Corollary 2

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

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

— Let E and F be two algebraic extensions of K such that $E \subset F$ . If F is quasi-Galois over K, then it is quasi-Galois over E.

We may assume that $F \subset \Omega$ . Let $u$ be an E-automorphism of $\Omega$ . Since $u$ is a K-automorphism of $\Omega$ and F is quasi-Galois over K, we have $u(F) = F$ , therefore F is quasi-Galois over E.

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

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