Let $R$ be a domain (ring without zero divisors) and $M$ be an $R$-module. I want to show that $M$ is torsion free iff the natural map $f: M \longrightarrow\operatorname{Hom}_ R\bigl(\operatorname{Hom}_R(M, R), R\bigr);
m \mapsto (\phi \mapsto \phi(m))$ is injective but I don't have any idea how to manage it.
I can probably prove "<=": $M$ is torsion, therefore there exist $r_0 \neq 0, m_0 \in M$ with
$r_0 m_0=0$. How can I show that then $f$ is not injective? A natural guess
is to show that $f(m_0) =0$.
$m_0$ is mapped by $f$ to $r$-module map $m_0^*:\phi \mapsto \phi(m_0)$.
Since $m_0$ is torsion and every $\phi\in M^*$ respects $R$-module
structure we have $r_0 \phi(m_0)= \phi(r_0 m_0)=0$. This imply that
$\phi(m_0)=0$ since $R$ domain. This argument seems to work.
As alternative way we kno that every submodule of a torsion free module
is torsion free. But is
$\operatorname{Hom}_ R\bigl(\operatorname{Hom}_R(M, R), R\bigr)$
torsion free?
"=>": is not clear. Assume $f$ is not injective, why is $M$ torsion?
Best Answer
The modules for which the natural map $f\colon M\rightarrow M^{\ast\ast}$ is injective are called torsion-less. This is not equivalent to being torsion-free. Over a domain, torsion-free implies torsion-less, as your argument correctly shows. To see the converse fails, note that $\mathbb{Q}$ is a torsion-free $\mathbb{Z}$-module, but $\mathbb{Q}^{\ast}=0$.
As an alternative argument, if $f$ is injective, then $M$ embeds as a submodule of $M^{\ast\ast}$, so if the latter is torsion-free, we are done. I will prove the following slightly more general assertion: if $N$ is a torsion-free $R$-module and $M$ is any $R$-module, then $\operatorname{Hom}_R(M,N)$ is torsion-free. To this end, assume that $f\in\operatorname{Hom}_R(M,N)$ is torsion, so there is a $0\neq r\in R$ such that $rf=0$. By definition, this means $rf(m)=(rf)(m)=0(m)=0$ for every $m\in M$. Since $r\neq0$ and $N$ is torsion-free, this implies $f(m)=0$ for all $m\in M$, but that just means $f=0$.