# Modules $M$ such that the automorphism of $M \otimes M \otimes M$ induced by the permutation $(123)$ is the identity

I've been struggling with the following problem for a couple of days and I don't seem to get any further:

Let $R$ be a commutative ring. I would like to get (something like) a classification of all finitely generated $R$-modules $M$ that satisfy the following condition:

When we look at $M \otimes_R M \otimes_R M$, the permutation (123) induces an automorphism of $M \otimes_R M \otimes_R M$ by sending $a \otimes b \otimes c$ to $c \otimes a \otimes b$. I demand that this automorphism be the identity map. In other words, in $M \otimes_R M \otimes_R M$, the elements $a \otimes b \otimes c, c \otimes a \otimes b, b \otimes c \otimes a$ should all be the same.

If $R$ is a field, it is easy to see that the only non-trivial finitely generated $R$-modules (i.e. finite-dimensional vector spaces) that satisfy this condition are the 1-dimensional ones. Furthermore, one sees more generally that all cyclic modules satisfy the condition, too. Up to now I've neither come up with an example of a non-cyclic module that satisfies the condition, nor was I able to prove that all modules that satisfy this condtion must be cyclic.

Can somebody help with this matter?

It is not true that a module that satisfies the above condition is cyclic. For instance, it is possible that $M \neq 0$ but $M \otimes M$ (and thus higher tensor powers) are zero. Consider for instance $M = \mathbb{Q}/\mathbb{Z}$ over the integers. Then $$M \otimes M = \mathbb{Q}/\mathbb{Z} \otimes \mathbb{Q}/\mathbb{Z} = 0$$ since to tensor with $\mathbb{Q}/\mathbb{Z}$ is to tensor over $\mathbb{Q}$ and take a quotient, and a torsion group tensored with $\mathbb{Q}$ is zero.

Now, suppose $M$ is finitely generated. The above example was not (and, in fact, the above phenomenon can't happen for a finitely generated module). Then it is easy to see that if $M$ satisfies your condition, so does the base-change $M \otimes_R R'$ for $R'$ any $R$-algebra (considered as an $R'$-module, that is!) because base extension is a monoidal functor with respect to the tensor product. It follows that for any prime ideal $\mathfrak{p}$, $M \otimes k(\mathfrak{p})$ (for $k(\mathfrak{p})$ the residue field, i.e. the quotient field of the residue ring) satisfies this property, so from what you have proved about vector spaces, it has rank at most one. Thus all the fibers of $M$ are of rank at most one. So at least $M$ is "close" to being cyclic.

More generally, the above observation on base-change reduces the question to the case of a local ring, because to check that two things are equal, it suffices to check on all the localizations. By Nakayama's lemma, it follows that for a local ring, and your observation for a field, any module satisfying your property is in fact cyclic. For local rings, we thus have an if and only if statement.

Putting the above observations all together, we find:

A module satisfies your property if and only if the localization at each prime is cyclic (as a module over the localized ring).

• This is a great answer, thank you very much! – Sebastian Jan 12 '11 at 14:50
• Notice that it is not true that «tensoring with $\mathbb Q$ does nothing to a divisible group», for the result of tensoring with $\mathbb Q$ is a vector space of $\mathbb Q$, so a uniquely divisible group, yet there do exist non-uniquely divisible abelian groups! – Mariano Suárez-Álvarez Nov 10 '11 at 5:30
• @Mariano: Thanks for the correction! – Akhil Mathew Nov 10 '11 at 7:09