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Let $F_2$ be the free group on two generators $x,y$ and let $R$ be the group algebra $\mathbf{Q}[F_2]$. Let $a,b,c$ be integers. Then define a right $R$-module

$M = R / (ax + by + c) R$.

I am interested in the question of whether $M$ is "detected" by finite quotients of $F_2$. That is; given $(a,b,c)$, does there exist a surjection $f$ from $F_2$ to a finite group $G$ such that $M \otimes_{\mathbf{Q}[F_2]} \mathbf{Q}[G]$ is nontrivial? There are some necessary conditions that aren't hard to check; for instance, a,b,c need to satisfy the triangle inequality with respect to any valuation on $\mathbf{Z}$. But are these conditions sufficient? For example, what if $(a,b,c) = (5,6,7)$; how would I check in this case whether $M$ is detected on some finite quotient of $F_2$?

Of course I am also interested in other elements of R, not just linear combinations of $x,y$, and $1$; but this seems like the simplest interesting case.

Remarks:

  1. This question has a lot in common with this one; but in the older question, we are really asking about integral group rings, where invertibility is harder to come by.

  2. The question is also related (at least in my mind) to questions about virtual positive Betti number for a finitely presented group $\Gamma$; in that case, Fox calculus gives a uniform presentation of the abelianization of the kernel of a quotient map $f: \Gamma \rightarrow G$ as a $\mathbf{Q}[G]$-module, which, if the virtual Betti number is positive, is nontrivial for some finite $G$.

  3. One might also think of this question as having something to do with a "non-abelian Manin-Mumford conjecture." Note that if $F_2$ were replaced with $\mathbf{Z}^2$, the question would ask about the intersection between torsion points in $\mathbf{G}_m^2$ and the line $ax+by+c=0$. Manin-Mumford tells us that this intersection is finite. Is there a similar finiteness statement in this case? I guess this might say something like: is there a finite set of finite quotients $G_1, ... G_k$ such that any finite quotient of $F_2$ detecting $M$ has some $G_i$ as a quotient?

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  • $\begingroup$ Let $G$ be a finite quotient of $F_2$. The $R=Q[F_2]$ maps onto both $Q[G]$ and $M$. What do you call "the projection of $M$ to $Q[G]$" (at first glance it would make sense if $M$ were a submodule rather than a quotient of $R$)? $\endgroup$
    – YCor
    Mar 25, 2016 at 20:14
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    $\begingroup$ @YCor. Maybe by "projection", Jordan means $M\otimes_{\mathbf{Q}[F_2]} \mathbf{Q}[G]$. Maybe Jordan is asking whether there exists a quotient $F_2\to G$ such that this tensor product module is a nonzero (right) $\mathbf{Q}[G]$-module. $\endgroup$
    – Jason Starr
    Mar 25, 2016 at 20:27
  • $\begingroup$ Yes, sorry, this is what I'm saying. I will modify the post to clarify. $\endgroup$
    – JSE
    Mar 25, 2016 at 22:01
  • $\begingroup$ OK thanks (btw, I don't think the tag gt is justified: most likely people subscribing to gt and neither gr nor ra will not be interested) $\endgroup$
    – YCor
    Mar 25, 2016 at 22:23
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    $\begingroup$ I only subscribe to gt and I'm interested. So for example, by taking $G$ to be the 3-element group, we get that triples of the form $a=b=c$ or $a+b+c=0$ are acceptable (those being irreducibles in $\mathbb{Q}[G]$.) In general, your condition says that, for some finite $G$ with two generators $x$ and $y$, $a+bx+cy$ is contained in a proper submodule of $\mathbb{Q}[G]$; that is, in some combination of subrepresentations of the regular representation which isn't all of them. (Just trying to reword the problem in lots of ways...) $\endgroup$
    – Fedya
    Mar 26, 2016 at 18:18

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