Bourbaki

Theorem 5

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

Algebra, chapter V, § 11. Abelian extensions

*a) The mapping A↦K(P−1(A))A \mapsto K(\mathscr{P}^{-1}(A)) is a bijection of the set of subgroups of KK containing P(K)\mathscr{P}(K) onto the set of abelian subextensions of exponent dividing pp in Ω\Omega. The inverse mapping is L↦P(L)∩KL \mapsto \mathscr{P}(L) \cap K. b) For every subgroup AA of KK containing P(K)\mathscr{P}(K), the homomorphism

a′:Gal(K(P−1(A))/K)→Hom(A/P(K),Fp) a' : \mathrm{Gal}(K(\mathscr{P}^{-1}(A))/K) \to \mathrm{Hom}(A/\mathscr{P}(K), \mathbf{F}_p)

is bijective, and it is a homeomorphism when Hom(A/P(K),Fp)\mathrm{Hom}(A/\mathscr{P}(K), \mathbf{F}_p) is equipped with the topology of simple convergence. c) Let AA be a subgroup of KK containing P(K)\mathscr{P}(K) and let BB be a basis of the vector Fp\mathbf{F}_p-space A/P(K)A/\mathscr{P}(K). For each a∈Ba \in B let xax_a be an element of Ω\Omega such that P(xa)\mathscr{P}(x_a) is a representative of aa in AA. Then the monomials xα=∏a∈Bxaα(a)x^\alpha = \prod_{a \in B} x_a^{\alpha(a)} with a = (\alpha(a)) in N^{(B)} such that 0 \leq \alpha(a) < p for each a \in B form a basis of the vector K-space K(\mathfrak{p}^{-1}(A)). In particular we have [K(\mathfrak{p}^{-1}(A)):K] = (A:\mathfrak{p}(K)).

Th. 5 is proved in the same way as Th. 4 (V, p. 89) mutatis mutandis.

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

Cited by

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Cites

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