TITLE: Why does free imply torsion-free?
QUESTION [7 upvotes]: I want to verify that if $R$ is an integral domain and $M$ is an $R$-module, that if $M$ is free, $M$ must also be torsion-free.
Where can I start with this? I feel like it is obvious but I can't see it. I am just getting started with my course.

REPLY [4 votes]: Fix a basis $\{b_i\}_{i \in I}$ of $M$. Take $m \in M$, $m \neq 0$, and $r\in R$ such that $rm = 0$. Write $m = m_{1}b_{i_1}+\cdots+m_k b_{i_k}$, for some scalars $m_j \in R$. Then $$rm = rm_1b_{i_1}+\cdots+ rm_k b_{i_k} = 0$$gives $rm_jb_{i_j} = 0$ for all $1 \leq j \leq k$. Since $m \neq 0$, there is $j^\ast$ with $m_{j^\ast} \neq 0$. Since $\{b_{i_{j^\ast}}\}$ is linearly independent, we have that $rm_{j^\ast}b_{i_{j^\ast}} = 0$ implies $rm_{j^\ast} = 0$. Since $R$ is an integral domain and $m_{j^\ast} \neq 0$, we get $r=0$ as wanted.