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The answer to this question says the following: "The general statement is if $A \to B$ is finite and injective, and $A$ is noetherian and regular, then $B$ is CM if and only if $A \to B$ is flat.

The if part is easy because $A$ is CM. Locally a regular sequence of $A$ is again a regular sequence in $B$. As a finite injective ring homomorphism preserves the dimensions, $B$ is CM.

The converse is harder but is standard. See for instance EGA IV 6.1.5.".

Questions (each question is independent of the others):

(1) I have not found this result in EGA iv 6.1.5 (there is no such iv 6.1.5). Could one please refer to this result in EGA or elsewhere?

(2) What if we weaken the condition that $B$ is a finitely generated $A$-module to the condition that $B$ is a finitely generated $A$-algebra? I guess the result does not hold anymore, but maybe a weaker version of it still holds?

(3) If, in addition to the assumptions, $A$ and $B$ are local rings and $A \to B$ is a local homomorphism, could we weaken '$B$ is a finitely generated $A$-module' to '$B$ is a finitely generated $A$-algebra' and obtain the same conclusion?

(3) Is it possible to replace '$A$ regular' in the assumptions by '$A$ is CM' and get the same conclusion or a similar one?

Thank you very much!

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  • $\begingroup$ For (1) I believe you should find the result in EGA IV, part 1, Corollaire (17.3.5). $\endgroup$
    – Matthieu Romagny
    Apr 21, 2021 at 18:03
  • $\begingroup$ Thank you very much! $\endgroup$
    – user237522
    Apr 21, 2021 at 18:06
  • $\begingroup$ For (2), think about $A = k$ a field. Then flatness is automatic, and there is of course no easy criterion to detect whether a $k$-algebra is CM. Maybe a natural hypothesis for the general case would be that all the fibres $B \otimes_A \kappa(\mathfrak p)$ for $\mathfrak p \subseteq A$ prime are CM, but I'm not sure if this is sufficient. (It is necessary when $A$ is regular, as $B \to B \otimes_A A/\mathfrak p$ is a regular immersion since $A \to A/\mathfrak p$ is .) $\endgroup$
    – R. van Dobben de Bruyn
    Apr 21, 2021 at 18:48
  • $\begingroup$ @R.vanDobbendeBruyn, thank you very much for your comment. Interesting. $\endgroup$
    – user237522
    Apr 21, 2021 at 19:11
  • $\begingroup$ What if $B$ equals $A \times (A/I)$, where $I$ is generated by a regular sequence? $\endgroup$
    – Jason Starr
    Apr 21, 2021 at 20:46

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