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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 that whether there exist $a'\in M$ or $b'\in N$ such that $$a\otimes b=0\otimes b'=a'\otimes 0?$$

We defined the tensor product to be the quotient group of the free abelian group on the set $M\times N$ divided by the subgroup generated by the usual relations for tensors.

If the question is true, I want to know the proof. If not, could you give me some counterexample?

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    The identity element in any group is unique, and both $ a \otimes b = 0 $ and $ 0 \otimes b' $ are identity elements, so they must always be equal no matter what $ b' $ is. – 2017-02-12
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    Yes: $0=0\otimes0$. – 2017-02-12

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Remember that the tensor product is $R$-balanced, meaning that $mr \otimes n = m \otimes rn$ for all $m \in M$, $n \in N$, $r \in R$. It follows that (choosing $r=0$)

$$a \otimes b = 0 = 0 \otimes 0 = 0 \otimes (0 \cdot b') = (0 \cdot 0) \otimes b' = 0 \otimes b'$$

for every $b' \in N$. A similar argument for $a' \in M$. This shows that every $a' \in M$ and $b' \in N$ satisfies your requirement.

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For any $b'\in N$, $0\otimes b'=0$. Therefore this is indeed equal to $a\otimes b$, if the latter is zero.