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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?

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    It is not true. Note $\mathbb{F}_p \bigotimes_{\mathbb{Z}} \mathbb{F}_q = 0$ when $p,q$ are distinct primes. – 2017-02-12
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    $r=0$, obviously. – 2017-02-12
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    @basket how is this a counter example? For any element $[x]_p\in\Bbb F_p$ you have that $p\cdot [x]_p=0$, so $[x]_p\otimes [y]_q = 0$ for any $[x]_p,[y]_q$ is not in contradiction to the statement. – 2017-02-12
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    @AlexM. I am sorry I forgot to write $r\neq 0$. – 2017-02-12
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    Is this question trivial? That's why a negative vote? Or is it because of my typo? – 2017-02-12
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    @AlexM. I am thinking the general situation but can you say something if $R=\mathbf{Z}$? – 2017-02-15

2 Answers 2

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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 $R$.

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The following argument works if $N$ is flat (in order for the embedding $aR \subseteq M$ to be preserved by the tensor product with $N$).

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 $M$. It follows that $aR \subseteq M$ is a free right $R$-module. Being free, it is also torsionless, so $\bigcap \limits _{f \in \hom _R (aR,R)} \ker f = 0$.

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 $b=0$.

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    This only works if $M=aR$. Note that the natural map $aR\otimes N\to M\otimes N$ may not be injective, you might have $a\otimes b=0$ in $M\otimes N$ but not in $aR\otimes N$. – 2017-02-15
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    @EricWofsey: What if $N$ were flat - would this fix the proof? – 2017-02-15
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    Yes, it should work if $N$ is flat. – 2017-02-15
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    The first version of this answer was wrong, because it was based on a misreading of the question; nevertheless, it got upvoted. Next, I completely rewrote the post, obtaining the current correct answer; the upvoter canceled his upvote... If I were an alien to judge the human species by just this behaviour, I would doubt it being intelligent. Finally, some other users decided to downvote without leaving any comment, which makes me ask: do they understand what's written above? – 2017-02-16