Is the tensor product of regular rings still regular

Question

An imprecise version of the question is that when $A$ and $B$ are regular rings, is $A \otimes B$ also regular? Please allow me to put more restrictions, here I am only interested in the case when $A$ and $B$ are finitely generated $k$-algebras. When $k$ is perfect, the answer is yes, see http://arxiv.org/abs/math/0210359. In fact, we can view $A$ and $B$ as coordinate rings of some affine $k$-varieties say $X$ and $Y$ respectively. Since $k$ is perfect, regularity is equivalent to being smooth and it can be shown easily $X \times_k Y$ is smooth. Hence the conclusion. Now when $k$ is not perfect, and assume in addition $X$ and $Y$ are both geometrically (absolutely) integral, moreover they contain some (regular but) not smooth point (so the above method doesn't apply), then is $X \times_k Y$ still regular?

Example, $k=\mathbb{F}_p(t)$, $p>2$, $A=k[x,y]/(x^p-x^{p-1}y-t)$, $X=\operatorname{Spec} A$. Then it is easy to show $X$ is geometrically integral, the maximal ideal generated by $(Y)$ in $A$ is regular but not smooth, and that $A$ is regular. The question is that is $X\times_k X$ regular?

Note, it is easy to produce a counter example when $X$ is not assumed to be geometrically integral, e.g. $A=k[x]/(x^p-t)$, then it is easy to show $X\times_k X$ is a $0$-dimension local ring but not a domain, therefore it can not be regular.

Answer 1

This is inspired by Tom Goodwillie's answer.

Let $X$ be an algebraic variety over a field $k$ (i.e. a scheme of finite type over $k$). If $X\times_k X$ is regular, then $X$ is smooth over $k$.

Proof. The first projection $X\times_k X\to X$ is faithfully flat, so the regularity of $X\times_k X$ implies that of $X$. To prove the smoothness of $X$, we can suppose $X$ is connected and affine. The diagonal morphism $\Delta: X\to X\times_k X$ is then a closed immersion from a regular scheme to a regular scheme. Therefore $\Delta$ is locally complete intersection. If $J$ is the ideal sheaf on $X\times_k X$ defining $\Delta(X)$, then $\Delta^*(J/J^2)$ is locally free of rank the codimension of $X$ in $X\times_k X$ which is equal to $\dim X$.

Now $\Delta^*(J/J^2)$ is isomorphic to the sheaf of differential forms $\Omega^1_{X/k}$ on $X$ (see Hartshorne), so the latter is locally free of rank $\dim X$. This implies that $X$ is smooth.

Answer 2

[EDIT: As shown in Ulrich's answer to this question, there must be some error here.]

It seems to me that if $A$ and $B\otimes B$ are regular then $A\otimes B$ is regular. Does this argument work? Use the fact that regularity is equivalent to finiteness of $Ext$-dimension. Fix an $(A\otimes B)$-module $M$ and factor the functor $Hom^{A\otimes B}(M,-)$ as $Hom^A(M,-)$ followed by $Hom^{B\otimes B}(B,-)$ $$ (A\otimes B)-Mod\ \to (B\otimes B)-Mod\to B-Mod. $$ This doesn't even seem to use that the ground ring is a field.

EDIT In more detail: Let $k$ be a commutative ring, and let $\otimes$ without a subscript mean $\otimes_k$. Let $A$ and $B$ be $k$-algebras. Fix an $(A\otimes B)$-module $M$.

There is the functor $F:X\mapsto Hom^{A\otimes B}(M,X)$ from $(A\otimes B)$-modules to $(A\otimes B)$-modules. This may be factored as the composition $H\circ G$ of two functors. The first is $G:X\mapsto Hom^A(M,X)$ from $(A\otimes B)$-modules to $(A\otimes B\otimes B)$-modules. Here $Hom^A(M,X)$ has two $B$-module structures: one from $M$ and one from $X$. The second is $H:Y\mapsto Hom^{B\otimes B}(B,Y)$ from $(A\otimes B\otimes B)$-modules to $(A\otimes B)$-modules.

All of these functors can be extended levelwise from modules ("discrete modules") to chain complexes of modules ("modules"), and then replaced by their left derived functors. I want to say that $LF=LH\circ LG$, the derived functor of the composition is the composition of the derived functors, but I haven't thought this through. If it's true, then the rest of the argument goes like this: If $A$ is regular then there is some $m\ge 0$ such that if $X$ is a discrete module then $LH(X)$ has its homology groups concentrated in dimensions $0$ through $-m$. If $B\otimes B$ is regular then there is some $n\ge 0$ such that if $Y$ is a discrete module then $LG(Y)$ has its homology groups concentrated in dimensions $0$ through $-n$, and if $Y$ is concentrated in $0$ through $-m$ then $LG(Y)$ is concentrated in $0$ through $-n$. So $LF(X)$ is concentrated in $0$ to $-m-n$ if $ X$ is discrete. So $A\otimes B$ is regular.

Answer 3

I will try to sketch something I know or collected from elsewhere. We know "regular$\Rightarrow$complete intersection$\Rightarrow$Gorestein$\Rightarrow$Cohen Macaulay". In thm 2 link text¹ it is shown when A and B are E algebras, A is flat over E, B is finitely generated over E, then:

1. If A is a complete intersection, B and E are regular, then $A\otimes_E B$ is a complete intersection.

2. If A, B, E are Gorestein, then $A\otimes_E B$ is Gorestein.

3. It is proved in EGA when A, B, (also E, but this may be superfluous) are Cohen Macaulay then so is $A\otimes_E B$.

When E is a field k, B is finitely generated, the above hypothesis is satisfied, therefore our question is "almost correct". Now in my old notation as in the original question, suppose X and Y are geometrically integral, therefore the function field of Y $K_Y=Frac(B)$ is linearly disjoint from the perfect closure of $k$, i.e. $k^{1\over {p^{\infty}}}$. Now by analogy with the "genus drop" phenomenon, see e.g.link text², I suspect (caveat!) $A\otimes_k Frac(B)$ is regular, in particular normal. Similarly $Frac(A)\otimes_k B$ is also regular, now $A\otimes_k B=A\otimes_k Frac(B)\cap Frac(A)\otimes_k B$ is normal.(Edit: fiber product of normal varieties is normal just by universal property, this part is superfluous) However we will show it is not regular.

Take the example as suggest by Qing Liu's comment, i.e. $A=B=k[S,T]/(T^2-S^p-t)$, $X=Spec A$, Y=$Spec B$, $x_0\in X$ is the maximal ideal generated by $(T)$, then there is a (in this case unique) maximal ideal in $A\otimes_k B$ containing $(T)\otimes_k B$ and $A\otimes_k (T)$, call it the point $(x_0, x_0)$.

(Note in general given $x\in X$, $y\in Y$, I suspect we do not get the point $(x,y)\in X\times_k Y$ for free by the universal property of fiber product. The reason is that a "closed point" in a scheme corresponds to a morphism from a field to the scheme, rather than the other way around.) The residue field at $(x,y)$ will be the composite field (ambiguity of Galois conjugation) of the corresponding residue field $k_{X,x}$ and $k_{Y,y}$, which may not be linearly disjoint even when $Frac(A)$ and $Frac(B)$ are.

Localize $A\otimes_k B$ at $(x_0, x_0)$, call it C with maximal ideal $m$. We need a lemma from EGA IV 17.1.8:

Lemma: Suppose C is Noetherian local ring, with maximal ideal $m$, $t\in m$, the following is equivalent:

1. C/tC is regular, and t is not a zero divisor of C;
2. C is regular, $t\notin m^2$.

Take t to be $1\otimes_k T\in m\setminus m^2$, then $C/tC=$ some localization of $A\otimes \mathbb{F}_p(t^{1\over p})$ which is not regular at the maximal ideal $m/tm$. Q.E.D

However, I don't know if there is a more direct way to see why $(x_0, x_0)$ is not a regular point of $X\times_k Y$, I would appreciate any comment.

¹ Kei-ichi Watanabe. Takeshi Ishikawa. Sadao Tachibana. Kayo Otsuka. "On tensor products of Gorenstein rings." J. Math. Kyoto Univ. 9 (3) 413 - 423, 1969. https://doi.org/10.1215/kjm/1250523903
²John Tate: Genus Change in Inseparable Extensions of Function Fields. Proceedings of the American Mathematical Society, Vol. 3, No. 3 (Jun., 1952), pp. 400-40