Timeline for When is a restricted enveloping algebra a domain? A finitely generated domain?
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| May 11, 2011 at 7:25 | comment | added | user91132 | 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? | |
| May 10, 2011 at 20:02 | comment | added | John Palmieri | 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. | |
| May 10, 2011 at 17:43 | history | edited | user91132 | CC BY-SA 3.0 |
added 5 characters in body
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| May 10, 2011 at 12:24 | comment | added | Bugs Bunny | 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? | |
| May 10, 2011 at 12:20 | comment | added | Bugs Bunny | An alternative proof is that $u(g)$ is a Hopf algebra, so it has a distinguished idempotent, called integral. | |
| May 10, 2011 at 11:02 | history | answered | user91132 | CC BY-SA 3.0 |