Torsion Free Quasi Coherent Module

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.

• Maybe try something by showing that if $am=0$ then $N(a)m=0$ where $N$ is the norm map. – Alex Youcis Apr 25 at 2:38
• 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? – KarlPeter Apr 27 at 1:40
• 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? – Alex Youcis Apr 27 at 2:15