Bourbaki

Remark

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

Algebra, V, p. 72, § 10. Galois extensions

Let KK and K′K' be two fields and uu a homomorphism of KK into K′K'. Let KsK_s (resp. Ks′K'_s) be a separable closure (V, p. 45, Prop. 14) of KK (resp. K′K') and Π\Pi (resp. Π′\Pi') the Galois group of KsK_s over KK (resp. Ks′K'_s over K′K'). Since KsK_s is a separable algebraic extension of KK and the extension (Ks′,u)(K'_s, u) of KK is separably closed, there exists (V, p. 45, Cor.) a homomorphism vv of KsK_s into Ks′K'_s extending uu. From vv we obtain a continuous homomorphism v∗v^* of Π′\Pi' into Π\Pi. Let v1v_1 be another extension of uu; since KsK_s is a quasi-Galois extension of KK, there exists an element asa_s of Π\Pi such that v1=v∘asv_1 = v \circ a_s. We conclude that v1∗(τ)=σ0−1v∗(τ)σ0v_1^*(\tau) = \sigma_0^{-1} v^*(\tau) \sigma_0 for all τ∈Π\tau \in \Pi.

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

Cited by

Nothing.

Cites

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