Timeline for Does $S_4$ inject into $SL(2,R)$ for some commutative ring $R$?
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
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| Nov 30, 2010 at 16:06 | comment | added | Jack Schmidt | Yes it does. Assuming R is local Artinian (i.e. taking Ai), its proper ideals are nilpotent, and so the congruence subgroup is nilpotent ( G(I) = GL(n,R,I) satisfies [ G(I), G(J) ] ≤ G(IJ), and I^n = 0 ), and so the kernel of the map from G into SL(2,R) to SL(2,R/I) is a normal nilpotent group, contained in the Fitting subgroup. I believe this shows "R" versus "product of fields" can only fix nilpotent normal subgroups. The abelian filtration is clearer, but I think only shows you cannot fix insoluble normal subgroups. | |
| Nov 30, 2010 at 15:56 | history | edited | user631 | CC BY-SA 2.5 |
reduction to Noetherian case added
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| Nov 30, 2010 at 15:28 | comment | added | Tim Dokchitser | So this does prove that $PSL_2({\mathbb F}_7)$ is not a subgroup of $SL_2(R)$ for any $R$, does it not? (This group has no 2-dim representations over any field.) This is what Jack Schmidt said he'd expect in his answer, and something I wondered about as well. | |
| Nov 30, 2010 at 11:01 | comment | added | Kevin Buzzard | In fact Step 1 itself is false without a Noetherian hypothesis: take for example the integers of Q_p-bar and set $x=p$. [Of course this isn't a problem because in Step 0 you replace $R$ by the subring of $R$ generated by the components of the image of $G$] | |
| Nov 30, 2010 at 10:34 | comment | added | Kevin Buzzard | This seems to deal with everything and in a way I understand too. Thanks! Thanks also to the other two answerers. To silence dogood: in step 1 you might want to put "wlog R is Noetherian" as the first line (before you apply Krull). | |
| Nov 30, 2010 at 10:30 | vote | accept | Kevin Buzzard | ||
| Nov 30, 2010 at 6:09 | history | answered | user631 | CC BY-SA 2.5 |