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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