Bourbaki

Lemma 1

Tag 00AQ, permanent. Cite it as [Bourbaki, Tag 00AQ].

Algebra, VIII, p. 159, § 9. Radical

Let aa be an element of a ring A such that a−a2a-a^2 is nilpotent. There exists a polynomial P in X+(X−X2)Z[X]X + (X-X^2)\mathbf{Z}[X] such that P(a)P(a) is idempotent in A.

Let nn be a strictly positive integer such that (a−a2)n=0(a-a^2)^n= 0. Set P(X) = 1−(1−Xn)n1-(1-X^n)^n. The polynomial P(X) is a multiple of XnX^n, and the polynomial 1−P(X)1-P(X) is a multiple of (1−X)n(1-X)^n, so P(X)−P(X)2P(X)-P(X)^2 is a multiple of (X−X2)n(X-X^2)^n, and we have P(a)=P(a)2P(a) = P(a)^2. Moreover, X−P(X)X-P(X) is a multiple of X and 1−X1-X, hence a multiple of X−X2X-X^2.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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Cites

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Transcribed from the printed edition, 2023, Springer Nature. Not translated.