Does $m\otimes n=0\in M\otimes N $ imply there is $ r\neq 0\in R$, $m'\in M$ such that $rm'=m$ and $rn=0$, or $n'\in N$ such that $rn'=n$ and $rm=0$?

Question

Let $R$ be a commutative ring with unit. Does $m\otimes n=0\in M\otimes N $ imply there exists $r\neq 0\in R$, $m'\in M$ such that $rm'=m$ and $rn=0$, or $n'\in N$ such that $rn'=n$ and $rm=0$?

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Phase in another way, if $m\otimes n=0$ when can I factor out a non-zero $r$ from $m$ such that it annilates $n$.

If $R$ is a zero ring, then there doesn't exists such $r$ obviously. But for general $R$, when can we factor out a $r$ from $m$, such that it annilates $n$?

Update:

• I proved below it's true if R is non-zero unital commutative ring.

• Some other related questions for non-commutative $R$: Does $a\otimes b=0$ in $M\otimes_R N$ imply $ar=0$ or $rb=0$ for some non-zero $r\in R$?.

Answer 1

I will answer myself, when $R$ is not the zero ring this will holds.

$\textbf{Lemma}$ : Let $R$ be a commutative ring, $M,N$ are $R$-modules. Then for any element $m\otimes n\in M\otimes N$, the annilator satisfies $$ann(m\otimes n)= ann(m)+ann(n).$$

Proof of Lemma: Consider the R-module map $R\rightarrow (m)$ whose kenerl is $ann(m)$. Then we have $R/ann(m)=(m)$ and $R/ann(n)=(n),$ and the submodule $(m\otimes n)\cdot R\subset M\otimes N $ satisfies $$R/ann(m\otimes n) = (m\otimes n)\cdot R = (m\cdot R)\otimes (n\cdot R) = R/ann(m)\otimes R/ann(m)$$ where $(m\cdot R)\subset M$, and $(n\cdot R)\subset N$. Since $$(R/I\otimes R/J)=R/(I+J)$$ it follows that $$R/ann(m\otimes n)=R/(ann(m)+ann(R))$$ which implies that the annilators satisfy $$ann(m\otimes n)=ann(m)+ann(n).$$

$\textbf{Proof of the claim}$: $m\otimes n=0$ implies there exists a non-zero $r$ such that $m\otimes n =m'\otimes rn $ and $rn=0.$

1. if $ann(n)=0$, then $ann(m\otimes n)=amm(m)$. Since $(m\otimes n)=0$ this implies $ann(m)=R$ so $m=0$. We can take $r=1$.
2. if $ann(n)\neq 0,$ we let $r\neq 0\in ann (n)$. If $m\in r\cdot M$ we are done since $m\otimes n = rm'\otimes n = m'\otimes rn$; If $m\notin r\cdot M$,we prove there is a contradiction by considering the following $R$ module $M/r\cdot M\times N$. Then the natural projection $f$ $M\times N \rightarrow M/r\cdot M\times N$ is a bilnear map which sends $(m,n)$ to $([m],n)$. It is not the zero map since $m\notin r\cdot M$. By the universal property of tensor product, it factors through $M\otimes N$ and it sends $(m\otimes n)$, which is zero, to $f(m,n)$ which is non-zero. So we have a contradiction and thus $m\in r\cdot M.$