Bourbaki

Remark

Tag 0A45, permanent. Cite it as [Bourbaki, Tag 0A45].

Lie Groups and Lie Algebras, chapter II, § 3. Enveloping algebra of the free Lie algebra

(1) Let K′K' be a commutative ring containing K. If A(X)A(X), L(X)L(X) and LK′(X)L_{K'}(X) are identified with subsets of AK′(X)A_{K'}(X), we deduce from part (a) of Theorem 1 the relation

L(X)=LK′(X)∩A(X). L(X) = L_{K'}(X) \cap A(X).

(2) Corollary 2 to Theorem 1 remains valid if it only assumes that the additive group of the ring K is torsion-free. For suppose first that K=ZK = \mathbf{Z}; every primitive element of A(X)A(X) is a primitive element of Aq(X)A_q(X) and hence is in Lq(X)∩A(X)=L(X)L_q(X) \cap A(X) = L(X) (Corollary 2 and formula (1)). In the general case, K is flat over Z\mathbf{Z} and we apply Remark 2 of § 1, no. 2 and Proposition 3 of § 2, no. 5.

(3) Let Δ\Delta be a commutative monoid, ϕ0\phi_0 a mapping of XX into Δ\Delta and ϕ:Mo(X)→Δ\phi : Mo(X) \to \Delta the homomorphism of the associated monoid; if A(X)A(X) is given the graduation (Aδ(X))δ∈Δ(A^\delta(X))_{\delta \in \Delta} defined in Algebra, Chapter III, § 3, no. 1, Example 3 and L(X)L(X) the graduation (Lδ(X))δ∈Δ(L^\delta(X))_{\delta \in \Delta} defined in § 2, no. 6, we have immediately, for δ∈Δ\delta \in \Delta, Lδ(X)⊂L(X)∩Aδ(X)L^\delta(X) \subset L(X) \cap A^\delta(X). As LL is the sum of the Lδ(X)L^\delta(X) for δ∈Δ\delta \in \Delta, and the sum of the L(X)∩Aδ(X)L(X) \cap A^\delta(X) for δ∈Δ\delta \in \Delta is direct, this implies

Lδ(X)=L(X)∩Aδ(X). L^\delta(X) = L(X) \cap A^\delta(X).

(4) Let AA be a unital associative algebra and t=(ti)i∈It = (t_i)_{i \in I} a family of elements of AA. We have a diagram

L(I)→ftAi↓↓gtA(I) \begin{array}{ccc} L(I) & \xrightarrow{f_t} & A \\ i \downarrow & & \downarrow g_t \\ A(I) \end{array}

where ii is the canonical injection, ftf_t is the Lie algebra homomorphism defined by tt and gtg_t is the unital algebra homomorphism such that gt(i)=tig_t(i) = t_i for i∈Ii \in I. The diagram is commutative for gt∘ig_t \circ i and ftf_t coincide on II. It follows that if P∈L(I)P \in L(I), the element P((ti)i∈I)P((t_i)_{i \in I}) defined in § 2, no. 4, coincides with the element P((ti)i∈I)P((t_i)_{i \in I}) defined in Algebra, Chapter III, § 2, no. 8, Example 2.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Cites

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