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?

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    $\begingroup$ For abelian $\mathfrak{g}$ the algebra $U(\mathfrak{g})$ is isomorphic to a polynomial ring in $\dim \mathfrak{g}$ variables, in particular never artinian (unless $\mathfrak{g} =0$). $\endgroup$ Apr 8, 2020 at 15:35
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    $\begingroup$ Also, cf. math.stackexchange.com/q/141437/96384, in particular the comments to the answer where @QiachuYuan links to the paper ams.org/journals/proc/1986-098-04/S0002-9939-1986-0861746-7/… which shows that the Jacobson radical of $U(\mathfrak g)$ only depends on the "basis ring (!)" of the Lie algebra; in particular, assuming we talk about Lie algebras over some field $K$ here, $Jac(U(\mathfrak{g})) = 0$ for all Lie algebras! $\endgroup$ Apr 8, 2020 at 15:42
  • $\begingroup$ @TorstenSchoeneberg this is a great comment, why don't you write an answer? (it doesn't sound to me redundant with the linked post) $\endgroup$
    – YCor
    Apr 8, 2020 at 18:51
  • $\begingroup$ @YCor: Done, please check. $\endgroup$ Apr 8, 2020 at 22:36
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    $\begingroup$ I'm pretty sure that in associative algebras you can define the Jacobson radical equivalently as the maximal nilpotent ideal and perhaps even as the maximal solvable ideal. So there is a way to make the parallel between the Lie and algebraic case more clear, but it doesn't run through the universal enveloping algebra. $\endgroup$
    – Vincent
    Apr 8, 2020 at 22:41

2 Answers 2


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.

  • $\begingroup$ Actually, I'm not sure what the exact sense of this "basis ring" statement. For if $\mathfrak{g}$ is a Lie algebra over a scalar (= associative commutative unital) ring $S$ and $R$ is a unital subring of $S$, then there is a canonical resulting $R$-algebra homomorphism $U_R(\mathfrak{g})\to U_S(\mathfrak{g})$, which is not injective in general (it is surjective, since the enveloping algebra is always generated by the image of the Lie algebra as a $\mathbf{Z}$-algebra). $\endgroup$
    – YCor
    Apr 8, 2020 at 22:52
  • $\begingroup$ @YCor: I've added the exact statement as I understand it from the paper. What you write about the situation of a subring $R \subset S$ is interesting but seems to not contradict the statement. $\endgroup$ Apr 9, 2020 at 3:19
  • $\begingroup$ So, two criteria are totally irrelevant... Interesting paper. Thanks! $\endgroup$
    – naughie
    Apr 9, 2020 at 5:10
  • $\begingroup$ My point is I don't guess the bare meaning of this statement. You write "as I understand it" but also you write the statement with quotation marks as if you don't understand it (and I don't either). $\endgroup$
    – YCor
    Apr 9, 2020 at 7:57
  • $\begingroup$ @YCor: Ah, my bad, I misquoted the paper: It states that it's the nilradical of the ground ring which (technically: via a canonical map $s: R \rightarrow U_R(\mathfrak g)$) generates the Jacobson radical of the universal enveloping algebra. Now that, unlike what I wrote before, has a chance to be compatible with the situation described by you, right? $\endgroup$ Apr 9, 2020 at 18:03

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, $\operatorname{Jac}(\mathfrak g) = 0$.


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