For this problem, suppose that $R$ is a commutative, unital ring. Further, for the sake of ease, suppose it is an integral domain.
Suppose also that $M$ is a finitely generated $R$ moduled, and $F$ is the field of fractions of $R$. Then $F \otimes M$ is naturally an $F$-vector space.
Then there is a cannonical map $f: M \to F \otimes M$ defined by $x \mapsto 1 \otimes x$.
My question is, how might one determine the kernel of the map $f$?
If we suppose that $R$ is a PID, then I believe we can use the structure theorem to show that $ker(f)=Tor(M)$ where $Tor(M)=\{ m \in M | rm=0 \text{ for some }r \in R \}$ is the torsion submodule of $M$.
My question is thus twofold:
1) Is there a way to show the above result without referring to the structure theorem?
2) Is there a result along the same lines for the case where $R$ is not a PID?
The module $F\otimes M$ is just the localization of $M$ with respect to the set $S$ of all nonzero elements of $R$, with the natural map $M\to F\otimes M$ being $m\mapsto\frac{m}{1}$. The fraction $\frac{m}{1}$ is equal to $0$ in the localization with respect to $S$ iff there is some $s\in S$ such that $sm=0$. That is, the kernel of $M\to F\otimes M$ is the set of torsion elements of $M$.