Bourbaki

Proposition 4

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

Algebra, chapter V, § 10. Galois extensions

— Let NN be a Galois extension of KK. Then the Galois group Γ=Gal(N/K)\Gamma = \mathrm{Gal}(N/K) is compact and totally disconnected.

Every element σ\sigma of Γ\Gamma has a base of neighbourhoods consisting of the open and closed sets UE(σ)U_E(\sigma), hence Γ\Gamma is totally disconnected (Gen. Top., I, p. 111). We have {σ}=⋂E∈ΛUE(σ)\{\sigma\} = \bigcap_{E \in \Lambda} U_E(\sigma), hence Γ\Gamma is separated. For each x∈Nx \in N the set of conjugates σ(x)\sigma(x) of xx, as σ\sigma ranges over Γ\Gamma, is finite because xx is algebraic over KK (V, p. Cor. 1); hence all the projections of Γ\Gamma on the factor spaces of NNN^N are finite sets, and this shows Γ\Gamma to be relatively compact in NNN^N (Gen. Top., I, p. 88). It remains to show that Γ\Gamma is closed in NNN^N. Now if uu is in the closure of Γ\Gamma in NNN^N, then for each pair of points (x,y)(x, y) of NN there exists a∈Γa \in \Gamma with u(x)=σ(x), u(y)=σ(y), u(x+y)=σ(x+y), u(xy)=σ(xy)u(x) = \sigma(x), \ u(y) = \sigma(y), \ u(x+y) = \sigma(x+y), \ u(xy) = \sigma(xy), whence u(x+y)=u(x)+u(y)u(x+y) = u(x) + u(y) and u(xy)=u(x)u(y)u(xy) = u(x)u(y). By the same reasoning we have u(x)=xu(x) = x for all x∈Kx \in K, hence uu is a KK-homomorphism of NN into NN; since NN is algebraic over KK, uu is a KK-automorphism of NN (V, p. 52, Prop. 1), hence u∈Tu \in T.

Let NN be a Galois extension of KK and (Ni)i∈I(N_i)_{i \in I} an increasing directed family of subextensions of NN. Suppose that NiN_i is Galois over KK for all i∈Ii \in I and that N=⋃i∈INiN = \bigcup_{i \in I} N_i. For each i∈Ii \in I denote by Γi\Gamma_i the Galois group of NiN_i over KK; for i≤ji \leq j in II we have Ni⊂NjN_i \subset N_j and the restriction homomorphism φij\varphi_{ij} of Γj\Gamma_j into Γi\Gamma_i is defined. It is continuous and so the family (Γi,φij)(\Gamma_i, \varphi_{ij}) is an inverse system of topological groups. Further, for each i∈Ii \in I denote by λi\lambda_i the restriction homomorphism of Gal(N/K)\mathrm{Gal}(N/K) into Gal(Ni/K)=Γi\mathrm{Gal}(N_i/K) = \Gamma_i; it is continuous and we have λi=φij∘λj\lambda_i = \varphi_{ij} \circ \lambda_j for i≤ji \leq j, hence the family (λi)i∈I(\lambda_i)_{i \in I} defines a continuous homomorphism Λ\Lambda of Gal(N/K)\mathrm{Gal}(N/K) into lim←⁡Γi\varprojlim \Gamma_i.

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

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Cites

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