In the course of working on an exercise in Atiyah-MacDonald (exercise 3 on p. 31), I've come to the belief that, for $A$ an arbitrary commutative ring and $M,N$ arbitrary $A$-modules,
$\operatorname{Ann}(M\otimes_A N)=\operatorname{Ann}M+\operatorname{Ann}N$
where $\operatorname{Ann}M$ is the annihilator of $M$, etc. While I think it is trivial that $\operatorname{Ann}(M\otimes_A N)\supset\operatorname{Ann}M+\operatorname{Ann}N$, I am having trouble establishing the reverse inclusion and would appreciate your help in doing so, if it is true, or if it is false I would appreciate a counterexample.
Here is what I have thought about so far:
On an intuitive level, because $M\otimes_A N$ is "the biggest/freest" bilinear image of $M\times N$, it should not be annihilated except by the smallest ideal of $A$ that annihilates either of $M$ and $N$, and this is $\operatorname{Ann}M+\operatorname{Ann}N$. I think this is why I believe the claim to be true.
My attempts to create an actual proof have focused on the universal properties of the relevant objects: (a) there is an $A$-bilinear map $M\times N\rightarrow M\otimes_A N$ such that any $A$-bilinear map from $M\times N$ factors through this one; and (b) any homomorphism from $A$ that factors through both $A/\operatorname{Ann}M$ and $A/\operatorname{Ann}N$ also factors through $A/(\operatorname{Ann}M+\operatorname{Ann}N)$. This is the forest I've gotten lost in. The homomorphisms from $A$ I can think of are into the endomorphism rings of $M,N,M\otimes_A N$. I haven't seen how to put them together with bilinear maps to show I'm aiming for.
Again, either an indication of how you'd complete the proof (preferably in terms of the universal properties, so I can see how to do what I've been trying to do) or a counterexample would be much appreciated.