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This question was asked earlier on math.stackexchange: click here. See the comments and the answer by Jack Schmidt there.

Let $M$ be a module over a commutative ring $R$.

It is possible that $M \otimes M = 0$ if $M$ is nonzero, for example when $R = \mathbb{Z}$ and $M = \mathbb{Q}/ \mathbb{Z}$.

What about when higher tensor powers of $M$ are zero? If $M \otimes M \otimes M = 0$, is it possible that $M \otimes M$ is nonzero? More generally if $M^{\otimes n} = 0$ for $n \geq 3$, is it possible that $M^{\otimes n-1}$ is nonzero?

The answer by Jack Schmidt in the math.stackexchange question shows that the answer to these questions is no, in the case of $R = \mathbb{Z}$, ie. in the case of abelian groups. Quote from his answer:

Theorem: If $M$ is an abelian group with $M \otimes M \otimes M = 0$, then $M$ is a divisible torsion abelian group and $M \otimes M = 0$.

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My previous attempt was completely wrong, as Jason Starr politely pointed out.

But I think the idea I was grasping for does work, in this example:

Let $R=k[x,y]$ for a field $k$, and let $$M=\frac{k[x,y,y^{-1}]}{k[x,y]}\oplus\frac{k[x,x^{-1},y]}{k[x,y]}.$$

[So that Jason's comment still makes sense, I'll leave my previous stupid wrong answer:

How about $R=\mathbb{Z}\times\mathbb{Z}$ and $M=(\mathbb{Q}/\mathbb{Z}\times\mathbb{Z})\oplus(\mathbb{Z}\times\mathbb{Q}/\mathbb{Z})$?]

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    $\begingroup$ That doesn't work. There is a surjective homomorphism to $R=\mathbb{Z}\times \mathbb{Z}$, namely multiplication by $(0,1)$ on the first summand plus multiplication by $(1,0)$ on the second summand. Since tensor product over $R$ is right exact, for every integer $n>0$, there is a surjectiction $M^{\otimes n} \to R^{\otimes n} = R$. $\endgroup$
    – Jason Starr
    Commented May 9, 2014 at 12:52
  • $\begingroup$ @JasonStarr: You're quite right, of course. I was being stupid. But I think the idea I was trying to make into an example does work, and I've edited my answer. $\endgroup$
    – Jeremy Rickard
    Commented May 9, 2014 at 13:43
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    $\begingroup$ I believe the new example works. By using $n$ variables instead of $2$ variables, you should also get a module $M$ such that $M^{\otimes n}$ is nonzero, yet $M^{\otimes(n+1)}$ is zero. $\endgroup$
    – Jason Starr
    Commented May 9, 2014 at 18:14

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