Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.


I have a doubt about this fact:

Let f:XS be a flat, proper and surjective morphism of complex spaces (or locally noetherian, excellent schemes) with n-pure dimensional fibers. Then f is Cohen-Macaulay if and only if the relative canonical sheaf

$\omega^{n}{X/S}=H^{-n}(f^{!}O{S})$ is $S$-flat.

Perhaps must we add the condition : $R^{n}f_{*}\omega^{n}_{X/S}\simeq {\cal O}{S}$ ?

Thank you very much.

share|improve this question

1 Answer 1

I don't think this is true. For example if you look at a scheme $X$ over a field $k$, and $f$ is the structure map $f : X \to Spec ( k)=:S$, then $\omega_{X/S} \cong \omega_X$, which is flat over $S$. However $X$ need not be Cohen-Macaulay. You can also soup up this example by pulling back over some other base. Also $R^n f_* \omega_{X/S} \cong O_S$ is satisfied in this example for any $X$, so that is still not enough for the equivalence (see Hartshorne section III.7., that the top duality is always satisfied for the dualizing sheaf). Probably you want to require duality on all cohomologies. That would give you an equivalent statement. You can check out the details in this article: Relative duality for quasicoherent sheaves, S Kleiman - Compositio math, 1980.

share|improve this answer
Also see Theorem 3.5.1 in Brian Conrad's book, I think that gives the desired answer. –  Karl Schwede Jun 30 '10 at 13:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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