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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.

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  • $\begingroup$ 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. $\endgroup$
    – Jason Starr
    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. $\endgroup$
    – Jason Starr
    May 3, 2015 at 19:07

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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 $A$.

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  • $\begingroup$ 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
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    $\begingroup$ 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

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