Bourbaki marks this one as harder than the rest.
Let g be a Lie algebra and U its enveloping algebra.
(a) Let g′ be an ideal of g, U′⊂U its enveloping algebra, x∈g and a1,…,an in U. Suppose that ai=x for the indices j1,…,jp and ai∈U′ for the other indices (let k1,…,kq be these indices, with k1<k2<⋯<kq). Then a1a2…an−xpak1ak2…akq is a sum of terms of the form xp′b, where b∈U′ and p′<p. (Argue by induction on p.)
(b) Suppose henceforth that g is nilpotent and K is of characteristic 0. Let g′ be an ideal of codimension 1 in g, U′ its enveloping algebra, x an element of g not belonging to g′ and U0 (resp. U0′) the subalgebra of U (resp. U′) annihilated by a set d of derivations of g (which extend to derivations of U) mapping g into g′. Suppose that U0 is contained in the centre of U and U0⊂U′. Show that there exist a1∈U0′, a2∈U′ such that a1=0 and a=xa1+a2∈U0. (Let xmbm+xm−1bm−1+⋯+b0∈U0 with bm,…,b0 in U′, m>0, bm=0. Show, using (a), that, for every derivation D∈d of g, Dbm=0, m(Dx)bm+Dbm−1=0 and hence D(mxbm+bm−1)=0.) Show that U0 is contained in the algebra K[a,a1−1,U0′] generated by a, a1−1, U0′ in the field of fractions of U0 (which can be formed by Corollary 7 to Theorem 1 of § 2). Deduce that the field of fractions of U0 is the field generated by a and U0′. Show that a is transcendental over K(U0′).
(c) Let {0}=g0⊂g1⊂…⊂gn=g be a sequence of ideals of g of dimensions 0,1,…,n. Let xj be an element of gj not belonging to gj−1. Let j1<j2<⋯<jq be the indices j such that there exists in Uj (enveloping algebra of gj) an element of the centre U0 of U not belonging to Uj−1. By (b) there exists aj1∈U0∩Uj and aj2∈Uj−1 such that aj1=0 and aj=xjaj1+aj2∈U0∩Uj. Show by induction on n that
U0⊂K[aj1,…,ajq,aj1−1,…,ajq−1] and that the field of fractions of
U0 is generated by the algebraically independent elements
aj1,…,ajq. In particular, the field of fractions of the centre of
U is a pure transcendental extension of
K.