{Q}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\F}{\mathbb{F}} \newcommand{\cG}{\mathcal{G}} \newcommand{\tG}{\widetilde{\Gamma}} \newcommand{\Cl}{\mathrm{Cl}} \newcommand{\rk}{\operatorname{rk}} \newcommand{\Gal}{\operatorname{Gal}} \newcommand{\Hom}{\operatorname{Hom}} \newcommand{\Tr}{\operatorname{Tr}} \newcommand{\cyc}{\mathrm{cyc}} \newcommand{\lb}{\bar\lambda} \DeclareFontFamily{U}{wncy}{} \DeclareFontShape{U}{wncy}{m}{n}{<->wncyr10}{} \DeclareSymbolFont{cyr}{U}{wncy}{m}{n} \DeclareMathSymbol{\Sha}{\mathord}{cyr}{`X} \newcommand{\eps}{\varepsilon} \newcommand{\sgn}{\mathrm{sgn}} \title{Forced $3$-adic anomalies of $S_3$ cubic fields} \author{Shawn Krisman} \email{krisman.shawn@gmail.com} \date{\today} \begin{document} \begin{abstract} Let $k$ be a totally real cubic field and $p$ a prime unramified in $k$. The $p$-adic regulator of $k$ is a $2\times2$ determinant of $p$-adic logarithms of units and is always divisible by $p^2$; call $k$ \emph{anomalous at $p$} if it is divisible by $p^3$. Among all $592\,421$ totally real cubic fields of discriminant up to $10^7$, the proportion of anomalous fields matches a random-matrix prediction at every prime $5\le p\le31$, but at $p=3$ it is $0.381$ instead of the predicted $0.407$. We explain this. For cubic fields of fundamental discriminant $D$ prime to $3$, we show that the $3$-class group and the ray class group modulo $9$ of the quadratic field $\Q(\sqrt D)$ can force every cubic field of discriminant $D$ to be anomalous at $3$. When the $3$-class group of $\Q(\sqrt D)$ is cyclic, we decide exactly when $k$ is anomalous, in terms of one invariant for which we give an explicit formula by cubic residue and Hilbert symbols. The phenomenon comes from the failure of semisimplicity of $\F_3[S_3]$ and has no analogue for $p\ge5$. The results fix one or two $3$-adic digits of the regulator and say nothing about Leopoldt's conjecture itself. \end{abstract} \maketitle % ====================================================================================================================== \section{Introduction} % ====================================================================================================================== \subsection{Leopoldt's conjecture for cubic fields as a $2\times2$ determinant} Let $k$ be a totally real cubic field with fundamental units $u_1,u_2$ and let $p$ be a prime unramified in $k$. Fix the three embeddings $\sigma_1,\sigma_2,\sigma_3\colon k\to\overline{\Q}_p$. The $p$-adic regulator \[ R_p(k)=\det\bigl(\log_p\sigma_i(u_j)\bigr)_{1\le i,j\le 2} \] is well defined up to sign, and Leopoldt's conjecture \cite{Leopoldt} for $k$ at $p$ is the statement $R_p(k)\neq0$. Leopoldt's conjecture is known for abelian fields by Brumer's $p$-adic Baker theorem \cite{Brumer}; for non-Galois cubic fields it is open. % [Lit] FJ intro; JN appendix Hofmann, Johnston and Nickel \cite[Appendix, Thm.~A.2]{JN} verified it at $p=3$ for every totally real Galois $S_3$ extension $E/\Q$ with $\operatorname{Disc}(O_E)<10^{20}$ (for an $S_3$-extension, Leopoldt's conjecture for $E$ and for its cubic subfield are equivalent \cite{FJ}), and Ferri and Johnston \cite{FJ} construct infinite families of totally real $S_3$-extensions satisfying it at any prescribed finite set of primes. Since $\log_p$ of a unit is divisible by $p$ in the unramified completions, $v_p(R_p(k))\ge2$. We call $\operatorname{depth}_p(k):=v_p(R_p(k))-2$ the \emph{depth} and say $k$ is \emph{anomalous at $p$} if the depth is positive. % [Lit] GrasKL, GrasPrat In Gras's language $R_p(k)/p^2$ is the normalised regulator, and (when $p\nmid h_k$) $k$ is anomalous at $p$ exactly when it is not $p$-rational \cite{GrasKL,GrasPrat}; see also \cite{MNQD,JNQD}. If the $2\times 2$ matrix of reduced logarithms were a Haar-random matrix over $\Z_p$, the anomaly probability would be \[ q(p)=1-(1-p^{-1})(1-p^{-2})=\tfrac1p+\tfrac1{p^2}-\tfrac1{p^3}. \] For cyclic cubic fields the Galois symmetry changes the model, and Hofmann and Zhang \cite{HZ} have studied the valuations of $R_p$ in that case; non-Galois cubic fields are not treated there. \subsection{The survey} % [Numerical] fe14_survey.py, runs/fe14_survey_D10000000.json; GRH-conditional unit groups (bnfinit without certification) We computed $R_p(k)$ modulo a high power of $p$ for every totally real cubic field of discriminant $d_k\le10^7$ (there are $592\,421$ of them) and every prime $3\le p\le 31$ unramified in $k$. The unit groups were computed with \textsc{PARI/GP} \cite{PARI} assuming the generalised Riemann hypothesis; all numerical statements in this paper inherit that caveat, and none of the theorems depend on it. Table~\ref{tab:survey} shows the result. \begin{table}[ht] \centering \begin{tabular}{rrccr} \toprule $p$ & fields & anomaly rate & $q(p)$ & $z$ \\ \midrule 3 & 330903 & 0.3814 & 0.4074 & -30.4 \\ 5 & 426455 & 0.2309 & 0.2320 & -1.6 \\ 7 & 477175 & 0.1597 & 0.1603 & -1.1 \\ 11 & 525213 & 0.0988 & 0.0984 & +0.9 \\ 13 & 538483 & 0.0820 & 0.0824 & -0.9 \\ 17 & 556937 & 0.0630 & 0.0621 & +2.9 \\ 19 & 561968 & 0.0551 & 0.0553 & -0.5 \\ 23 & 567513 & 0.0449 & 0.0453 & -1.2 \\ 29 & 572657 & 0.0355 & 0.0356 & -0.6 \\ 31 & 574070 & 0.0330 & 0.0333 & -1.0 \\ \bottomrule \end{tabular} \caption{Anomaly rates over all totally real cubic fields with $d_k\le10^7$ and $p$ unramified ($z$ = deviation from $q(p)$ in standard errors).} \label{tab:survey} \end{table} At every $p\ge5$ the Haar value fits within the expected fluctuation. At $p=3$ the rate is $0.3814$ against $11/27=0.4074$, a deviation of $30$ standard errors. The deficit is not a small-discriminant effect: the rate stabilises near $0.385$ (Table~\ref{tab:p3bins}). The purpose of this paper is to explain it. \subsection{Results} Let $D>0$ be a fundamental discriminant with $3\nmid D$, let $F=\Q(\sqrt D)$ and let $k$ be a cubic field with $d_k=D$. Then the Galois closure $K$ of $k$ is a cyclic cubic extension of $F$ unramified at every place, so $h:=\rk_3\Cl(F)\ge1$, and there are $(3^h-1)/2$ such fields $k$. Put $d:=\rk_3\Cl_9(F)$, the $3$-rank of the ray class group of $F$ modulo $9$. Let $\cG=\Gal(F_S(3)/F)$ be the Galois group of the maximal pro-$3$ extension of $F$ unramified outside $S=\{3,\infty\}$; then $d=\dim_{\F_3}H^1(\cG,\F_3)$. Let $\chi_K,\chi_\cyc\in H^1(\cG,\F_3)$ cut out $K$ and $F\cdot\Q(\zeta_9)^+$. \begin{theorem}[forced anomaly]\label{thmA} % [Proved] One has $d\in\{h+1,h+2\}$, and $d=h+2$ if and only if every $x\in F^\times$ such that $(x)$ is the cube of an ideal is a cube in $F\otimes\Q_3$. If $d=h+2$, then every cubic field $k$ with $d_k=D$ satisfies $v_3(R_3(k))\ge3$. \end{theorem} For $h=1$ the condition reads: the fundamental unit $\eps_F$ and an element $\alpha$ with $(\alpha)=\mathfrak b^3$, $\mathfrak b$ of order $3$ in $\Cl(F)$, are both cubes in $F\otimes\Q_3$; equivalently the torsion $T_3(F)$ of the Galois group of the maximal abelian pro-$3$ extension of $F$ unramified outside $3$ is not cyclic (Proposition~\ref{prop:resolvent}). For $h=1$ we can say exactly what happens. Let $\lb(E_k)\subset O_k/3$ be the image of the units of $k$ under the reduced logarithm $u\mapsto\log_3(u)/3 \bmod 3$; it lies in the trace-zero plane, which contains the line $\F_3\cdot1$. \begin{theorem}[$h=1$]\label{thmB} % [Proved] Suppose $h=1$ and put $c=1$ if $\chi_K\cup\chi_\cyc=0$ in $H^2(\cG,\F_3)$ and $c=0$ otherwise. Then \[ \rk_3\Cl_9(k)=d-1+c,\qquad \dim_{\F_3}\lb(E_k)=4-d-c,\qquad 1\in\lb(E_k)\iff c=0 . \] Consequently: \begin{center} \begin{tabular}{cccc} \toprule $d$ & $c$ & $\lb(E_k)$ & $\operatorname{depth}_3(k)$\\ \midrule $2$ & $0$ & the trace-zero plane & $0$\\ $2$ & $1$ & a line $\neq\F_3\cdot1$ & $\ge1$\\ $3$ & $0$ & $\F_3\cdot1$ & $\ge1$\\ $3$ & $1$ & $0$ & $\ge2$\\ \bottomrule \end{tabular} \end{center} \end{theorem} In words: in the forced stratum the reduced logarithms of all units of $k$ are \emph{rational} modulo $3$, i.e.\ every unit of $k$ is, at every place above $3$, a rational $1$-unit times a local cube. The invariant $c$ can be computed explicitly (Theorem~\ref{thm:cform}). Write a generator $x$ of the relevant Kummer group of $F$ as a norm $N_{K/F}(Y)$. Then $c$ is the vanishing of a product of two symbols: a cubic residue symbol at an ideal recording how the $3$-class group of $F$ capitulates in $K$, and cubic Hilbert symbols at the primes above $3$ of the local Hilbert-90 cofactors of $Y$. When $|\Cl(F)[3^\infty]|=3$ and $\eps_F$ is a local cube, the first symbol is trivial and $c$ is a local condition at $3$ on one unit of $K$. The proof of Theorem~\ref{thm:cform} is given in outline (one cochain identification is sketched); the formula agrees with $c$ on all $2500$ fields on which we tested it. We conjecture that $c=1$ with probability $1/3$ for $d=2$ and $1/9$ for $d=3$ (Conjecture~\ref{heur:third}), and explain why this seems out of reach of current methods. The reason is a parity phenomenon. By Poitou--Tate duality $H^2(\cG,\F_3)$ is dual to a group of Kummer classes of $F$ on which $\Gal(F/\Q)$ acts by $-1$, so $H^2(\cG,\F_3)$ has no invariants under $\Gal(F/\Q)$ (Proposition~\ref{prop:resolvent}). Hence the cup product of two sign-type characters of $\cG$ vanishes. When $d=h+2$ there is a sign-type character $\chi_E$ independent of the class-group characters, and the vanishing of $\chi_K\cup\chi_E$ lets $\chi_E$ lift to the non-semisimple $\F_3[S_3]$-module $[\sgn;1]$, the reduction modulo $3$ of the reflection representation. Through Shapiro's lemma the lift becomes a cyclic cubic extension of $k$ unramified outside $3$ and independent of $k\cdot\Q(\zeta_9)^+$, and two such extensions are exactly what anomaly means (Lemma~\ref{lem:depth}). For $p\ge5$ the module $\F_p[S_3/S_2]$ is semisimple and the mechanism disappears (Section~\ref{sec:pge5}), in accordance with Table~\ref{tab:survey}. \subsection{The mixture law} % [Numerical] fe16b_scan.py, fe16c_analyze.py Among the $259\,403$ fields of the survey with fundamental discriminant prime to $3$, the strata have weights $0.879$ ($h=1$, $d=2$), $0.092$ ($h=1$, $d=3$), $0.028$ ($h=2$, $d=3$) and $0.0004$ ($h=2$, $d=4$). Theorem~\ref{thmA} accounts for the second and fourth stratum (all $24\,009$ fields are anomalous), and Theorem~\ref{thmB} reduces the first to the vanishing of one element of $\F_3$, which Theorem~\ref{thm:cform} computes explicitly. If that element is equidistributed (Conjecture~\ref{heur:third}), the anomaly rate over fundamental discriminants should be close to $0.879\cdot\frac13+0.092+0.028\cdot\frac{11}{27}+0.0004\approx0.397$; the observed value is $0.390$, the difference coming from a small deficit in the first stratum that decreases with $D$ (Table~\ref{tab:bins}). Fields of non-fundamental discriminant $d_k=f^2D$, $f>1$, are not covered by our results; their observed rate is $0.351$. \subsection{Context and organisation} The ingredients --- the ray class group modulo $p^2$ as a test of $p$-rationality \cite{GrasPrat}, the Shafarevich--Koch formula, Poitou--Tate duality and Shapiro's lemma \cite{NSW,Koch,GrasCFT} --- are standard. We have not found Theorems~\ref{thmA} or~\ref{thmB} in the literature on $p$-rational fields \cite{MNQD,JNQD,GrasPrat,LQ} or on $p$-adic regulators of cubic fields \cite{HZ,JN,FJ}, but a statement of this kind may well be implicit in that literature and we make no strong priority claim. In a much deeper setting, $p$-adic regulators of units --- determinants built from $p$-adic logarithms and valuations of units --- give the leading terms of $p$-adic $L$-functions, as in the Gross--Stark conjecture proved by Dasgupta, Kakde and Ventullo \cite{DKV}; here only the reduction modulo $3$ of such a determinant matters. Section~\ref{sec:setup} fixes notation and proves the depth criterion. Section~\ref{sec:resolvent} studies $\cG$, and Section~\ref{sec:proof} proves Theorems~\ref{thmA} and~\ref{thmB}. Section~\ref{sec:cform} gives an explicit formula for $c$. Section~\ref{sec:d2} discusses the generic stratum, Section~\ref{sec:pge5} the primes $p\ge5$, Section~\ref{sec:numerics} the computations, and Section~\ref{sec:leopoldt} the relation with Leopoldt's conjecture and with the function-field analogue. % ====================================================================================================================== \section{Set-up and the depth criterion}\label{sec:setup} % ====================================================================================================================== Throughout, $D$, $F$, $h$, $d$, $k$, $K$ are as in the introduction. Write $\sigma$ for the generator of $\Gal(F/\Q)$ and $\Gal(K/\Q)=S_3=\langle\tau,s\rangle$ with $\tau^3=s^2=1$, $k=K^{s}$, $F=K^{\tau}$. The prime $3$ is unramified in $K$. \subsection{Reduced logarithms} For $u\in O_k$ prime to $3$ put $\lambda(u)=\frac13\log_3(u)\in O_k\otimes\Z_3$, where $\log_3(u)=\log_3(u^N)/N$ with $N=\operatorname{lcm}(3^f-1:f\le3)=104$, and let $\lb(u)\in O_k/3$ be its reduction. Since $\log(1+3y)\equiv 3y\pmod 9$, \[ \lb(u)=N^{-1}\,\frac{u^N-1}{3}\bmod 3 . \] The same definitions apply to $F$ and $K$. Let $C_k=\ker\bigl(\Cl_9(k)[3^\infty]\to\Cl(k)[3^\infty]\bigr)$. \begin{lemma}\label{lem:depth} % [Proved] Let $k$ be a totally real cubic field in which $3$ is unramified, with unit group $E_k$. \begin{enumerate} \item $\lb(E_k)$ lies in the trace-zero plane of $O_k/3$, and $C_k\cong(O_k/3)/\lb(E_k)$. \item $\operatorname{depth}_3(k)\ge1$ if and only if $\dim\lb(E_k)\le1$, if and only if $\dim C_k\ge2$. \item If $\lb(E_k)=0$ then $\operatorname{depth}_3(k)\ge2$. \end{enumerate} \end{lemma} \begin{proof} Let $M=(\sigma_i\lambda(u_j))$, a $3\times2$ matrix over the ring of integers of an unramified extension of $\Q_3$; then $R_3(k)=9\det(M_{i,j\le2})$. Each column of $M$ sums to $\Tr\lambda(u_j)=\lambda(N_{k/\Q}u_j)=\lambda(\pm1)=0$, so the three $2\times2$ minors of $M$ agree up to sign, and $3\mid\det M_{\le2}$ if and only if the reduction $\overline M$ has rank at most $1$. As $k\otimes\Q_3$ is unramified, the embeddings induce an isomorphism $(O_k/3)\otimes\overline{\F}_3\cong\overline{\F}_3^3$ carrying $\lb(u_j)$ to the $j$-th column of $\overline M$; linear dependence over $\overline\F_3$ and over $\F_3$ coincide. This proves the first equivalence in (2), the trace statement in (1), and (3) (if $M\equiv0\bmod3$ then $9\mid\det M_{\le2}$). Taking $3$-primary parts in the exact sequence $E_k\to(O_k/9)^\times\to\Cl_9(k)\to\Cl(k)\to1$, note that the $3$-Sylow subgroup of $(O_k/9)^\times$ is $(1+3O_k)/(1+9O_k)$, isomorphic to $O_k/3$ via $1+3y\mapsto y$ because $(1+3y)(1+3y')\equiv1+3(y+y')\pmod 9$. The $3$-component of the image of a unit $u$ is $u^{Nm}$ for any $m$ with $Nm\equiv1\bmod 3$, and corresponds to $\lb(u)$. Hence $C_k\cong(O_k/3)/\lb(E_k)$, and $\dim C_k=3-\dim\lb(E_k)$. \end{proof} No hypothesis on Leopoldt's conjecture is used. Lemma~\ref{lem:depth} is the cubic case of Gras's ray-class test for $p$-rationality \cite{GrasPrat}. \subsection{Ray class groups as Galois cohomology} Let $S=\{3,\infty\}$ and let $F_S(3)$ be the maximal pro-$3$ extension of $F$ unramified outside $S$. Since $K/F$ is unramified, $K\subset F_S(3)$, and $F_S(3)/\Q$ is Galois. Put \[ \cG=\Gal(F_S(3)/F),\quad \tG=\Gal(F_S(3)/\Q),\quad \Gamma_k=\Gal(F_S(3)/k),\quad H=\Gal(F_S(3)/K)\lhd\tG, \] so that $\tG/H\cong S_3$ and $[\tG:\cG]=2$. Let $F^{\rm ur}(3)$ be the maximal unramified pro-$3$ extension of $F$ and define $\tG'$, $\Gamma_k'$ in the same way. \begin{lemma}\label{lem:raycoh} % [Proved] For $L\in\{F,k,K\}$, $\rk_3\Cl_9(L)=\dim H^1(\Gal(F_S(3)/L),\F_3)$; for $L\in\{F,k\}$, $\rk_3\Cl(L)=\dim H^1(\Gal(F^{\rm ur}(3)/L),\F_3)$. \end{lemma} \begin{proof} Let $v\mid3$ be a place of $L$. As $L_v/\Q_3$ is unramified, $\log$ maps $1+3O_v$ isomorphically onto $3O_v$, so $(1+3O_v)^3=1+9O_v$ and every character of order $3$ of $L_v^\times$ is trivial on $1+9O_v$. Real places do not ramify in extensions of odd degree. Hence the cyclic cubic extensions of $L$ unramified outside $S$ correspond to the characters of order $3$ of $\Cl_9(L)$. Each such extension $M/L$ lies in $F_S(3)$: the Galois closure of $MK$ over $F$ is a compositum of cyclic cubic extensions of $K$ unramified outside $S$, hence a $3$-extension of $F$ unramified outside $S$. The unramified statement is proved in the same way. \end{proof} \subsection{Some $\F_3[S_3]$-modules} Write $\mathbf 1$ and $\sgn$ for the trivial and sign characters of $S_3$ over $\F_3$, viewed as $\tG$-modules through $\tG\to S_3$. Let $P=\F_3[S_3/\langle s\rangle]$ be the permutation module on the three cosets, with basis $e_1,e_2,e_3$, where $e_1$ is the coset fixed by $s$. Its submodules are $0\subset\F_3(e_1+e_2+e_3)\subset N\subset P$, where $N=\{\sum a_ie_i:\sum a_i=0\}$; the successive quotients are $\mathbf 1$, $\sgn$, $\mathbf 1$ and $P$ is uniserial. Thus there are exact sequences \begin{equation}\label{eq:seqs} 0\to\mathbf1\to N\to\sgn\to0,\qquad 0\to N\to P\xrightarrow{\ \Sigma\ }\mathbf1\to0,\qquad 0\to\sgn\to P/\mathbf1\to\mathbf1\to0, \end{equation} none of which splits. By Shapiro's lemma, $H^i(\Gamma_k,\F_3)\cong H^i(\tG,P)$, and the map induced by the augmentation $\Sigma$ is corestriction $H^i(\Gamma_k,\F_3)\to H^i(\tG,\F_3)$. Since $[\tG:\cG]=2$ is prime to $3$, for every $\F_3[\tG]$-module $M$ we have $H^i(\tG,M)=H^i(\cG,M)^{\sigma}$ (invariants under $\tG/\cG$). We write $H^i(\cG,\F_3)^{\pm}$ for the $\pm1$-eigenspaces of $\sigma$; then $H^i(\tG,\mathbf 1)=H^i(\cG,\F_3)^+$ and $H^i(\tG,\sgn)=H^i(\cG,\F_3)^-$. % ====================================================================================================================== \section{The resolvent field}\label{sec:resolvent} % ====================================================================================================================== Let $W=\{x\in F^\times:(x)=\mathfrak b^3\text{ for an ideal }\mathfrak b\}/F^{\times3}$; then $\dim W=1+h$, spanned by $\eps_F$ and elements $\alpha_1,\dots,\alpha_h$ attached to a basis of $\Cl(F)[3]$. Let \[ V_S=\{x\in W: x\in F_v^{\times3}\text{ for all }v\mid 3\}. \] \begin{prop}\label{prop:resolvent} % [Proved] \begin{enumerate} \item $\sigma$ acts on $W$, and hence on $V_S$, by $-1$. \item $H^2(\cG,\F_3)^+=0$; that is, $H^2(\tG,\mathbf1)=0$. \item $d=1+\dim V_S$, and $V_S$ is the kernel of $\lb\colon W\to(O_F/3)^-$, a map to a line. Hence $d\in\{h+1,h+2\}$, and $d=h+2$ if and only if every element of $W$ is a cube in $F\otimes\Q_3$. \item $\rk_3T_3(F)=d-1$. For $h=1$: $\eps_F$ is a cube in $F\otimes\Q_3$ if and only if the local part of $T_3(F)$ is nonzero, and $d=3$ if and only if $\eps_F$ and $\alpha_1$ are both cubes in $F\otimes\Q_3$, if and only if $T_3(F)$ is not cyclic. \end{enumerate} \end{prop} \begin{proof} (1) If $x\in W$ is $\sigma$-invariant modulo cubes, then $x^2\equiv x\sigma(x)=N_{F/\Q}(x)$, so $x\equiv r:=N_{F/\Q}(x)^2$ modulo cubes, with $r\in\Q^\times$ and $(r)O_F$ a cube. For a prime $p$ unramified in $F$ this gives $3\mid v_p(r)$; for $p$ ramified it gives $3\mid 2v_p(r)$. So $r=\pm m^3$ with $m\in\Q$, a cube in $F$. As $2$ is invertible, $W=W^-$. (2) Inflation $H^2(\cG,\F_3)\to H^2(G_S(F),\F_3)$ is injective and $\tG$-equivariant: in the five-term exact sequence for $\Gal(\overline F_S/F_S(3))\lhd G_S(F)$ the obstruction $H^1(\Gal(\overline F_S/F_S(3)),\F_3)^{\cG}$ vanishes, because a nonzero invariant homomorphism would cut out a cyclic cubic extension of $F_S(3)$ that is Galois and pro-$3$ over $F$ and unramified outside $S$. For $v\in S$ the local groups $H^2(F_v,\F_3)$ vanish: for $v\mid3$ they are dual to $H^0(F_v,\mu_3)=0$ since $F_v/\Q_3$ is unramified, and for $v$ real they are killed by $2$. Hence $H^2(G_S(F),\F_3)=\Sha^2_S(\F_3)$, which by Poitou--Tate duality \cite[Ch.~I, Thm.~4.10]{Milne}, \cite[Ch.~VIII]{NSW} is dual to $\Sha^1_S(\mu_3)$. By Kummer theory $H^1(F,\mu_3)=F^\times/F^{\times3}$, and the classes unramified outside $S$ and locally trivial at $S$ form exactly $V_S$. The duality pairing is invariant under conjugation by a lift of $\sigma$, since the local invariant maps satisfy $\operatorname{inv}_{\sigma v}\circ\sigma=\operatorname{inv}_v$. So $\Sha^2_S(\F_3)^+$ is orthogonal to $V_S^-=V_S$, hence zero. (3) Every Galois extension of $F$ of degree $3$ unramified outside $S$ lies in $F_S(3)$, so $H^1(\cG,\F_3)=H^1(G_S(F),\F_3)$. Tate's global Euler characteristic formula \cite[Ch.~I, Thm.~5.1]{Milne}, \cite[Ch.~VIII]{NSW} gives $h^0-h^1+h^2=\sum_{v\text{ real}}h^0(F_v,\F_3)-[F:\Q]=0$ for $G_S(F)$ with coefficients $\F_3$, so $d=1+h^2(G_S(F),\F_3)=1+\dim V_S$; this is the Shafarevich--Koch formula in our case \cite{Koch,NSW}. Every class of $W$ has a representative $x$ prime to $3$ (move $\mathfrak b$ in its class), and such an $x$ is a cube in $F_v$ exactly when $x^N\in1+9O_v$, i.e.\ when $\lb(x)$ vanishes at $v$. Since $N_{F/\Q}(x)=\pm m^3$ with $3\nmid m$, we get $\Tr\lambda(x)=3\lambda(m)\equiv0\pmod 3$, so $\lb(x)\in(O_F/3)^{\Tr=0}=(O_F/3)^-$, which is a line. Hence $h\le\dim V_S\le h+1$. (4) The Galois group $X_F$ of the maximal abelian pro-$3$ extension of $F$ unramified outside $3$ is $\Z_3\oplus T_3(F)$ (Leopoldt's conjecture holds for $F$), so $\rk_3 T_3(F)=d-1$. The local part of $T_3(F)$ is the torsion of $U^1_F/\overline{E}_F$; under $\lambda$ it is $I_F/\Z_3\lambda(\eps_F)$ with $I_F$ the trace-zero line of $O_F\otimes\Z_3$, which is nonzero exactly when $\lambda(\eps_F)\in3I_F$. The remaining statements follow from (3) with $W=\langle\eps_F,\alpha_1\rangle$. \end{proof} \begin{remark} In the language of the introduction, the forced stratum is the case where ``the unit part of $T_3(F)$ splits off its class-group part''; precisely, it means $\rk_3T_3(F)=h+1$. When $9\mid h_F$ this need not be a direct sum decomposition, and indeed $2428$ of the $23\,914$ forced fields in our data have $9\mid h_F$. \end{remark} % ====================================================================================================================== \section{Proof of the theorems}\label{sec:proof} % ====================================================================================================================== \begin{thm}[rank formula]\label{thm:rank} % [Proved] For every cubic field $k$ with $d_k=D$ one has $\rk_3\Cl_9(k)=d-1+c$, where $c=1$ if $\chi_K\cup\chi_\cyc=0$ in $H^2(\cG,\F_3)$ and $c=0$ otherwise. \end{thm} \begin{proof} By Lemma~\ref{lem:raycoh} and Shapiro's lemma, $\rk_3\Cl_9(k)=\dim H^1(\tG,P)$. We use: \begin{itemize} \item $H^0(\tG,\mathbf1)=\F_3$ and $H^0(\tG,\sgn)=0$; \item $H^1(\tG,\mathbf1)=\Hom(\tG,\F_3)=\F_3\chi_\cyc$, because $\Q(\zeta_9)^+$ is the only cyclic cubic field unramified outside $3$; \item $H^1(\tG,\sgn)=H^1(\cG,\F_3)^-$, of dimension $d-1$, containing $\chi_K$ (because $\sigma$ acts on $\Gal(K/F)$ by $-1$); \item $H^2(\tG,\mathbf1)=0$ (Proposition~\ref{prop:resolvent}(2)) and $H^2(\tG,\sgn)=H^2(\cG,\F_3)$. \end{itemize} The first sequence of \eqref{eq:seqs} gives, since $H^0(\tG,\mathbf 1)\to H^0(\tG,N)=N^{S_3}=\F_3(e_1+e_2+e_3)$ is an isomorphism, \[ 0\to H^1(\tG,\mathbf1)\to H^1(\tG,N)\to H^1(\tG,\sgn)\to H^2(\tG,\mathbf 1)=0, \] so $\dim H^1(\tG,N)=d$, and also that $H^2(\tG,N)\to H^2(\tG,\sgn)$ is injective. In the second sequence, $H^0(\tG,P)=\F_3(e_1+e_2+e_3)$ maps to $3=0$ in $H^0(\tG,\mathbf1)$, so the connecting map $\F_3=H^0(\tG,\mathbf 1)\to H^1(\tG,N)$ is injective and \[ 0\to H^1(\tG,N)/\F_3\to H^1(\tG,P)\xrightarrow{\ \operatorname{cor}\ } H^1(\tG,\mathbf1)\xrightarrow{\ \delta\ }H^2(\tG,N). \] Thus $\dim H^1(\tG,P)=d-1+\dim\ker\delta$. Composing $\delta$ with the injection $H^2(\tG,N)\to H^2(\tG,\sgn)$ gives, by naturality, the connecting map of the third sequence of \eqref{eq:seqs}, which is the cup product with its extension class in $H^1(\tG,\Hom(\mathbf1,\sgn))=H^1(\cG,\F_3)^-$. With the section $1\mapsto e_1$ that class is the cocycle $g\mapsto(g-1)e_1$; it vanishes on a lift of $s$, and for $g\in\cG$ acting as $\tau^{\chi_K(g)}$ it equals $\chi_K(g)(e_2-e_1)$ modulo constants, because $e_3-e_1\equiv-(e_2-e_1)$ there. So the class is $\chi_K$, and $\delta(\chi_\cyc)=0$ if and only if $\chi_K\cup\chi_\cyc=0$. \end{proof} \begin{remark}[the witness]\label{rem:witness} The proof also says where the extensions come from. Modulo the class of the extension $P$ (which maps to $\chi_K$), $H^1(\tG,N)$ is spanned by $\chi_\cyc$ and by lifts to $N=[\sgn;\mathbf1]$ of the sign-type characters other than $\chi_K$; such lifts exist because their obstruction $\chi_K\cup\chi$ lies in $H^2(\tG,\mathbf1)=0$. When $h=1$ and $d=3$ the extra sign-type character $\chi_E$ cuts out a cyclic cubic extension $E/F$, ramified at $3$, with $E/\Q$ an $S_3$-extension. Its lift gives a cyclic cubic extension $L_E/k$ unramified outside $3$ and different from $k\cdot\Q(\zeta_9)^+$, whose Galois closure over $F$ is a non-abelian group of order $27$ containing $EK$. This field is the explicit cause of the anomaly. \end{remark} \begin{remark}[$c$ as a transfer]\label{rem:transfer} % [Proved] The image of $\operatorname{cor}$ is $\ker\delta$, so $c=1$ if and only if corestriction $H^1(\Gamma_k,\F_3)\to H^1(\tG,\F_3)$ is nonzero, i.e.\ if and only if the transfer $\Gal(\Q(\zeta_9)^+/\Q)\to\Gamma_k^{\rm ab}/3$ is nonzero. By class field theory the transfer is the extension of ideals $\Cl_9(\Q)[3]=\langle[2]\rangle\to\Cl_9(k)/3$, so $c=0$ if and only if the class of $2O_k$ is trivial in $\Cl_9(k)/3$. \end{remark} \begin{lemma}\label{lem:clk} % [Proved] $\rk_3\Cl(k)\le h-1$. In particular $\Cl(k)[3]=0$ when $h=1$. \end{lemma} % [Lit, NOT READ] Gerth 1976: bibliographic data checked on Crossref only; the attribution below is a presumption % from the title, to be verified against the paper before submission. This bound is presumably classical: Gerth computes $3$-ranks of class groups of non-Galois cubic fields in terms of the quadratic resolvent \cite{Gerth}. We have not consulted that paper, so we include a proof. \begin{proof} Run the argument of Theorem~\ref{thm:rank} with $\tG'=\Gal(F^{\rm ur}(3)/\Q)$ in place of $\tG$ (Lemma~\ref{lem:raycoh}). Now $H^1(\tG',\mathbf 1)=0$ (there is no unramified cyclic cubic extension of $\Q$) and $H^1(\tG',\sgn)=\Hom(\Cl(F),\F_3)$ has dimension $h$. The second sequence gives $\rk_3\Cl(k)=\dim H^1(\tG',N)-1$, and the first gives $H^1(\tG',N)\hookrightarrow H^1(\tG',\sgn)$. \end{proof} \begin{proof}[Proof of Theorem~\ref{thmA}] The first two statements are Proposition~\ref{prop:resolvent}(3). Since $C_k$ is the kernel of $\Cl_9(k)[3^\infty]\to\Cl(k)[3^\infty]$, Theorem~\ref{thm:rank} and Lemma~\ref{lem:clk} give \[ \dim C_k\ \ge\ \rk_3\Cl_9(k)-\rk_3\Cl(k)\ \ge\ (d-1+c)-(h-1)\ =\ d-h+c . \] If $d=h+2$ this is at least $2$, and Lemma~\ref{lem:depth} gives $\operatorname{depth}_3(k)\ge1$. \end{proof} The same inequality shows that $k$ is anomalous whenever $d=h+1$ and $c=1$. \begin{proof}[Proof of Theorem~\ref{thmB}] For $h=1$, Lemma~\ref{lem:clk} gives $\Cl(k)[3]=0$, so $C_k=\Cl_9(k)[3^\infty]$ is elementary abelian of dimension $d-1+c$, and $\dim\lb(E_k)=3-(d-1+c)$ by Lemma~\ref{lem:depth}. The class of the principal ideal $2O_k$ in $\Cl_9(k)$ is the image of $2\in(O_k/9)^\times$, whose $3$-component is that of $-2=7=1+3\cdot2$, i.e.\ the class of $2\cdot1$ in $(O_k/3)/\lb(E_k)$. By Remark~\ref{rem:transfer}, $c=0$ if and only if $1\in\lb(E_k)$. The table follows: for $d=3$, $c=0$ the space $\lb(E_k)$ is a line containing $1$; for $d=3$, $c=1$ it is $0$ and Lemma~\ref{lem:depth}(3) applies; for $d=2$ it is the plane ($c=0$) or a line not containing $1$ ($c=1$). \end{proof} \begin{remark}[units of $K$] The classical route through the unit group of $K$ gives only part of this. For $h=1$ and $|\Cl(F)[3^\infty]|=3$, the Herbrand quotient of $E_K$ for the unramified extension $K/F$ is $1/3$, so $H^1(\Gal(K/F),E_K)\cong\Z/3$ and $\mathfrak bO_K=(\beta)$ is principal. In the forced stratum $\eps_F$ and $\eta=\alpha/\beta^3$ are units of $K$ that are local cubes at $3$ and independent modulo cubes. Both are of sign type, however, and by themselves impose no dependence on the reduced logarithms of $E_k$; the additional input is the vanishing $H^2(\tG,\mathbf 1)=0$. \end{remark} % ====================================================================================================================== \section{An explicit formula for $c$}\label{sec:cform} % ====================================================================================================================== % [Proof sketch + Numerical] F5_P3_LOCAL.md; fe26_cform.gp, fe26b_test.py, fe26c_analyze.py Theorem~\ref{thmB} reduces the first free $3$-adic digit to the invariant $c$, defined by a cup product. Here we evaluate that cup product. One might expect a formula in terms of the cubic Hilbert symbols $(\kappa_K,\zeta_3)_{\mathfrak P}$, $\mathfrak P\mid3$, of the Kummer generators of $K$ and of $\Q(\zeta_9)^+$ over $L=F(\zeta_3)$. Those symbols are the local components of $\chi_K\cup\chi_\cyc$, and they vanish identically, because $H^2(\cG,\F_3)$ consists of $\Sha^2$ only (Proposition~\ref{prop:resolvent}). All the information is in the Poitou--Tate pairing with $V_S$, which mixes a local term at $3$ with a class-group term. Let $\lambda_\Q\colon\Z_3^\times\to\F_3$, $u\mapsto\frac13\log_3(u)\bmod 3$, extended to rational numbers prime to $3$. \begin{definition}\label{def:I} Let $x\in V_S$, and let $\tau$ generate $\Gal(K/F)$. \begin{enumerate} \item By the Hasse norm theorem $x=N_{K/F}(Y)$ for some $Y\in K^\times$: $x$ is a local norm everywhere, since $K/F$ is unramified and $x$ is a local cube at $3$. Replacing $Y$ by $Y\,W/\tau(W)$ for a suitable $W$, we may assume $Y$ is a unit at every prime above $3$. \item Write $(x)=\mathfrak b^3$. Since $(Y)\mathfrak b^{-1}O_K$ has norm $1$ and the ideal group of $K$ is a permutation module, $(Y)=\mathfrak bO_K\cdot\tau(\mathfrak A)/\mathfrak A$ for an ideal $\mathfrak A$ of $K$ prime to $3$. \item For $v\mid3$ let $\xi_v\in F_v$ be the cube root of $x$, and write $Y=\xi_v\,\tau(Z_v)/Z_v$ with $Z_v\in(K\otimes_FF_v)^\times$ a $1$-unit (Hilbert~90). \end{enumerate} Put \[ I(x)=\lambda_\Q\bigl(N_{K/\Q}\mathfrak A\bigr)-\lambda_\Q\Bigl(\prod_{v\mid3}N_{K\otimes F_v/\Q_3}(Z_v)\Bigr)\in\F_3 . \] \end{definition} Over $L$, the first term is the cubic residue symbol $(\zeta_3/N_{K/F}\mathfrak A)_3$, which records how $\mathfrak b$ capitulates in $K$. The second is the product of the cubic Hilbert symbols $(N Z_v,\zeta_3)_{\mathfrak P}$ at the primes above $3$. \begin{thm}\label{thm:cform} % [Proof sketch; Numerical: 2500/2500 fields, 600/600 invariance checks] $I(x)$ does not depend on the choices made, and $\langle\chi_K\cup\chi_\cyc,x\rangle=\pm I(x)$ under $\Gal(\Q(\zeta_9)^+/\Q)\cong\F_3$. Hence for $h=1$: \begin{itemize} \item if $d=2$, then $c=1\iff I(x)=0$, where $x$ generates $V_S$; \item if $d=3$, then $c=1\iff I(\eps_F)=I(\alpha_1)=0$. \end{itemize} \end{thm} \begin{proof}[Proof (outline)] \emph{Independence of the choices.} \begin{itemize} \item Replacing $Y$ by $Y\,W/\tau(W)$ replaces $\mathfrak A$ by $\mathfrak A(W)$ and $Z_v$ by $Z_vW$. This changes $I$ by $\lambda_\Q$ of the principal id\`ele $N_{K/\Q}(W)$, which is $0$ by Artin reciprocity for $\Q(\zeta_9)^+/\Q$. \item Changing $\mathfrak A$ by an ideal of $F$, or $Z_v$ by an element of $F_v^\times$, changes the norms by cubes. \end{itemize} \emph{The pairing.} Use the explicit cochain description of the pairing $\Sha^1\times\Sha^2\to\Q/\Z$ in \cite[Ch.~I, \S4, after Remark~4.19]{Milne}, with $\alpha=\chi_\cyc\cup\chi_K$ and $\beta=x$. There one trivialises $\beta$ and $\alpha$ locally at each $v\in S$, writes $\beta\cup\alpha=d\epsilon$ for a global $2$-cochain $\epsilon$ with values in the multiplicative group (the $S$-units), and sums the local Brauer invariants of the resulting local $2$-cocycles. At $v\mid3$ trivialise $x$ locally by its cube root $\xi_v$. Globally take $\epsilon=-\chi_\cyc\cup\theta$, where $\theta$ is a $1$-cochain with values in $\overline F^\times$ and $d\theta=\chi_K\cup x$; it exists because the cyclic algebra $(K/F,x)$ is split by $Y$. \begin{itemize} \item The local term at every finite place takes the form $\operatorname{inv}_v(\chi_\cyc\cup y_v)$, where $y_v\in F_v^\times/F_v^{\times3}$ is the difference of the global splitting (from $Y$) and the local one (from $\xi_v$ for $v\mid3$, from a local generator of $\mathfrak b$ for $v\nmid3$). Apart from this term, the local contributions at $v\notin S$ are unramified classes of degree $2$, which vanish. \item Comparing two splittings of a cyclic algebra gives $y_v=N(Z_v)$, where $Y$ is the local splitting element times $\tau(Z_v)/Z_v$. At $v\nmid3$ this is a generator of $(N_{K/F}\mathfrak A)_v$. \item Hence $\langle\alpha,\beta\rangle=\chi_\cyc(\operatorname{rec}_F(y))$ for the id\`ele $y=(y_v)$. Artin's map for $\Q(\zeta_9)^+/\Q$ sends $\mathrm{Frob}_\ell$ to $\zeta\mapsto\zeta^\ell$ and $u\in\Z_3^\times$ to $\zeta\mapsto\zeta^{u^{-1}}$, which gives $\pm I(x)$. \end{itemize} We have not written out the cochain identification $y_v=N(Z_v)$ in full. If it failed, $I(x)$ would depend on the choice of $Y$; it is invariant in every randomised test below. For $d=3$, use that $H^2(\cG,\F_3)\cong V_S^\vee$ and $V_S=\langle\eps_F,\alpha_1\rangle$. \end{proof} \begin{cor}\label{cor:local} % [Proved, given Theorem cform] Suppose $h=1$, $d=2$, $|\Cl(F)[3^\infty]|=3$ and $\eps_F$ is a cube in $F\otimes\Q_3$. Then $\eps_F=N_{K/F}(\eta)$ for a unit $\eta$ of $K$, up to cubes of units of $F$, and $c=1$ if and only if $\lambda_\Q\bigl(\prod_{v\mid3}N_{K\otimes F_v/\Q_3}(Z_v)\bigr)=0$, where $\eta=\xi_v\tau(Z_v)/Z_v$. So in this substratum $c$ is a purely local condition at $3$ on one unit of $K$. \end{cor} \begin{proof} Here the capitulation kernel of $K/F$ has order $3=|H^1(\Gal(K/F),E_K)|=3\,[E_F:N_{K/F}E_K]_3$, so $\eps_F\in N(E_K)E_F^3$. Take $Y=\eta e$ with $e\in E_F$. Then $(Y)=(1)$, so $\mathfrak A=1$. \end{proof} \emph{Numerical check.} We evaluated $I$ with \textsc{PARI/GP} on random fields of the scan of Section~\ref{sec:numerics} and compared with $c=\rk_3\Cl_9(k)-d+1$: \begin{itemize} \item Theorem~\ref{thm:cform} held in every case: $1000$ fields with $h=1$, $d=2$ and $\eps_F$ not a local cube; $1000$ glued fields; $500$ forced fields with $d=3$. \item $I(x)$ was unchanged under $600$ random replacements $Y\mapsto Y\,W/\tau(W)$, while its two terms changed. \item The alternative sign $\lambda_\Q(N\mathfrak A)+\lambda_\Q(\prod N Z_v)$ disagrees with $c$ on $43$ of these fields. \item In the substratum of Corollary~\ref{cor:local} the class-group term vanished in all $898$ tested fields. Elsewhere it was nonzero in $3$--$25\%$ of them, so both terms are needed in general. \end{itemize} % ====================================================================================================================== \section{The generic stratum}\label{sec:d2} % ====================================================================================================================== Suppose $h=1$ and $d=2$, which is the case for $88\%$ of our fields. By Theorem~\ref{thmB}, $k$ is anomalous if and only if $\chi_K\cup\chi_\cyc=0$ in the line $H^2(\cG,\F_3)\cong V_S^\vee$, i.e.\ if and only if one element of $\F_3$ vanishes. The line $V_S$ is spanned by $\alpha_1\eps_F^{j}$ for the unique $j$ making it a local cube when $\eps_F$ is not a local cube, and by $\eps_F$ when $\eps_F$ is a local cube but $\alpha_1$ is not (the ``glued'' case, $T_3(F)$ cyclic). In both cases $\cG$ has two generators and one relation, and nothing distinguishes the two cases at the level of this invariant. \begin{heur}\label{heur:third} % [Conjecture] Order the real quadratic fields $F$ with $3\nmid D$ by discriminant. Among those with $h=1$ and $d=2$, the proportion with $c=1$ tends to $1/3$, both when $\eps_F$ is a local cube and when it is not. Among those with $h=1$ and $d=3$, it tends to $1/9$. \end{heur} By Theorem~\ref{thmB}, Theorem~\ref{thm:cform} and Corollary~\ref{cor:local}, the following are equivalent forms of the first statement: \begin{itemize} \item a cubic field of the stratum fails to be $3$-rational ($T_3(k)\neq0$) with probability $1/3$; \item the symbol $I(x)$ of Definition~\ref{def:I} vanishes with probability $1/3$; \item in the substratum of Corollary~\ref{cor:local}, the local symbol at $3$ of the norm unit $\eta$ vanishes with probability $1/3$. \end{itemize} We do not know how to prove any of them. Three obstacles stand in the way: \begin{itemize} \item The density of the stratum itself ($3$-rank of $\Cl(F)$ exactly one) is a Cohen--Lenstra prediction, not a theorem; Davenport--Heilbronn controls only the mean of $|\Cl(F)[3]|$. \item $c=1$ is equivalent to the existence of a second cyclic cubic extension of $k$ unramified outside $3$. Counting such extensions means counting nonic fields, for which no parametrisation is known. \item The symbol $I(x)$ is built from a norm from $K$, so it is not a Frobenius condition in a field that stays fixed as $F$ varies, and Chebotarev's theorem does not apply. \end{itemize} A naive count misled us at first. Writing the reduced log matrix of $k$ with one difference-type linear relation coming from the unit index of $K$ and an inhomogeneous term $e$ equal to the reduced logarithm of $\eps_F$, a uniform model gives anomaly probability $1/3$ for $e\neq0$ and $5/9$ for $e=0$, predicting $5/9$ in the glued case. The data give $0.324$. The correct inhomogeneous term is the image $\lb(W)\subset(O_F/3)^-$ of the whole group $W$, which by Proposition~\ref{prop:resolvent} is nonzero exactly when $d=h+1$: in the glued case it is carried by $\alpha_1$, through the capitulation relation $\alpha_1=\eta\beta^3$. It vanishes exactly in the forced case, where the answer is $1$ rather than $5/9$. Observed: the anomaly rate is $0.3257$ when $\eps_F$ is not a local cube ($171\,138$ fields) and $0.3241$ in the glued case ($56\,912$ fields). Overall $P(c=1\mid h=1,d=2)=0.3253\pm0.0010$, with a deficit relative to $1/3$ that is largest for small $D$ (Table~\ref{tab:bins}); above $4\cdot10^6$ the pooled rate is $0.328$ ($4.5$ standard errors below $1/3$), so we cannot yet decide whether the limit is exactly $1/3$. For $d=3$ we observe $P(c=1)=0.101\pm0.002$ against $1/9$. In the forced stratum the probability of depth $\ge2$ given $c=0$ is $0.338$, so the model ``$c=1$ with probability $1/9$, otherwise one further uniform digit'' predicts $P(\operatorname{depth}\ge2)\approx\frac19+\frac89\cdot\frac13=\frac{11}{27}$, against $0.405$ observed. % ====================================================================================================================== \section{Why nothing is forced for $p\ge5$}\label{sec:pge5} % ====================================================================================================================== Let $p\ge5$ be unramified in $K$. Lemma~\ref{lem:depth} holds with $3$ replaced by $p$ and $\Cl_9$ by $\Cl_{p^2}$, with the same proof. The forcing mechanism of Section~\ref{sec:proof} used two features specific to $p=3$: $K$ lies in a pro-$3$ extension of $F$, so the arithmetic of $F$ (the characters $\chi_K$, $\chi_E$) and that of $k$ live in the cohomology of one group; and the permutation module $\F_3[S_3/\langle s\rangle]$ glues the sign character to the trivial one. For $p\ge5$ neither holds. \begin{prop}\label{prop:pge5} % [Proved] Let $p\ge5$ be unramified in $K$. As an $\F_p[S_3]$-module, $\Cl_{p^2}(K)/p\cong\mathbf 1^{a}\oplus\sgn^{b}\oplus\bar\rho^{m}$, with $\bar\rho$ the two-dimensional irreducible representation. Here $a=\rk_p\Cl_{p^2}(\Q)=1$, $b=\rk_p\Cl_{p^2}(F)-1$, and $\rk_p\Cl_{p^2}(k)=1+m$. In particular $\rk_p\Cl_{p^2}(k)$ does not depend on the sign-type part, which is the only part through which invariants of $F$ enter. \end{prop} \begin{proof} $\F_p[S_3]$ is semisimple, so the decomposition exists. Since $[K:k]$, $[K:F]$ and $[K:\Q]$ are prime to $p$, taking invariants under $\langle s\rangle$, $\langle\tau\rangle$ and $S_3$ is exact and identifies $\Cl_{p^2}(k)/p$, $\Cl_{p^2}(F)/p$ and $\Cl_{p^2}(\Q)/p$ with the corresponding invariants. Now $\mathbf1^{\langle s\rangle}=\mathbf 1$, $\sgn^{\langle s\rangle}=0$ and $\dim\bar\rho^{\langle s\rangle}=1$, while $\mathbf1^{\tau}=\mathbf1$, $\sgn^{\tau}=\sgn$ and $\bar\rho^{\tau}=0$. Finally $\Cl_{p^2}(\Q)[p]\cong\Z/p$ corresponds to the degree-$p$ subfield of $\Q(\zeta_{p^2})$. \end{proof} So for $p\ge5$ the anomaly of $k$ is a statement about the $\bar\rho$-multiplicity alone, and no condition on the resolvent can force it. This matches Table~\ref{tab:survey}, where the Haar law holds at every $p\ge5$ (we do not claim that the Haar law is a theorem). % ====================================================================================================================== \section{Computations}\label{sec:numerics} % ====================================================================================================================== All computations use \textsc{PARI/GP} 2.17.3 \cite{PARI} with \texttt{bnfinit} assuming GRH; the class and unit groups are not certified. Scripts and data are in the directory of this paper. \subsection{The survey} \texttt{fe14\_survey.py} enumerates the totally real cubic fields with $d_k\le10^7$ ($592\,421$ fields), computes the units, and computes the reduced logarithm matrix modulo $p^{8}$ for $3\le p\le31$, exactly in $\Z_p[x]/(f)$. Table~\ref{tab:p3bins} shows the rate at $p=3$ by discriminant range. \begin{table}[ht] \centering \begin{tabular}{lrcr} \toprule $d_k$ range ($10^6$) & fields & rate & $z$ vs $11/27$\\ \midrule $(0,1]$ & 31435 & 0.3666 & -14.7 \\ $(1,2]$ & 32531 & 0.3791 & -10.4 \\ $(2,4]$ & 66076 & 0.3798 & -14.5 \\ $(4,6]$ & 66600 & 0.3854 & -11.5 \\ $(6,8]$ & 67043 & 0.3834 & -12.6 \\ $(8,10]$ & 67218 & 0.3852 & -11.7 \\ \bottomrule \end{tabular} \caption{Anomaly rate at $p=3$ by discriminant.} \label{tab:p3bins} \end{table} \subsection{The strata} \texttt{fe15\_strata.py} and \texttt{fe15b\_cube.py} compute $d$, $h$ and whether $\eps_F$ is a local cube for the $254\,176$ resolvents of the $259\,403$ fields with fundamental $d_k$ prime to $3$ (Table~\ref{tab:strata}). The cell ($h=1$, $d=3$, $\eps_F$ not a local cube) is empty, as Proposition~\ref{prop:resolvent}(4) requires. \begin{table}[ht] \centering \begin{tabular}{cccrcc} \toprule $h$ & $d$ & $\eps_F$ local cube & fields & anomaly rate & depth $0$ / $1$ / $\ge2$\\ \midrule 1 & 2 & no & 171138 & 0.3257 & 0.674 / 0.217 / 0.109 \\ 1 & 2 & yes & 56912 & 0.3241 & 0.676 / 0.218 / 0.106 \\ 1 & 3 & yes & 23914 & 1.0000 & 0.000 / 0.595 / 0.405 \\ 2 & 3 & no & 5146 & 0.3984 & 0.602 / 0.255 / 0.143 \\ 2 & 3 & yes & 2198 & 0.3922 & 0.608 / 0.258 / 0.134 \\ 2 & 4 & yes & 95 & 1.0000 & 0.000 / 0.568 / 0.432 \\ \bottomrule \end{tabular} \caption{Fields with fundamental discriminant prime to $3$, by stratum.} \label{tab:strata} \end{table} \subsection{The full check} \texttt{fe16b\_scan.py} recomputes, for each of the $259\,403$ fields, $\rk_3\Cl(k)$, $\rk_3\Cl_9(k)$, $\lb(E_k)$, whether $1\in\lb(E_k)$, whether $\lb(E_k)\subset\F_3\cdot1$, and the class of $2O_k$ in $\Cl_9(k)/3$; for each resolvent it recomputes $\dim V_S$ from $\lb(\eps_F)$ and $\lb(\alpha_i)$. \texttt{fe16c\_analyze.py} checks every statement of Proposition~\ref{prop:resolvent}(3), Theorem~\ref{thm:rank}, Lemma~\ref{lem:clk}, Theorems~\ref{thmA} and~\ref{thmB}, Remark~\ref{rem:transfer} and Lemma~\ref{lem:depth} (against the independently computed depth of the survey). There were no exceptions and no computational errors (run time $1901$\,s on $24$ processes). Table~\ref{tab:scan} gives the distribution of $c$. \begin{table}[ht] \centering \begin{tabular}{cccrcccc} \toprule $h$ & $d$ & $c$ & fields & anomaly & depth $\ge2$ & $1\in\lb(E_k)$ & $\lb(E_k)\subset\F_3\cdot1$\\ \midrule 1 & 2 & 0 & 153859 & 0.0000 & 0.0000 & 1.0000 & 0.0000 \\ 1 & 2 & 1 & 74191 & 1.0000 & 0.3321 & 0.0000 & 0.0000 \\ 1 & 3 & 0 & 21493 & 1.0000 & 0.3376 & 1.0000 & 1.0000 \\ 1 & 3 & 1 & 2421 & 1.0000 & 1.0000 & 0.0000 & 1.0000 \\ 2 & 3 & 0 & 6589 & 0.3274 & 0.1131 & 0.7534 & 0.0807 \\ 2 & 3 & 1 & 755 & 1.0000 & 0.3788 & 0.0000 & 0.0596 \\ 2 & 4 & 0 & 90 & 1.0000 & 0.4000 & 0.9556 & 1.0000 \\ 2 & 4 & 1 & 5 & 1.0000 & 1.0000 & 0.0000 & 1.0000 \\ \bottomrule \end{tabular} \caption{The invariant $c$ of Theorem~\ref{thm:rank} on all $259\,403$ fields (proportions per row).} \label{tab:scan} \end{table} For $h=2$ Lemma~\ref{lem:clk} is only an inequality and Theorem~\ref{thmB} does not apply; the table shows that $c=1$ still forces the anomaly there (as Theorem~\ref{thmA} predicts via $\dim C_k\ge d-h+c$), while $d=3$, $c=0$ gives a rate of about $1/3$. \begin{table}[ht] \centering \begin{minipage}{0.45\textwidth} \centering \begin{tabular}{lrcr} \toprule $D$ ($10^6$) & fields & $P(c=1)$ & $z$ vs $\frac13$\\ \midrule $[0,1)$ & 21939 & 0.3146 & -5.89 \\ $[1,2)$ & 22659 & 0.3242 & -2.92 \\ $[2,3)$ & 22773 & 0.3218 & -3.70 \\ $[3,4)$ & 22860 & 0.3263 & -2.26 \\ $[4,5)$ & 22942 & 0.3284 & -1.59 \\ $[5,6)$ & 22962 & 0.3297 & -1.18 \\ $[6,7)$ & 23027 & 0.3287 & -1.51 \\ $[7,8)$ & 22942 & 0.3251 & -2.65 \\ $[8,9)$ & 23010 & 0.3312 & -0.70 \\ $[9,10)$ & 22936 & 0.3229 & -3.35 \\ \bottomrule \end{tabular} \end{minipage}\hfill \begin{minipage}{0.5\textwidth} \centering \begin{tikzpicture} \begin{axis}[width=0.92\textwidth,height=5cm,xlabel={$D/10^6$},ylabel={$P(c=1)$},ymin=0.30,ymax=0.345, xmin=0,xmax=10,grid=major,tick label style={font=\small},label style={font=\small}] \addplot[only marks,mark=*,mark size=1.5pt,error bars/.cd,y dir=both,y explicit] table[row sep=\\,x=x,y=p,y error plus expr=\thisrow{hi}-\thisrow{p},y error minus expr=\thisrow{p}-\thisrow{lo}]{x p lo hi\\ 0.5 0.31460 0.30845 0.32074\\ 1.5 0.32420 0.31810 0.33029\\ 2.5 0.32178 0.31572 0.32785\\ 3.5 0.32629 0.32021 0.33237\\ 4.5 0.32839 0.32232 0.33447\\ 5.5 0.32968 0.32359 0.33576\\ 6.5 0.32866 0.32259 0.33472\\ 7.5 0.32508 0.31902 0.33114\\ 8.5 0.33116 0.32508 0.33724\\ 9.5 0.32290 0.31685 0.32895\\}; \addplot[dashed,domain=0:10] {1/3}; \end{axis} \end{tikzpicture} \end{minipage} \caption{The generic stratum $h=1$, $d=2$: $P(c=1)$ (the anomaly rate) by discriminant, with $95\%$ intervals; dashed line $1/3$.} \label{tab:bins} \end{table} \subsection{Reproducibility and formal verification} Every table is regenerated from the stored JSON outputs by \texttt{make\_tables.py}. The intermediate identities were in addition checked by hand on ten fields from each stratum (\texttt{fe16\_check.py}), including three forced fields with $9\mid h_F$. We do not offer a formal (Lean) certificate. The weight of the proof is carried by Poitou--Tate duality, Tate's Euler characteristic formula and class field theory (Lemma~\ref{lem:raycoh}, Proposition~\ref{prop:resolvent}), none of which is available in a proof assistant library at present. What remains --- the structure of the permutation module $P$ and the bookkeeping of the long exact sequences in the proof of Theorem~\ref{thm:rank} --- is elementary, and formalising it alone would certify the least doubtful part of the argument while leaving the load-bearing inputs untouched. The \textsc{PARI/GP} checks instead test every link of the chain on every field of the survey, including the inputs we cite (for instance $d=1+\dim V_S$ and $\rk_3\Cl(k)\le h-1$), with the caveat that they are GRH-conditional. % ====================================================================================================================== \section{Relation to Leopoldt's conjecture and to function fields}\label{sec:leopoldt} % ====================================================================================================================== Theorems~\ref{thmA} and~\ref{thmB} force the first $3$-adic digit of $R_3(k)/9$ (two digits when $d=3$ and $c=1$), and say nothing about $R_3(k)\neq0$. Nothing in the argument controls the further digits, and the data are consistent with the next digit being random once the forced one is accounted for (depth exactly $1$ in $59.5\%$ of the forced fields). Writing $O_K\otimes\Z_3\cong\Z_3[S_3]$, the regulator of $k$ is the $\rho$-factor of the group determinant of the reduced logarithms of one unit of $K$, an irreducible quadratic form in $3$-adic logarithms, and the non-vanishing of such forms is the open ($p$-adic four exponentials) part of Leopoldt's conjecture for $S_3$ cubic fields. The non-maximality of $\Z_3[S_3]$ changes the distribution of the reduction of this form, but not its transcendence content. There is a parallel in positive characteristic. For the Carlitz module over $\F_q(\theta)$ the analogue of Leopoldt's conjecture can fail, and the failure is forced by the same Brauer reduction: in characteristic $3$ the reflection representation of $S_3$ reduces to the non-split extension $[\sgn;\mathbf1]$, and for a real cubic extension of $\F_3(\theta)$ whose Galois closure is imaginary and wildly ramified at $\infty$ the Goss zeta function factors through the non-real quadratic resolvent, which forces a positive Leopoldt defect at every prime \cite{KrismanCarlitz}. Over number fields the group algebra $\Q[S_3]$ is semisimple and no rank defect can be forced; what survives of the mechanism is the forced digit of Theorem~\ref{thmA}. % ====================================================================================================================== \section{Open questions}\label{sec:open} % ====================================================================================================================== \begin{enumerate} \item \emph{Equidistribution of $c$.} Theorem~\ref{thm:cform} makes $c$ an explicit symbol, and in the substratum of Corollary~\ref{cor:local} a purely local condition on one unit of $K$. Prove Conjecture~\ref{heur:third}, even conditionally on a Cohen--Lenstra--Martinet-type hypothesis for $T_3$ of $S_3$ cubic fields that accounts for the Galois data of Theorems~\ref{thmA} and~\ref{thmB}, and explain the small-discriminant deficit of Table~\ref{tab:bins}. \item \emph{The cochain identification.} Complete the outline proof of Theorem~\ref{thm:cform}, for instance by identifying $\langle\chi_K\cup\chi_\cyc,x\rangle$ with a R\'edei-type triple symbol. \item \emph{The strata with $h\ge2$.} For $h=2$, $d=3$ and $c=0$ the anomaly rate is about $1/3$; a description analogous to Theorem~\ref{thmB} requires controlling $\Cl(k)[3]$ and its interaction with $C_k$. \item \emph{Non-fundamental discriminants.} When $d_k=f^2D$ with $f>1$, $K/F$ is ramified and $K$ need not lie in a pro-$3$ extension of $F$ unramified outside $3$; the observed rate $0.351$ calls for a separate analysis. \item \emph{Limits.} Is the anomaly rate in the generic stratum exactly $1/3$ in the limit, and is $P(c=1\mid h=1,d=3)$ exactly $1/9$? \end{enumerate} % bibliography inlined from main_src.bbl (amsplain style; source refs.bib) \providecommand{\bysame}{\leavevmode\hbox to3em{\hrulefill}\thinspace} \providecommand{\MR}{\relax\ifhmode\unskip\space\fi MR } % \MRhref is called by the amsart/book/proc definition of \MR. \providecommand{\MRhref}[2]{% \href{http://www.ams.org/mathscinet-getitem?mr=#1}{#2} } \providecommand{\href}[2]{#2} \begin{thebibliography}{10} \bibitem{Brumer} Armand Brumer, \emph{On the units of algebraic number fields}, Mathematika \textbf{14} (1967), 121--124. \bibitem{DKV} Samit Dasgupta, Mahesh Kakde, and Kevin Ventullo, \emph{On the {G}ross--{S}tark conjecture}, Ann. of Math. 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