Timeline for Does $S_4$ inject into $SL(2,R)$ for some commutative ring $R$?
Current License: CC BY-SA 2.5
16 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Nov 30, 2010 at 11:02 | history | edited | Kevin Buzzard | CC BY-SA 2.5 |
fixed mathematical error
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| Nov 30, 2010 at 10:30 | vote | accept | Kevin Buzzard | ||
| Nov 30, 2010 at 7:11 | comment | added | Kevin Buzzard | ...ideal in a polynomial ring with about 20 generators over the integers and then asking if this ideal contains 1, and this is apparently too much for current algorithms/computers. The non-conceptual approach to this question looked like a similar, but easier, problem. | |
| Nov 30, 2010 at 7:10 | comment | added | Kevin Buzzard | @JSE: any commutative group scheme of order $n$ is killed by $n$: the "old-fashioned" proof that any commutative group of order $n$ is killed by $n$ (multiply all the elements together, call the result $x$, and note that $gx=x$) generalises very nicely. But there are non-commutative group schemes of order 4. The issue of course is a base with 2 locally nilpotent. If no conceptual proof is known one can try writing down the universal group scheme of order 4 and checking it on this, but the bottom line is that this "non-conceptual" approach involves writing down about 30 generators for an... | |
| Nov 30, 2010 at 6:09 | answer | added | user631 | timeline score: 16 | |
| Nov 30, 2010 at 1:25 | comment | added | JSE | "the open problem of whether every finite flat group scheme of order 4 was killed by 4" Holy crow, this isn't known? | |
| Nov 30, 2010 at 1:12 | answer | added | user6976 | timeline score: 9 | |
| Nov 30, 2010 at 0:05 | answer | added | Jack Schmidt | timeline score: 14 | |
| Nov 29, 2010 at 23:04 | answer | added | Pace Nielsen | timeline score: 4 | |
| Nov 29, 2010 at 22:44 | comment | added | Ian Agol | This book might be useful: books.google.com/… In particular, try to compute H[S_4] (in the book's notation). Any SL(2) rep. will have to factor through this quotient algebra. If S_4 doesn't inject in H[S_4], then the answer to question 2 is no. | |
| Nov 29, 2010 at 20:03 | history | edited | Kevin Buzzard | CC BY-SA 2.5 |
typo/clarification
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| Nov 29, 2010 at 19:53 | comment | added | Kevin Buzzard | By CRT this is just $SL(2,2)\times SL(2,3)$ which is $S_3\times\tilde{S}_4$. Now $\tilde{S}_4$ is (by definition) the non-split central extension of $S_4$ by $C_2$ and contains no $S_4$, so $SL(2,6)$ can't contain an $S_4$ either (indeed any map from $S_4$ to $SL(2,6)$ will contain $V_4$ in the kernel, this being the minimal normal subgroup). | |
| Nov 29, 2010 at 19:39 | comment | added | Qiaochu Yuan | Kevin, if you want to check more examples with computer algebra I would really like it if you checked SL_2(Z/6Z). | |
| Nov 29, 2010 at 19:37 | answer | added | Qiaochu Yuan | timeline score: 0 | |
| Nov 29, 2010 at 19:21 | history | asked | Kevin Buzzard | CC BY-SA 2.5 |