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This is lemma 1.15. of Deligne, Mumford's paper. Let $X$ be an irreducible stable curve over an algebraically closed field, $\phi$ an automorphism on $X$ which induces the identity on $\text{Pic}^0X$. Then is $\phi$ the identity on $X$?

Let $X'$ be the normalization and $\phi'$ the action of $\phi$ on $X'$. (induced by the universal property of the normalization) The author says that choosing a singular point of $X$ and an prdering of its inverse image in $X'$ defines a distinct morphism from $\mathbb{G}_m$ to $\text{Pic}^0X$, and therefore the inverse image $S$ of the singular locus of $X$ is pointwise fixed by $\phi'$. But I can't understand this at all. (And I don't know what $\mathbb{G}_m$ is.)

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  • $\begingroup$ @abx Is this a group scheme? If so, I know what it is, but I have no idea about relation between $\text{Pic}$ and it. And I think that this is not the group scheme because in lemma 1.16. the author says an abstract group is isomorphic to $H^1 \otimes \mathbb{G}_m$. ($H^1$ is the cohomology group of a graph.) $\endgroup$ – k.j. Mar 31 '19 at 13:00
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    $\begingroup$ Of course $\Bbb{G}_m$ is the group scheme, this is a universally adopted notation. See my answer below. $\endgroup$ – abx Mar 31 '19 at 13:18
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Let $s$ be a double point of $X$, and let $p,q$ be the two points of $\pi ^{-1}(s)$ (ordering chosen). Let $\pi :X_s\rightarrow X$ be the partial normalization of $X$ at $s$. There is an exact sequence of sheaves $$1\rightarrow \mathcal{O}_{X}^*\rightarrow \pi _*\mathcal{O}_{X_s}^*\xrightarrow{\ \varphi \ } \kappa (s)^*\rightarrow 1$$here $\kappa (s)^*$ is the skyscrapper sheaf over $s$ with stalk $k^*$, and $\varphi $ maps a function $f$ to $f(p)/f(q)$. The coboundary of the associated long exact sequence gives an injective homomorphism of algebraic groups $h_s:\mathbb{G}_m\rightarrow \operatorname{Pic}^{\mathrm{o}}(X) $. If $\operatorname{Sing}(X)=\Sigma $, the homomorphisms $h_s \ (s \in \Sigma )$ define an injective homomorphism $\mathbb{G}_m^{\Sigma }\rightarrow \operatorname{Pic}^{\mathrm{o}}(X) $; if $\phi$ induces the identity on $\operatorname{Pic}^{\mathrm{o}}(X) $, it must therefore preserve $\Sigma $ pointwise, and also $S=\pi ^{-1}(\Sigma )$ because exchanging $p$ and $q$ exchanges $h$ and $h^{-1}$.

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  • $\begingroup$ Thank you very much, this is so helpful! $\endgroup$ – k.j. Mar 31 '19 at 16:56
  • $\begingroup$ You are welcome. $\endgroup$ – abx Mar 31 '19 at 18:13

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