Free rank 0 vs torsion?

Question

For $R$ a PID and M a finitely generated R-Module, is saying $M$ has free rank 0 and $M$ is torsion the same thing? Free rank 0 means the only linearly independent set is {0} and $M$ being torsion means each element $m \in M$ has a nonzero $r \in R$ such that $mr = 0.$ So I think these are synonymous...

Answer 1

As $R$ is a pid and $M$ is finitely generated, we can decompose $M$ into a direct sum. That is, $M = M^{r} \oplus M/(a_{1}) \oplus ... \oplus M/(a_{n})$ where $a_{i} | a_{i+1}$. Here the first part, $M^{r}$ is the free part of the decompositon, and the rest is the torsion part. So if the free rank is 0, that is $r = 0$ we have that $M$ is torsion, so yes they are equivalent. Also be wary, as Bernard said above ${0}$ is not linearly independent.