Bourbaki

Proposition 5

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

Algebra, chapter V, § 4. Algebraically closed extensions

— *Let K be a field, (fi)i∈I(f_i)_{i \in I} a family of non-constant polynomials in K[X]K[X], E an extension of K, and F, F' subextensions of E which are each a splitting extension of (fi)i∈I(f_i)_{i \in I}. Then F=F′F = F'.

Let RiR_i be the set of roots of fif_i in E and R=⋃i∈IRiR = \bigcup_{i \in I} R_i. Since fif_i is a product of polynomials of the first degree lying in F[X]F[X], we have Ri⊂FR_i \subset F. By Def. 2, we have F=K(R)F = K(R); in the same way we find that F=K(R)F = K(R).

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

Cited by

Cites

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