# When is a restricted enveloping algebra a domain? A finitely generated domain?

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 $L$?

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.

• 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? – user91132 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.