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This question is inspired, of course, by this question, and I don't know enough commutative algebra to know whether it's answered by silence dogood's answer to this follow-up question. If the answer is "no," then the original problem will be solved. If the answer is "yes," then the strategy that several people (including myself) were pursuing can't work.

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Kevin Buzzard didn't think it was answered, last I heard. I don't think it's answered, but also don't know enough commutative algebra to meaningfully contribute any more. – Jonathan Kiehlmann Dec 3 '10 at 16:13
You probably want R to have a unit. Otherwise take $R$ to be a nilpotent ring, say, with $xy=0$ for all $x,y$. Then $SL_3(R)=SL_2(R)=\emptyset$. Also finite subgroups are not the best testers for the embedding of $SL_3$ into $SL_2$. It is better to consider the maximal solvable or even nilpotent subgroups. – Mark Sapir Dec 4 '10 at 1:33
Yes, rings are unital here. Mark, if you have better ideas about the original question I encourage you to post an answer there; I would love to see the original problem solved. – Qiaochu Yuan Dec 4 '10 at 1:56
@Qiaochu: It is not an easy problem, it would certainly require more time than I can spend. But I would consider the Hirsh lengths of the maximal nilpotent subgroup in $SL_3$ and $SL_2$. That may fail, of course. Perhaps somebody else can do it. – Mark Sapir Dec 4 '10 at 2:24

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