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Let $f\colon A \rightarrow S$ be an abelian scheme of dimension $d$. I would like a reference or an argument for the fact that $R^1f_* \mathcal{O}_A$ is locally free of dimension $d$ and that its formation commutes with arbitrary base change. Using usual limit arguments one may reduce to the case when $S$ is the spectrum of a Noetherian ring. I would be especially happy if the answer could mention whether similar local freeness and cohomological flatness claims hold true for $R^if_* \mathcal{O}_A$ for arbitrary $i \ge 0$.

I think that the key input should be a theorem of Grauert reducing the claim to the fibers, but I couldn't isolate a precise reference that would apply in the desired generality. I know that there is a slew of related results in EGA III, section 7 but I have not studied them in detail, so due to heavy notation used there it is difficult for me to gauge their applicability. There are also improvements to this part of EGA in SGA 6, Expose III, but those are mostly formulated in terms of total derived functors, so I couldn't see how to extract information about a single $R^i f_*$.

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    $\begingroup$ Use cohomology and base change + backwards induction on $i$. The point is that $R^df_*\mathcal{O}_A$ is a vector bundle, so cohomology commutes with base change for $R^{d-1}f_*\mathcal{O}_A$, which is thus a vector bundle (as the rank is constant on fibers), etc... $\endgroup$ Commented Dec 17, 2014 at 22:35
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    $\begingroup$ Berthelot, Breen and Messing, Théorie de Dieudonné cristalline, II (LNM 930), Prop. 2.5.2 compute the De Rham cohomology of an abelian scheme - the proof uses a bit of spectral sequences. $\endgroup$
    – ACL
    Commented Dec 17, 2014 at 23:17
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    $\begingroup$ @DanielLitt: Your argument with fiber-ranks is insufficient, since it doesn't say anything when the base is artinian. You may have had in mind the case of a reduced base, for which Grauert's result directly gives the answer since we know the fiber-ranks. $\endgroup$
    – user74230
    Commented Dec 17, 2014 at 23:44
  • $\begingroup$ Thank you both. @DanielLitt: Could you provide a reference which "cohomology and base change" statement I should be using? Some of them have assumptions that the base be reduced (in addition to Noetherian), which is troubling me. $\endgroup$
    – Lisa S.
    Commented Dec 17, 2014 at 23:44
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    $\begingroup$ @DanielLitt: OK, I was being dumb, since of course ${\rm{R}}^pf_{\ast}(\Omega^q)$ includes the case of interest as $q=0$; sorry about that! So Deligne's Hodge degeneration paper takes care of it in char. 0 in general (any base, reduced or not). $\endgroup$
    – user74230
    Commented Dec 18, 2014 at 6:02

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It suffices to prove the vector bundle property for all cohomological degrees when the base is an artin local ring $R$ (by considerations with direct limits to pass to the noetherian case and then using the standard base change formalism). We can make a flat local extension $R \rightarrow R'$ to an artin local ring with algebraically closed residue field, so we may assume the residue field $k$ of $R$ is algebraically closed. If $k$ has characteristic 0 then we can make $R$ into a $k$-algebra and pass to the case $k = \mathbf{C}$, so the self-contained slick argument of Deligne from his Hodge degeneration paper can be applied to conclude the vector bundle property (as $\Omega^1$ is globally free for an abelian scheme over a local ring).

If $k$ has characteristic $p > 0$ then by picking an ample line bundle on the special fiber and raising it to a sufficiently high $p$-power (depending on the nilpotence order of the maximal ideal of $R$) we can lift it to the abelian $R$-scheme so as to make a polarization over $R$. (This trick doesn't work for residue characteristic 0.) It is harmless to pass to $(A \times A^{\vee})^4$ (as a direct summand of a finite free $R$-module is finite free), so by Zarhin's trick we can assume $A$ is principally polarized over $R$. But as explained in Theorem 2.4.1 of Oort's 1970 article "Finite group schemes, local moduli for abelian varieties, and lifting problems", the formal deformation ring of an abelian variety equipped with a polarization of degree prime to the characteristic is formally smooth. (This is a theorem of Grothendieck.) Hence, by formal GAGA we get a lift over a formal power series ring over $W(k)$. Now the base is reduced, so we can use Grauert's theorem to conclude.

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    $\begingroup$ Woah, surprisingly tricky to get it for non-reduced bases. Perhaps there's a simpler argument using Hodge-to-de Rham degeneration? (But I only know how to prove degeneration for Abelian schemes over Artin bases via lifting, which requires a similar argument to the one you give.) $\endgroup$ Commented Dec 18, 2014 at 0:06
  • $\begingroup$ @DanielLitt: Oda published a proof of degeneration of Hodge-to-dR for abelian varieties over fields via induction on cohomological degree with a spectral sequence, but in that proof doesn't seem to address the base case of the induction (degree 1) which can however be handled by getting the dimension of degree-1 dR cohomology via Oda's comparison with Dieudonne theory for $A[p]$ in his (published) thesis. The Deligne-Illusie lifting method is perhaps the simplest way to go (avoids algebraization of formal deformations). $\endgroup$
    – user74230
    Commented Dec 18, 2014 at 1:58
  • $\begingroup$ @user74230: Thank you very much! In the last sentence of the first paragraph, how do you pass to the case $k = \mathbb{C}$? Even though the base is Artinian, $A$ need not be "constant" (i.e., come from $k$); e.g., how do you handle elliptic schemes whose $j$-invariant is not in your chosen lift of $k$? Would you perhaps instead need to make a lifting to a reduced from an Artinian base again (after reducing to the principally polarized case via Zarhin's trick)? Also, for the direct summand reduction, is there some Kunneth under the hood? $\endgroup$
    – Lisa S.
    Commented Dec 18, 2014 at 4:46
  • $\begingroup$ @user74230: also, could you elaborate on how to lift the polarization (to get Zarhin going)? I realize that in my previous comment I overlooked that a polarization over the Artinian base is necessary. $\endgroup$
    – Lisa S.
    Commented Dec 18, 2014 at 4:52
  • $\begingroup$ @LisaS.: Talk in person with local experts in algebraic geometry to learn these (standard) methods; it is always better to talk with someone in person when possible. Every equi-characteristic complete local noetherian ring (e.g., equi-char. artin local) admits a structure of algebra over its residue field -- this is part of the Cohen Structure Theorem (but equi-char. 0 is an exercise: lift transcendence basis, use separability, etc.). Don't think about $j$-invariants; that's a red herring. Kunneth has nothing to do with these things either. Treat polarization lift as an exercise for yourself. $\endgroup$
    – user74230
    Commented Dec 18, 2014 at 5:07

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