1
$\begingroup$

Suppose $f: X \rightarrow Y$ is a finite, flat (hence locally free) morphism of curves (i.e. schemes of dimension 1, not smooth or even reduced). Suppose $L$ is a reflexive sheaf on $X$, locally free of rank 1 at each generic point of $X$.

Is the direct image $f_* L$ still reflexive on $Y$? (Better its top exterior power). What if $Y$ is smooth?

$\endgroup$

1 Answer 1

3
$\begingroup$

Here is a (probably non-optimal) statement that may apply in your situation. In your situation with curves, the hypothesis says that you need $X$ and $Y$ to be Gorenstein.

Claim. Let $X$ and $Y$ be noetherian schemes satisfying $G_1$ and $S_2$. If $f\colon X \to Y$ is a finite surjective morphism and $\mathscr{F}$ is a coherent reflexive sheaf on $X$, then $f_*\mathscr{F}$ is a coherent reflexive sheaf on $Y$.

Proof. On noetherian schemes satisfying $G_1$ and $S_2$, reflexivity is equivalent to being $S_2$ (in Hartshorne's sense) [Hartshorne 1994, Thm. 1.9]. The claim then follows since the $S_r$ property is preserved under pushforward by finite surjective morphisms by [EGAIV$_2$, Prop. 5.7.9]. $\blacksquare$

I wanted to prove a statement for non-finite morphisms as well, and for integral schemes, you can say a bit more:

Claim. Let $X$ and $Y$ be integral noetherian schemes satisfying $G_1$ and $S_2$. If $f\colon X \to Y$ is a proper dominant morphism with all fibers of the same dimension. If $\mathscr{F}$ is a coherent reflexive sheaf on $X$, then $f_*\mathscr{F}$ is a coherent reflexive sheaf on $Y$.

Proof. The fact that $f_*\mathscr{F}$ is coherent and normal follows from the proof of [Hartshorne 1980, Cor. 1.7]. By [Hartshorne 1994, Rem. 1.11], to show $f_*\mathscr{F}$ is reflexive, it therefore suffices to show that it satisfies $S_1$. But being $S_1$ is equivalent to torsion-freeness for integral noetherian schemes [Hartshorne 1994, Lem. 1.5], hence the claim follows by the fact that torsion-freeness is preserved under pushforwards by dominant morphisms. $\blacksquare$

Edit. Added the hypothesis that $f$ is surjective in the first claim.

$\endgroup$
3
  • $\begingroup$ Thanks! This is interesting. If the sheaf on $X$ is locally free of rank 1 at the generic points, is this true also for the pushforward? $\endgroup$
    – Raffaele C
    Sep 12, 2018 at 8:55
  • $\begingroup$ @Ramac I apologize but I am a bit confused by your comment. What are your assumptions on $X$ and the morphism by which you are pushing forward? Even for a morphism like $X \amalg X \to X$, where $X$ is a connected regular curve, it seems like $\mathcal{O}_{X \amalg X}$ is locally free of rank 1 at the generic points, but the pushforward will be of rank 2 at the generic point, so perhaps I am misunderstanding your question. $\endgroup$ Sep 17, 2018 at 1:42
  • $\begingroup$ Sorry, I was confused about it. My comment is nonsense. $\endgroup$
    – Raffaele C
    Sep 24, 2018 at 9:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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