For $A$ a not necessarily commutative algebra over a field $\mathbb{k}$, can there exist two bi-modules $M$ and $N$ over $A$ such that, for some $m \in M$, and $n \in N$, $$ m \otimes_A n = 0? $$
I can see that this cannot happen if $N$ is projective as a left $A$-module, which is the case I happen to work with. However I am curious to know if projectivity is necessary to prevent $m \otimes_A n = 0$. I know that such examples exist over general rings, but for the special case of algebras over a field it is not clear to me what happens.
Yes, it can certainly happen. For example if $A = \Bbbk[X]$ and $M = N = A/(X^2)$, then $X \otimes X$ is an element of $M \otimes_A N$ which vanishes (because $X \otimes X = 1 \cdot X \otimes X = 1 \otimes X^2 = 0$).