Bourbaki

Corollary 1 (binomial formula)

Tag 05H4, permanent. Cite it as [Bourbaki, Tag 05H4].

Algebra, chapter I, § 8. Rings

Let x and y be two elements of a commutative ring A. Then:

(x+y)n=∑p=0n(np)xpyn−p. (x + y)^n = \sum_{p=0}^n \binom{n}{p} x^p y^{n-p}.

Formula (14) applied to L={1,2}L = \{1, 2\}, x1=xx_1 = x and x2=yx_2 = y gives

(x+y)n=∑p+q=nn!p!q!xpyq, (x + y)^n = \sum_{p+q=n} \frac{n!}{p! q!} x^p y^q,

the sum extending over ordered pairs of positive integers p,qp, q with p+q=np + q = n. The binomial formula follows immediately from this (Set Theory, III, § 5, no. 8).

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.