Push forward of the structure sheaf along covering

Question

Let $f: X \to Y$ be a covering (proper, surjective, finite regular map) of smooth projective varieties of degree $d$. How one can show that in this case $f_* \mathcal{O}_X$ is a locally free sheaf of rank $d$?

Answer 1

The answer relies on the following incredibly general and simple result:

Given a finite morphism of schemes $f:X\to Y$ with $Y$ locally noetherian, the sheaf $f_*\mathcal O_X$ is locally trivial iff $f$ is flat.

The proof consists in quoting a result in commutative algebra: a module $M$ over a ring $R$ is flat of finite presentation iff it is finitely generated and projective
[cf. Bourbaki, Commutative Algebra, Chapter II, §5.3, Corollary 2, page 111]

Once more Mumford's brilliant aphorism applies:

Flatness is a riddle that comes out of algebra , but which technically is the answer to many prayers.

If $Y$ is smooth, any finite surjective morphism is flat and the above applies, so that $f_*\mathcal O_X$ is locally free, just as you wished.

Edit
The last assertion is a particular case of a wonderful result, aptly named by some geometers miracle flatness. It goes like this:

Let $f:X\to Y$ be a surjective morphism between varieties over a field, with $X$ Cohen-Macaulay (for example regular) and $Y$ regular. If all fibers of $f$ are of the same dimension, then $f$ is flat.

A reference for this theorem is GÖRTZ-WEDHORN, Corollary 14.128, page 475.

Answer 2

With your assumptions, $f$ is a flat morhism, and $H^1(X_y,\mathscr O_{X_y})=0$ for all $y\in Y$. Then the theory of cohomology and base change says that $f_\ast\mathscr O_X$ is locally free. It is rank $d$ because $d=h^0(X_y,\mathscr O_{X_y})$.

If you want a precise statement, here it is:

Suppose you have a proper morphism $f:X\to Y$, with $Y$ locally Noetherian, and a coherent $\mathscr O_X$-module $\mathscr F$, flat over $Y$. If $R^if_\ast\mathscr F=0$ for all $i\neq 0$, then $f_\ast\mathscr F$ is locally free.