# $I\otimes I$ is torsion free for a principal ideal $I$ in domain $R$

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 $m=pa$ 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....

• Any principal ideal in a domain $R$ is $\cong R$ as an $R$-module. – darij grinberg Jul 23 '15 at 19:31
• Oh Oh oh..... So, $I\cong_R R$ so, $I\otimes_R I\cong_R\otimes_R R=R$.. Thus, $I\otimes_R I \cong_R$. As $R$ is torsion free as $R$ module, then so is $I\otimes I$?? @darijgrinberg – user87543 Jul 23 '15 at 19:34
• For understanding purposes, I am still interested in what @PraphullaKoushik asked, namely: if you can show that every simple tensor is torsion free, can you conclude that every element in $I\otimes_R I$ is torsion free? Does anyone have an answer? – Pawel Aug 6 '16 at 23:00

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.