# Direct image of reflexive sheaf via finite, flat map

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?

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.

• 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? – Ramac Sep 12 '18 at 8:55
• @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. – Takumi Murayama Sep 17 '18 at 1:42
• Sorry, I was confused about it. My comment is nonsense. – Ramac Sep 24 '18 at 9:40