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

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.

  • 1
    $\begingroup$ Are you over a field? Smooth over a field implies regular ( stacks.math.columbia.edu/tag/056S ), so if you require $A$ to be smooth over $k$ (stronger than regular, in general), then $B$ will be also be smooth over $k$, and in particular regular. $\endgroup$ May 26 '15 at 6:52
  • $\begingroup$ Great! Actually, I have a specific example in mind in which $A$ is smooth over a field $k$. $\endgroup$
    – user237522
    May 26 '15 at 14:32

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.

  • $\begingroup$ See mathoverflow.net/questions/175766/…. $\endgroup$
    – Vinteuil
    May 26 '15 at 12:01
  • $\begingroup$ Cool! But now I'm curious, is there a more 'elementary' way to prove it, that doesn't use André-Quillen homology? $\endgroup$ May 27 '15 at 4:00
  • $\begingroup$ I also wanted to ask @Vinteuil the same question... $\endgroup$
    – user237522
    May 27 '15 at 8:47
  • $\begingroup$ I do not know any reference. I feel in the finite type case there should be a more elementary proof, and maybe you can find one in some place of EGA IV, part 4. In the general case (for noetherian rings), maybe completing, then using relative Cohen factorizations (similar to those in Avramov-Foxby-Herzog "Structure of local homomorphisms") and then Bourbaki, Algèbre Commutative, chap. X, $7 n.5 Thèoréme 2, we can deduce a proof, but it would not be very 'elementary'. $\endgroup$
    – Vinteuil
    May 27 '15 at 12:01
  • $\begingroup$ Ok thanks! I just thought it would be nice if for example there was some argument using the characterization of regular rings given by the Auslander-Buchsbaum formula.. but nevermind :) $\endgroup$ May 28 '15 at 3:46

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 $B_\mathfrak{q}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.