Bourbaki

Corollary 2

Tag 05H5, permanent. Cite it as [Bourbaki, Tag 05H5].

Algebra, chapter I, § 8. Rings

*Let A be a commutative ring, X a set, u=(ux)x∈X\mathbf{u} = (u_x)_{x \in X} and v=(vx)x∈X\mathbf{v} = (v_x)_{x \in X} two families of elements of A. Let u+v\mathbf{u} + \mathbf{v} denote the family (ux+vx)x∈X(u_x + v_x)_{x \in X}. For all λ∈N(X)\lambda \in \mathbf{N}^{(X)} we write λ!=∏x∈Xλ(x)!\lambda! = \prod_{x \in X} \lambda(x)!. Then for all α∈N(X)\alpha \in \mathbf{N}^{(X)}, in the notation of § 7, no. 8,

(u+v)α=∑β+γ=αα!β!γ!uβvγ. (\mathbf{u} + \mathbf{v})^\alpha = \sum_{\beta + \gamma = \alpha} \frac{\alpha!}{\beta! \gamma!} \mathbf{u}^\beta \mathbf{v}^\gamma.

For x∈Xx \in X,

(ux+vx)α(x)=∑m+n=α(x)α(x)!m!n!uxmvxn (u_x + v_x)^{\alpha(x)} = \sum_{m+n=\alpha(x)} \frac{\alpha(x)!}{m! n!} u_x^m v_x^n

by Corollary 1. Taking the product of these equations for x∈Xx \in X and using (13), we obtain the corollary.

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 edition, 1998, Springer. Not translated.