Skip to main content
4 of 4
deleted 157 characters in body
Yosemite Sam
  • 1.9k
  • 1
  • 14
  • 27

If $f: X \to Y$ is a finite flat morphism of schemes, $g: Y \to Z$ is a proper morphism of relative dimension one, $Z$ is affine and $E$ is a vector bundle on $Y$ with $R^1g_*E=0$ then $H^1(X,f^*E)=0$?

Let $f: X \to Y$ and $g: Y \to Z$ be morphisms of schemes* such that f is flat and finite, g is proper and $R^{> 1}g_*E=0$ for all sheaves and Z is affine.

Let E be a vector bundle on Y such that $R^1 g_* E=0$. Can we say anything about $H^1 (X,f^* E)$? By the projection formula this is the same as $R^1 g_* ( E \otimes f_* \mathcal{O} ) $ and as f is flat and finite $f_* \mathcal{O}_X $ is a vector bundle. But I can't seem to be able to say much else.

* There are many hypotheses I'd be happy to make: everything is finite type over a field, the field is algebraically closed of characteristic zero, all the schemes involved are integral, Y is regular, g is actually the restriction of a morphism of projective varieties $g': Y' \to Z'$ to an open affine patch $Z$ of $Z'$. The dual of $E$ is globally generated. $X,Y,Z$ have all the same dimension (equal to 3). g is birational. $Rg_* \mathcal{O}_Y = \mathcal{O}_Z$ and $Z$ is Gorenstein.

Yosemite Sam
  • 1.9k
  • 1
  • 14
  • 27