Bourbaki

Proposition 7

Tag 06B0, permanent. Cite it as [Bourbaki, Tag 06B0].

Algebra, chapter IV, § 2. Zeros of polynomials

— Let f∈A[X]f \in \mathbf{A}[X] and let a∈Aa \in \mathbf{A} be a root off: For aa to be a simple root off it is necessary and sufficient that aa should not be a root of the derivative DfDf of ff.

By hypothesis we have f=(X−a) gf = (X - a)\ g, where g∈A[X]g \in \mathbf{A}[X]. For aa to be a simple root of ff it is necessary and sufficient that g(α)≠0g(\alpha) \neq 0. Now we have Df=g+(X−α) DgDf = g + (X - \alpha)\ Dg, whence (Df)(α)=g(α)(Df)(\alpha) = g(\alpha).

More generally:

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

Cited by

Cites

Nothing.

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