# 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.

• Awesome question! I tried in abelian groups and I'm pretty sure it can't happen there. I would check a local ring of dim 2, like a localization of k[x,y]. – Jack Schmidt Apr 2 '14 at 22:32
• Do you know other examples of abelian groups $M_i \neq 0$ with $M_i \otimes M_i = 0$ such that $M_i \otimes M_j \neq 0$ for $i \neq j$? These will provide examples, e.g. let $M := M_1 \oplus M_2$ then $M \otimes M = M_1^{\otimes 2} \oplus 2 M_1 \otimes M_2 \oplus M_2^{\otimes 2} = 2 M_1 \otimes M_2 \neq 0$. Nevertheless $M^{\otimes 3} = 0$. This still works for more than 2 summands since sums of nilpotents are nilpotent. – Dune Apr 3 '14 at 20:43
• Just a remark when looking for examples (I'm not sure if this is clear): finitely generated modules can't work, as for finitely generated modules over a commutative ring $A$ it holds $\operatorname{supp}(M \otimes N) = \operatorname{supp}(M) \cap \operatorname{supp}(N)$ where $\operatorname{supp}(M) = \left\{ \mathfrak{p} \in \operatorname{Spec}(A) | M_{\mathfrak{p}} \neq 0\right\}$ – Louis Apr 4 '14 at 0:08
• @Louis: I guess $M = N = \mathbb{Q}/\mathbb{Z}$ is a counterexample to that statement if we drop the finitely generated hypothesis? (I'm confused about this because I found a few statements online that this holds without the finitely generated hypothesis.) – Qiaochu Yuan Apr 4 '14 at 4:19
• I just found a paper ("Vanishing Tensor Powers of Modules", R.Wiegand & S.Wiegand, Math.Zeit. 129, (1972), 351-358) that might be of interest. Among other things, they prove that if $R$ is a commutative Noetherian ring and $d$ is the largest integer such that there is a module $M$ with $M^{\otimes d+1}=0$ but $M^{\otimes d}\neq0$, then $d$ is less than or equal to the Krull dimension of $R$. – Jeremy Rickard May 11 '14 at 10:21

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 $M\otimes M\otimes M=0$.

• Thanks! Three questions: (1) Is $M_1 \otimes M_2 \cong k[x,x^{-1},y,y^{-1}]/k[x,y]$? (2) Everything still works if we localize at $(x,y)$, right? Then $M_1 \otimes M_2$ is just the field of fractions. (3) We can use more variables to get larger powers, since we get something like the ring $k[ M_1, M_2, \ldots, M_n ]/(M_1^2,M_2^2,\ldots,M_n^2)$ with $\oplus$ the sum and $\otimes$ the product, right? – Jack Schmidt May 10 '14 at 14:56
• @JackSchmidt: (1) Not quite: $M_1\otimes M_2$ is $k[x,x^{-1},y,y^{-1}]/(k[x,x^{-1},y]+k[x,y,y^{-1}])$. It has a basis consisting of those $x^iyy^j$ where both $i<0$ and $j<0$. (2) I think it works if you localize, but I think that then $M_1\otimes M_2$ is the quotient of the field of fractions by the submodule consisting of those rational functions in $x$ and $y$ which don't have both $x$ and $y$ in the denominator. (3) Yes. – Jeremy Rickard May 11 '14 at 9:56
• So what is the general case? $R=k[x_1,\ldots,x_n]$, $M_i=\frac{k[x_1,\ldots,x_i^{\pm1},\ldots,x_n]}{k[x_1,\ldots,x_n]}$, $M=M_1\!\oplus\!\ldots\!\oplus\!M_n$, so that $M_i\otimes M_j\cong0$ iff $i=j$, and $M^{\otimes n}\ncong0\cong M^{\otimes n+1}$? – Leo Jul 17 '14 at 18:24

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.

• The theorem shows that if $k \geq 1$ and $M^{\otimes 3k} = 0$, then $M^{\otimes 2k} = 0$. Thus if $k \geq 1$ and $M^{\otimes 3k} = 0$, then $M^{\otimes (3k-1)} = 0$ since $2k \leq 3k-1$. Similarly if $k \geq 1$ and $M^{\otimes (3k+2)} = 0$, then $M^{\otimes 3(k+1)} = 0$ so $M^{\otimes 3k+1} = 0$ since $2(k+1) \leq 3k+1$. Finally if $k \geq 2$ and $M^{\otimes (3k+1)} = 0$, then $M^{\otimes 3k} = 0$. But what about the case $M^{\otimes 4} = 0$? How can we apply the theorem to prove $M^{\otimes 3} = 0$? – spin May 8 '14 at 13:34
• If (M (x) M)^((x)2) = 0, then M(x)M is a divisible DSC abelian group, so is 0. This sort of thing works for any Dedekind domain: the result of a tensor product is actually simpler than the ingredients, so the more tensoring there is, the less weird things can actually happen. – Jack Schmidt May 8 '14 at 14:43