Bourbaki

Proposition 4

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

Algebra, chapter V, § 9. Conjugate elements and quasi-Galois extensions

— Let (Ni)i∈I(N_i)_{i \in I} be a family of quasi-Galois extensions of KK contained in Ω\Omega. Let N=⋂i∈INiN = \bigcap_{i \in I} N_i and M=K(⋃i∈INi)M = K\left( \bigcup_{i \in I} N_i \right); then the extensions NN and MM of KK are quasi-Galois.

Let uu be a KK-automorphism of Ω\Omega. We have u(Ni)=Niu(N_i) = N_i for all i∈Ii \in I, hence we obtain the equalities u(N)=Nu(N) = N and u(M)=Mu(M) = M; now the proposition follows from Cor. 1 (V, p. 54).

Let AA be a set of elements of Ω\Omega, and let BB be the set of elements of Ω\Omega which are conjugates of elements of AA; expressed differently, if GG is the group of KK-automorphisms of Ω\Omega, we have B=⋃u∈Gu(A)B = \bigcup_{u \in G} u(A). For each u∈Gu \in G we have u(B)=Bu(B) = B whence u(K(B))=K(B)u(K(B)) = K(B). Therefore K(B)K(B) is a quasi-Galois extension of KK containing AA and it is immediate that every quasi-Galois extension NN of KK containing AA contains BB, hence K(B)K(B). We shall say that K(B)K(B) is the quasi-Galois extension generated by AA. This definition applies in particular when AA is an extension of KK.

The next proposition follows directly from the preceding remarks.

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, 2003, Springer. Not translated.