MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $\mathfrak{g}$ is a restricted Lie algebra over a field of characteristic $p>0$. Are there conditions on $\mathfrak{g}$ and its restriction which ensure that its restricted enveloping algebra is a domain? That it is a finitely generated domain? I really care about the graded case, so what if we assume further that $\mathfrak{g}$ is graded, concentrated in positive even degrees, and finite-dimensional in each degree?

Edit: for example, if the restriction is injective, does that mean that $u(\mathfrak{g})$ is a domain? What if you only know that if $x^{[p]}=0$, then $x=0$ – is that good enough? In the graded case, if the restriction is also surjective in all sufficiently large degrees, is $u(\mathfrak{g})$ a finitely generated domain?

Are there conditions on $\mathfrak{g}$ and its restriction which make $u(\mathfrak{g})$ isomorphic to an ordinary enveloping algebra $U(L)$ of some Lie algebra $L$?

share|cite|improve this question

When $\mathfrak{g}$ is non-zero and finite dimensional over $k$, its restricted enveloping algebra $u(\mathfrak{g})$ is never a domain. To see this, note that $u(\mathfrak{g})$ is itself finite dimensional over $k$ (of dimension $p^{\dim \mathfrak{g}}$ by Jacobson's PBW theorem) so if it's a domain it must be a division ring. But the definition of $u(\mathfrak{g})$ by generators and relations shows that $u(\mathfrak{g})$ always has a proper ideal --- the augmentation ideal $\mathfrak{g} u(\mathfrak{g})$. This is a contradiction.

When $\mathfrak{g}$ is infinite dimensional over $k$, it can happen that $u(\mathfrak{g})$ is a finitely generated domain. Here is an example: let $\mathfrak{g}$ be the $k$-linear span of the monomials $X, X^p, X^{p^2}, X^{p^3}, \ldots$ inside the polynomial algebra $k[X]$ and view it as an abelian restricted Lie algebra with the obvious restricted structure $x^{[p]} = x^p$ for all $x \in \mathfrak{g}$ (the latter calculated inside $k[X]$). It turns out that actually $u(\mathfrak{g})$ is isomorphic to $k[X]$ in this case, hence $u(\mathfrak{g})$ is a domain.

Note that if $p = 2$ and we view $k[X]$ as a graded ring with $\deg X = 2$, then $\mathfrak{g}$ is a graded restricted Lie algebra concentrated in even positive degrees and finite dimensional in each degree --- i.e. it satisfies your conditions.

Starting from an arbitrary group $G$, it is possible to cook up a restricted graded Lie algebra $\mathfrak{g} = \bigoplus_{n=1}^\infty (D_n/D_{n+1}) \otimes_{\mathbb{F}_p}k$. Here $D_1 \geq D_2 \geq D_3 \geq \cdots $ is the so-called modular dimension series of $G$, defined by Lazard's closed formula

$D_n = \prod_{ip^j \geq n} \gamma_i(G)^{p^j}$

where $\gamma_i(G)$ is the lower central series of $G$ defined by $\gamma_1(G) = G$ and $\gamma_i(G) = [G, \gamma_{i-1}(G)]$ for all $i \geq 2$. The $p$-power operation on $\mathfrak{g}$ is induced by the $p$-power map on the group. Then the theorem of S.A.Jennings asserts that $u(\mathfrak{g})$ is isomorphic to the associated graded ring of the group algebra $k[G]$ with respect to the filtration by powers of the augmentation ideal $I$ of $k[G]$:

$u(\mathfrak{g}) \cong \bigoplus_{n=0}^\infty \frac{I^n}{I^{n+1}}$.

The example I gave above is obtained by taking $G = \mathbb{Z}$. Similar examples can be obtained by taking $G$ to be a uniform pro-$p$ group of rank $d$; then $\mathfrak{g}$ will be an abelian, restricted graded Lie algebra with $u(\mathfrak{g})$ isomorphic to the polynomial algebra $k[X_1,\ldots, X_d]$.

You can find a good account of this material in Chapters 11 and 12 of the book "Analytic pro-$p$ groups" by Dixon, Du Sautoy, Mann and Segal.

share|cite|improve this answer
An alternative proof is that $u(g)$ is a Hopf algebra, so it has a distinguished idempotent, called integral. – Bugs Bunny May 10 '11 at 12:20
Still no answer to the question!! Say $\gamma (x)=0$ implies $x=0$ and $g$ contains no non-trivial finite dimensional restricted subalgebras. Does it imply that $u(g)$ is a domain? – Bugs Bunny May 10 '11 at 12:24
I've edited the original question to add something like Bugs Bunny's question, among other things. Note also that in the graded case, it is even clearer that $\mathfrak{g}$ must be infinite-dimensional, if $u(\mathfrak{g})$ is a domain. Finally, if $R$ is a $k$-algebra, where $k$ has characteristic $p$, then if you view $R$ as a restricted Lie algebra, it will be its own enveloping algebra, and you can get lots of examples this way, just by choosing $R$ to be a domain. – John Palmieri May 10 '11 at 20:02
It's not clear to me why $R$ will be its own enveloping algebra. Certainly there's a map $u(R) \to R$ by the universal property but it isn't an injection in general. For example if $\dim R < \infty$ then $\dim u(R) = p^{\dim R}$ so this map is never an injection in this case, and I see no reason for it to be injective in the infinite dimensional case either. And then why must $u(R)$ be a domain? – Konstantin Ardakov May 11 '11 at 7:25

Let $g$ be a restricted Lie algebra over a field of positive characteristic $p$. It is not difficult to see that a necessary condition such that the restricted enveloping algebra $u(g)$ of $g$ is a domain is that $g$ has no nonzero $p$-algebraic elements. (An element $x \in g$ is said to be $p$-algebraic if the restricted subalgebra generated by $x$ is finite-dimensional.) The converse of this property is a well-known open question posed by V. Petrogradsky (see "DNIESTER NOTEBOOK: Unsolved Problems in the Theory of Rings and Modules", Problem 3.59). Apparently, this is the Lie theoretical analog of the Kaplansky Problem about zero-divisors of group algebras of torsion-free groups.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.