If the tensor power $M^{\otimes n} = 0$, is it possible that $M^{\otimes n-1}$ is nonzero? 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? Can we find examples among $\mathbb{Z}$-modules (abelian groups)?
Here $M^{\otimes n} = M \otimes \cdots \otimes M$ denotes the tensor product of $M$ with itself $n$ times.
 A: I posted this example earlier on MathOverflow.
Let $R=k[x,y]$ for a field $k$, and let
$$M=\frac{k[x,y,y^{-1}]}{k[x,y]}\oplus\frac{k[x,x^{-1},y]}{k[x,y]}.$$
Then $M$ is a direct sum $M_1\oplus M_2$ of two modules for which $M_1\otimes M_1=0$, $M_2\otimes M_2=0$, but $M_1\otimes M_2\neq0$, so that $M\otimes M\cong(M_1\otimes M_2)\oplus (M_1\otimes M_2)\neq0$ but $M\otimes M\otimes M=0$. 
A: Here is the case for abelian groups:
Theorem: If $M$ is an abelian group with $M \otimes M \otimes M = 0$, then $M$ is a divisible torsion abelian group and $M \otimes M = 0$.
The proof is pretty standard abelian group theory. Basic subgroups are probably not well known outside of that theory (at least I never learned about them over rings that weren't Dedekind domains). Fuchs's Infinite Abelian Groups volume 1 has all the details (chapter V.27, VI.32-33, and X.61 should be everything needed beyond a basic course on modules). I suspect the results depend on $R$ being a Dedekind domain, since otherwise the notion of DSC subgroup is completely broken (affecting the proof in several spots). If you want to salvage it, the categorical version of DSC is called pure projective.
At any rate, I don't think anything like this works for $R=k[x,y]_{(x,y)}$, hence my suggestion to look there.
$p$-groups
Proposition: If $M$ and $N$ are abelian $p$-groups then $M \otimes N=0$ iff at least one of $M$ or $N$ is divisible.
Suppose $A$ is an abelian $p$-group. Then there exists a (so called basic subgroup) $B \leq A$ with the properties: (1) $B$ is a direct sum of cyclic groups, (2) $B$ is pure in $A$, and (3) $A/B$ is divisible.
Consider $A \otimes M$ for an abelian $p$-group $M$. By (2), we get that $0 \to B \otimes M \to A \otimes M \to (A/B) \otimes M \to 0$ is exact, and by (3) we get that $(A/B) \otimes M = 0$, so that $B \otimes M \cong A \otimes M$. By (1) we get that $B \cong \bigoplus_{i=1}^\infty \left(\mathbb{Z}/p^i\mathbb{Z}\right)^{(n_i)}$ for cardinals $n_i$. Hence $$A\otimes M \cong B \otimes M \cong \bigoplus_{i=1}^\infty \left(M/p^iM\right)^{(n_i)}$$
This is zero iff for each $i$, $M=p^iM$ or $n_i=0$. Note that $M=p^iM$ iff $M=pM$ so the dependence on $i$ is misleading: either $B=0$ or $M=pM$. By (3), we get either that $B$ is divisible or $M$ is divisible. This proves the proposition.
Proposition: If $M$ and $N$ are abelian $p$-groups, then $M \otimes N$ is divisible iff it is zero.
Note that $M/p^iM$ is a bounded $p$-group, so a direct sum of cyclic groups, hence tensor products of abelian $p$-groups are direct sums of cyclic groups, hence, reduced.
Corollary: If $M$ is an abelian $p$-group with $M \otimes M \otimes M =0$, then $M$ is divisible and $M \otimes M=0$.
From $M \otimes M \otimes M = 0$ we get that $M \otimes M$ is divisible, and so $M \otimes M = 0$, and so $M$ is divisible.
torsion groups
Torsion abelian groups are direct sums of $p$-groups, written $T = \oplus T_p$. If $P$ is an abelian $p$-group and $Q$ is an abelian $q$-group for $p\neq q$, then $P \otimes Q=0$. If $M$ and $N$ are torsion abelian groups, then so is $M \otimes N$, and $(M \otimes N)_p = M_p \otimes N_p$. Hence the propositions and corollary hold with “$p$-group” replaced with “torsion group”.
torsion-free groups
Proposition: If $M$ and $N$ are torsion-free abelian groups, then $M \otimes N = 0$ iff $M=0$ or $N=0$.
Corollary: If $M$ is a torsion-free abelian group with $M \otimes M \otimes M = 0$ then $M = 0$.
mixed groups
Consider the pure-sequence $0 \to t(M) \to M \to M/t(M) \to 0$. Hence $0 \to t(M) \otimes N \to M \otimes N \to M/t(M) \otimes N \to 0$ is exact. If $M \otimes N = 0$, then $t(M) \otimes N = M/t(M) \otimes N = 0$ as well.
Suppose $M \otimes M \otimes M = 0$. Then $t(M) \otimes (M \otimes M) = 0$ so $t(M) \otimes t(M) \otimes t(M) = 0$, so $t(M)$ is divisible by the torsion corollary. Similarly, abbreviating $X/t(X)$ as $tf(X)$, we get $tf(M) \otimes tf(M) \otimes tf(M) = 0$ so $tf(M) = 0$. Hence $M =t(M)$, and the theorem is proven.
