Why does free imply torsion-free? 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.
 A: 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.
A: Show that a direct sum of torsion-free modules is torsion-free, and that the module $R$ is torsion-free.
A: Any free $R$-module $M$ is a direct sum of copies of $R$.  That is, there is some isomorphism of $R$-modules $$M \cong \bigoplus_{i \in X} R$$ and since $R$ is an integral domain, you can check that direct sum is torsion-free.
A: Since $M$ is free $M$ has a basis $E$.
Take a non-zero element $m\in M$.
Then $m=a_1e_1+a_2e_2+\dots +a_ne_n$ with some $a_i\neq 0$.
then for any non-zero $r\in R$ we have:
$rm=(ra_1)e_1+(ra_2)e_2+\dots + (ra_n)e_n$. Notice $ra_i\neq 0$ because $a_i\neq 0$ and $R$ is a domain. therefore at least one coefficient is non-zero and $rm\neq 0$. Proving $M$ is torsion-free.
