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.

  • $\begingroup$ Much as I wanted an answer to the above question, I wonder in general if we can have A, B regular domains, $A\otimes B$ also a domain, but not regular? $\endgroup$
    – Ying Zhang
    Dec 21, 2010 at 17:27
  • $\begingroup$ related: mathoverflow.net/questions/2039/… $\endgroup$
    – J.C. Ottem
    Dec 21, 2010 at 19:49
  • 2
    $\begingroup$ Thanks for the link. I just noticed EGA IV comment 6.7.5 gave a counter example to my question in the comment. Consider $A=k[S,T]/(T^2-S^p-t)$, $B=\mathbb{F}_p(t^{1\over p})$, A is geometrically integral, regular, $A\otimes B$ is a domain but not normal. It is similiar in flavour to Taisong's example in the above link. However in both examples B is not geometrically integral, (it is a finite purely inseparable extensions of the base field.) EGA also has an interesting comment that it is related to the "genus drop" phenomenon. I wonder what can we say when both X and Y are absolutely integral? $\endgroup$
    – Ying Zhang
    Dec 21, 2010 at 20:29
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    $\begingroup$ @Ying Zhang: Just take $X=Y={\mathrm Spec}A$ with your $A$, then $X\times_k Y$ is integral, normal, with one singular point $(x_0, x_0)$, where $x_0$ corresponds to the maximal ideal of $A$ generated by $T$. $\endgroup$
    – Qing Liu
    Dec 21, 2010 at 21:48

3 Answers 3


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.


[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.

  • $\begingroup$ Thanks, but I have a confusion. Here by tensor you always mean over a fixed base ring C, right? I think we need some assumptions: I only know "regularity $\Leftrightarrow$ finiteness of $Ext$-dimension" when the ring is Noetherian. (Is it true for non Noetherian?) In general if the rings A, B and base ring C is Noetherian, $A\otimes_C B$ may not be Noetherian, see mathoverflow.net/questions/29764/pushouts-of-noetherian-rings $\endgroup$
    – Ying Zhang
    Dec 22, 2010 at 21:32
  • $\begingroup$ Now if we assume all the rings and tensor products involved are Noetherian, could you pls provide some more details for your argument? E.g. do you mean $Hom_{A\otimes_C B}(M,-)=Hom_{B\otimes_C B}(B, Hom_A(M,-))$ for any $A\otimes_C B$ module N? (This looks to me like an adjointness argument, but I don't know how to show it...) Besides, I guess there is a typo, first you used $Hom_A(M,-)$, then in the rightmost of the line below it is $B-Mod$, (well this line is not an exact sequence, is it?) If these however are standard facts provided somewhere, I would appreciate any reference. $\endgroup$
    – Ying Zhang
    Dec 22, 2010 at 21:43
  • 2
    $\begingroup$ For a $B\otimes B$-module $X$, a $B\otimes B$-linear map $B\to X$ corresponds precisely to an element $x\in X$ satisfying the condition that for every $b\in B$ we have $(b\otimes 1)x=(1\otimes b)x$. Taking $X$ to be the group of $A$-module maps $M\to N$ and defining the module structure by $((b_1\otimes b_2)f)(m)=b_1f(b_2m)$, we find that this condition becomes $f(bm)=bf(m)$, or in other words, $f$ is not just $A$-linear but $(A\otimes B)$-linear. $\endgroup$ Dec 23, 2010 at 4:15
  • $\begingroup$ @Tom Goodwillie: Thanks very much, it is much clearer now. I was still a little worried about "the derived functors of the composition is the composition of the derived functors". I was trying to show the finiteness directly. Suppose for an arbitrary $A\otimes B$ module M, find an injective resolution $I^{\cdot}$, which are also (at least acyclic) $A$ and $B\otimes B$ modules, could this always be achieved? Then, we have finite homological dimension when applying the functor $G$. $\endgroup$
    – Ying Zhang
    Dec 23, 2010 at 23:15
  • $\begingroup$ (continuation of above)Then, to show $G(I^{\cdot})$ is also an injective(acyclic?) sequence of $B\otimes B$ modules, I think we need sth like $M$ to be a flat $B\otimes B$ module, which is not always true, since we choose $M$ arbitrarily. Maybe there are some ways to get around with this, or that I was trying to show sth not necessary. Anyway, the core question can be stated very clearly: is the composition of two functors each of whose derived functor has finite homological dimension also the derived functor for itself with finite homological dimension? $\endgroup$
    – Ying Zhang
    Dec 23, 2010 at 23:22

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

  • $\begingroup$ @Ying Zhang: In general, the fiber product of normal varieties is not normal (consider $X=Y$ equal to the spectrum of a finite inseparable extension of $k$). On the other hand, my answer inspired by Tom Goodwillie's one shows that $X\times_k X$ is always singular when $X$ is not smooth. And the examples you consider are not smooth as you said. $\endgroup$
    – Qing Liu
    Dec 22, 2010 at 18:52
  • $\begingroup$ @Qing Liu: Thanks, I got the part where $X\times_k X$ is regular implies X is smooth. In general, if in addition to X and Y being normal, we require $X\times_k Y$ to be integral, then $X\times_k Y$ is normal. The reason is that its normalization is itself: any normal variety Z mapping to $X\times_k Y$ mapping to X and Y by composing with the naturual projections. Then they uniquely factor through the normalization of X and Y, (but the normalizations $\widetilde{X}=X$, $\widetilde{Y}=Y$), so the map $Z\rightarrow X\times_k Y$ uniquely factors through $\widetilde{X}\times_k \widetilde{Y}$. $\endgroup$
    – Ying Zhang
    Dec 22, 2010 at 20:11
  • $\begingroup$ (continuation of above)Therefore $\widetilde{X\times_k Y}=\widetilde{X}\times_k\widetilde{Y}=X\times_k Y.$ $\endgroup$
    – Ying Zhang
    Dec 22, 2010 at 20:13
  • $\begingroup$ Sorry Ying Zhang, I don't understand your proof. You showed that any morphism from a normal variety $Z$ to $X\times_k Y$ factors through $X\times_k Y$, but this is just tautological. You can prove the normality with Serre's criterion R$_1$ and S$_2$. $\endgroup$
    – Qing Liu
    Dec 24, 2010 at 15:52
  • $\begingroup$ @Qing Liu: You are right, I did miss sth in my proof. Afterall, suppose a corrected proof of "fiber product over k of normal varieties is still normal (if integral)", then I think my argument just shows $\widetilde{X\times_k Y}=\widetilde{X}\times_k\widetilde{Y}$. $\endgroup$
    – Ying Zhang
    Dec 25, 2010 at 1:11

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