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I am trying to understand why finite dimensional lie algebras have a unique maximal solvable ideal. This is from Humphrey's textbook.

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I have a question about the statement: "Let S be a maximal solvable ideal.." . So the author did not elaborate much on this statement but is the reason why such a maximal solvable ideal exist due to Zorn's lemma.

Also, part c) of the proposition states that the sum of solvable ideals is solvable. Now the author states that "$S+I=S$ or $I \subset S$. Was there any part of this statement that made use of the fact that $I$ and $S$ is solvable. If $I$ were not solvable, would $I \subset S$ still be true.

Edit: one point about this proof that is making me confused is because it seems like we did not had to use the fact that our ideals were solvable. Could we have said the same thing about $I$ and $S$ if we did not specify they were solvable?

Bill
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2 Answers2

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You posed two questions and let me try to answer them.

First, you ask about the existence of a maximal solvable ideal. This is really a consequence of the finite-dimensionality of your Lie algebra (and in particular Zorn's Lemma has no use). Indeed, just try to add elements such that we still have a solvable ideal. If this is not possible, then we have reached a maximal solvable ideal.

Then, you ask about $I \subseteq S$. As stated in the book, this is by maximality of $S$: Since $I$ and $S$ are solvable, also $I + S$ is solvable by (c). On the other hand, $S$ is a maximal solvable ideal, so maximality implies $I + S \subseteq S$.

Qi Zhu
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First of all, the misleading may be comes from the term "arbitrary" in the statement

let $L$ be an arbitrary Lie algebra ...

There, the author means that $L$ is a Lie algebra that is not necessarily solvable or of any kind.

For Zorn's lemma there is no need for it here because Humphreys' book deals only with finite dimensional Lie algebras.

Besides, in the infinite case, there is no unique maximal solvable ideal, as pointed by @Dietrich Burde in his comment here.

Finally, for your question in "Edit" note. The answer is that you can say nothing if you didn't use the solvability of $I$ and $S$ (and of course the point $(c)$). Instead, you can substitute solvability by nilpotency (cf. Nilradical's definition). But, if you don't consider any caracterization then what exactely would you prove, other than $S$ is a maximal vector space?

Youness EL KHARRAF
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