Bourbaki

Lemma 3

Tag 0A5F, permanent. Cite it as [Bourbaki, Tag 0A5F].

Lie Groups and Lie Algebras, chapter II, § 5. Magnus algebras

*Let x1,…,xsx_1, \ldots, x_s be elements of X\mathbf{X} such that s≥1s \geq 1 and xi≠xi+1x_i \neq x_{i+1} for 1≤i≤s−11 \leq i \leq s-1; let n1,…,nsn_1, \ldots, n_s be non-zero rational integers. Then the element ∏i=1s(1+xi)ni\prod_{i=1}^s (1+x_i)^{n_i} of A^(X)\hat{\mathbf{A}}(\mathbf{X}) is ≠1\neq 1.

Let mm be a maximal ideal of KK and kk the field K/mK/m; let p:A^K(X)→A^k(X)p : \hat{\mathbf{A}}_K(\mathbf{X}) \to \hat{\mathbf{A}}_k(\mathbf{X}) be the unique continuous homomorphism of unital KK-algebras such that p(x)=xp(x) = x for x∈Xx \in \mathbf{X} (no. 1, Proposition 1). It suffices to prove that p(∏i=1s(1+xi)ni)≠1p\left( \prod_{i=1}^s (1+x_i)^{n_i} \right) \neq 1 and the problem is reduced to the case where KK is a field.

In the notation of Lemma 2:

∏i=1s(1+xi)ni=∑b1≥0⋯∑bs≥0cb1,n1⋯cbs,nsx1b1⋯xsbs. \prod_{i=1}^s (1+x_i)^{n_i} = \sum_{b_1 \geq 0} \cdots \sum_{b_s \geq 0} c_{b_1, n_1} \cdots c_{b_s, n_s} x_1^{b_1} \cdots x_s^{b_s}.

By Lemma 2, there exist integers ai>0a_i > 0 such that cai,ni≠0c_{a_i, n_i} \neq 0 (1≤i≤s1 \leq i \leq s). By Algebra, Chapter I, § 7, no. 4, Proposition 6, no monomial x1b1⋯xsbsx_1^{b_1} \cdots x_s^{b_s} such that bi≥0b_i \geq 0 and (b1,…,bs)≠(a1,…,as)(b_1, \ldots, b_s) \neq (a_1, \ldots, a_s) can be equal to x1a1…xsasx_1^{a_1} \ldots x_s^{a_s}. It follows that the coefficient of x1a1…xsasx_1^{a_1} \ldots x_s^{a_s} in ∏i=1s(1+xi)ni\prod_{i=1}^s (1 + x_i)^{n_i} is ca1,n1…cas,ns≠0c_{a_1, n_1} \ldots c_{a_s, n_s} \neq 0, which implies the result.

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

Cited by

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Cites

Transcribed from the printed volume. Not translated.