Bourbaki

Exercise 19, § 1

Commutative Algebra, IX

Let K be a field of characteristic pp, P its prime subfield, Ω\Omega an algebraic closure of K. We denote by ℘\wp the endomorphism x↦Fx−xx \mapsto Fx - x of the group W(Ω)W(\Omega), and also, for every integer n⩾1n \geqslant 1, the endomorphism of Wn(Ω)W_n(\Omega) which it induces by passing to quotients.

a) Let us fix an integer n⩾1n \geqslant 1. The endomorphism ℘\wp of W(Ω)W(\Omega) (resp. Wn(Ω)W_n(\Omega)) leaves W(K)W(K) (resp. Wn(K)W_n(K)) stable and its kernel is W(P)W(P) (resp. Wn(P)W_n(P)).

In the sequel to this exercise, W(P)W(P) and Zp\mathbf{Z}_p, Wn(P)W_n(P) and Z/pnZ\mathbf{Z}/p^n\mathbf{Z} will be identified (§ 1, No. 7, Example 3).

b) Let a∈Wn(K)a \in W_n(K). Denoting by V the shift homomorphism Wn(K)→Wn+1(K)W_n(K) \to W_{n+1}(K), prove that one has a∈pWn(K)a \in \mathfrak{p} W_n(K) if and only if one has Va∈pWn+1(K)Va \in \mathfrak{p} W_{n+1}(K). If K is separably closed, one has pWn(K)=Wn(K)\mathfrak{p} W_n(K) = W_n(K).

c) Let a∈Wn(Ω)a \in W_n(\Omega) be such that pa∈Wn(K)\mathfrak{p} a \in W_n(K). We denote by K(a)K(a) the subfield of Ω\Omega generated by K and the components a0,...,an−1a_0, ..., a_{n-1} of aa. If A is a subset of Wn(K)W_n(K), we denote by K(p−1(A))K(\mathfrak{p}^{-1}(A)) the subextension of Ω\Omega generated by the fields K(a)K(a), for all elements aa of Wn(Ω)W_n(\Omega) satisfying pa∈A\mathfrak{p} a \in A.

Reasoning as in A, V, p. 87 and using the preceding exercise, prove the following assertions:

(i) Let L be a Galois extension of K in Ω\Omega. There exists a unique mapping (σ,a)↦[σ,a](\sigma, a) \mapsto [\sigma, a] of Gal(L/K)×(pWn(L)∩Wn(K))/pWn(K)\mathrm{Gal}(L/K) \times (\mathfrak{p} W_n(L) \cap W_n(K))/\mathfrak{p} W_n(K) into Z/pnZ\mathbf{Z}/p^n\mathbf{Z} such that for every σ∈Gal(L/K)\sigma \in \mathrm{Gal}(L/K) and every x∈Wn(L)x \in W_n(L) such that p(x)∈Wn(K)\mathfrak{p}(x) \in W_n(K), one has, denoting by p(x)‾\overline{\mathfrak{p}(x)} the class of p(x)\mathfrak{p}(x) modulo pWn(K)\mathfrak{p} W_n(K)

[σ,p(x)‾]=σx−x. [\sigma, \overline{\mathfrak{p}(x)}] = \sigma x - x.

If σ,σ′∈Gal(L/K)\sigma, \sigma' \in \mathrm{Gal}(L/K) and a,a′∈(pWn(L)∩Wn(K))/pWn(K)a, a' \in (\mathfrak{p} W_n(L) \cap W_n(K))/\mathfrak{p} W_n(K), one has

[σσ′,a]=[σ,a]+[σ′,a] [\sigma \sigma', a] = [\sigma, a] + [\sigma', a]

and

[σ,a+a′]=[σ,a]+[σ,a′]. [\sigma, a + a'] = [\sigma, a] + [\sigma, a'].

(ii) Let us denote, for every Galois extension L of K in Ω\Omega,

aL:(pWn(L)∩Wn(K))/pWn(K)→Hom(Gal(L/K),Z/pnZ) a_L : (\mathfrak{p} W_n(L) \cap W_n(K))/\mathfrak{p} W_n(K) \to \mathrm{Hom}(\mathrm{Gal}(L/K), \mathbf{Z}/p^n\mathbf{Z})

and

aL′:Gal(L/K)→Hom((pWn(L)∩Wn(K))/pWn(K),Z/pnZ) a'_L : \mathrm{Gal}(L/K) \to \mathrm{Hom}((\mathfrak{p} W_n(L) \cap W_n(K))/\mathfrak{p} W_n(K), \mathbf{Z}/p^n\mathbf{Z})

the homomorphisms deduced from the mapping (σ,a)↦[σ,a](\sigma, a) \mapsto [\sigma, a] constructed in (i). For every Galois extension L of K in Ω\Omega, the homomorphism aLa_L is injective, and its image is the group of continuous homomorphisms of the topological group Gal(L/K)\mathrm{Gal}(L/K) into the discrete group Z/pnZ\mathbf{Z}/p^n\mathbf{Z}.

(iii) The mapping A↦K(p−1(A))A \mapsto K(\mathfrak{p}^{-1}(A)) is a bijection of the set of subgroups of Wn(K)W_n(K) containing pWn(K)\mathfrak{p} W_n(K) onto the set of extensions of K in Ω\Omega, abelian over K and of exponent dividing pnp^n. The inverse mapping is L↦pWn(L)∩Wn(K)L \mapsto \mathfrak{p} W_n(L) \cap W_n(K).

(iv) For every subgroup A of Wn(K)W_n(K) containing pWn(K)\mathfrak{p} W_n(K), the homomorphism

aK(p−1(A))′:Gal(K(p−1(A))/K)→Hom(A/pWn(K),Z/pnZ) a'_{K(\mathfrak{p}^{-1}(A))} : \mathrm{Gal}(K(\mathfrak{p}^{-1}(A))/K) \to \mathrm{Hom}(A/\mathfrak{p} W_n(K), \mathbf{Z}/p^n\mathbf{Z})

is bijective. When one endows Hom(A/pWn(K),Z/pnZ)\mathrm{Hom}(A/\mathfrak{p} W_n(K), \mathbf{Z}/p^n\mathbf{Z}) with the topology of simple convergence, it is a homeomorphism.

Solution

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Machine translation of the English, by gpt-5.4. Not checked by a person.