What does a zero tensor product imply? 
I'm trying to prove that for two finitely generated $A$-modules $M,N$ ($A$ being any cmmutative ring), the tensor product $M\otimes_A N$ is zero iff $\operatorname{Ann}(M)+\operatorname{Ann}(N)=A$.

The if direction is of course easy- just show $1$ as a sum of two annihilating elements, $r\in Ann(M)$ and $s\in Ann(N)$, and for any $m\otimes n\in M\otimes_A N$ we have that 
$$m\otimes n=(r+s)(m\otimes n)=rm\otimes n+m\otimes sn=0$$
The only if directions is what got me baffled. By now I know that it suffices to show that  $$M\otimes_A N=0\Rightarrow N=Ann(M)N\text{ or } M=Ann(N)M\tag{$**$}$$ 
since both modules are fin.gen, the claim will follow (either trivially, if one of the annihilators is zero, or by Nakayama's Lemma, if both are non-zero).
But I'm stuck on showing that, and I'm not even sure if it is always the case that $(**)$ holds... Does anybody have any idea? Maybe a hint in the case where it is true?
(It could be that the ring $A$ should be noetherian, I'm not sure about that...)
In any case- I would very much appreciate if someone can suggest some intuitions on how to prove when a tensor product is non-zero, or equivalently, what can be entailed from a zero tensor product.
Thanks in advance
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ADDED: In response to @Dylan Moreland's question on how I intend on using Nakayama's Lemma:
Once we've seen that (for example) $N=Ann(M)N$, since $N$ is a finitely generated module (and after seeing that $Ann(M)\subsetneq A$) we have by NL that there exists some $\alpha\equiv 1\mod Ann(M)$ such that $\alpha N=0$. In paticular $\alpha\in Ann(N)$, and since $\alpha\equiv 1\mod Ann(M)$ we have some $\beta\in Ann(M)$ such that $\alpha+\beta=1$. This implies that $1\in Ann(N)+Ann(M)$ and so the only if direction is proved (at least I think this proof is sound, if someone sees a flaw, I'd be happy to hear about it)
 A: Here is an alternative proof, using the facts


*

*$\mathrm{supp}(M \otimes N) = \mathrm{supp}(M) \cap \mathrm{supp}(N)$ (Hint: Nakayama and Linear algebra)

*$\mathrm{supp}(M) = V(\mathrm{Ann}(M))$


This easily implies $V(\mathrm{Ann}(M \otimes N)) = V(\mathrm{Ann}(M) + \mathrm{Ann}(N))$. Hence, $M \otimes N=0$ iff the set is empty iff $\mathrm{Ann}(M) + \mathrm{Ann}(N)=A$.
A: Try the contrapositive.  If $Ann(M) + Ann(N)$ is a proper ideal, what interesting kind of ideal can you choose that contains it?  Try using that ideal to help.
Also, as a general rule, to show that $M\otimes N$ is non-zero, try to find a map to some quotient that is simpler to understand (and so simpler to show is non-zero).
A: Just a summary,
let $M,N$ fin. gen. $A$-modules. TFAE:
i) $M\otimes N=0$
ii) $Ann(M) + Ann(N)=A$. 
iii)$Ann(M) N = N$
Proof: $i)\Leftrightarrow ii)$ Done above.
$ii)\Rightarrow iii)$ Multiply $ii)$ by $N$.
$iii) \Rightarrow i)$ $M\otimes N = M\otimes  Ann(M) N =Ann(M) M\otimes  N =0$.
In particular $(**)$ holds.
RH & eduard
