Semisimple Lie algebra and Jacobson radical

Question

In the theory of Lie algebras, the radical $\mathrm{rad} (\mathfrak{g})$ of a Lie algebra $\mathfrak{g}$ is defined to be a (the) maximal solvable ideal of $\mathfrak{g}$, and the Lie algebra $\mathfrak{g}$ is said to be semisimple if $\mathrm{rad} (\mathfrak{g}) = 0$.

On the other hand, in the theory of associative algebras, the Jacobson radical $\mathrm{rad} (A)$ of an algebra $A$ is the intersection of all maximal (left) ideal of $A$, and the algebra $A$ is semisimple if $A$ is artinian and $\mathrm{rad} (A) = 0$.

(A semisimple algebra is a semisimple module (direct sum of simple modules) over itself.)

Then, there rises two questions to me:

• Is the universal enveloping algebra $U (\mathfrak{g})$ artinian?
• Do this two kinds of semisimplicity coincide; that is, $\mathfrak{g}$ is semisimple iff $U (\mathfrak{g})$ is semisimple?

If not, under which circumstances can we deduce the equivalence of semisimplicity?

Answer 1

In a comment to his answer to the related question Ties between Lie algebras and ring theory, Qiaochu Yuan points out the very short paper by Erazm J. Behr, Jacobson Radical of Filtered Algebras (Proc. AMS 98(4), 1986) which shows that the Jacobson radical of the universal enveloping algebra $U(\mathfrak{g})$ only depends on the "basis ring (!)" of the Lie algebra $\mathfrak{g}$. More precisely, it shows:

Let $L$ be a Lie algebra over a commutative unital ring $R$. Then the Jacobson radical of $U_R(L)$ (the universal enveloping algebra of $L$ viewed as an $R$-algebra) is generated by $Nil(R)$, the nilradical of $R$.

In particular, if our Lie algebra is defined over some field (or just integral domain) $K$, then $Jac(U(\mathfrak{g}))=0$ automatically! (The paper gives credit for this result to an earlier paper by R. S. Irving.)

This in particular implies that the Jacobson radical of $U(\mathfrak{g})$ has virtually nothing to do with the Lie-theoretic radical of $\mathfrak{g}$.

As for the property of being artinian, one obvious counterexample is given by abelian Lie algebras $\mathfrak{a}$ (over a field $K$): For these, it is well-known that $U(\mathfrak{a})$ is isomorphic to a polynomial ring in $\dim_K \mathfrak{g}$ variables over $K$. In particular, these rings are non-artinian unless $\mathfrak{g} = 0$.

If I'm not mistaken, we don't need the abelianness here, but with a bigger gun we get a much stronger result: If $0 \neq x \in \mathfrak{g}$, then by Poincare-Birkhoff-Witt $R:=U(\mathfrak g)$ contains the infinite chain of left ideals $Rx \supsetneq Rx^2 \supsetneq Rx^3 \supsetneq ...$.

Upshot: If $\mathfrak{g} \neq 0$ is any Lie algebra over a field, $U(\mathfrak g)$ is never artinian but always Jacobson-semisimple.

Answer 2

You may also be interested in the relation between the solvable (usual) radical $\operatorname{rad}(\mathfrak g)$ and the Jacobson radical $\operatorname{Jac}(\mathfrak g)$ (the intersection of maximal ideals).

Theorem (Marshall). One has $\operatorname{Jac}(\mathfrak g) = [\mathfrak g, \operatorname{rad}(\mathfrak g)]$.

E. I. Marshall, The Frattini Subalgebra of a Lie Algebra, Journal of the London Mathematical Society, Volume s1-42, Issue 1, 1967, Pages 416–422. https://academic.oup.com/jlms/article-abstract/s1-42/1/416/844116.

$\operatorname{Jac}(\mathfrak g)$ is nilpotent but need not be the maximal nilpotent ideal. For example when $\mathfrak g$ is abelian, $\operatorname{Jac}(\mathfrak g) = 0$.