0

I've this doubt from proof in book of Humphreys Lie Algebras, chapter 2, first theorem: we have a solvable algebra $\;L\le \mathfrak g\mathfrak l(V)\;,\;\;\dim V<\infty\;$ and $\;V\neq0\;$ . Then V contains a common eigenvector for all elements in $\;L\;$ .

Question: I've problems in second parraph of proof: Why is it true that if I have a subspace of $\;L/[LL]\;$ of codimension $\;1\;$ , then its inverse image (under the canonical homomorphism, I think. It isn't said) is ideal of codimension $\;1\;$ , too?

By correspondence theorem (of rings, groups, not the one between Lie groups and algebras), I know index of ideal in $\;L/LL]\;$ is the same as index of inverse image of that ideal in $\;L\;$ , but I am unable to connect this with codimension.

Any help is greatly appreciated.

  • Just go through the proof of the statement about index and see that it also works for codimension (the index is not a very useful thing to consider in these cases as it will always be either infinite or $0$) – Tobias Kildetoft Mar 25 '17 at 12:48
  • @TobiasKildetoft Thank you very much, but I've some doubts: not every subspace is ideal, but I think the other way is true (right?). Now, with proof of index I take representatives in quotients, lift them to inverse image and show they are full set of representatives (this is very similar as with groups, I believe). But here: if $;\hat T\le L/[LL];$ is subspace it doesn't need to be ideal (even that $;L/[LL];$ abelian...or am I wrong here?), so how I lift basis of it to basis of $;T=\pi^{-1}(\hat T);$ in $;L;$ with number of elements one less than $;\dim L;$ ? –  Mar 25 '17 at 13:16
  • Inverse image of an ideal is an ideal (this is part of the correspondence theorem). And just go through the same argument using a basis instead of a set of coset representatives in order to see that it preserves codimensions. – Tobias Kildetoft Mar 25 '17 at 13:49

0 Answers0