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I failed to understand the radical of finite dimensional Lie algebra $\mathfrak{g}$. There are two definitions for radical:

  1. It is the sum of all solvable ideals of $\mathfrak{g}$.

  2. It is the unique maximal solvable ideal of $\mathfrak{g}$.

For the first definition, we know that if $\mathfrak{a}$ and $\mathfrak{b}$ are solvable ideals, then so if $\mathfrak{a}+\mathfrak{b}$. However, to ensure that the sum of all solvable ideals are solvable, it seems that we need that there are only finitely many solvable ideals to apply induction, but I failed to see why.

For the second definition, how can we ensure the existence and uniqueness of maximal solvable ideal? For rings, we can use Zorn's Lemma to deduce the existence of maximal ideal, but I don't know how to prove the existence of maximal solvable ideal in this case.

Any help would be appreciated. Thanks in advance.

jialong zhang
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  • The first question can be answered by the finite dimensionality of your Lie-algebra. If you sum up infinitely many ideals which have non-trivial intersection (pairwise), the resulting ideal is of infinite dimension. With this in mind, the uniqueness in 2. stems from adding up all of the finitely many solvable ideals. In finite dimension, there is no need for a Zorn's Lemma/Axiom of Choice type of argument I think. – Nuffie Feb 27 '24 at 12:24
  • I agree, @Nuffie, that finite dimension is the key here, but your second sentence is not true. – Torsten Schoeneberg Feb 28 '24 at 01:45
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    Agreed, I was a bit hasty in formulating "If you find infinitely many summands that change the sum of ideals, the dimension has to be infinite". Your approach is the elegant version of what came to my mind. – Nuffie Feb 28 '24 at 14:01

1 Answers1

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There might be various ways to go about this, but I recommend the following:

Because $\mathfrak g$ has finite dimension, among all its solvable ideals there exists one that has maximal dimension. Now show:

If $M$ is such an ideal (of maximal dimension among the solvable ones), and $I \subseteq \mathfrak g$ is any solvable ideal, then $I \subseteq M$.

From this one can conclude that such $M$ contains (hence equals) the sum of all solvable ideals, that it is unique, and that it is in fact maximal among the solvable ideals.

Torsten Schoeneberg
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