Counterexample involving tensor product of $A$-modules equals zero

Question

I have proved the following statement and I'm looking for a counterexample where $M$ or $N$ is not finitely generated so that the implication does not hold. I would apppreciate any tips.

Let $A$ be a local ring, $M$ and $N$ finitely generated $A$-modules. Show that if $M \otimes_A N = 0$ then $M=0$ or $N=0.$

Answer 1

Consider $M=N=\mathbb{Q}/\mathbb{Z}$, then $M\otimes_{\mathbb{Z}} N = 0$, now take a localization: $\mathbb{Q}/\mathbb{Z}_{(p)}\otimes_{\mathbb{Z}_{(p)}} \mathbb{Q}/\mathbb{Z}_{(p)} =0 $.

Answer 2

For a simple example, let $A$ be any local ring with an element $a\in A$ that is neither nilpotent nor a unit. Then the localization $M=A[a^{-1}]$ is nonzero (since $a$ is not nilpotent) and the quotient $N=A/(a)$ is nonzero (since $a$ is not a unit), but $M\otimes_A N=0$ since $a$ must both annihilate it and act as a unit on it.