Are abelian varieties unobstructed? That is, given an abelian scheme $X \to \mathrm{Spec}R $ for $R$ a local artinian ring and $\mathrm{Spec} R \to \mathrm{Spec} S $ a nilpotent thickening, can we extend $X$ to an abelian scheme over $\mathrm {Spec }S $?

This assertion was made somewhat quickly in an article of Katz I was trying to read, at the bottom of p. 144. Mumford shows in his GIT book that it's enough to deform $X$ together with the identity section $\mathrm{Spec} R \to \mathrm{Spec } X$, and the abelian structure follows. I don't see how to get that, though.

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    $\begingroup$ This follows from a theorem of Grothendieck. See, for example, Theorem 2.2.1 of the article by F. Oort "Finite group schemes, local moduli for abelian varieties ...", Compositio Math. 1971. $\endgroup$ – ulrich Jun 15 '12 at 3:38
  • $\begingroup$ One idea: use limit theorems and reduction mod primes to reduce to the case of char p residue field. Now the Abelian scheme is projective: for an ample on the closed fiber, a sufficiently high $\text{p}^{\text{th}}$ power will extend to your deformation over $R$. Take a general complete intersection curve $C$ of sufficiently ample divisors (on the dual) to embed into $\text{Jac}(C)$. Since $C$ deforms over $S$, also $\text{Jac}(C)$ deforms. Now you have the simpler problem of showing that for some deformation of $C$, also $X$ deforms as a subvariety of $\text{Jac}(C)$ (sheafy arguments). $\endgroup$ – Jason Starr Jun 15 '12 at 12:58
  • $\begingroup$ Why does a general curve which is a complete intersection of very ample divisors let you imbed an abelian variety into the Jacobian? $\endgroup$ – Akhil Mathew Jun 15 '12 at 14:42
  • $\begingroup$ @Akhil -- The question is whether the pullback map from Picard schemes of the (dual) Abelian variety to the Picard scheme of the curve is a closed immersion. First, the pullback map on $H^1(\mathcal{O})$ is injective using Serre vanishing, Serre duality and the usual (Koszul-type) resolution of the structure sheaf of a complete intersection subvariety. So the pullback map is unramified, hence has finite, 'etale kernel. Since elements in the kernel are torsion, vanishing of the kernel reduces to a problem about surjectivity of pushforward on $\pi_1$, which follows by Bertini's theorem. $\endgroup$ – Jason Starr Jun 15 '12 at 15:10
  • $\begingroup$ @Akhil -- Actually the last step about injectivity for restriction of torsion invertible sheaves, which is the same as surjectivity for the restriction map on global sections of the invertible sheaf, can also be proved by "Serre vanishing - Serre duality - resolution" computation (it does also follow by Bertini's connectedness theorem / lemma of Enriques-Severi-Zariski). $\endgroup$ – Jason Starr Jun 15 '12 at 16:26

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