Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $d$?

share|improve this question

2 Answers 2

up vote 6 down vote accepted

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.

share|improve this answer
    
Thanks! Could you please add explanation or reference to "if $Y$ is smooth, any finite surjective morphism is flat"? –  Alex Aug 26 at 10:28
    
Dear Alex, I have added an edit with a generalization and a reference. –  Georges Elencwajg Sep 2 at 14:28

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.