A group $G$ is said to satisfy the Tits alternative if any finitely generated subgroup of $G$ is either virtually solvable or contains a nonabelian free subgroup. Tits proved this for linear groups over a field. I want to ask the following natural question: Is it true that any linear group over a commutative ring satisfies the Tits alternative? I guess the answer is no. Can someone give a counterexample?
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2$\begingroup$ For anybody wondering about matrices over skew fields, the Tits alternative fails for $1 \times 1$ matrices: Lichtman, A. On subgroups of the multiplicative group of skew fields. Proc. Amer. Math. Soc. 63 (1977). zbmath.org/0352.20026 $\endgroup$– Giles GardamCommented Oct 14 at 7:15
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Yes it's true and follows from the field case.
We can suppose the ring $R$ is finitely generated. First, in the reduced case ($r^2=0$ implies $r=0$ in $r$), $R$, being also noetherian, embeds into a finite product of fields and the result follows.
In general, let $N$ be the radical of $R$, i.e. its set of nilpotent elements, so $R/N$ is reduced. Since $R$ is noetherian, there exists $m$ such that $r_1\dots r_m=0$ for all $r_1,\dots,r_m\in N$. It follows without difficulty that the kernel of $\mathrm{GL}_n(R)\to\mathrm{GL}_n(R/N)$ is nilpotent, and in turn the Tits alternative then passes from $\mathrm{GL}_n(R/N)$ to $\mathrm{GL}_n(R)$.
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$\begingroup$ You are right! Thank you for your wonderful explanation! $\endgroup$– NobodyCommented Oct 12 at 7:02