When a smooth algebra is regular?

Question

Let $A \subseteq B$ be noetherian integral domains, $A$ regular (=every localization at maximal ideal is a regular local ring) and $B$ is a smooth $A$-algebra. For the definition of a smooth algebra, please see the first page of

Robert A Morris, Stuart Sui-Sheng Wang, A Jacobian criterion for smoothness
Journal of Algebra, Volume 69, Issue 2, April 1981, Pages 483–486
doi:10.1016/0021-8693(81)90217-9

which says that $B$ is a smooth $A$-algebra if the following two conditions are satisfied:

(1) For each $A$-algebra $C$, and each ideal $J$ in $C$ with $J^2=0$, the canonical homomorphism $Hom_{A-alg}(B,C) \to Hom_{A-alg}(B,C/J)$ is surjective.

(2) $B$ is finitely presented as an $A$-algebra.

My question: Is it true that $B$ must be regular too? If not, what additional conditions should we assume in order that $B$ will be regular?

I really apologize if this question is trivial; it's just that only recently I have started to study regular rings/smooth extensions.

Answer 1

Yes. More generally, let $A \to B$ be a homomorphism of noetherian rings satisfying the condition (1) of your question (that is, B is formally smooth in the sense of [EGA IV.17.1.1]). Let $\mathfrak q$ be a prime ideal of $B$ and $\mathfrak p$ its contraction in $A$, $k(\mathfrak p)$, $k(\mathfrak p)$ the residue fields. I will use André-Quillen homology modules and references are to André's book. We have

$$H^1(A_{k(\mathfrak p)},B_{k(\mathfrak q)},k(\mathfrak q))=H^1(A,B,k(\mathfrak q))=0$$

by 5.27 and 16.17. By 5.1, 7.4, 3.21, 6.26 we have an exact sequence

$$ H^1(A_{k(\mathfrak p)},B_{k(\mathfrak q)},k(\mathfrak q)) \to H^2(B_{k(\mathfrak q)},k(\mathfrak q),k(\mathfrak q))\to H^2(A_{k(\mathfrak p)},k(\mathfrak q),k(\mathfrak q))=Hom_{k(\mathfrak p)}(H_2(A_{k(\mathfrak p)},k(\mathfrak p),k(\mathfrak p)),k(\mathfrak q))=0$$

Therefore $H^2(B_{k(\mathfrak q)},k(\mathfrak q),k(\mathfrak q))=0$ and again by 6.26 $B$ is regular.

Answer 2

Here's a proof not using André-Quillen homology. It uses that $\varphi\colon A\to B$ is flat with regular fibers (which is the case if $A\to B$ is smooth).

Let $\mathfrak{q}\subseteq B$ be a prime ideal and $\mathfrak{p}=\varphi^{-1}(\mathfrak{q})$. Pick a regular sequence $f_1,f_2,\dots,f_n$ in $A_\mathfrak{p}$ that generates $\mathfrak{p}A_\mathfrak{p}$. Since $A\to B$ is flat, $\varphi(f_1),\dots,\varphi(f_n)$ is a regular sequence in $B_\mathfrak{p}$. The quotient $B_\mathfrak{p}/(\varphi(f_1),\dots,\varphi(f_n))$ is a fiber, hence regular. It follows that $B_\mathfrak{p}$ is regular, hence also $B_\mathfrak{q}$.