5
$\begingroup$

For simplicity, all varieties in this question are quasiprojective varieties over an algebraically closed field of characteristic $0$.

The coherent cohomological dimension $cd(X)$ of a variety $X$ is the maximal dimension $n$ such that there exists a quasicoherent sheaf $\mathcal{F}$ such that $H^n(X;\mathcal{F}) \neq 0$. Since cohomology commutes with direct limits, it is enough to consider coherent $\mathcal{F}$. As an example, $X$ is affine if and only if $cd(X)=0$.

Question: Let $f\colon X \rightarrow Y$ be a surjective affine morphism. Is it the case that $cd(X) = cd(Y)$?

We have $cd(X) \leq cd(Y)$; indeed, if $\mathcal{F}$ is a quasicoherent sheaf on $X$, then since $f$ is affine the Leray spectral sequence degenerates to show that $H^n(X;\mathcal{F}) = H^n(Y;f_{\ast}(\mathcal{F}))$. Moreover, in the article

R. Hartshorne, Cohomological Dimension of Algebraic Varieties, Ann. Math. 88 (3), 1968, 403-450.

it is proven that $cd(X) = cd(Y)$ if $f$ is a surjective finite morphism.

Improve this question
$\endgroup$
2
  • 12
    $\begingroup$ By the Jounalou trick, any quasi-projective variety admits a surjective map from an affine one. A morphism from an affine scheme to a separated scheme is affine, so for any quasi-projective variety $Y$ there is a morphism $f:X\to Y$ with $cd(X)=0$. $\endgroup$
    – SashaP
    Sep 21, 2019 at 3:21
  • $\begingroup$ @SashaP: Thanks! If you posted that as an answer, I would be delighted to accept it. $\endgroup$
    – Linda
    Sep 27, 2019 at 0:02

0

Reset to default

Your Answer

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