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...

  • $\begingroup$ If $M$ is finitely generated, yes.. B.t.w., $\{0\}$ is not linearly independent. $\endgroup$ – Bernard Mar 16 '17 at 0:26
  • $\begingroup$ What sort of problems would we run into if $M$ were not finitely generated? $\endgroup$ – green frog Mar 16 '17 at 0:28
  • $\begingroup$ The torsion part of $M$ is not necessarily a direct summand of $M$, if I remember well. $\endgroup$ – Bernard Mar 16 '17 at 0:29

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.

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  • $\begingroup$ Thank you! Totally forgot that we can decompose $M$ like that... $\endgroup$ – green frog Mar 16 '17 at 0:39

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