Does $M^{*}$ contain an injective function?

Question

Let $R$ be a PID and $M$ a finitely generated $R$-module. Let $F$ be the field of fractions of $R$ and $P = F/R.$ Let $M^{*}$ = $\mathrm{Hom}_R(M, P).$ Suppose $M$ is torsion. Is there an injective function in $M^*?$

What I've done so far was let $M =\left <e_1, ..., e_n\right>$ and let $\phi(e_i) = 1/r_{e_i} + R$ where $r_{e_i}$ is the nonzero annihilator of $e_i.$ I don't know if these works though...any hints?

Answer 1

The answer is:

$M^*$ contains an injective morphism if and only if $M$ is a cyclic torsion $R$-module, i.e. $M \cong R/(a)$ for some $0 \neq a \in R$.

The backward direction is trivial, because $R/(a) \cong \langle \frac{1}{a} \rangle \subset F/R$.

The forward direction goes as follows: It is well known that any finitely generated submodule of $F$ is cyclic, hence the same is true for $P$. In particular an injection $M \to P$ with $M$ finitely generated implies that $M$ is cyclic.

The main ingredient, that

any finitely generated submodule of $F$ is cyclic

is proven here.