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.