2
$\begingroup$

I wonder if the miracle flatness theorem Generalizing miracle flatness (Matsumura 23.1) via finite Tor-dimension still works if the rings involved are not local (and the dimension condition is deleted)?

In other words, is it true that $f: A \to B$, $A$ regular, $B$ Cohen-Macaulay, implies that $f$ is flat? I guess the answer is no (maybe taking $A=k[x,y]$, $B=A/I$, with a right choice of an ideal $I$ of $A$ will serve as a counterexample? or maybe no).

But what if we further assume that: $f$ is injective, $A$ is Noetherian and $B$ is a finitely generated $A$-module? Is $f$ flat in this case?

I ask this question since I have seen here this claim as a fact https://math.stackexchange.com/questions/296971/what-has-projectiveness-to-do-with-cohen-macaulay-rings/297320#297320, with reference to EGA IV 6.1.5, though it seems (I do not know French) that Grothendieck talks about the local case.

$\endgroup$
2
  • $\begingroup$ Certainly there are counterexamples if you drop the dimension hypothesis, even with all of the other hypotheses you propose. For instance, the ring homomorphism $k[x,y] \mapsto k[x,y/x]$ is a counterexample. $\endgroup$ May 3, 2015 at 18:54
  • $\begingroup$ In my previous comment I missed the hypothesis that $B$ is a finitely generated $A$-module. Obviously this implies that the fibers are finite, hence zero-dimensional. So the new hypothesis is simply replicating the "usual" hypothesis that the fiber dimension be constant. $\endgroup$ May 3, 2015 at 19:07

1 Answer 1

4
$\begingroup$

The answer to your first question is obviously no, as you point out: the homomorphism $A\rightarrow A/I$ is essentially never flat. It is OK under your hypotheses ($f$ injective, $A$ noetherian, $B$ finitely generated) if you assume moreover that $B$ is a domain. A reference (in french, sorry) is Bourbaki, Algèbre Commutative X, §4, no. 3, Corollaire.

The hypothesis that $B$ is a domain is necessary for stupid reasons: just take $B=A\times A/I$ with $A/I$ Cohen-Macaulay, again this is not flat over $A$.

$\endgroup$
8
  • $\begingroup$ Thank you very much! I will try to read Bourbaki's Corollary. $\endgroup$
    – user237522
    May 3, 2015 at 19:46
  • $\begingroup$ Bourbaki's corollary even says that $B$ is a projective $A$-module. Precisely, if $A \subseteq B$, $A$ is regular, $B$ is a noetherian integral domain which is finitely generated as an $A$-module, then: $B$ is Macaulay iff $B$ is a projective $A$-module. $\endgroup$
    – user237522
    May 10, 2015 at 16:35
  • 1
    $\begingroup$ This is not stronger: a flat, finitely generated module over a noetherian ring is projective. $\endgroup$
    – abx
    May 10, 2015 at 19:27
  • $\begingroup$ Yes, you are right (a regular ring is, by definition, noetherian). Actually, I now notice that in my question: $\endgroup$
    – user237522
    May 11, 2015 at 19:43
  • $\begingroup$ "But what if we further assume that...$A$ is noetherian", I should delete "$A$ is noetherian", since $A$ is already noetherian (since I have already mentioned the assumption that $A$ is regular). $\endgroup$
    – user237522
    May 11, 2015 at 20:00

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.