$\operatorname{Ann}(M\otimes_A N)=\operatorname{Ann}M+\operatorname{Ann}N$? 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.
 A: Consider the $\mathbb{Z}$-modules $M:=\mathbb{Z}/m\mathbb{Z}$ and $N:=\mathbb{Q}$. Then if I am not terribly mistaking:
$\text{ann}M=m\mathbb{Z}$,
$\text{ann}N=0$,
$M\otimes_\mathbb{Z} N=0$.
A: This is an excellent question, and your intuition is not too far off.
Note that the counterexamples given in the other answers involve a non-finitely generated module ($\mathbb Q$ thought of as a $\mathbb Z$-module), and constructions like annihilators behave much better for finitely generated modules than for non-finitely generated ones (as a general rule).
E.g. Suppose that $M = R/I$ and $N = R/J$ are cyclic.  Then $M\otimes N = R/(I+J),$ and so your conjectured formula is true in that case.
As you observe, always $\operatorname{Ann} M + \operatorname{Ann} N \subset 
\operatorname{Ann} M\otimes N$.  Regarding the converse,
what is true for general finitely generated $R$-modules $M$ and $N$ is that $\operatorname{Ann}M\otimes N$ is contained in the radical of $\operatorname{Ann} M +  \operatorname{Ann} N$.  So your conjecture becomes correct if you restrict to f.g. modules, and take radicals of both sides.
[To see this, use the fact that for a finitely generated module, the radical of $\operatorname{Ann} M$ is equal to the intersection of those prime ideals $\mathfrak p$ for which $\kappa(\mathfrak p) \otimes_R M \neq 0$, where $\kappa(\mathfrak p)$ denotes the fraction field of $R/\mathfrak p$.]
I don't think that we can do much better than this though, in the finitely generated but non-cyclic case.  
To see the kind of things that can happen, choose three ideals $I_1,I_2,J,$
and let $M = R/I_1 \oplus R/I_2$ and $N = R/J$.  Then $\operatorname{Ann} M = I_1\cap I_2$, $\operatorname{Ann} N = J$, and $M\otimes N = R/I_1+J \oplus R/I_2 + J$, so that $\operatorname{Ann} M\otimes N = (I_1 + J) \cap (I_2  +J).$
Now we always have $(I_1\cap I_2) + J \subset (I_1 + J) \cap (I_2 + J) 
\subset \operatorname{rad}\bigl((I_1 \cap I_2) + J\bigr),$ but the first inclusion is typically not an equality (because the lattice of ideals in $R$ is always modular, but typically not distributive).
So if you take a counterexample to the distributive property for some lattice of ideals, feeding it into the above example will give a counterexample to the precise form of your conjecture (i.e. equality on the nose, rather than up to taking radicals) involving only finitely generated modules.
A: The containment $\mathrm{Ann}_A(M)+\mathrm{Ann}_A(N)\subseteq\mathrm{Ann}_A(M\otimes_AN)$ can be proper. For an example, take $A=\mathbf{Z}$, $M=\mathbf{Z}/2\mathbf{Z}$, and $N=\mathbf{Q}$. Then $\mathrm{Ann}_A(M)=2\mathbf{Z}$, and $\mathrm{Ann}_A(N)=0$, but $M\otimes_AN=\mathbf{Z}/2\mathbf{Z}\otimes_{\mathbf{Z}}\mathbf{Q}=0$ because the left tensor factor is killed by $2$ but $\mathbf{Q}$ is divisible. So $\mathrm{Ann}_A(M\otimes_AN)=\mathbf{Z}$. 
A: Let $R= k[x, y]$. We consider $M = I :=(x, y)^2$ and $N = R/J$ with $J = (x^2, y^2)$. It is easy to check that $I^2 = IJ$. We have $M \otimes_R N = I/IJ$. So we we have $I \subseteq \mathrm{Ann}(M \otimes_R N) $ (if fact the equality holds clearly). But note that $\mathrm{Ann}M = 0$ and $\mathrm{Ann}N = J$.
