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Let $C$ be regular curve. Consider a finite and locally free morphism $f: C \to \mathbb{P}^1$. (the latter mean that $f_* \mathcal{O}_{C}$ is a free $\mathcal{O}_{P^1}$ module)

Let $\mathcal{F}$ be a quasicoherent sheaf of finite type on $C$.

Assume that we know that the pushforward $f_*\mathcal{F}$ is torsion free. How to deduce that then $\mathcal{F}$ is also torsion free?

My ideas: the problem is local and $f$ is affine (because $f$ finite). Therefore $f$ reduces to a morphism $\phi:= f^{\#}: R \to A$ of Dedekind rings $R = \Gamma(U, \mathcal{O}_{P^1}),R = \Gamma(f^{-1}(U), \mathcal{O}_{C}) $.

In this setting $\mathcal{F}= \widetilde{M}$ for $A$-module $M$.

The pushforward is $f_*\mathcal{F}= f_*\widetilde{M}= \widetilde{M \vert_R}$.

By assumption classification theorem for finitely generated modules over Dedekind rings $M \vert_R$ is free $R$-module since it has no torsion.

So it suffice to show that $M$ is also free as $A$-module.

But how? The $A$-module structure of $M$ is fixed at the beginning. The naive approach by tensoring $M \vert_R$ by $A$ induces a (free)$A$-module structure on $M \vert_R$ but it doesn't "bring" the previous $A$-module structure back.

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  • $\begingroup$ Maybe try something by showing that if $am=0$ then $N(a)m=0$ where $N$ is the norm map. $\endgroup$ – Alex Youcis Apr 25 at 2:38
  • $\begingroup$ what do you mean by the norm map $N(-): A \to R$? I know only a norm map in case of field extensions. So do you mean that in the light of the extension $Frac(R) \subset Frac(A)$? If yes, then - considering $a$ as element of $Frac(A)$ - the image $N(a)$ lies in $Frac(R)$ and not generally in $R$. Or do you mean that in the sense that (probably) $N$ maps integers to integers? $\endgroup$ – KarlPeter Apr 27 at 1:40
  • $\begingroup$ let $\mathrm{Spec}(B)$ be a small enough open in $\mathbb{P}^1$ such that $f^{-1}(\Spec(A))$ is such that $B$ is a free $A$-module. You then naturally get a map of multiplicative monoids $N:B\to A$ such that $N^{-1}(0)=0$ by literally taking the determinant of the multiplication by $b$ map on $B$ thought of as a free $A$-module. The point is that if $b\in B$ is such that $bm=0$ then $N(b)m=0$ I believe which implies that $N(b)=0$ so $b=0$. Something like that? $\endgroup$ – Alex Youcis Apr 27 at 2:15

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