This question was asked earlier on math.stackexchange: [click here](https://math.stackexchange.com/questions/737003/if-the-tensor-power-m-otimes-n-0-is-it-possible-that-m-otimes-n-1-i). 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 $M \otimes M = 0$.
> 
> 
> 



If the tensor power $M^{\otimes n} = 0$, is it possible that $M^{\otimes n-1}$ is nonzero?
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 $M=(\mathbb{Q}/\mathbb{Z}\times\mathbb{Z})\oplus(\mathbb{Z}\times\mathbb{Q}/\mathbb{Z})$?]


