Timeline for When is the Levi subalgebra an ideal?
Current License: CC BY-SA 3.0
12 events
when toggle format | what | by | license | comment | |
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Apr 18, 2011 at 12:31 | vote | accept | user14312 | ||
Apr 18, 2011 at 11:12 | answer | added | Jim Humphreys | timeline score: 6 | |
Apr 18, 2011 at 5:22 | vote | accept | user14312 | ||
Apr 18, 2011 at 12:31 | |||||
Apr 18, 2011 at 5:10 | answer | added | Ben Webster♦ | timeline score: 3 | |
Apr 18, 2011 at 4:58 | history | edited | user14312 | CC BY-SA 3.0 |
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Apr 18, 2011 at 4:49 | history | edited | user14312 | CC BY-SA 3.0 |
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Apr 18, 2011 at 4:45 | comment | added | user14312 | I will edit the question. | |
Apr 18, 2011 at 4:44 | comment | added | user14312 | As direct sum I mean direct sum of vector subspaces, not submodules. | |
Apr 18, 2011 at 4:43 | comment | added | Kevin Ventullo | The answer is no. For instance, if $L$ is a non-abelian solvable Lie algebra, then $S=0$ is certainly an ideal, but $L=rad(L)\neq Z(L)$. | |
Apr 18, 2011 at 4:36 | comment | added | Ben Webster♦ | I'm not sure what the question here is. S is an ideal if and only if L is the direct sum of S and rad(L); this is essentially the definition of "direct sum." | |
Apr 18, 2011 at 4:13 | history | edited | user14312 | CC BY-SA 3.0 |
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Apr 18, 2011 at 3:41 | history | asked | user14312 | CC BY-SA 3.0 |