Essentially, I believe my question boils down to this: does the correspondence theorem at least partially hold for Lie algebras (or, indeed, any algebra)?
Lie algebras, . Let be some homomorphism of Lie algebras and let be an ideal of , . Then is it true that ?
This seems to be used in a proof in my lecture notes, but I can't work out why it holds (if, indeed, it does)!
Thanks in advance.
Hmm. I suppose I should probably stop asking questions in abstraction and just state the Lemma...
"For every finite dimensional Lie algebra the quotient algebra is semisimple."
That is to say, for the ideal containing every soluble ideal of then the quotient contains no proper soluble ideals.
Clearly, if is soluble then is soluble in , and so it is sufficient to prove that . Your proof requires , and I can't quite see why that is. My thought is that there exists some such that mod with but I'm not sure if that is true. (unless...would the canonical map not give the result? and taking to be ? Does the set of coset representative of form an ideal of ?)
Thanks for giving me/this problem so much time, by the way!
this is just the "correspondence theorem" that you're usually taught right after introducing quotient (Lie) algebras and you should find it in your notes:
suppose L is an algebra and I is an ideal of L. define the map by let be an ideal of and let
1) is an ideal of and
proof: if and then thus and so is an ideal of also if then thus i.e.
2)
proof: if then by definition of thus conversely, if then and hence i.e.
it's also clear that if are ideals of then is an ideal of thus we proved that ideals of are exactly in the form where are ideals of