Vanishing of $R^1f_*\mathcal O_X$

Question

I am probably missing something obvious here, but none the less, here goes: Is the following statement (or perhaps some minor modification of it) true and if so, why: $R^1f_*\mathcal O_X = 0$ for a finite morphism of curves $f\colon X\to Y$. The question is inspired by exercise 4.2.6 in Hartshorne, where it seems to be necessary to use some sort of similar result. Thanks in advance for any help!

Answer 1

Here's one way to see it. This is probably overkill with the machinery. Finite morphisms are proper, and the higher direct images of a coherent sheaf under a proper morphism are still coherent. Thus $R^1f_*\mathcal{O}_X$ is coherent.

But now by the Theorem on Formal Functions, the completion of the stalks are just cohomology of the fibers: $\widehat{(R^1f_*\mathcal{O}_X)_y}\simeq \varprojlim H^1((X_y)_n, \mathcal{O}_{(X_y)_n})$. Since the fibers of a finite morphism are finite, the $X_y$ are zero dimensional and hence the $H^1=0$ and hence $(R^1f_*\mathcal{O}_X)_y=0$ for all $y\in Y$ which shows $R^1f_*\mathcal{O}_X=0$.