22
$\begingroup$

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

.................................................

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)

$\endgroup$
10
  • 2
    $\begingroup$ $M \otimes N$ being zero means that every $A$-bilinear map emanating from $M \times N$ is identically zero. Does this help as a hint? $\endgroup$
    – Dactyl
    Dec 21, 2011 at 17:39
  • $\begingroup$ @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? $\endgroup$
    – kneidell
    Dec 21, 2011 at 18:11
  • 1
    $\begingroup$ I asked a similar question, about what can we say about a zero tensor product, here: math.stackexchange.com/questions/81640/… $\endgroup$
    – Klaus
    Dec 21, 2011 at 19:16
  • $\begingroup$ @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 $x\in A$ $\endgroup$
    – kneidell
    Dec 21, 2011 at 20:08
  • 2
    $\begingroup$ 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, $\endgroup$
    – Matt E
    Dec 21, 2011 at 20:11

3 Answers 3

12
$\begingroup$

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).

$\endgroup$
2
  • $\begingroup$ How do we proceed? The natural candidate for an ideal containing $J=Ann(M)+Ann(N)$ is taking a maximal ideal $I$ containing $J$. Then do we claim that the bilinear map $M \times N \to M \otimes N / I (M \otimes N)$ given by $f(m,n)=m\otimes n + I M \otimes N$ is nonzero? And why? $\endgroup$
    – Emolga
    Feb 15, 2016 at 20:20
  • 1
    $\begingroup$ What I did (about $\aleph_0$ years ago, when I posted this question), was to show that $M/IM$ and $N/IN$ are non-zero vector spaces over $A/I$ (you'll have to use Nakayama for this one), and that there's a surjective map $M\otimes_A N\to (M/IM)\otimes_(A/I) (N/I N)$. In particular, we get that the domain of the map cannot be zero. $\endgroup$
    – kneidell
    Feb 22, 2016 at 7:23
7
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.