Let $R$ be a (unital) ring and let $A$ be a right $R$-module and let $B$ be a left $R$-module. Then we can define the tensor product $A\otimes_{R} B$, with the expected relations. In particular, we have that
$$a\cdot r\otimes b=a\otimes r\cdot b$$
for all $a\in A$, $r\in R$ and $b\in B$, where $\cdot$ is used to denote the appropriate module action.
Inspired by the question [https://math.stackexchange.com/questions/81640/proving-that-the-tensor-product-is-right-exact], I started wondering about when the equality $$a\otimes b=\tilde{a}\otimes\tilde{b}$$ holds in general. In essence, this seems to come down to the question of what can be said in general about $a\in A$ and $b\in B$ if $$a\otimes b=0.$$ Clearly, if either $a=0$ or $b=0$, then the statement is trivially true, so let us assume that both $a$ and $b$ are nonzero. Then what are necessary and sufficient conditions for the tensor product $a\otimes b$ to be zero? Right now I'm thinking that it should be necessary for at least one of $a$ and $b$ to be a zero divisor (i.e., there is an $r\neq 0$ such that $a\cdot r=0$ or $r\cdot b=0$).
Assuming that $a$ is a zero divisor, then my logic goes as follows: the expression $a\otimes b=0$ iff $b=r\cdot c$ for a $c\in B$ and $r\in R$ such that $a\cdot r=0$.
Likewise, if we assume that $b$ is a zero divisor then my logic is that $a\otimes b=0$ iff $a=c\cdot r$ for a $c\in A$ and $r\in R$ such that $r\cdot b=0$.
What I'm interested in knowing is whether my conclusion is correct or if I'm mistaken and there is a counterexample that breaks my argument. If so, what are the real necessary and sufficient conditions for an expression $a\otimes b$ to be zero? And more broadly, what can be said about $a,\tilde{a},b$ and $\tilde{b}$ if $a\otimes b=\tilde{a}\otimes\tilde{b}$? Because right now I'm thinking that this is true iff there is an $r\in R$ such that either $a=\tilde{a}\cdot r$ and $\tilde{b}=r\cdot b$ or $b=r\cdot\tilde{b}$ and $\tilde{a}=a\cdot r$.
