Diagonal morphism of regular variety is a regular embedding

Question

Let $X$ be a regular $k-$variety (i.e. all of its local rings are regular) of pure dimension $d$. Then I would like to show that the diagonal morphism $X\rightarrow X\times_k X$ is a regular embedding of codimension $d$, to be able to prove exercise $12.2.M$ in Ravi Vakil's "Foundations of Algebraic Geometry".

The preceding exercise shows that any closed embedding $Y \rightarrow Z$ with $Y$ regular and $Z$ regular at all points in the image of $Y$ is a regular embedding. I'd like to use this (particularly because Vakil lists $12.2.M$ as a consequence of the preceding exercise), only I'm having trouble proving that $X \times_k X$ is regular at points on the diagonal. Thus what I would really like to know is:

Let $X$ be a regular $k-$variety. Then $X \times_k X$ is regular at points on the diagonal.

Since $X$ is finite type, I can explicitly describe the local rings of $X$ as quotients of local rings of affine space, and use the kernels involved to explicitly describe the local rings of $X \times_k X$ as quotients of the local rings of $X$, but I can't then do very much with this, and I'm not sure that this is even the right approach.

Edit: Some of the answers suggest that regularity is not enough to show that the diagonal is a regular embedding. One provides an example of a regular scheme $W$ where $W \times W$ isn't regular at the diagonal, so the approach taken in this question is certainly flawed.

Answer 1

Without some hypothesis on $k$ I don't think it's true that the diagonal embedding $X \to X \times_k X$ is regular. Like take everybody's favorite regular but non-smooth scheme over the non-perfect field $k:= \Bbb{F}_p(t)$. Namely, the scheme $\operatorname{Spec} \Bbb{F}_p(t)[x]/(x^p - t)$. We first compute the tensor product

$$\begin{eqnarray*} \Bbb{F}_p(t)[x]/(x^p - t) \otimes_{\Bbb{F}_p(t)} \Bbb{F}_p(t)[y]/(y^p - t) &\cong& \Bbb{F}_p(t)[x,y]/(x^p - t, y^p - t) \\ &\cong& L[y]/(y-\alpha)^p \end{eqnarray*}$$ where $L := \Bbb{F}_p(t)[x]/(x^p-t)$ and $\alpha$ is the image of $x$ in $L$.

We see that the tensor product is a local non-reduced of dimension $0$. Hence the diagonal $X \to X \times_k X$ is a map from a point to a fat point and so is not regular.

Edit: Ravi told me via email that there is indeed an error in the notes with regards to this question.

Answer 2

If $X$ is a flat locally noetherian scheme over $S$ (and in particular in your case), it is (formally) smooth over $S$ if and only if the diagonal is a regular embedding. So you are trying to prove that your variety is smooth over $k$. This is true for instance if $k$ is algebraically closed or of characteristic zero, but not in general.

For a proof of the first claim, see Bourbaki, Algèbre Commutative chap. 10, § 7 n. 10 Théorème 4, (i) ⟺ (v). As in most of this section of that book, there are some finiteness hypotheses on the morphisms, but they are not necessary if instead of "smoothness" you want to obtain "formal smoothness for the discrete (and so for any other) topology" (though the proof without them is more dificult). If you are interested in the proof in general I can sketch one here, but using deep homological methods.

For "regular does not imply smooth" you have examples in https://en.wikipedia.org/wiki/Geometrically_regular_ring

Answer 3

Sadly I cannot comment. In any case:

(1) make clear what $k$ is (any field?) It depends greatly on the properties of $k$, as was already mentioned by A.G.

(2) unlike what some people think, there is no standard definition of "variety". E.g. everyone assumes *of finite type" (I think), but some assume irreducible, geometrically integral, whatever. For that reason you should always state what you mean by it.

(3) Over any field "smoothness" is equivalent to geometric regularity (see EGA IV), and thus "smoothness" is equivalent to regularity over a perfect field.

(4) The notion of smoothness is superior to regularity, exactly because it behaves very well with respect to base change. Over nonperfect fields I think regularity is less interesting.

Answer 4

This is an answer to a question in Hoot's comments (I'm sorry not been able to post it as a comment by lack of reputation): "I wonder if demanding that they be geometrically reduced patches things up?"

No, $X\times_kX$ is regular (if and) only if $X$ is smooth over $k$. More generally, if $A$ is a noetherian ring, $B$ a local flat $A$-algebra such that $B\otimes_AB$ is regular, then $B$ is an essentially of finite type smooth $A$-algebra. The finiteness condition follows from Corollaire (3.6) in Ferrand, Monomorphismes et morphismes absolument plats. An then, for the smoothness, if $B\otimes_AB$ is regular, then $B$ is regular since it is a retract, and therefore the surjective map $B\otimes_AB \to B$ is a closed regular embedding. We apply now Bourbaki, Algèbre Commutative chap. 10, § 7 n. 10 Théorème 4, (i) ⟺ (v).