\section{The group \texorpdfstring{$\Sel$}{L24(S,5)}}\label{sec:selmer}

The sieve of \cref{sec:sieve} is carried out in coordinates on $\Sel$.  Since the class number of $\Lt$ is $1$, $\Sel$ is the group of $S$-units of $\Lt$ modulo fifth powers, with an explicit basis (\cref{prop:basis}).  The norm from $\Lt$ to $\Lo$ is given by an explicit matrix (\cref{lem:normdata}).

\subsection{The class number of \texorpdfstring{$\Lt$}{L24}}

\begin{theorem}\label{thm:classnumber}
The class number of $\Lt$ is $1$.
\end{theorem}

\Cref{app:classnumber} proves it without assuming the generalized Riemann hypothesis.  The descent uses only the consequence that $5$ does not divide the class number of $\Lt$, through \cref{prop:basis}(ii) below, and nothing about the class group of $\Lo$.

\subsection{A basis}

Let $\OS$ be the ring of $S$-integers of $\Lt$.  If $\alpha\in\Lt^\times$ represents a class in $\Sel$, then $\alpha\,\OS=\mathfrak a^5$ for a fractional ideal $\mathfrak a$ of $\OS$, and the class of $\mathfrak a$ in the class group $\Cl(\OS)$ depends only on the class of $\alpha$.  This gives an exact sequence
\begin{equation}\label{eq:exact}
0\longrightarrow\OS^\times/\OS^{\times5}\longrightarrow\Sel\longrightarrow\Cl(\OS)[5]\longrightarrow0,
\end{equation}
where the first map is injective because an $S$-unit that is a fifth power in $\Lt$ is the fifth power of an $S$-unit, and the kernel of the second consists of the classes of $S$-units: if $\mathfrak a=w\OS$ then $\alpha w^{-5}\in\OS^\times$.  The second map is surjective: if $\mathfrak a^5=\alpha\OS$, then the class of $\alpha$ lies in $\Sel$ and maps to that of $\mathfrak a$.  The group $\Cl(\OS)$ is a quotient of the class group of $\Lt$.  By Dirichlet's $S$-unit theorem, $\OS^\times\isom\mu(\Lt)\times\Z^{r}$ with $r=r_1+r_2-1+\#S=2+11-1+12=24$, and $\mu(\Lt)=\{\pm1\}$ because $\Lt$ has a real embedding.  So $\OS^\times/\OS^{\times5}\isom\F_5^{24}$.

The elements $B_1,\dots,B_{24}$ of $\Lt$ are listed in \cref{tab:B}, as polynomials in $a$ and $b$.  For $i\le12$, $B_i$ generates the prime $\PT_i$ of \cref{tab:Sprimes}, and $B_{13},\dots,B_{24}$ are units.

\begin{proposition}\label{prop:basis}
\begin{enumerate}[label=\textup{(\roman*)},leftmargin=2.2em]
\item The elements $B_1,\dots,B_{24}$ are $S$-units, $v_{\PT_i}(B_j)=1$ if $i=j\le12$ and $v_{\PT_i}(B_j)=0$ otherwise, and their classes in $\Lt^\times/\Lt^{\times5}$ are linearly independent over $\F_5$.
\item  The  classes of $B_1,\dots,B_{24}$ form a basis of $\Sel\isom\F_5^{24}$.
\end{enumerate}
\end{proposition}

\begin{proof}
(i) Each $B_j$ has a characteristic polynomial over $\Q$ whose coefficients have denominators divisible only by $2$, $5$ and $7$, and its norm $\Norm{\Lt/\Q}{B_j}$ is $-2$ ($j=1$), $5$ ($2\le j\le8$), $-7$ ($9\le j\le12$) or $1$ ($j\ge13$).  So $B_j$ is integral at every prime of $\Lt$ outside $S$, and its norm is prime to every rational prime other than $2$, $5$, $7$.  Hence $B_j$ is an $S$-unit.  The valuations are computed with \texttt{nfeltval}.  For independence, let $\qQ$ run over the $97$ primes of $\Lt$ above the sixteen primes of $\cQ$.  The $97\times24$ matrix $\big(\sym_{\qQ}(B_j)\big)$ over $\F_5$ has rank $24$ (the rows for $q\in\{11,101,131,181\}$ already have rank $24$).  Each $\sym_{\qQ}$ is a homomorphism on the $\qQ$-units that vanishes on fifth powers.  If $\prod_jB_j^{c_j}=\gamma^5$ with $\gamma\in\Lt^\times$, then $\gamma$ is an $S$-unit, hence a unit at every such $\qQ$, and applying the symbols gives $\sum_jc_j\sym_{\qQ}(B_j)=0$ for all $\qQ$.  As the matrix has rank $24$, $c\equiv0\pmod5$.  These computations are listed in \cref{app:computations}.

(ii)  Since $\Cl(\OS)$ is a quotient of the class group of $\Lt$, which is trivial by \cref{thm:classnumber}, we have  $\Cl(\OS)[5]=0$ and~\eqref{eq:exact} gives $\Sel=\OS^\times/\OS^{\times5}\isom\F_5^{24}$, in which the $24$ independent classes of (i) form a basis.
\end{proof}

If the class of $\alpha\in\Lt^\times$ lies in $\Sel$, we write $c=(c_1,\dots,c_{24})\in\F_5^{24}$ for its coordinate vector on this basis.  Since the valuation matrix $\big(v_{\PT_i}(B_j)\big)_{i\le12,\,j\le24}$ is $(I_{12}\mid0)$ by \cref{prop:basis}(i), $c_i\equiv v_{\PT_i}(\alpha)\pmod5$ for $i\le12$.

\subsection{The norm map}\label{sec:normdata}

Let $u_1,\dots,u_{10}\in\Lo$ be the elements of \cref{tab:u}: $u_1,\dots,u_6$ generate the primes of $\Lo$ above $2$, $5$, $7$ (the valuation matrix is the identity), and $u_7,\dots,u_{10}$ are units.

\begin{lemma}\label{lem:normdata}
\begin{enumerate}[label=\textup{(\roman*)},leftmargin=2.2em]
\item The classes of $u_1,\dots,u_{10}$ in $\Lo^\times/\Lo^{\times5}$ are linearly independent over $\F_5$.
\item There are exact identities
\[
\Norm{\Lt/\Lo}{B_j}=\pm\prod_{i=1}^{10}u_i^{\NM_{ij}}\quad(1\le j\le24),\qquad 2=\pm\prod_{i=1}^{10}u_i^{\TT_i},
\]
with the integer matrix $\NM\in M_{10\times24}(\Z)$ and vector $\TT\in\Z^{10}$ printed in~\eqref{eq:NM}.
\end{enumerate}
\end{lemma}

\begin{proof}
(i) The fifth-power residue symbols of $u_1,\dots,u_{10}$ at the $47$ primes of $\Lo$ above the primes of $\cQ$ form a matrix of rank $10$ over $\F_5$.  Since the $u_i$ are units at these primes, the argument of \cref{prop:basis}(i) shows that their classes are independent.  This matrix can be recomputed from \cref{tab:u,tab:primesq}: the primes of $\Lo$ above $q$ are the ideals $(q,h_k(a))$ of \cref{tab:primesq}, with residue fields $\Fq q[x]/(h_k)$.

(ii) For an element $B(a,b)$ of $\Lt$ written as a polynomial in $a$ and $b$, $\Norm{\Lt/\Lo}{B}$ is the resultant $\operatorname{Res}_t\big(\psi(t)/25,\,B(a,t)\big)$, computed exactly in $\Q[a]/(h(a))$.  It equals the stated product up to sign, and the identity for $2$ is checked by computing the product $\prod_iu_i^{\TT_i}$ in the same ring.  See \cref{app:computations}.
\end{proof}