TITLE: Does $a\otimes b=0$ in $M\otimes_R N$ imply $ar=0$ or $rb=0$ for some non-zero $r\in R$?
QUESTION [6 upvotes]: Let $R$ be a ring and let $M$ be a right $R$-module and $N$ be a left $R$-module. Then $M\otimes_R N$ is the tensor product over $R$.
Consider $a\in M$ and $b\in N$ such that the pure tensor $a\otimes b=0$.
Then I understand that this does not imply that $a=0$ or $b=0$.
My question is: is there $r\in R$, $r\ne0$ such that $ar=0$ or $rb=0$?
If the question is true, I want to know the proof. If not, could you give me some counterexample?

REPLY [2 votes]: Here's a silly counterexample.  Let $R$ be the zero ring.  Then there are no nonzero elements of $R$, so such an $r$ trivially can never exist.  (Note that the only $R$-module is the $0$ module, so there is only one version of this example: $M=N=0$ and $a=b=0$.)
If $R$ is an integral domain, then the answer is yes.  Let us suppose $a\in M$ and $b\in N$ are not torsion (i.e., they are not annihilated by any nonzero element of $R$).  Let $K$ be the field of fractions of $R$ and let $T(M)$ be the submodule of torsion elements of $M$.  Note that $M/T(M)$ is torsion-free, and so the inclusion map $M/T(M)\to M/T(M)\otimes_R K$ is injective (you can prove this by identifiying $M/T(M)\otimes_R K$ with the localization of $M/T(M)$ with respect to all nonzero elements of $R$).  In particular, since $a\in M$ is not a torsion element, then the image of $a$ under the composition $M\to M/T(M)\to M/T(M)\otimes_R K$ is nonzero.  Since $M/T(M)\otimes_R K$ is a $K$-vector space, there exists a $K$-linear map $M/T(M)\otimes_R K\to K$ which sends the image of $a$ to $1$.  Composing all these maps together, we get a homomorphism $f:M\to K$ such that $f(a)=1$.
Since $b\in N$ is also not torsion, then by the same argument there is a homomorphism $g:N\to K$ such that $g(b)=1$.  There is then a unique homomorphism $h:M\otimes N\to K$ such that $h(m\otimes n)=f(m)g(n)$ for all $m$ and $n$.  We then have $h(a\otimes b)=1$, so $a\otimes b\neq 0$.
I don't know what happens for arbitrary rings $R$.