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For a morphism of schemes $f:X\to S$ with $S=\text{Spec}(R)$ affine, let's write $A(X)=H^0(X,\mathcal{O}_X)$. I'm interested in the morphism of $R$-algebras $$ c:A(X)\otimes_R A(Y)\to A(X\times_SY) $$ for some $X,Y$ over $S$ which I don't want to assume affine or proper. With no affineness condition, the classical case where $c$ is shown to be an isomorphism is when $X\to S$ is flat and $A(Y)$ is flat (SGA3 I, Lemme 11.1).

I want to prove that $c$ is an isomorphism in the following case: $R$ is an artinian ring with algebraically closed residue field and $X$ and $Y$ are smooth, of finite type, connected if that helps, over $R$.

In fact in my intended application I have $X=Y=G$ a smooth connected group scheme. Over the residue field, by Chevalley we know $G$ is an extension of an abelian variety by a an affine group scheme, or by the 'dual' Chevalley also an extension of an affine by an anti-affine. But I'm not sure this is relevant in any way.

My attempts to prove this so far were as follows. If $X,Y$ are both affine, or both proper, then the claim is true. I tried dévissage, that is, $X\to S$ factors as an affine morphism $X\to Z$ followed by a proper morphism $Z\to S$ (for schemes, Temkin 2011; for group schemes this is Chevalley) but was unable to really go further.

Another approach I tried was to use the Künneth formula (say in the form stacks project Tag 0FLQ. Taking $H^0$ in the Künneth formula, in the RHS I get exactly $A(X\times_SY)$, while in the LHS the algebra $A(X)\otimes_R A(Y)$ should appear as a graded piece of the abutment of the second hypercohomology spectral sequence, if I'm right. The problem here is that I'm not sure how to write that spectral sequence, because instead of a functor, in this case I have a bifunctor (namely $\otimes$) and then I get lost.

Note that although I'd be quite surprised, it is still possible that my guess for the isomorphism is wrong, and I'd be happy (well... would I?) to be disproved.

Thank you for any help or suggestion!

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  • $\begingroup$ Do you know an example of $Y/S$ smooth and of finite type (with $S=\operatorname{Spec} R$ as in your question) such that $A(Y)$ is not flat over $R$? $\endgroup$
    – Piotr Achinger
    Commented Jan 21, 2020 at 9:14
  • $\begingroup$ @Piotr: Excellent question. I must think about it. $\endgroup$
    – Matthieu Romagny
    Commented Jan 21, 2020 at 11:50
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    $\begingroup$ @PiotrAchinger We probably can construct an example using deformation theory. Take $Y$ to be the deformation of $\mathbb{A}^2\setminus 0$ over $S=k[e]/e^2$ given by gluing the charts $k[x^{\pm 1},y][e]$ and $k[x,y^{\pm 1}][e]$ using the infinitesimal automorphism $id+e \frac{1}{xy}\frac{\partial}{\partial x}$ of the intersection $k[x^{\pm 1},y^{\pm 1}][e]$. Then no function of the form $x+e f,f\in k[x^{\pm},y]$ extends to a global function on $Y$ so $e x\in H^0(Y,\mathcal{O})$ is an elmement killed by $e$ that is not divisible by $e$, this module is thus not flat. $\endgroup$
    – SashaP
    Commented Jan 21, 2020 at 16:21
  • $\begingroup$ @SashaP: your example is nice, it's of course a very natural one. I'll try to compute the map $c$ in case $X=Y=$ your example, and see how it's like. Thanks! $\endgroup$
    – Matthieu Romagny
    Commented Jan 22, 2020 at 22:41

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