Does $a\otimes b=0$ in $M\otimes_R N$ imply $ar=0$ or $rb=0$ for some non-zero $r\in R$? 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?
 A: 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 $R$.
A: The following argument works if $N$ is flat (in order for the embedding $aR \subseteq M$ to be preserved by the tensor product with $N$) and $R$ is a reduced ring.
If $a$ is a torsion element of $M$, then there exist a regular $r \in R$ such that $ar = 0$, regardless of whether $a \otimes b = 0$ or not. In this case, your conclusion is trivially true.
Let us asume now that $a$ is not a torsion element of $M$. Just by definition, it follows that $aR \subseteq M$ is a torsionless $R$-module: every $f \in (aR)^* = \hom_R (aR,R)$ is given by its value in $a$, say $f(a)=q \in R$, so if $r \in R$ is such that $f(ar) = 0$ for all $f \in (aR)^*$ it follows that $qr = 0$ for all $q \in R$, and in particular $r^2 = 0$, whence $r=0$ (because $R$ is reduced). We conclude that $\bigcap \limits _{f \in \hom _R (aR,R)} \ker f = 0$.
If $f \in \hom _R (aR,R)$ then
$$0 = (f \otimes \text{id}_N) (0) = (f \otimes \text{id}_N) (a \otimes b) = f(a)b .$$
There are two possibilities:

*

*if $a \in \bigcap \limits _{f \in \hom _R (aR,R)} \ker f$ then $a=0$, so you may just take $r=1$ (assuming $R$ to be unital);


*if $a \notin \bigcap \limits _{f \in \hom _R (aR,R)} \ker f$ then there exist $f_0 \in \hom _R (aR,R)$ such that $f_0 (a) \ne 0$, so letting $r_0 = f_0 (a)$ implies that $r_0 b = 0$.
If $R$ is unital we can do even better: considering the form $f(ar)=r$, we get $r_0 =f(a)=1$, which implies $b=0$.
