Nonsolvable finite quotients of matrix groups

Question

Suppose that $\Gamma$ is a finitely generated nonsolvable subgroup of $GL(n, R)$. Is it in the literature that $\Gamma$ has a nonsolvable finite quotient? I know how to prove it (the hardest ingredient is due to Nori), but would prefer to give a reference instead of a proof.

Answer 1

It is a straightforward consequence of the combination of two old results of Malcev and most likely Malcev was aware of this consequence.

1) The first is the well-known Malcev's residual finiteness result (every finitely generated linear group over, say, a field, is residually finite), or rather the stronger result that follows from the proof. Indeed, the main ingredient is the lemma saying that every finitely generated domain is residually a finite field. As a corollary, we have that for every $n$, every field $K$, every finitely generated subgroup of $\mathrm{GL}_n(K)$ is residually a subgroup of $\mathrm{GL}_n(F)$ where $F$ ranges over finite fields (with the same $n$).

2) The second is Malcev's result (1951, English translation 1956) that solvable subgroups of $\mathrm{GL}_n(K)$ have solvability length $\le r_n$ for some $r_n$, independently of the field $K$. This can be found (Theorem 3.21) in the book of D. Robinson Finiteness conditions and generalized soluble groups I, Theorem 3.21. Actually, Malcev proved the stronger result that for some $s_n$ independent of the field, every solvable subgroup of $\mathrm{GL}_n(K)$ has a subgroup of index $\le s_n$ that is conjugate, in $\mathrm{GL}_n(\hat{K})$, to the upper triangular group ($\hat{K}$ denoting an algebraic closure).

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3) The combination of these two facts entails the result for linear groups over fields. Now suppose we have a non-solvable subgroup $\Gamma$ of $\mathrm{GL}_n(R)$ where $R$ is an arbitrary (associative unital) commutative ring. Let $N$ be the nilradical of $R$. It is no restriction to assume that $R$ is finitely generated. So the kernel of $\mathrm{GL}_n(R)\to\mathrm{GL}_n(R/N)$ is nilpotent; thus the image of $\Gamma$ in $\mathrm{GL}_n(R/N)$ is not solvable. Thus, we can suppose that $R$ is reduced (i.e., $N=0$). So $\{0\}$ is the intersection of the finitely many minimal prime ideals $P_i$ of $R$. It follows that the image of $\Gamma$ in at least one of the $\mathrm{GL}_n(R/P_i)$ is non-solvable. This reduces to the case of a field.