Suppose $\Gamma$ is a (finitely presented, but this is probably irrelevant) group, and $M$ is a finitely generated (EDIT: finitely presented) module over $\mathbb{Q}\Gamma$ which is infinite-dimensional (as a vector space over $\mathbb{Q}$.) Define a module $M \wedge_{\mathbb{Q}} M$ as either the symmetric or the alternating tensor product over $\mathbb{Q}$, with the module structure given by $\gamma(m \wedge n)=\gamma m \wedge \gamma n$. Is it true that $M \wedge_{\mathbb{Q}} M$ cannot be finitely generated?

My intuition is that there will always be some $\alpha \in M$ and a sequence of $\gamma_i \in \Gamma$ such that every element of the form $\alpha \wedge \gamma_i\alpha$ can't be generated by a finite set of elements. The simplest example is when $\Gamma=\mathbb{Z}$ and $M=\mathbb{Q}[\mathbb{Z}]$. Then it's easy to show that a finite set in $M \wedge_{\mathbb{Q}} M$ can only generate a finite-dimensional set of elements of the form $\alpha \wedge m$, where $\alpha$ generates $M$. [In fact, this argument generalizes to any cyclic module over any group. This also means that the statement is true for groups $\Gamma$ for which $\mathbb{Q}\Gamma$ is Noetherian, but these are rather rare.] <- the argument I had in mind here doesn't actually work, never mind

To prove the general statement it's enough to prove it for modules over the free group rings, $\mathbb{Q}F_n$, though this is unlikely to make it any easier.

This question, though reduced to pure algebra, came up in a topological setting. I'm wondering if there's a ring-theoretic reason that the statement is true.


1 Answer 1


If $\Gamma$ acts $2$-transitively on an infinite set $X$, then the permutation module $\mathbb{Q}[X]$ will be a counterexample.

For example, take an action of the free group of rank $2$ on a countable set, where one generator acts transitively and the other generator has one fixed point and acts transitively on the remaining elements.

Edit: In fact, $2$-transitivity is not necessary for this to work: it's sufficient that a point stabilizer acts on $X$ with only finitely many orbits. Also, if the point stabilizer is finitely generated then $\mathbb{Q}[X]$ will be a finitely presented $\mathbb{Q}[\Gamma]$-module.

An example is Thomson's group $F$ acting on the set $X$ of dyadic rationals strictly between $0$ and $1$. It is a finitely presented group, as is a point stabilizer (which is isomorphic to $F\times F$), and a point stabilizer acts on $X$ with three orbits.

[I have no reason to think there couldn't be a far simpler example, except for the fact that I didn't think of one.]

  • $\begingroup$ OK, I'm going to change the goalposts a bit. Is there a finitely presented counterexample? $\endgroup$
    – Fedya
    Oct 10, 2014 at 0:59
  • $\begingroup$ @Fedya I've edited my answer to give a finitely presented example (which might well be absurdly over-complicated). $\endgroup$ Oct 10, 2014 at 8:59

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