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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?

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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 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 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. $\blacksquare$

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

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