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 $M \otimes M = 0$.

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 $M \otimes M = 0$.

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 $M^{\otimes n-1} = 0$?

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 $M \otimes M = 0$.

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 $M \otimes M = 0$.