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Let $g$ be a Lie ring (Lie algebra over $\mathbb Z$), and let $U(g)$ and $S(g)$ denote the universal enveloping and symmetric algebra of $g$. The Poincaré-Birkhoff-Witt theorem (in the form proved by Lazard, see "Sur les algebres enveloppantes universelles de certaines algebres de Lie, M Lazard - Publ. Sci. Univ. Alger. Ser. A, 1954) yields a ring isomorphism between $S(g)$ and an associated graded of $U(g)$.

I can prove that $S(g)$ and $U(g)$ are isomorphic as $\mathbb Z$-modules; this essentially follows from the proof by Lazard. Is this already known?

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Just remark isomorphism fails for char = p… – Alexander Chervov Aug 6 '12 at 10:43
Doc, you make my head spin. Surely, PBW holds over any field. – Bugs Bunny Aug 7 '12 at 12:53
Clarification: I consider algebras over $\mathbb Z$, in which case Lazard shows that there is a ring isomorphism between $gr(U(g))$ and $S(g)$. Indeed $\mathbb Z$ is a PID. The typical "counterexamples" to PBW involve algebras over non-domains such as $\mathbb F_p[a,b,c]/(a^p,b^p,c^p)$. My question is about isomorphism of $S(g)$ and $U(g)$ as abelian groups, not as rings. – grok Aug 8 '12 at 11:29

Earlier Lazard's result (Sur les algèbres enveloppantes universelles de certaines algèbres de Lie. C. R. Acad. Sci. Paris 234, (1952). 788–791. ) does it for Lie rings over PID.

For general Lie rings the first counterexample was constructed by Shirshov (On the representation of Lie rings as associative rings. Uspehi Matem. Nauk (N.S.) 8, (1953). no. 5(57), 173–175.)

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It is known if $g$ is free as a $\mathbb{Z}$-module, and fails otherwise, see e.g. P.M.Cohn, A Remark on the Birkhoff-Witt Theorem J. London Math. Soc. (1963) s1-38(1): 197-203

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I don't think this is claimed anywhere in Cohn. – darij grinberg Aug 6 '12 at 15:02
Oh I mean, in Cohn it is shown that it fails. Under the $\mathbb{Z}$-module freeness, it is an easy exercise which is done in way too many textbooks. – Vladimir Dotsenko Aug 6 '12 at 15:13
It fails in the category of "algebras over $\Phi$", for some rings $\Phi$ that contain zero divisors; however, this is not exactly the question I wanted to ask; I added a clarification. – grok Aug 8 '12 at 11:30
@grok: I see what you're saying - that in PM Cohn's example you will say that both are, as abelian groups, countably dimensional vector spaces over $\mathbb{F}_p$, and hence isomorphic? – Vladimir Dotsenko Aug 8 '12 at 12:29
@Dotsenko: yes indeed. Most people seek an algebra isomorphism, so replace $U(g)$ by an associated graded; I'm interested in less, but don't want to pass to an associated graded. My question only makes sense for Lie rings with $\mathbb Z$ or $\mathbb Z/q$ additive factors with $q$ non-prime. – grok Aug 9 '12 at 12:03

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