2
$\begingroup$

I would like to fill a detail in the answer given by Francesco Polizzi to the question "Is the dualizing sheaf on a Cohen-Macaulay scheme reflexive?".

Let $X$ be a normal, Cohen-Macaulay scheme of dimension $n$. Let $U$ be the smooth locus of $X$ and let $i : U \rightarrow X$ be the natural inclusion. Now, let $$ \omega_X = i_*{\Omega_U^n} $$ I would like to prove that $\omega_X$ coincides with the dualizing sheaf on $X$ in the sense of Serre duality. Basically, I would like to show for any locally free sheaf $F$ on $X$, we have $$ H^i(X, F) \cong H^{n-i}(X, F^\vee\otimes\omega_X)' $$

My immediate thought is to use projection formula for $F^\vee\otimes\omega_X$. Use projection formula, we have $$ H^{n-i}(X, F^\vee\otimes\omega_X) = H^{n-i}(X, i_*(F|_U\otimes\Omega_U^n)) $$ But I don't know how to get back to $H^i(X, F)$ from the right side of the above equation.

$\endgroup$

1 Answer 1

2
$\begingroup$

This is more or less standard material on Grothendieck and Serre Duality Theory, and there are several references available.

For instance, the statement that you want is a consequence of the results in Chapter 5 of S.Ishii's book Introduction to Singularities. Look in particular at Theorem 5.3.6, Theorem 5.3.8 and Corollary 5.3.9.

Note that the last corollary also provides an answer to your previous question, since it shows that for any normal, $n$-dimensional variety $X$ the sheaf $\omega_X$ is divisorial, namely reflexive of rank one.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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