# Prove that $M/\mathrm{Tor}(M)$ is an R-module with zero torsion

I need to prove that $M/\mathrm{Tor}(M)$ is an $R$-module with zero torsion. ($R$ is assumed to be an integral domain.)

I already prove that $M/Tor(M)$ is an R-module, but in the part of zero torsion, I am lost. This is my attempt:

To prove that $M/Tor(M)$ has zero torsion, we have: $$Tor(M/Tor(M)) = \{\widehat{x} \in M/Tor(M) | r*\widehat{x} = 0, r\neq0,r \in R\}$$ where: $\widehat{x} = x + Tor(M).$ In other words, we have: $$r*(x + Tor(M)) = 0,$$

$$(r*x) + Tor(M) = 0,$$Since $r \neq 0 \Rightarrow r*x \in Tor(M) \Rightarrow Tor(M) = 0$, hence $M/Tor(M)$ has zero torsion.

Basically you want to show that if $r\hat x =0$, then $\hat x =0$. Equivalently, if $rx\in\operatorname{Tor}(M)$, then $x\in\operatorname{Tor}(M)$.
• @MatthewLeingangSo, what this means is that since $\hat x =0$, we end with $x\in\operatorname{Tor}(M)$, so basically all the elements in the form $\hat x$ are in $\operatorname{Tor}(M)$,then $\operatorname{Tor}(M/\operatorname{Tor}(M)) = 0$. – richitesenpai Feb 7 '15 at 3:25