Let $A \subseteq B$ and $A \subseteq C$ be commutative noetherian domains. Assume that $A$ and $C$ are regular rings (=every localization at a maximal ideal is a regular local ring). Assume that $B$ is a free $A$-module of finite rank $r$, so $B \cong \oplus_{i=1}^{r} A$ as $A$-modules. Is it possible to decide whether the ring $C \otimes_A B$ is regular or not?
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3$\begingroup$ Take $A=C$ to be the ring of a smooth affine variety over some field. Then $C\otimes_A B=B$, and you're just asking for an irreducible affine finite flat cover of $\mathrm{Spec}(A)$ that is not regular, and there should be plenty of those. For example take $A=k[x]$ and $B=k[x,y]/(y^2-x^3+x^2)$. $\endgroup$– Mattia TalpoCommented May 10, 2015 at 17:47
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$\begingroup$ Thank you. Actually, I am interested in a specific case where $A \neq C$ (maybe I should have written it in first place). Anyway your example for a non-regular tensor is nice, though I am not sure I know why $B$ is a free $A$-module. $\endgroup$– user237522Commented May 10, 2015 at 18:31
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$\begingroup$ There are examples where the inclusion $A\subseteq C$ is strict, for example take $A=B=C=k[x]$ and both $A\subseteq B$ and $A\subseteq C$ to be $x\mapsto x^n$. Freeness of $B$ follows (for example) from the fact that $f_*\mathcal{O}_X$ is locally free of finite rank if $f\colon X\to Y$ is finite and flat, and that over $k[x]$ every locally free module of finite rank is free. Maybe you can write down the specific example you have in mind? $\endgroup$– Mattia TalpoCommented May 10, 2015 at 19:45
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$\begingroup$ Interesting, thanks! As for my specific example: $A=K[p,q] \subseteq K[x,y]$, where $K[x,y]$ is the polynomial ring in two indeterminates, and $p,q \in K[x,y]$ have invertible Jacobian (so $K[p,q] \cong K[x,y]$). $B=K[p,q][w] \subseteq K[x,y]$ and further assume that $w$ is integral over $K[p,q]$ (without assuming $w$ is integral over $K[p,q]$ we only know that it is algebraic over $K[p,q]$). $C=A_m/m_m$ with $m=A\cap M$ where $M$ is a maximal ideal of $B$. I have just noticed that in my question I should have demanded $A \to C$ and not $A \subseteq C$, since $A \subseteq A_m \to A_m/m_m$. $\endgroup$– user237522Commented May 10, 2015 at 22:08
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$\begingroup$ Since you answered my original question in two different counterexamples, you can write it as an answer, if you like. $\endgroup$– user237522Commented May 10, 2015 at 22:23
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2 Answers
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Alright, I'm posting this as an answer.
As in the comments above, there are counterexamples to your general question.
As for the specific situation, unless you want to assume something more on $p$, $q$ and $w$ I guess it's still false: assume $p=x$ and $q=y$, so that $A=k[x,y]$, and take $B=k[x,y,w]/(w^2-x)$, $M=(x,y,w)$, so $m=(x,y)$. So $C$ is the residue field of $(x,y)$, and the (spectrum of the) tensor product $B\otimes_A C$ is the fiber of the map $\mathrm{Spec}(B)\to \mathrm{Spec}(A)$ over $(x,y)$. You can easily see that the fiber is $\mathrm{Spec}(k[w]/(w^2))$, which is not regular.
Since you're basically looking at regularity of fibers, it seems that what you really want is smoothness of $\mathrm{Spec}(B)\to \mathrm{Spec}(A)$ to get your conclusion.
answered May 10, 2015 at 22:43
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Another well known example: it is clear that the tensor product of two field extensions need not be regular. So this answers your original question.
Regarding the one raised in the comments, if $A$ and $B\otimes_AC$ are regular, then $B$ is regular, so the first thing we have to check is if $B$ is regular. In that case, some smoothness conditions on any of the two algebras will to be sufficient, as indicated by Mattia Talpo. For the proofs and precise statements you can see for instance Majadas, On tensor products of complete intersections, Bull. London Math.Soc. and the references in that paper.
