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 finite morphism of curves. Let $L$ be a line bundle on $X$. Why is $f_\ast L$ a line bundle and is the degree of $f_\ast L$ equal to $\deg f$ or $\deg f+ \deg L$?

Here is my strategy and some related questions.

Firstly, since $f$ is proper, we have that $f_\ast L$ is coherent on $Y$. I want to see that it is locally free. I have read that this follows from the flatness, but this is too abstract for me.

Can't we do something more explicit? The problem is local on $Y$. Suppose that $V_1,\ldots,V_r$ are trivializing opens for $L$ on $X$. Then, how does one trivialize $f_\ast L$. Problems might occur at the branch point, but therefore I ask: if I show that $(f_\ast L)_y$ is free for all non-branch points $y$ of $f:X\to Y$, does it follow that $f_\ast L$ is locally free?

I don't want to use Grothendieck-Riemann-Roch or anything similar.

share|improve this question
add comment

1 Answer

up vote 6 down vote accepted

No, in general $f_*L$ is not a line bundle.

Suppose $X, Y$ are smooth with fields of functions $K(X), K(Y)$. Then $f$ is flat, so $L$ is flat over $Y$. As $f$ is affine, $f_*L$ is flat over $Y$. As you said it is coherent, so $f_*L$ is locally free. At the generic point of $Y$, the stalk of $f_*L$ is just $L\otimes k(X)\simeq K(X)$ viewed as a vector space over $K(Y)$, its dimension is $[K(X) : K(Y)]=\deg f$. So $f_*L$ is locally free of rank $\deg f$.

To say something more, suppose $X,Y$ are projective. We have to use Riemann-Roch which says that if $V$ is locally free of rank $r$ on $Y$, then the degree of the determinant of $V$ can be computed with the Euler-Poincaré characteristics $$\deg\det (V)= \chi(V)-r \chi(O_Y).$$ So $$\deg\det(f_*L)=\chi(f_*L)-r\chi(O_Y)=\chi(L)-r\chi(O_Y)=\deg L + \chi(O_X)-r\chi(O_Y).$$

If $L$ is associated to a divisor $D$ on $X$, then one can consider the pushforward $f_*D$ as a divisor on $Y$ and compare the locally free sheaf $f_*L$ with the divisor $f_*D$. We have $$ \det(f_*L)\simeq \det(f_*O_X) \otimes O_Y(f_*D).$$
See Hartshorne, Exercise IV.2.6.

share|improve this answer
add comment

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.