Let $\phi: R \to A$ be a finite morphism of Dedekind rings (so $A$ is a finitely generated $R$-module) and $M$ a finitely generated $A$-module. Obviously, if we restrict the action of $A$ on $M$ to $R$ via $(r,m) \mapsto \phi(r) \cdot m$ we obtain an $R$-module $M_R$.
My question is:
If the restriction $M_R$ is a torsion-free $R$-module how to show that $M$ is a torsion-free $A$-module?
My ideas:
By classification theorem for finitely generated modules over Dedekind rings $M= A^r \oplus T_A$ with torsion $T_A$ and $M_R = R^s \oplus T_R$. since $M_R$ torsionfree $M_R= R^s$.
Why does this already imply that $M$ torsionfree as $A$-module (therefore $T_A=0$)?
Background: this question arises from following problem: Consider a finite and locally free morphism $f:C→\mathbb{P^1}$ where $C$ a curve. Furthermore let $F$ be a quasicoherent sheaf of finite type on $C$. The aim is to show that if the pushforward $f_*F$ is locally free then $F$ is already locally free.
Best Answer
Let me keep the discussion with a general ring homomorphism $\varphi: R \rightarrow A$ between Dedekind rings (which corresponds to the originally asked question).
1) The implication "$M_R$ torsion-free $\Rightarrow M$ torsion-free":
The implication holds if and only if maximal ideals of $A$ contract to maximal ideals of $R$. A module $M$ has torsion iff $M$ has an associated maximal ideal $\mathfrak{P}$ of $A$, but then $M_R$ has an associated prime $\mathfrak{p}=\varphi^{-1}(\mathfrak{P})$, showing that $M_R$ has torsion in this case as well. On the other hand, if there is a maximal ideal $\mathfrak{P} \subseteq A$ such that $\mathfrak{P} \cap R=0,$ then $M=A/\mathfrak{P}$ is a module that becomes torsion-free over $R$, but it is a torsion $A$-module.
For example, when $R \rightarrow A$ is $\mathbb{Z} \rightarrow \mathbb{C}[X]$ and $\mathfrak{P}=(X)$, we see that $\mathbb{C}=\mathbb{C}[X]/(X)$ is torsion-free $\mathbb{Z}$-module which is a torsion $\mathbb{C}[X]$-module.
2) The implication "$M_R$ torsion-free $\Leftarrow M$ torsion-free":
The implication holds if and only if the map $\varphi: R \rightarrow A$ is injective. If it is, and $M$ is a torsion-free $A$-module, then it is torsion-free as $R$-module since for any $m \in M$, $\mathrm{Ann}_R(m)=\varphi^{-1}(\mathrm{Ann}_A(m))=\varphi^{-1}(0)=0$. On the other hand, by the same reasoning any $A$-module $M$ is always annihilated by $\mathrm{Ker}\,\varphi$ as an $R$-module, so if $\varphi$ is not injective, every $A$-module $M$ will have torsion as an $R$-module.
(A concrete example is, similarly, $\mathbb{Z}\rightarrow \mathbb{F}_p[X]$. Then all $\mathbb{F}_p[X]$-modules are obviously $p$-torsion Abelian groups.)
3} Regarding the assumption that $\varphi$ is finite:
The condition that $R \rightarrow A$ is finite (or, more generally, integral) is sufficient for both the implications 1} and 2) to hold. Such morphism is necessarily injective (note that any non-trivial quotient $R/I$ is $0$ dimensional, so $R/I \subseteq A$ finite implies $A$ is $0$-dimensional), showing that 2) holds. Also the fact that maximal ideals contract to maximal ideals follows by the going-up property of integral (e.g. finite) extensions, showing that 1) holds as well.