I have the following exercise:
Let R be a commutative non-trivial ring. Show that R is torsion-free (as a R-module) iff R is an integral domain.
So I assumed first R is torsion-free, so I take a non-zero element $r \in R$, then choose two elements $a,b \in R$ such that $ra = rb$ then we have that $r(a-b) = 0 \space$ (because $r \neq 0$ and as R is torsion free if we have that $ a \in R$ non-zero and $r \in R$ then $ra = 0 \Rightarrow r=0$) $\Rightarrow a-b = 0 \Rightarrow a=b$ so we have that R is an integral domain.
For the converse we have that $ \forall a,b \in R \space a\neq 0 \neq b$ we have $ab \neq 0$ this because R is assumed an integral domain; so we take any $r \in R$ non-zero and $a \in R$ with $ar = 0$ so as $r \neq 0$ this implies (because R is a domain) that $a=0$, so we conclude R is torsion-free
Is my proof correct? As we are seeing modules in my algebra class this problems seems to be trivial, and I think that I must be skipping something. Please clarify.