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\newcommand{\M}{M_{0,n}}
\newcommand{\Sn}{\mathfrak S}
\newcommand{\Pic}{\operatorname{Pic}}
\newcommand{\Aff}{\mathrm{Aff}}
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\title[The tangent line at a marked point]{The tangent line at a marked point on $M_{0,n}/\mathfrak S_{n-1}$,\\ after a table of Grothendieck}
\author{Michel Hua}
\address{Commutator, Paris, France}
\email{michel@commutator.io}
\thanks{ORCID 0009-0002-8304-0683.}
\date{7 November 2026}
\subjclass[2020]{14H10; 14C22, 14D23}
\keywords{moduli of pointed rational curves, Picard group of a quotient stack, binary forms, discriminant, tangent line bundle, Lean}

\begin{document}
\begin{abstract}
Let $L_n$ be the line bundle on $M_{0,n}$ whose fibre is the tangent line to $\mathbb P^1$ at the last marked point, and $L$ the tensor product of the tangent lines at all of them. We show that, in characteristic zero, the Picard group of the quotient stack $[M_{0,n}/\mathfrak S_{n-1}]$ is cyclic of order $(n-1)(n-2)$ for every $n \ge 3$, generated by $L_n$, and that $L$ has order $n-1$ in $\operatorname{Pic}[M_{0,n}/\mathfrak S_n]$. For even $n \ge 6$ the group was known, through the integral Chow ring computed by Edidin and Hu; the odd case does not seem to be in print. The question comes from a table in Grothendieck's working notes, which gives these orders for four points and only guesses for five and six. The proof takes a few lines once the stack is written as monic squarefree polynomials modulo the affine group. An elementary version, with explicit trivialisations and fixed configurations, is formalised in Lean.
\end{abstract}

\maketitle

\section{Introduction}

Let $k$ be a field of characteristic zero and $n \ge 3$. Let $\M$ be the moduli space of $n$ distinct ordered points $x_1, \dots, x_n$ on $\PP$, up to $\mathrm{PGL}_2$. For each $i$ the tangent lines $T_{x_i}\PP$ form a line bundle $L_i$ on $\M$, and we put $L = L_1 \otimes \dots \otimes L_n$. The symmetric group $\Sn_n$ acts on $\M$ by renumbering the points and acts compatibly on $L$; the subgroup $\Sn_{n-1}$ fixing the last point acts on $L_n$. This defines classes
\[
[L] \in \Pic[\M/\Sn_n], \qquad [L_n] \in \Pic[\M/\Sn_{n-1}],
\]
and the question is their order: the least $e \ge 1$ for which $L^{\otimes e}$, or $L_n^{\otimes e}$, has a nowhere-vanishing section compatible with the action.

\begin{theorem}\label{thm:main}
For every $n \ge 3$:
\begin{enumerate}
\item $\Pic[\M/\Sn_{n-1}]$ is cyclic of order $(n-1)(n-2)$, generated by $[L_n]$;
\item $[L]$ has order $n-1$ in $\Pic[\M/\Sn_n]$.
\end{enumerate}
\end{theorem}

For $n = 4, 5, 6$ the orders are $3, 4, 5$ for $L$ and $6, 12, 20$ for $L_n$.

\subsection*{The source}
The University of Montpellier holds the mathematical papers that Alexandre Grothendieck left to Jean Malgoire, and publishes their facsimiles. Page~70 of its \fac{146}, \emph{Groupo\"ides de Teichm\"uller}, has a table which gives, for $n = 4, 5, 6$, the best exponent that makes $L$ and the $L_i$ trivial compatibly with the symmetric group. Its first row gives $(n-1)(n-2) = 6, 12, 20$. In the column $n = 4$ it gives $3$ for $L$ and $6$ for $L_i$, and both are right. The columns $n = 5$ and $n = 6$ hold only queried values: \emph{12\,?} for both bundles when $n = 5$, and \emph{20\,?} for $L$ and \emph{60\,?} for $L_i$ when $n = 6$. Below the table, a computation for $n = 5$ ends with $(L^4)^3 \simeq \mathcal O$, that is, $L^{\otimes 12}$ is trivial. Page~72 builds sections indexed by the three-element subsets of the points. For $n = 4$ it writes $L^{\otimes 6} \simeq 1$, which is true and weaker than the order~$3$. For $n = 5$ it writes $L^{\otimes 6} \simeq 1$ again (our transcription marks the words before the formula as an uncertain reading); this is false, since by Theorem~\ref{thm:main} $L$ then has order~$4$. The theorem completes the table. For $n = 5$ the queried $12$ is right for $L_5$, but it is not minimal for $L$, whose order is $4$. For $n = 6$ the orders are $5$ and $20$, not $20$ and $60$. In every column the order of $L_n$ is the number in the first row.

\begin{table}[h]\centering
\caption{The table of \fac{146}, page~70, and the orders given by Theorem~\ref{thm:main}.}\label{tab:table}
\begin{tabular}{lcccccc}
\toprule
& \multicolumn{2}{c}{$n = 4$} & \multicolumn{2}{c}{$n = 5$} & \multicolumn{2}{c}{$n = 6$}\\
& $L$ & $L_n$ & $L$ & $L_n$ & $L$ & $L_n$\\
\midrule
Page 70 & 3 & 6 & 12\,? & 12\,? & 20\,? & 60\,?\\
Theorem~\ref{thm:main} & 3 & 6 & 4 & 12 & 5 & 20\\
\bottomrule
\end{tabular}
\end{table}

\subsection*{What was known}
The stack $[\M/\Sn_n]$ is the stack of smooth binary forms of degree $n$ up to scalars, modulo $\mathrm{PGL}_2$. Arsie and Vistoli computed the Picard groups of stacks of smooth cyclic covers of projective spaces from such presentations by forms \cite[Th.~5.1]{arsievistoli2004}. For $n = 2g+2$ with $g \ge 2$, Gorchinskiy and Viviani identify $[\M/\Sn_n]$ with the stack $\mathcal D_{2g+2}$ of $\PP$-bundles with an \'etale divisor of degree $2g+2$, and find $\Pic \mathcal D_{2g+2} \cong \mathbb Z/2(n-1)$ \cite[Prop.~3.4, Th.~3.6]{gorchinskiyviviani2008}. The order of $[L]$ follows from these methods in a few lines (Remark~\ref{rem:sn}), but we have not found it stated. For $n = 2g+2$ and $g \ge 2$, Edidin and Hu compute the integral Chow ring of $[M_{0,2g+2}/\Sn_{2g+1}]$, which they present as binary forms with one marked root, as $\mathbb Z[l]/(2g(2g+1)\,l)$ \cite[Th.~1.3]{edidinhu2025}, building on Pernice's description of the stack of $1$-pointed hyperelliptic curves \cite{pernice2022}. So $\Pic[\M/\Sn_{n-1}]$ is cyclic of order $(n-1)(n-2)$ for even $n \ge 6$. Their generator is written in terms of characters of a torus \cite[Th.~5.8]{edidinhu2025}, and its identification with $L_n$ is not stated. For $n = 4$, sending an elliptic curve $E$ to $E/\{\pm1\} \cong \PP$, with the origin as last point, makes $\mathcal M_{1,1}$ a $\mu_2$-gerbe over $[M_{0,4}/\Sn_3]$. Pulling back along it is injective on Picard groups, and its image consists of the classes on which $\mu_2$ acts trivially. Since $\Pic \mathcal M_{1,1} = \mathbb Z/12$ is generated by the Hodge bundle $\lambda$, on which $-1$ acts by $-1$ \cite{mumford1965,fultonolsson2010}, the image is $2\mathbb Z/12 \cong \mathbb Z/6$, and $L_4$ pulls back to $T_0E^{\otimes 2} \cong \lambda^{-2}$, of order~$6$. For odd $n$ we found no printed statement. D.~Edidin confirms that their paper treats only the even case and that he knows no reference for the odd one \cite{edidin2026}. The author asked the question on MathOverflow (\href{https://mathoverflow.net/q/516243}{question 516243}) and posted the argument of Section~\ref{sec:stack} as an answer.

\section{The stack as polynomials modulo the affine group}\label{sec:stack}

Put $m = n - 1 \ge 2$. Since $\mathrm{PGL}_2$ acts transitively on $\PP$, we may send the last point to $\infty$; its stabiliser is the affine group $\Aff = \Ga \rtimes \Gm$, acting on $\mathbb A^1$ by $x \mapsto ax + b$, and the other $m$ points, unordered, are the roots of a monic squarefree polynomial of degree $m$. Taking the slice $x_n = \infty$ gives an isomorphism of stacks
\[
[\M/\Sn_{n-1}] \;\cong\; [U_m/\Aff],
\]
where $U_m \subset \mathbb A^m$ is the open set of monic squarefree polynomials $x^m + c_1x^{m-1} + \dots + c_m$, on which $\Aff$ acts by substitution (Figure~\ref{fig:slice}).

The translations $\Ga$ act freely, and every orbit meets the closed set $c_1 = 0$ exactly once, since $m$ is invertible in $k$: replacing $f(x)$ by $f(x - c_1/m)$ centres the roots. So
\[
[U_m/\Aff] \;\cong\; [V_m/\Gm], \qquad V_m = \bigl\{x^m + c_2x^{m-2} + \dots + c_m \text{ squarefree}\bigr\} \subset \mathbb A^{m-1},
\]
where $t \in \Gm$ acts by $x \mapsto tx$, that is, on the coefficient $c_j$ with weight $j$. The complement of $V_m$ in $\mathbb A^{m-1}$ is the discriminant hypersurface $\disc = 0$. As a function of the roots, $\disc = \prod_{i<j}(r_i - r_j)^2$ is homogeneous of degree $m(m-1)$, so it is a semi-invariant of weight $m(m-1)$. It is irreducible in $k[c_2, \dots, c_m]$. Indeed, the discriminant of the generic monic polynomial is irreducible in $k[c_1, \dots, c_m]$ and invariant under translation of the roots. Since $m$ is invertible, $k[c_1, \dots, c_m] = k[c_1, c'_2, \dots, c'_m]$, where the $c'_j$ are the coefficients of the centred polynomial $f(x - c_1/m)$, and the discriminant is the same polynomial in $c'_2, \dots, c'_m$, so it stays irreducible there.

\begin{figure}[ht]
\centering
\begin{tikzpicture}[scale=1.05, >=stealth]
\begin{scope}
\draw[black!35] (0,1) circle (1);
\draw[black!70] (-3.2,0) -- (2.2,0) node[right] {\footnotesize $\mathbb A^1$};
\fill (0,2) circle (1.8pt); \node[above=1pt] at (0,2) {\small $x_n = \infty$};
\foreach \px/\py/\r/\l in {-0.94/1.342/-2.856/1, -0.819/0.426/-1.041/2, 0.342/0.06/0.353/3, 0.94/0.658/1.4/4} {
  \draw[black!30] (0,2) -- (\r,0);
  \fill (\px,\py) circle (1.4pt);
  \fill (\r,0) circle (1.6pt);
  \node[below=2pt] at (\r,0) {\footnotesize $r_{\l}$};
}
\node at (-0.45,-0.9) {\small (a) $\PP$, projected from $x_n$};
\end{scope}
\begin{scope}[xshift=6.9cm]
\draw[black!70] (-2.6,0) -- (2.6,0) node[right] {\footnotesize $\mathbb A^1$};
\draw (0,0.12) -- (0,-0.12); \node[below=3pt] at (0,0) {\footnotesize $0$};
\foreach \r/\l in {-2.32/1, -0.505/2, 0.889/3, 1.936/4} {
  \fill (\r,0) circle (1.6pt);
  \draw[->,thick] (\r,0) -- (\r+0.42,0.42);
}
\node[align=left, anchor=west] at (-2.6,1.45) {\footnotesize $x \mapsto x + b$: \ $c_1 = -\textstyle\sum r_i \mapsto c_1 - mb$};
\node[align=left, anchor=west] at (-2.6,0.95) {\footnotesize $x \mapsto tx$: \ $c_j \mapsto t^j c_j$, \ $T_{r_i}\PP \mapsto t$, \ $T_\infty\PP \mapsto t^{-1}$};
\node at (0,-0.9) {\small (b) centred roots, $\sum r_i = 0$};
\end{scope}
\end{tikzpicture}
\caption{The slice of Section~\ref{sec:stack} for $n = 5$. (a) With the last point at $\infty$, the other points become four roots on the affine line. (b) Translations move $c_1$ freely, so one may centre the roots; the scalings that remain act on the coefficient $c_j$ with weight $j$, on each tangent line at a root (arrows) with weight $1$, and on the tangent line at $\infty$ with weight $-1$.}\label{fig:slice}
\end{figure}

\begin{lemma}\label{lem:pic}
$\Pic[V_m/\Gm] \cong \mathbb Z/m(m-1)$, generated by the line bundle attached to the character $t \mapsto t$.
\end{lemma}

\begin{proof}
Since $V_m$ is open in an affine space, $\Pic V_m = 0$, so a $\Gm$-linearised line bundle on $V_m$ is the trivial bundle with a linearisation. A linearisation is a unit $\phi(t, x)$ of $\mathcal O(\Gm \times V_m) = \mathcal O(V_m)[t^{\pm 1}]$ satisfying the cocycle rule. As $\disc$ is irreducible, the units of $\mathcal O(V_m)$ are the $c\,\disc^j$ with $c \in k^\times$, so $\phi = c\,\disc(x)^j\,t^a$, and $\phi(1, x) = 1$ forces $c = 1$ and $j = 0$. So every linearisation is a character $t \mapsto t^a$. Two characters $t^a$ and $t^b$ give isomorphic bundles exactly when there is a unit $u$ on $V_m$ with $u(t\cdot x) = t^{b-a} u(x)$, that is, a semi-invariant unit of weight $b - a$. The units $c\,\disc^j$ have weights $j\,m(m-1)$. Hence $\Pic[V_m/\Gm] = \Pic^{\Gm}(V_m) \cong \mathbb Z/m(m-1)$, generated by the character of weight~$1$.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:main}]
In the model $[U_m/\Aff]$ the last point is $\infty$. In the coordinate $u = 1/x$ at $\infty$, the map $x \mapsto ax + b$ is $u \mapsto u/(a + bu)$, whose derivative at $u = 0$ is $a^{-1}$; so $\Gm$ acts on $T_\infty\PP$ through the character of weight $-1$, and $[L_n]$ corresponds to $-1 \in \mathbb Z/m(m-1)$. It generates the group, which proves~(1).

At a finite root $r$, the derivative of $x \mapsto ax + b$ is $a$, so each $T_{r_i}\PP$ has weight $+1$, and the pull-back of $L$ to $[\M/\Sn_{n-1}]$ has weight $m - 1 = n - 2$. Its order in $\mathbb Z/(n-1)(n-2)$ is $n - 1$, so the order of $[L]$ is a multiple of $n - 1$. Conversely, write the points as vectors $v_i \in k^2 \smallsetminus \{0\}$ and $[v_iv_j] = \det(v_i, v_j)$. Since $T_{\langle v\rangle}\PP \cong \operatorname{Hom}(\langle v\rangle, k^2/\langle v\rangle) \cong \langle v\rangle^{\otimes -2} \otimes \det k^2$, and since the centre of $\mathrm{GL}_2$ acts trivially on $\langle v\rangle^{\otimes -2} \otimes \det k^2$, descent along the $\Gm^n$-torsor of vectors over configurations shows that trivialising $L^{\otimes(n-1)}$ compatibly with the action amounts to giving a nowhere-vanishing function of the $v_i$, homogeneous of degree $2(n-1)$ in each $v_i$, transforming under $g \in \mathrm{GL}_2$ by $\det(g)^{n(n-1)}$ and invariant under renumbering. The function
\[
\prod_{i \ne j}[v_iv_j] \;=\; (-1)^{n(n-1)/2}\prod_{i<j}[v_iv_j]^2
\]
is such a function: each $v_i$ occurs in $2(n-1)$ brackets, it vanishes nowhere on $\M$, and the right-hand side is invariant under $\Sn_n$. This proves~(2).
\end{proof}

\begin{remark}\label{rem:sn}
For $n = 3$ the stack $[M_{0,3}/\Sn_2]$ is $B(\mathbb Z/2)$, whose Picard group is $\mathbb Z/2$, as the theorem says. The same method computes $\Pic[\M/\Sn_n]$. Let $B_n$ be the space of binary forms of degree $n$, and let $H = (\mathrm{GL}_2 \times \Gm)/\{(t, t^n)\}$ act on it by $(g, s)\cdot f = s\,(f \circ g^{-1})$; then $[\M/\Sn_n] \cong [B_n^{\mathrm{sm}}/H]$. The characters of $H$ are the $\det^a s^b$ with $2a + nb = 0$, a group isomorphic to $\mathbb Z$. The discriminant, irreducible of degree $2(n-1)$ \cite[\S5]{arsievistoli2004}, has character $\det^{-n(n-1)} s^{2(n-1)}$, which is $2(n-1)$ times a generator when $n$ is even and $n-1$ times a generator when $n$ is odd. The argument of Lemma~\ref{lem:pic} then shows that $\Pic[\M/\Sn_n]$ is cyclic of order $2(n-1)$ for even $n$, as in \cite[Th.~3.6]{gorchinskiyviviani2008}, and of order $n-1$ for odd $n$. By Theorem~\ref{thm:main}(2), $[L]$ generates this group when $n$ is odd, and generates its subgroup of index~$2$ when $n$ is even.
\end{remark}

\section{An elementary version}

The argument can be phrased without stacks, in the vector model used above. In the same way, a trivialisation of $L_n^{\otimes e}$ compatible with $\Sn_{n-1}$ is a nowhere-vanishing function $F(v_1, \dots, v_n)$ which is homogeneous of degree $2e$ in $v_n$ and of degree $0$ in the other $v_i$, transforms by $\det(g)^e$ under $\mathrm{GL}_2$, and is symmetric in $v_1, \dots, v_{n-1}$.

\emph{Upper bound.} The function
\[
F = \Bigl(\prod_{i<n}[v_iv_n]\Bigr)^{2(n-2)} \Bigm/ \prod_{\substack{i \ne j\\ i,j<n}}[v_iv_j]
\]
has these properties for $e = (n-1)(n-2)$.

\emph{Lower bound.} A trivialisation over $k$ gives one over an algebraic closure, so we may assume that $k$ contains the roots of unity used below. Let $\omega$ be a primitive $(n-1)$-th root of unity, and consider the configuration with $x_n = 0$ and $x_1, \dots, x_{n-1}$ the $(n-1)$-th roots of unity. Write $0 = \langle(0,1)\rangle$ and the other points as $\langle(z,1)\rangle$. The matrix $g = \mathrm{diag}(\omega, 1)$ fixes $(0,1)$ and permutes the vectors $(z,1)$, so $F(gv) = F(v)$ by symmetry, while $F(gv) = \det(g)^e F(v) = \omega^e F(v)$. Hence $\omega^e = 1$. With $\zeta$ a primitive $(n-2)$-th root of unity and the configuration formed by $x_n = 0$, $\infty$ and the $(n-2)$-th roots of unity, the same argument with $\mathrm{diag}(\zeta, 1)$ gives $\zeta^e = 1$. This matrix also multiplies the vector $(1,0)$ of $\infty$ by $\zeta$, which $F$ does not see, since it has degree $0$ in that variable. As $n - 1$ and $n - 2$ are coprime, $(n-1)(n-2)$ divides $e$ (Figure~\ref{fig:rot}).

\begin{figure}[ht]
\centering
\begin{tikzpicture}[scale=1.4, >=stealth]
\begin{scope}
\draw[black!35] (0,0) circle (1);
\foreach \k in {0,...,4} { \fill (\k*72:1) circle (1.6pt); }
\node[right=1pt] at (0:1) {\footnotesize $1$};
\fill (0,0) circle (1.8pt); \node[below left=1pt] at (0,0) {\small $0$};
\draw[->,thick] (0,0) -- (0:0.62) node[pos=0.75, below=1pt] {\footnotesize $v$};
\draw[->,thick,dashed] (0,0) -- (72:0.62) node[pos=0.8, left=2pt] {\footnotesize $\omega v$};
\draw[->,black!60] (12:0.36) arc (12:60:0.36);
\node[align=center] at (0,-1.45) {\small $x_6 = 0$, $x_1, \dots, x_5 \in \mu_5$\\ \small rotated by $z \mapsto \omega z$};
\end{scope}
\begin{scope}[xshift=4.2cm]
\draw[black!35] (0,0) circle (1);
\foreach \k in {0,...,3} { \fill (\k*90:1) circle (1.6pt); }
\node[right=1pt] at (0:1) {\footnotesize $1$};
\fill (0,0) circle (1.8pt); \node[below left=1pt] at (0,0) {\small $0$};
\draw[->,thick] (0,0) -- (45:0.62) node[pos=0.8, right=2pt] {\footnotesize $v$};
\draw[->,thick,dashed] (0,0) -- (135:0.62) node[pos=0.8, left=2pt] {\footnotesize $\zeta v$};
\draw[->,black!60] (57:0.36) arc (57:123:0.36);
\fill (1.3,0.9) circle (1.6pt); \node[right=1pt] at (1.3,0.9) {\footnotesize $\infty$};
\node[align=center] at (0,-1.45) {\small $x_6 = 0$, $x_1 = \infty$, $x_2, \dots, x_5 \in \mu_4$\\ \small rotated by $z \mapsto \zeta z$};
\end{scope}
\end{tikzpicture}
\caption{The two configurations of the lower bound for $n = 6$. The last point is $0$. A rotation permutes the other points and turns the tangent vector $v$ at $0$ into $\omega v$ or $\zeta v$, with $\omega = e^{2\pi i/5}$ and $\zeta = i$. A trivialisation compatible with the renumbering must take the same value at both, so $5 \mid e$ and $4 \mid e$, hence $20 \mid e$ for $L_6$.}\label{fig:rot}
\end{figure}

Both directions, for every $n$, and the corresponding statements for $L$, are formalised in Lean~4 with Mathlib, over any field containing the relevant roots of unity, with $F$ allowed to be any function on the points over that field (theorems \texttt{trivL\_iff} and \texttt{trivLi\_iff} of \texttt{Grothendieck.Folder146}, in the repositories \href{https://github.com/Commutator-IO/grothendieck-project}{Commutator-IO/grothendieck-project} and \href{https://github.com/Commutator-IO/grothendieck-project-palomar}{Commutator-IO/grothendieck-project-palomar}; they belong to the Lean appendix of \cite{hua2611}). The dictionary between equivariant line bundles on $\M$ and such functions is not formalised, that is, the identification of the equivariant Picard group with the first cohomology of $\Sn$ with values in the units; Mathlib does not have the spaces $\M$.

\section*{Statement on the use of artificial intelligence}
The transcription of folder~146, the search of the literature, the general formula, its proofs, the Lean formalisation and a first draft of this text were produced with language models (Claude Opus~5 and Opus~5.5, Anthropic), under the author's direction. The author checked the proofs and the references, and takes full responsibility for the content. The formal proofs are checked mechanically by Lean.

\section*{Acknowledgements}
The author thanks Dan Edidin for his answer about the odd case. The facsimiles are published by the University of Montpellier, \href{https://grothendieck.umontpellier.fr/}{grothendieck.umontpellier.fr}.

\begin{thebibliography}{9}
\bibitem{arsievistoli2004} A.~Arsie, A.~Vistoli, Stacks of cyclic covers of projective spaces, \emph{Compos.\ Math.} 140 (2004), no.~3, 647--666, \href{https://doi.org/10.1112/S0010437X03000253}{doi:10.1112/S0010437X03000253}.
\bibitem{edidin2026} D.~Edidin, personal communication, e-mail of 28 October 2026.
\bibitem{edidinhu2025} D.~Edidin, Z.~Hu, The integral Chow rings of the stacks of hyperelliptic Weierstrass points, \emph{Michigan Math.\ J.} 76 (2026), no.~5, 1091--1113, \href{https://doi.org/10.1307/mmj/20236479}{doi:10.1307/mmj/20236479}, arXiv:2208.00556.
\bibitem{fultonolsson2010} W.~Fulton, M.~Olsson, The Picard group of $\mathcal M_{1,1}$, \emph{Algebra Number Theory} 4 (2010), no.~1, 87--104, \href{https://doi.org/10.2140/ant.2010.4.87}{doi:10.2140/ant.2010.4.87}.
\bibitem{gorchinskiyviviani2008} S.~Gorchinskiy, F.~Viviani, Picard group of moduli of hyperelliptic curves, \emph{Math.\ Z.} 258 (2008), no.~2, 319--331, \href{https://doi.org/10.1007/s00209-007-0173-9}{doi:10.1007/s00209-007-0173-9}.
\bibitem{hua2611} M.~Hua, Grothendieck Montpellier. Énoncés des lectures modernisées démontrés, Hexagon preprint \href{https://hexagonmath.org/2611.00125v1}{hexagon:2611.00125v1}, 2026.
\bibitem{mumford1965} D.~Mumford, Picard groups of moduli problems, in \emph{Arithmetical Algebraic Geometry} (Purdue, 1963), Harper \& Row, New York, 1965, 33--81.
\bibitem{pernice2022} M.~Pernice, The integral Chow ring of the stack of 1-pointed hyperelliptic curves, \emph{Int.\ Math.\ Res.\ Not.} 2022, no.~15, 11539--11574, \href{https://doi.org/10.1093/imrn/rnab072}{doi:10.1093/imrn/rnab072}.
\end{thebibliography}

\end{document}