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Let $X_{K}$ be a curve over a complete DVR $R$, $R/m:=k$ an algebraically closed field. We suppose the minimal field extension $L$ of $K$ such that $X_{L}$ has stable model $X_{R_{L}}$, and the special fiber is $X_{k}$. We obtain an action of $Gal(L/K)$ on $X_{k}$.

My question is:

Does $Gal(L/K) \longrightarrow Aut(X_{k})$ is an injection ?

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up vote 4 down vote accepted

This appears in the literature, see e.g. Theorem 2.2 in Lehr and Matignon, Wild monodromy and automorphisms of curves, Duke Math. 2006. It is attributed there to Deligne and Mumford.

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@Matthieu: The reference to DM within that paper is for the existence of a stable model over some finite extension of the ground field. No doubt DM knew how to deduce the injectivity from what they'd done, but it isn't in the DM paper, right? ([DM, 1.13] is suggestive, but not quite enough, as one does have to bring in the orthogonality theorem and the inertial unipotence criterion more directly, it seems.) Does the other reference (called [D]) in the Lehr-Matignon paper give a statement and proof of the injectivity? Just curious if the proof is in the literature. – user30379 Jan 27 '13 at 20:25
Pranavk, you're right that the proof is not in Deligne-Mumford and I don't know if it appears in the literature. Your answer is certainly very useful and I just wanted to point out that the statement appears. – Matthieu Romagny Jan 28 '13 at 20:20
It is proved with full details in Mireille Deschamps's talk in Asteristisque vol. 86 (I guess this is [D] in Green-Matignon). – Qing Liu Feb 15 '13 at 17:57

The answer is affirmative when $X_K$ is smooth and geometrically connected with genus $\ge 2$ (which I am guessing are implicit hypotheses), without restriction on the residual characteristic nor on the generic characteristic. The key ingredients are theorems of Grothendieck concerning semistable reduction and the link between stable reduction of curves and semistable reduction of Jacobians (proved by Deligne and Mumford).

By increasing $K$ if necessary, the "minimality" hypothesis on $L$ reduces the problem to showing that if the ${\rm{Gal}}(L/K)$-action on the special fiber $X_k$ of $X_{R_L}$ is trivial then $X$ has stable reduction over $R$. By Deligne-Mumford, it is equivalent to deduce that the Jacobian $J$ of $X_K$ has semistable reduction. By Grothendieck's inertial criterion for semistable reduction, this in turn is equivalent to deducing that the action of $G_K = {\rm{Gal}}(K_s/K)$ on $V_{\ell}(J)$ is unipotent for some (equivalently, any) prime $\ell \ne {\rm{char}}(k)$.

By a theorem of Raynaud, $P := {\rm{Pic}}^0_{X_{R_L}/R_L}$ is a scheme, and more specifically is semi-abelian with generic fiber $J_L$. The subgroup $G_L = {\rm{Gal}}(K_s/L)$ of $G_K$ acts unipotently on $V_{\ell}(J)$, and Grothendieck's orthogonality theorem (applied over $R_L$) gives much more: $V_{\ell}(J)^{G_L} = V_{\ell}(P_k)$ with $V_{\ell}(J)/V_{\ell}(J)^{G_L}$ canonically dual to $V_{\ell}(T)$ where $T$ is the maximal torus of $P_k$ (here using the auto-duality of $J$ over $K$). This "canonicity" includes $G_K$-equivariance, so to prove the unipotence of the $G_K$-action on $V_{\ell}(J)$ it suffices to prove the triviality of the ${\rm{Gal}}(L/K)$-action on $V_{\ell}(P_k)$. But this latter Galois action is just the composition of functoriality of $V_{\ell}$ applied to the ${\rm{Gal}}(L/K)$-action on $P_k = {\rm{Pic}}^0_{X_k/k}$ arising from the ${\rm{Gal}}(L/K)$-action on $X_k$. (Various implicit compatibility verifications are left as an exercise.) This latter action is assumed to be trivial, so we are done.

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This is correct in characteristic $0$. Let $G$ be the kernel of the action; then one has to show that $X_{R_L}/G$ is a stable curve over $R_L^G$. This is Lemma 3.2 in my paper with Dan Abramovich Complete moduli for fibered surfaces,

I am pretty sure that it is false in positive characteristic, though.

[Edit:] I guess I was wrong about positive characteristic.

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