Bourbaki

Remark 1

Tag 06AS, permanent. Cite it as [Bourbaki, Tag 06AS].

Algebra, IV, p. 15, § 2. Zeros of polynomials

If f=0f = 0 we agree to say that aa has order ≥h\geq h relative to ff, whatever a∈Aa \in \mathbf{A} and the integer h≥0h \geq 0. For any f∈A[X]f \in \mathbf{A}[X] and a∈Aa \in \mathbf{A}, to say that aa has order ≥h\geq h relative to ff means that (X−α)h(X - \alpha)^h divides ff.

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

Cited by

Nothing.

Cites

Nothing.

Transcribed from the printed edition, 2003, Springer. Not translated.