\section{The class number of \texorpdfstring{$\Lt$}{L24}}\label{app:classnumber}

This appendix proves \cref{thm:classnumber}.  Minkowski's bound for $\Lt$ is about $1.42\cdot10^{15}$, so a proof that examines every prime ideal below it is out of reach.  We use instead a criterion of Belabas, Diaz y Diaz and Friedman~\cite[Theorem~5.1]{BDF08}, deduced from Weil's explicit formula, under which it is enough to examine the prime ideals of norm at most $10^7$.  In this appendix the letters $h$, $g$, $c$, $m$, $n$, $f$, $F$, $S_i$, $T$, $L$, $\chi$, $\Phi$ and $\pe$ denote the objects defined below, not those of the same or similar names in the body of the paper.

\subsection{The criterion}\label{app:criterion}

Let $K$ be a number field of degree $n$, with $r_1$ real places, $r_2$ complex places and discriminant $d_K$.  For a prime ideal $\pe$ of $K$ let $\Nm\pe$ be its absolute norm, and let $\gamma=0.5772\ldots$ be Euler's constant.  Let $c>0$, let $J\ge1$, and let $a_0,\dots,a_{J-1}$ be real numbers, not all zero.  Put
\begin{equation}\label{eq:testh}
h(y)=\sum_{j=0}^{J-1}a_j\cos\Big(\frac{2\pi jy}{c}\Big)\ \text{ for }|y|\le\frac c2,\qquad h(y)=0\ \text{ for }|y|>\frac c2,
\end{equation}
and let $g=h*h$, that is, $g(x)=\int_{\R}h(y)h(x-y)\,dy$.  Then $g$ is real, even and continuous, $g(x)=0$ for $|x|\ge c$, and $g(0)=\int_\R h^2>0$.  Put $\tilde g=g/g(0)$, let $\kappa(x)=r_1/(1+\cosh x)+n/\sinh x$ for $x>0$, and
\begin{multline}\label{eq:R}
R=r_1+n(\gamma+\log4\pi)-\log|d_K|-\int_0^\infty\big(1-\tilde g(x)\big)\kappa(x)\,dx\\
+\sum_{\pe}\ \sum_{\substack{m\ge1\\ \Nm\pe^m<e^c}}\frac{4\log\Nm\pe}{\Nm\pe^m+1}\,\tilde g(m\log\Nm\pe),
\end{multline}
where $\pe$ runs over the prime ideals of $K$.  The integral converges: its integrand is bounded near $0$, because $g$ is Lipschitz, and equals $\kappa(x)$ for $x\ge c$.  The number $R$ depends only on $K$, $c$ and the $a_j$, not on the class group of $K$.

\begin{proposition}[Belabas, Diaz y Diaz and Friedman]\label{prop:criterion}
If $R>0$, then the classes of the prime ideals of $K$ of norm less than $e^c$ generate the class group of $K$.
\end{proposition}

\begin{proof}
This is~\cite[Theorem~5.1]{BDF08} with $T=e^c$ and $f$ the restriction of $\tilde g$ to $[0,\infty)$: since $2\cosh^2(x/2)=1+\cosh x$, the inequality~(16) of~\cite{BDF08} is $R>0$.  The hypotheses of that theorem hold.  The function $f$ is supported on $[0,c]$ and $f(0)=1$.  Its Fourier cosine transform is $2\int_0^\infty\tilde g(x)\cos(xt)\,dx=\hat h(t)^2/g(0)\ge0$, where $\hat h(t)=\int_\R h(y)e^{-ity}\,dy$ is real because $h$ is real and even.  And $F(x)=\tilde g(x)/\cosh(x/2)$ is even, continuous, compactly supported, and real-analytic on $[0,c]$ (by~\eqref{eq:gclosed} below) and on $[c,\infty)$, so it satisfies the conditions of Weil's formula in the form of Poitou~\cite[Th\'eor\`eme, p.~6-06]{Poitou}.

For the reader's convenience we recall the argument.  Let $H$ be the subgroup of the class group generated by the classes of the prime ideals of norm less than $e^c$, and suppose that $H$ is a proper subgroup.  Then there is a nontrivial character $\chi$ of the class group that is trivial on $H$.  Put $\Gamma_\R(s)=\pi^{-s/2}\Gamma(s/2)$ and $\Gamma_{\mathbb C}(s)=2(2\pi)^{-s}\Gamma(s)$, and let $\Lambda(s,\chi)=|d_K|^{s/2}\Gamma_\R(s)^{r_1}\Gamma_{\mathbb C}(s)^{r_2}L(s,\chi)$ be the completed $L$-function.  Since $\chi$ is unramified at every place, $\Lambda(s,\chi)$ has the same gamma factors as the Dedekind zeta function of $K$.  It is entire, of order $1$, and satisfies $\Lambda(s,\chi)=W(\chi)\Lambda(1-s,\bar\chi)$ with $|W(\chi)|=1$.  Poitou's proof of Weil's formula for the Dedekind zeta function~\cite[\S1]{Poitou}, including his bound on the horizontal segments of the contour (his Proposition~1), carries over to $\Lambda(s,\chi)$ with two changes: the functional equation relates $\Lambda(s,\chi)$ to $\Lambda(1-s,\bar\chi)$, and there is no pole.  With $\Phi(s)=\int_\R F(x)e^{(s-1/2)x}\,dx$ it gives
\begin{equation}\label{eq:EF}
\sum_\rho\Phi(\rho)=\log|d_K|-A(F)-2\sum_{\pe}\sum_{m\ge1}\operatorname{Re}\big(\chi(\pe)^m\big)\frac{\log\Nm\pe}{\Nm\pe^{m/2}}F(m\log\Nm\pe),
\end{equation}
where $\rho$ runs over the zeros of $L(s,\chi)$ with $0<\operatorname{Re}\rho<1$, counted with multiplicity and summed symmetrically in $\operatorname{Im}\rho$, and
\[
A(F)=\frac{\pi r_1}{2}+n(\gamma+\log8\pi)-r_1\int_0^\infty\frac{1-F(x)}{2\cosh(x/2)}\,dx-n\int_0^\infty\frac{1-F(x)}{2\sinh(x/2)}\,dx
\]
(see also~\cite{Weil52}, \cite[Chapter~XVII]{Lang} and~\cite[(3)]{BDF08}).  A term of the prime sum with $F(m\log\Nm\pe)\ne0$ has $\Nm\pe\le\Nm\pe^m<e^c$, so $\chi(\pe)=1$ there.  On the other side, $\operatorname{Re}\Phi(\rho)\ge0$ whenever $0<\operatorname{Re}\rho<1$.  Indeed, write $\rho=\frac12+\sigma+i\tau$ with $|\sigma|<\frac12$.  Then
\[
\operatorname{Re}\Phi(\rho)=\int_\R\tilde g(x)\mu_\sigma(x)\cos(\tau x)\,dx\qquad\text{with }\mu_\sigma(x)=\frac{\cosh(\sigma x)}{\cosh(x/2)}.
\]
The Fourier transform of $\mu_\sigma$ is $2\pi\cos(\pi\sigma)\cosh(\pi\xi)/|\cosh(\pi(\xi+i\sigma))|^2>0$, and $\hat g=\hat h^2\ge0$, so Parseval's formula gives $\operatorname{Re}\Phi(\rho)\ge0$.  (Alternatively, positivity on the strip follows from positivity on the line $\operatorname{Re}s=1$ by the maximum principle~\cite[p.~6-08]{Poitou}, \cite[\S5]{BDF08}.)  Finally, $2N^{-m/2}/\cosh(\frac m2\log N)=4/(N^m+1)$ for $N=\Nm\pe$, and the identities
\[
\int_0^\infty\big(1-\operatorname{sech}u\big)\operatorname{sech}u\,du=\frac\pi2-1,\qquad\int_0^\infty\frac{1-\operatorname{sech}u}{\sinh u}\,du=\log2
\]
give $A(F)=r_1+n(\gamma+\log4\pi)-\int_0^\infty(1-\tilde g)\kappa$.  So the real part of the right side of~\eqref{eq:EF} is $-R$, and $0\le-R$, which contradicts $R>0$.
\end{proof}

For the computation we use the following closed form, which is elementary (substitute $y=u+x/2$ and use the product-to-sum formulas).  Put $\omega_j=2\pi j/c$.  For $0\le x\le c$,
\begin{equation}\label{eq:gclosed}
g(x)=(c-x)\sum_{j=0}^{J-1}\beta_j\cos(\omega_jx)+\sum_{i=1}^{J-1}S_i\sin(\omega_ix),
\end{equation}
with $\beta_0=a_0^2$, $\beta_j=a_j^2/2$ for $j\ge1$, and
\[
S_i=-\frac{a_i^2}{2\omega_i}-2a_i\sum_{\substack{0\le k<J\\ k\ne i}}(-1)^{i+k}a_k\frac{\omega_i}{\omega_i^2-\omega_k^2}.
\]

\subsection{The computation for \texorpdfstring{$\Lt$}{L24}}\label{app:cn-computation}

We apply \cref{prop:criterion} to $K=\Lt$, with $n=24$, $r_1=2$, $r_2=11$ and $d_K=-2^{44}\cdot5^{20}\cdot7^{23}$ (\cref{lem:fields}), with $T=10^7$ and $c=\log T$, and with $J=20$ and the coefficients $a_j$ of \cref{tab:aj}.  The $a_j$ were found as follows.  Both $g(0)$ and $R\,g(0)$ are quadratic forms in $(a_0,\dots,a_{J-1})$, so the coefficients that maximize $R$ form the top eigenvector of a generalized symmetric eigenvalue problem.  We solved it in floating point for $J=20$ and $c=\log10^7$, normalized the eigenvector to $a_0=1$ and rounded it to eight decimal places.  The proof uses only the printed values, read as exact rational numbers.  The same floating-point computation shows that for $J=20$ the maximum of $R$ changes sign between $c=15.75$ and $c=16$, which is why we took the round bound $T=10^7$, with $c=16.118\ldots$.

\begin{table}[ht]
\centering
\small
\begin{tabular}{@{}r@{\quad}l@{\qquad\qquad}r@{\quad}l@{\qquad\qquad}r@{\quad}l@{\qquad\qquad}r@{\quad}l@{}}
\toprule
$j$ & $a_j$ & $j$ & $a_j$ & $j$ & $a_j$ & $j$ & $a_j$\\
\midrule
$0$ & $\phantom{-}1$ & $5$ & $\phantom{-}0.02000542$ & $10$ & $-0.01007476$ & $15$ & $\phantom{-}0.00413216$\\
$1$ & $\phantom{-}0.15211830$ & $6$ & $-0.01334957$ & $11$ & $\phantom{-}0.00896303$ & $16$ & $-0.00403606$\\
$2$ & $-0.08506698$ & $7$ & $\phantom{-}0.00970009$ & $12$ & $-0.00736544$ & $17$ & $\phantom{-}0.00363395$\\
$3$ & $\phantom{-}0.04674648$ & $8$ & $-0.00852986$ & $13$ & $\phantom{-}0.00616254$ & $18$ & $-0.00327830$\\
$4$ & $-0.02962281$ & $9$ & $\phantom{-}0.00863041$ & $14$ & $-0.00513556$ & $19$ & $\phantom{-}0.00335562$\\
\bottomrule
\end{tabular}
\vspace{3pt}
\caption{The coefficients $a_j$ of the function $h$ used for $\Lt$ ($J=20$, $c=\log10^7$).}\label{tab:aj}
\end{table}

\emph{(a) The prime ideals.}  Since $10^7$ is not a prime power, the condition $\Nm\pe^m<e^c=10^7$ in~\eqref{eq:R} selects the same terms as $\Nm\pe^m\le10^7$, and the prime ideals in \cref{prop:criterion} are those of norm at most $10^7$.  We listed them in two ways.  First, with PARI's \texttt{idealprimedec} on the maximal order of $\Lt$, which is certified with \texttt{nfcertify}.  Second, from the decompositions in $\Lo$ and in $\Q(b)$.  The polynomial $x^3+2x-2$ defines $\Q(b)$: if $y$ is a root of it, then $b=(-2y^2+y-4)/5$ is a root of $\psi$.  The discriminant $-140$ of $x^3+2x-2$ equals that of $\Q(b)$, and the index of $\Z[a]$ in the ring of integers of $\Lo$ is $2^7\cdot5\cdot41$ (\cref{lem:fields}).  So for $p\notin\{2,5,7,41\}$ the prime $p$ is unramified in $\Lt=\Lo\otimes_\Q\Q(b)$, and, by the Dedekind--Kummer theorem, its residue degrees in $\Lo$ and in $\Q(b)$ are the degrees of the irreducible factors modulo $p$ of the octic polynomial in~\eqref{eq:L8} and of $x^3+2x-2$.  Each pair of primes of $\Lo$ and of $\Q(b)$ above $p$, of residue degrees $d$ and $d'$, gives $\gcd(d,d')$ primes of $\Lt$ of residue degree $\operatorname{lcm}(d,d')$.  For $p\in\{2,5,7,41\}$ we used \texttt{idealprimedec}.  Both lists consist of the same $665\,770$ prime ideals, above $280\,528$ rational primes, and the sum in~\eqref{eq:R} has $666\,361$ terms.\looseness=-1

\emph{(b) The value of $R$.}  We evaluated~\eqref{eq:R} in ball arithmetic (Arb~\cite{Arb}, through python-flint), in which every real number is represented by an interval that is guaranteed to contain it.  The function $g$ is evaluated by~\eqref{eq:gclosed}.  The integral is split at $\delta=2^{-20}$ and at $c$.  On $[\delta,c]$ it is computed by Arb's certified integration of the analytic integrand.  On $[0,\delta]$ we use $\kappa(x)\le r_1/2+n/x$ and $0\le g(0)-g(x)\le Lx$, where the lower bound is the Cauchy--Schwarz inequality and $L=\sum_j|\beta_j|(1+c\,\omega_j)+\sum_i|S_i|\,\omega_i$ bounds $|g'|$ on $[0,c]$ by~\eqref{eq:gclosed}.  So this part of the integral lies between $0$ and $\delta(nL+r_1L\delta/4)/g(0)$.  On $[c,\infty)$, where $g=0$, the integral equals $r_1(1-\tanh\frac c2)-n\log\tanh\frac c2$.  At $128$-bit precision this gives $R\in[0.629536,\,0.629544]$, the computed interval rounded outward.  At $200$-bit precision and with $\delta=2^{-30}$ it gives $R=0.62954314\pm3.72\cdot10^{-9}$.  The three parts of~\eqref{eq:R} are $r_1+n(\gamma+\log4\pi)-\log|d_K|=-30.84540973\ldots$, $\int_0^\infty(1-\tilde g)\kappa=2.37838643\ldots$ and the prime sum $33.85333930\ldots$, with $g(0)=16.39716346\ldots$.  In particular $R>0$.

\emph{(c) Principality.}  For each of the $665\,770$ prime ideals $\pe$ of (a) we found an element $\alpha\in\Lt$ with $\alpha\mathcal O_{\Lt}=\pe$.  A candidate $\alpha$ was proposed by PARI's \texttt{bnfisprincipal}, from a class group computation that is proved correct only under the generalized Riemann hypothesis.  This computation serves only to find $\alpha$.  Each candidate was then checked by exact arithmetic:
\begin{enumerate}[label=(\arabic*),leftmargin=2em]
\item the coordinates of $\alpha$ on an integral basis of the certified maximal order of $\Lt$ are integers, so $\alpha$ is an algebraic integer;
\item $\lvert\Norm{\Lt/\Q}{\alpha}\rvert=\Nm\pe$;
\item $v_\pe(\alpha)\ge1$.
\end{enumerate}
By (1) and (3), $\alpha\mathcal O_{\Lt}=\pe\mathfrak a$ with $\mathfrak a$ an integral ideal, and by (2), $\Nm\mathfrak a=1$, so $\alpha\mathcal O_{\Lt}=\pe$.  All $665\,770$ prime ideals pass these checks.  The generators are not stored: the program deposited in~\cite{Companion257} finds them again and repeats these exact checks in $3765$ seconds on one core.

\begin{proof}[Proof of \cref{thm:classnumber}]
By (b), $R>0$, so by \cref{prop:criterion} and (a) the classes of the prime ideals of $\Lt$ of norm at most $10^7$ generate the class group of $\Lt$.  By (c) all these classes are trivial.  Hence the class group of $\Lt$ is trivial.
\end{proof}

These computations were repeated by separately written programs (\cref{app:independent}).

\begin{remark}\label{rem:classnumberL8}
As a smaller example of the criterion, with the simplest test function, the same method shows without the generalized Riemann hypothesis that $\Lo$ has class number $1$.  Take $K=\Lo$ ($n=8$, $r_1=2$, $d_K=-2^{14}\cdot5^6\cdot7^7$), $J=1$ and $a_0=1$, so that $h$ is the indicator function of $[-c/2,c/2]$ and $g(x)=\max(c-|x|,0)$, and take $T=e^c=300$.  The sum in~\eqref{eq:R} then runs over the $61$ prime ideals of $\Lo$ of norm less than $300$ ($87$ terms), and $R=1.45103\ldots>0$.  Every prime ideal of $\Lo$ of norm at most $1000$, more than the bound $300$ requires, is principal, as checked in the manner of (c).
\end{remark}