\section{The member at 5: ramification, discriminants, groups and fields}\label{sec:bounds} The members above $5$ of a compatible system as in \cref{sec:S} are crystalline at $5$ with Hodge--Tate weights in the Fontaine--Laffaille range, which bounds the ramification of their reductions modulo $5$ (\cref{prop:ram}). With this bound, the lower bounds for discriminants, the facts on finite groups and the class numbers given below, \cref{sec:reductions} shows that their monodromy at $\p$ is not zero (step~(1) of \cref{sec:intro-uncond}). The \emph{root discriminant} of a number field $L$ of degree $n$ and discriminant $d_L$ is $\rd(L)=|d_L|^{1/n}$. Upper ramification groups $G^y$ ($y\ge-1$ real) are numbered as in \cite[Chapitre~IV]{SerreCL}; so $G^y=1$ for $y>0$ in a tamely ramified extension. \subsection{Notation}\label{sec:notation5} From now on in this part, $\lambda$ is a prime of $M$ above $5$ and $R=R_\lambda$. The prime $5$ is inert in $K$. The completion of $K$ at $5$ is $\Qtf$, the unramified quadratic extension of $\Q_5$, and $G_{\Qtf}\subset G_K$ is a decomposition group at $5$. Let $E\subset\overline M_\lambda$ be a finite extension of $\Q_5$ over which $R$ is defined, with ring of integers $\OO_E$, maximal ideal $\mE$ and residue field $\FE$. A \emph{stable lattice} is a $G_K$-stable $\OO_E$-lattice $\Lambda$ in the space of $R$, and its reduction is $\Rb_\Lambda=\Lambda/\mE\Lambda$, a representation of $G_K$ over $\FE$. The semisimplification $\Rb^{\mathrm{ss}}$ of $\Rb_\Lambda$ is independent of $\Lambda$. We enlarge $E$ when needed, so that every constituent of $\Rb^{\mathrm{ss}}$ is absolutely irreducible and $E$ contains $\Qtf$. As $5\notin\{3,7,p\}$, \cref{prop:members,prop:member-irreducible} apply: $R$ is irreducible, and it has the properties (M1)--(M3). By (M3) there is a unipotent automorphism $U$ of the space of $R$, the image under $R\otimes\kappa_\lambda^{-1}$ of a topological generator of the tame inertia group at $\p$, such that every $\sigma\in I_\p$ acts by $\kappa_\lambda(\sigma)U^{t_5(\sigma)}$, where $t_5(\sigma)\in\Z_5$ is defined by that generator. The representation $R$ is polarized (\cref{sec:polar}): there is a nondegenerate symmetric pairing $\Psi$ on its space with $\Psi(\sigma x,(\tilde\cc\sigma\tilde\cc)y)=\varepsilon_5(\sigma)^{-2}\Psi(x,y)$; in particular $R^\cc\isom R^\vee\otimes\varepsilon_5^{-2}$. \subsection{Ramification} \begin{theorem}[Fontaine; Abrashkin]\label{thm:FA} Let $\ell$ be a prime, $F$ a finite unramified extension of $\Q_\ell$, and $h$ an integer with $0\le h\le\ell-2$. Let $V$ be a crystalline representation of $G_F$ on a finite-dimensional $\Q_\ell$-vector space with Hodge--Tate weights in $[0,h]$, and let $T'\subset T$ be $G_F$-stable $\Z_\ell$-lattices in $V$ with $\ell T\subset T'$. Then the ramification groups $G_F^y$, $y>h/(\ell-1)$, act trivially on $T/T'$. \end{theorem} This is \cite[\S2, 8.1]{Abrashkin89} (generalized in \cite[Theorem~2.3]{Abrashkin15}); for $h=1$ it is Fontaine's theorem on finite flat group schemes \cite{Fontaine85}. It is also established in the course of the proof of \cite[Th\'eor\`eme~2]{Fontaine93}, and \cite[Theorem~2.19, Corollary~2.13 and \S4.4]{Hattori19} gives a proof (there $G^{(j)}=G^{j-1}$). The convention for the sign of the weights does not matter, and the interval $[0,h]$ may be replaced by any interval of length $h$: twisting $V$ by a power of $\varepsilon_\ell$ twists $T/T'$ by a power of $\bar\varepsilon_\ell$, which is tamely ramified and so trivial on $G_F^y$ for $y>0$. \begin{proposition}\label{prop:ram} Let $\Lambda$ be a stable lattice and $\Theta$ a subquotient of the $\FE[G_K]$-module $\Rb_\Lambda$. \begin{enumerate}[(a)] \item The ramification groups $G_{\Qtf}^y$, $y>1/2$, act trivially on $\Theta$. \item $U$ stabilizes $\Lambda$; let $\bar U_\Theta$ be the automorphism of $\Theta$ that it induces. It is unipotent, of order $1$ or $5$, and $\sigma\in I_\p$ acts on $\Theta$ by $\bar\kappa_\lambda(\sigma)\bar U_\Theta^{t_5(\sigma)}$. So the image of $I_\p$ in the projective linear group of $\Theta$ is generated by the image of $\bar U_\Theta$; it is tame, of order $1$ or $5$. \item $\Theta$ is unramified outside $\{\p,5\}$. \item Let $L/K$ be a finite Galois extension contained in the field cut out by the projective representation attached to $\Theta$, let $e_\p\in\{1,5\}$ be the order of its inertia groups at $\p$, and let $5^m$ be the order of its wild inertia groups at $5$. Then \[ \rd(L)<3^{\,1-1/(2e_\p)}\cdot5^{\,1+(1-5^{-m})/2}. \] In particular $\rd(L)<3^{1/2}\cdot5^{3/2}$ if $e_\p=1$; $\rd(L)<3^{1/2}\cdot5^{7/5}$ if $e_\p=1$ and $25\nmid[L:K]$; and $\rd(L)<3^{1/2}\cdot5$ if $e_\p=1$ and $5\nmid[L:K]$. \end{enumerate} \end{proposition} \begin{proof} (a) By (M2), $R|_{G_{\Qtf}}$, viewed as a representation over $\Q_5$, is crystalline with Hodge--Tate weights in $[0,2]$, and $\Lambda\supset\mE\Lambda\supset5\Lambda$ are $G_{\Qtf}$-stable $\Z_5$-lattices in it. Apply \cref{thm:FA} with $\ell=5$, $F=\Qtf$ and $h=2\le\ell-2$: the groups $G_{\Qtf}^y$ with $y>2/4$ act trivially on $\Lambda/\mE\Lambda$, hence on $\Theta$. (b) $U=R(\sigma_0)\kappa_\lambda(\sigma_0)^{-1}$ for some $\sigma_0\in I_\p$, so $U$ stabilizes $\Lambda$; as $(U-1)^3=0$, its reduction satisfies $(\bar U-1)^5=\bar U^5-1=0$. The character $\bar\kappa_\lambda$ is a scalar, and the wild inertia group at $\p$, a pro-$3$ group, acts through it. (c) is (M1). (d) By (c), $L$ is unramified outside $\{3,5\}$. The primes of $L$ above $3$ are conjugate, and so are those above $5$. At $3$, $L/\Q$ is tamely ramified with ramification index $2e_\p$, which contributes $3^{1-1/(2e_\p)}$ to $\rd(L)$. At $5$, let $\Delta_5\subset\Gal(L/K)$ be a decomposition group, with ramification groups $\Delta_5^y$ in the upper numbering. Since $\Qtf/\Q_5$ is unramified, the exponent of $5$ in $\rd(L)$ is the valuation, normalized by $v(5)=1$, of the different of the corresponding extension of $\Qtf$, which is \[ \int_{-1}^\infty\Big(1-\frac1{|\Delta_5^y|}\Big)\,dy \] (\cite[Chapitre~IV, \S1, Proposition~4]{SerreCL} and the definition of the upper numbering). By (a), $\Delta_5^y=1$ for $y>1/2$; for $y>0$, $\Delta_5^y$ is contained in the wild inertia subgroup, of order $5^m$; and for $-11$ \cite[(9)]{Poitou77}. It is twice continuously differentiable and nonnegative, and it is twice the convolution square of the function equal to $\cos(\pi x)$ on $[-\frac12,\frac12]$ and to $0$ elsewhere, so its Fourier transform is nonnegative. For $b>0$ we use the test function $\Phi(x)=\Phi_0(x/b)/{\cosh(x/2)}$. For a zero $\varrho=\beta+it$, $\operatorname{Re}\widehat\Phi(\varrho)$ is the Fourier transform at $t$ of $\Phi_0(x/b)\cosh((\beta-\frac12)x)/{\cosh(x/2)}$, a product of two functions with nonnegative Fourier transforms (that of $\cosh(\alpha x)/{\cosh(x/2)}$, for $|\alpha|<\frac12$, is $4\pi\cosh(\pi t)\cos(\pi\alpha)/(\cosh2\pi t+\cos2\pi\alpha)$); so it is nonnegative, and \eqref{eq:rdbound} holds unconditionally \cite[Proposition~5]{Poitou77}. Here $I_2(\Phi)=4b/\pi^2$. For fixed $b$, and for $r_1=0$ or $r_1=n$, the right side of \eqref{eq:rdbound} increases with $n$. This is the method of Odlyzko. For totally complex fields the supremum over $b$ and $n$ of these bounds is $4\pi e^\gamma<22.39$. \begin{proposition}\label{prop:odlyzko} Let $L$ be a totally complex number field of degree $n$. If $n\ge18$ then $\rd(L)>9.2709$; if $n\ge120$ then $\rd(L)>16.9886$; if $n\ge2002$ then $\rd(L)>21.4136$. On the other hand \[ 3^{1/2}\cdot5<8.6603,\qquad 3^{1/2}\cdot5^{7/5}<16.4862,\qquad 3^{1/2}\cdot5^{3/2}<19.3650. \] \end{proposition} \begin{proof} The right side of \eqref{eq:rdbound}, with $r_1=0$, was evaluated with certified error bounds (\cref{app:computations}, item~C4) at $(n,b)=(18,\frac{23}4)$, $(120,13)$ and $(2002,36)$; any $b>0$ is admissible, and these values of $b$ are close to optimal. The exponential of each value exceeds the number stated, and the bounds increase with $n$. \end{proof} The proof of \cref{thm:cases} needs these bounds for $[L:\Q]\ge18$, $[L:\Q]\ge120$ and $[L:\Q]\ge7200$; in the last case any degree up to $7200$ at which the bound exceeds $19.3650$ would serve, and $2002$ is one. The exponent $3/2$ at $5$ in \cref{prop:ram}~(d) is what makes the method work: $3^{1/2}\cdot5^{3/2}<19.37$ lies below the supremum $4\pi e^\gamma$ of these bounds. A crystalline representation with Hodge--Tate weights in an interval of length $3$ is still covered by \cref{thm:FA} at $\ell=5$, since $3\le\ell-2$, but only for $y>3/4$. The argument of \cref{prop:ram}~(d) then gives the exponent $1+\frac34=\frac74$ at $5$ in place of $\frac32$, and $3^{1/2}\cdot5^{7/4}>28.9$, far above that supremum. \subsection{Finite groups} For a finite group $\Delta$, $O_5(\Delta)$ is its largest normal $5$-subgroup. \begin{lemma}\label{lem:no-abelian} Let $d\in\{2,3\}$ and let $\Delta\subset\PGL_d(\Fbar_5)$ be a finite subgroup whose preimage in $\GL_d(\Fbar_5)$ acts irreducibly on $\Fbar_5^d$. If $5$ divides $|\Delta|$, then $\Delta$ has no nontrivial abelian normal subgroup. In particular $\Delta$ is not solvable, $|\Delta|\ge60$, and $O_5(\Delta)=1$. \end{lemma} \begin{proof} Let $\tilde\Delta\subset\SL_d(\Fbar_5)$ be the preimage of $\Delta$; it is finite and irreducible. Let $A\ne1$ be an abelian normal subgroup of $\Delta$ and $\tilde A$ its preimage, a normal subgroup of $\tilde\Delta$ whose commutator subgroup lies in the center $\mu_d$; so $\tilde A$ is nilpotent. By Clifford's theorem the restriction of $\Fbar_5^d$ to $\tilde A$ is semisimple and $\tilde\Delta$ permutes its isotypic components transitively. As $d$ is prime there are three cases. (i) There are $d$ components. They are lines, $\tilde\Delta$ permutes them, and the kernel of $\tilde\Delta\to S_d$ consists of diagonal matrices, so its order is prime to $5$; as $|S_d|$ divides $6$, $5\nmid|\Delta|$. (ii) There is one component, a multiple of a character. Then $\tilde A$ consists of scalars and $A=1$. (iii) $\tilde A$ acts irreducibly. The Sylow $5$-subgroup of the nilpotent group $\tilde A$ is normal in $\tilde\Delta$; its fixed vectors form a nonzero $\tilde\Delta$-stable subspace, which is therefore the whole space, so this Sylow subgroup is trivial. An irreducible representation of a nilpotent group of order prime to $5$ is a tensor product of irreducible representations of its Sylow subgroups, whose degrees are powers of the corresponding primes. As $d$ is prime, the Sylow $d$-subgroup acts irreducibly and the other Sylow subgroups act by scalars. Scalars of order prime to $d$ in $\SL_d$ are trivial, so $\tilde A$ is a $d$-group. Its center consists of scalars, so it is $\mu_d$, and $\tilde A/\mu_d=A$ is abelian. The character of a faithful irreducible representation of such a group vanishes outside the center, so $d^2=[\tilde A:\mu_d]=|A|$; and $A$ is not cyclic, since otherwise $\tilde A$ would be abelian. So $A\isom(\Z/d)^2$. An element $x$ of $\GL_d(\Fbar_5)$ whose image commutes with $A$ satisfies $xqx^{-1}=\eta(q)q$ for $q\in\tilde A$, with $\eta$ a character of $A$ with values in $\mu_d$; the commutator pairing identifies $A$ with the group of such characters, so $x$ differs from an element of $\tilde A$ by an element commuting with $\tilde A$, a scalar. Hence $A$ is its own centralizer in $\PGL_d(\Fbar_5)$, and $\Delta/A$ embeds in $\Aut(A)=\GL_2(\Z/d)$, of order $6$ or $48$. So $|\Delta|$ divides $24$ or $432$, and $5\nmid|\Delta|$. Each case contradicts $5\mid|\Delta|$ or $A\ne1$. A nontrivial solvable group has a nontrivial abelian normal subgroup (the last nontrivial term of its derived series), and the smallest non-solvable group has order $60$. Finally, if $O_5(\Delta)\ne1$ its center would be a nontrivial abelian normal subgroup. \end{proof} \pagebreak[3]\begin{lemma}\label{lem:groups} The following statements on finite groups hold. \begin{enumerate}[(a)] \item A finite group with no nontrivial abelian normal subgroup and of order divisible by $25$ has order at least $3600$. \item A finite subgroup of $\PGL_3(\Fbar_5)$ whose preimage in $\GL_3(\Fbar_5)$ is irreducible has order at least $9$. \item A finite subgroup of $\PGL_2(\Fbar_5)$ of order prime to $5$ whose preimage in $\GL_2(\Fbar_5)$ is irreducible is dihedral of order $2m\ge4$, or isomorphic to $A_4$ or $S_4$. \end{enumerate} \end{lemma} \begin{proof} We use two standard facts. First, if a finite group $\Delta$ has no nontrivial abelian normal subgroup, then its socle is a direct product of non-abelian simple groups and has trivial centralizer, so $\Delta$ embeds in the automorphism group of its socle. Second, the non-abelian simple groups of order less than $3600$ are $A_5$, $\mathrm{PSL}_2(\F_7)$, $A_6$, $\mathrm{PSL}_2(\F_8)$, $\mathrm{PSL}_2(\F_{11})$, $\mathrm{PSL}_2(\F_{13})$, $\mathrm{PSL}_2(\F_{17})$, $A_7$, $\mathrm{PSL}_2(\F_{19})$, of orders $60$, $168$, $360$, $504$, $660$, $1092$, $2448$, $2520$, $3420$, and their automorphism groups have orders $120$, $336$, $1440$, $1512$, $1320$, $2184$, $4896$, $5040$, $6840$ \cite{ATLAS}; none of these is divisible by $25$. (a) If the socle has at least two simple factors, its order is at least $60^2$; if it is a simple group $S$ of order less than $3600$, then $|\Delta|$ divides $\lvert\Aut(S)\rvert$, which is not divisible by $25$. (b) If the order of the group is prime to $5$, the representation of its preimage $\tilde\Delta$ in $\SL_3(\Fbar_5)$ lifts to an irreducible representation in characteristic $0$, of degree $3$, and $9\le[\tilde\Delta:\mu_3]$. Otherwise \cref{lem:no-abelian} gives order at least $60$. (c) The preimage in $\SL_2(\Fbar_5)$ has order prime to $5$, so it lifts to characteristic $0$, and the group is isomorphic to a finite subgroup of $\PGL_2(\C)$: cyclic, dihedral, $A_4$, $S_4$ or $A_5$. A cyclic group has an abelian, hence reducible, preimage, and $|A_5|$ is divisible by $5$. \end{proof} \subsection{Class numbers and abelian extensions} \begin{lemma}\label{lem:classfields} The following statements hold. \begin{enumerate}[(a)] \item The fields $K$, $K(\sqrt5)$ and $K(\zeta_5)=\Q(\zeta_{15})$ have class number one. \item The abelian extensions of $K$ of degree prime to $5$ that are unramified outside $5$ are $K$, $K(\sqrt5)$ and $K(\zeta_5)$; and $\Gal(K(\zeta_5)/K)\isom\Z/4$. \item An abelian extension of $K(\sqrt5)$ of degree prime to $5$ that is unramified outside $5$ has degree at most $2$ over $K(\sqrt5)$. \end{enumerate} \end{lemma} \begin{proof} (a) The three fields are totally complex, of degrees $2$, $4$, $8$ and discriminants $-3$, $3^2\cdot5^2$, $3^4\cdot5^6$, so their Minkowski bounds are smaller than $1.11$, $2.28$ and $7.11$. No prime ideal has norm below the bound: in $K(\sqrt5)$ the primes above $2$ have norm $4$ ($2$ is inert in $K$ and in $\Q(\sqrt5)$); in $\Q(\zeta_{15})$ the primes above $2$, $3$, $5$, $7$ have norms $16$, $81$, $25$, $2401$. So every ideal class contains the unit ideal. (b) Such an extension is tamely ramified at the prime $5\OO$, so its conductor divides $5\OO$. As $K$ has class number one, no real place and unit group $\mu_6$, the ray class group of conductor $5\OO$ is $\F_{25}^\times/\mu_6\isom\Z/4$. The extension $K(\zeta_5)/K$ is abelian of degree $4$, unramified outside $5$ and tamely ramified, so it is the ray class field, and its subextensions are the three fields~named. (c) The field $K(\sqrt5)$ has one prime $\mathfrak Q$ above $5$, with residue field $\F_{25}$, and the conductor of the extension divides $\mathfrak Q$. By (a) the ray class group of conductor $\mathfrak Q$ is the quotient of $\F_{25}^\times$ by the image of the units. The image of $-\omega$ has order $6$, and the image of $(1+\sqrt5)/2$ is $1/2=3$, of order $4$; together they generate a subgroup of order $12$ of the cyclic group $\F_{25}^\times$. The image of the full unit group contains this subgroup, so the quotient has order at most~$2$. \end{proof}