# 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.

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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)

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$M \otimes N$ being zero means that every $A$-bilinear map emanating from $M \times N$ is identically zero. Does this help as a hint? –  Dactyl Dec 21 '11 at 17:39
@Dactyl: Thanks, It might- any chance of fishing for a little more? I'm thinking of finding a surjective bilinear map $M\times N\to N/Ann(M)N$, am I in the right direction? –  kneidell Dec 21 '11 at 18:11
I asked a similar question, about what can we say about a zero tensor product, here: math.stackexchange.com/questions/81640/… –  Klaus Dec 21 '11 at 19:16
@Klaus: We can look at the tensor produce $A\beta(b)\otimes_AAm$. If my assertion is correct, then $1=s+r\in Ann(\beta(b))+Ann(m)$, and $(s+r)(b\otimes m)=sb\otimes m+0$. since we know that $s\beta(b)=\beta(sb)=0$ this imples that $sb\in\ker(\beta)=Im(\alpha)$ and $b\otimes m=\alpha(x)\otimes m$ for some $x\in A$ –  kneidell Dec 21 '11 at 20:08
Dear kneidell, 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. (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.) Regards, –  Matt E Dec 21 '11 at 20:11

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).

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Here is an alternative proof, using the facts

1. $\mathrm{supp}(M \otimes N) = \mathrm{supp}(M) \cap \mathrm{supp}(N)$ (Hint: Nakayama and Linear algebra)
2. $\mathrm{supp}(M) = V(\mathrm{Ann}(M))$

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 $\mathrm{Ann}(M) + \mathrm{Ann}(N)=A$.

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