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 $H^1(X,f^*E)=0$?

Question

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.

Answer 1

Actually, for any such projective $g$ and any $E$ there exists an $f$ such that this fails. In fact you may even assume that $f$ is a double cover. (I suspect that it also fails without the projective assumption, but this seems convincing enough).

Let $g:Y\to Z$ be a projective morphism, $Y$ smooth, $Z$ quasi-projective and $E$ a coherent sheaf on $Y$. Assume that $R^ig_*(E\otimes N)=0$ for all $N$ line bundles and $i>1$. Then there exists a finite, flat $f:X\to Y$ such that $X$ is smooth and $H^1(X,f^*E)\neq 0$. Notice that this is less than assumed and probably even less is enough for the above claim.

Claim Let $L$ be a $g$-ample line bundle on $Y$. Then $g_*(E\otimes L^{-m}) = 0$, but $Rg_*(E\otimes L^{-m})\neq 0$ for $m\gg 0$.
(Here $Rg_*$ stands for the total push-forward and this statement is equivalent to saying that there exists an $i$ such that $R^ig_*(E\otimes L^{-m})\neq 0$).

Proof The first claim is obvious, we can "kill" every section in $E$ by "dividing" with sections of $L$ enough times. The second one is relatively easy using Grothendieck duality and observing that $g_*(F\otimes L^m)\neq 0$ for $m\gg 0$ and any $F$ (and you just have to use an appropriate $F$ that comes from GD). I am sure this can be proved by alternative means. $\square$

So, now take $N=L^m$ such that $g_*(E\otimes N^{-1}) = 0$ and $Rg_*(E\otimes N^{-1})\neq 0$. By the assumption that $R^ig_*(E\otimes N)=0$ for all $i>1$ it follows that $R^1g_*(E\otimes N^{-1})\neq 0$. We may assume that $N^2$ is very ample and choose a general section $s\in H^0(Y,N^2)$. Let $\mathscr A=\mathscr O_Y\oplus N^{-1}$ and make it an $\mathscr O_Y$ algebra using $s$ to "wrap back" from $N^{-2}$ to $\mathscr O_Y$ as usual. Finally, let $X=\mathrm{Spec}_Y\mathscr A$ and $f$ the associated morphism. From the construction it is clear that $f$ is flat and finite of degree $2$. Again by the construction $f_*\mathscr O_X\simeq \mathscr A$ and then by the above, $R^1g_*(E\otimes f_*\mathscr O_X)=R^1g_*E\oplus R^1g_*(E\otimes N^{-1})\neq 0$.

Answer 2

This doesn't seem to be true without additional hypotheses. Here is a counterexample:

Take $Z = \mathrm{Spec}(\mathbb{C})$ , $X = Y = \mathbb{P}^1_\mathbb{C}$, $f:X\to Y$ the map sending $[x: y]$ to $[x^2:y^2]$ (then $f_* \mathcal{O}_X = \mathcal{O}_Y\oplus\mathcal{O}_Y(-1)$, in particular $f$ is finite flat). Take $E = \mathcal{O}_Y(-1)$. Then the dual of $E$ is globally generated and $H^1(Y, E) = 0$. We have $f^* E = \mathcal{O}_X(-2)$ which has one-dimensional $H^1$.