# Torsion-free module morphism

I am trying to prove the statement:

Let $$R$$ be a PID but not a field and let $$M$$ be an $$R$$-module. Then $$M \space \text{is torsion-free R-module} \space \text{iff} \space \operatorname{Hom}_R(S,M)=0 \space \text{for all simple R-modules } S.$$

I could show (or at least I think I could) the implication $$M$$ is torsion-free then the only morphism from a simple $$R$$-module to $$M$$ is the trivial one:

Take $$S$$ simple $$R$$-module and $$f \in \operatorname{Hom}_R(S,M)$$, one can show that $$S$$ is simple iff $$S \cong R/ \langle p \rangle$$ for $$p$$ irreducible. Using this property, we have an isomorphism of $$R$$-modules $$g: R/ \langle p \rangle \to S$$. Take $$s \in S$$, there is $$\overline{a} \in R/ \langle p \rangle : g(\overline{a})=s$$, then $$0=g(\overline{0})=g(p\overline{a})=pg(\overline{a})=ps$$. Now, for any $$0 \neq s \in S$$, we have $$0=f(ps)=pf(s)$$, and since $$f(s) \in M$$ which is torsion-free and $$p \neq 0$$, one must thus have $$f(s)=0$$. It follows $$f$$ is exactly $$0$$.

I don't know what to do to show the other implication, I would appreciate suggestions to complete the solution and also if someone could check if what I wrote is correct, feel free to make any corrections and/or add an alternative proof for that part.

If $M$ is simple, then $M\simeq R/(p)$ with $p\in R$ irreducible. Let $f\in \operatorname{Hom}(R/(p),M)$. We have $f(\bar 1)\in M$ and $pf(\bar 1)=f(p\cdot\bar 1)=f(\bar 0)=0$. Since $M$ is torsion-free we get $f(\bar 1)=0$, so $f(\bar a)=0$ for all $a\in R$, that is, $f=0$.
Conversely, $\operatorname{Hom}(R/(p),M)=0$ for all $p\in R$ irreducible. Let $x\in M$ such that $px=0$ for some $p$. Then define $f:R/(p)\to M$ by $f(\bar a)=ax$. This is well defined (why?) and an $R$-module homomorphism, so $f=0$. In particular, $f(\bar 1)=0$, so $x=0$.
Now let $x\in M$ and $a\in R$ such that $ax=0$. If $a$ is invertible then $x=0$. If not, then $a=p_1\cdots p_n$ with $p_i$ irreducible. By induction on $n$ one gets $x=0$.
• Just checking well definition: take two representatives of the same equivalence class $\overline{a}$ and $\overline{b}$. Then $a-b \in (p)$, so $0=f(\overline{0})=(a-b)x$, it follows $f(\overline{a})=ax=bx=f(\overline{a})$. – user16924 Nov 13 '14 at 14:19
Any simple module is torsion, so if $\operatorname{Hom}_R(S,M)\ne0$ for a simple module $S$, then $M$ contains a submodule isomorphic to $S$ and so it is not torsionfree.
Conversely, suppose $M$ is not torsionfree; if $x\in M$ is a torsion element, $x\ne0$, then $xR$ is isomorphic to $R/aR$ for some $a\ne0$, $a$ non invertible. The ring $R/aR$ is artinian, because it is noetherian and has Krull dimension zero (every prime ideal is maximal), so $xR\cong R/aR$ is artinian, hence it has essential socle. In particular it has at least a simple submodule.
I assume $$M$$ to be finitely generated. So I can use the fundamental decomposition theorem for module over PID. The theorem states that $$M = R^n \oplus (\bigoplus_i R/p_i^{a_i})$$, where $$n$$ is the rank of $$M$$. If $$\operatorname{Hom}(S,M)=0$$ for all simple modules $$S$$, then $$M$$ is torsion-free, as there would be no $$R/p_i$$ parts. Hence statement is proved.