Bourbaki

Corollary 1

Tag 0A2S, permanent. Cite it as [Bourbaki, Tag 0A2S].

Lie Groups and Lie Algebras, chapter II, § 2. Free Lie algebras

The family (ϕ(x))x∈X(\phi(x))_{x \in X} is free over K in L(X)L(X).

Let x1,x2,…,xnx_1, x_2, \ldots, x_n be distinct elements of X and λ1,…,λn\lambda_1, \ldots, \lambda_n be elements of K such that

λ1⋅ϕ(x1)+⋯+λn⋅ϕ(xn)=0. \lambda_1 \cdot \phi(x_1) + \cdots + \lambda_n \cdot \phi(x_n) = 0.

Let gg be the commutative Lie algebra with K as underlying module. For i=1,2,…,ni = 1, 2, \ldots, n, there exists a homomorphism FiF_i of L(X)L(X) into gg such that Fi(ϕ(xi))=1F_i(\phi(x_i)) = 1 and Fi(ϕ(x))=0F_i(\phi(x)) = 0 for x≠xix \neq x_i (Proposition 1); applying FiF_i to relation (3), we obtain λi=0\lambda_i = 0.

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

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Transcribed from the printed volume. Not translated.