9
$\begingroup$

As it was shown by Malcev, unlike the commutative case, in which every domain can be embedded in a field, there are noncommutative domains that can't be embedded in a division ring.

For noncommutative domains, the nicest way they can be embedded in a division ring is if they satisfie Ore's conditions. In this case, a noncommutative analogue of the field of fractions exist.

As an example, we have enveloping algebras o finite dimensional Lie algebras in zero characteristic.

By Jategaonkar's Lemma, the enveloping algebra of infinite dimensional algebra can still be an Ore domain, if their growth is sufficiently slow as to make their enveloping algebras having subexponential growth - and there are many of these examples.

However, in general the enveloping algebra will not be a an Ore domain.

I remember seeing, a long time ago, a result by P. M. Cohn using methods from ring theory and universal algebra, showing that, in the characteristic 0 case and any Lie algebra $\mathfrak{g}$, $U(\mathfrak{g})$ can be embedded in a division ring.

Has anyone heard about this result, and has a reference for it?

Thank you.

$\endgroup$

2 Answers 2

10
$\begingroup$

This is in P. M. Cohn, "On the Embedding of Rings in Skew Fields", Proceedings of the London Mathematical Society, Volume s3-11, Issue 1 (1961), Pages 511-530. I do not think that the zero characteristic is important: he proves that any algebra having a filtration for which the associated graded algebra satisfies the Ore conditions can be embedded in a division algebra.

$\endgroup$
2
$\begingroup$

I think Tamari has also shown this result using a combinatorial trick.

Tamari, Dov, On the embedding of Birkhoff-Witt rings in quotient fields, Proc. Am. Math. Soc. 4, 197-202 (1953). ZBL0053.21504.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .