Question is :
Suppose $I$ is a principal ideal in a domain $R$. Prove that the $R$ module $I\otimes_R I$ is torsion-free.
Suppose we have $r(m\otimes n)=0$. Just for simplicity assume that $m\otimes n$ is a simple tensor so we have $m,n\in I$.
As $I$ is principal ideal, we have $I = aR$ for some $a \in R$, and thus $m=pa$ and $n=qa$ for some $p,q\in R$.
So, we have $r(pa\otimes qa)=0$ i.e, $rpq(a\otimes a)=0$.
We have $\varphi: I\times I\rightarrow I$ with $(xa,ya)\rightarrow xy$. This $\varphi$ is a bilinear map so we have $\varphi: I\otimes I\rightarrow I$.
Suppose we have $r(m\otimes n)=0$ then we should have $\varphi(r(m\otimes n))=0$ i.e, $rpq=0$...
As $R$ is an integral domain and $r\neq 0$ we should have $p=0$ or $q=0$ in which case we have $m=0$ or $n=0$ thus, $m\otimes n=0$...
So, this seem to be fine, but if i assume $m\otimes n$ is not a simple tensor then we have difficulty..
Suppose $m\otimes n=p_1a\otimes p_2a+p_3a\otimes p_4a$ then by previous observation this would mean $p_1p_2p_3p_4=0$ which says one of $p_i$ is zero which does not mean $m\otimes n=0$...
I am not sure how to proceed from here... Please suggest something....
Since $I$ is a principal ideal, it is isomorphic to $R$ as an $R$-module. Hence $$I \otimes_R I \cong R \otimes_R R \cong R$$ which is obviously torsionfree.