Bourbaki

Lemma 8

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

Algebra, chapter V, § 11. Abelian extensions

Let L be a Galois extension of K. There exists a unique mapping (σ,a)↦[a,a′)(\sigma, a) \mapsto [a, a') of Gal(L/K)×((φ(L)∩K)/φ(K))\mathrm{Gal}(L/K) \times ((\varphi(L) \cap K)/\varphi(K)) into Fp\mathbf{F}_p such that for all σ∈Gal(L/K)\sigma \in \mathrm{Gal}(L/K) and every element x∈Lx \in L such that φ(x)∈K\varphi(x) \in K we have, on denoting by φ(x)‾\overline{\varphi(x)} the residue class of φ(x)\varphi(x) mod φ(K)\varphi(K),

[σ,φ(x)‾]=σ(x)−x. [\sigma, \overline{\varphi(x)}] = \sigma(x) - x.

This mapping is Z-bilinear (for σ,τ∈Gal(L/K)\sigma, \tau \in \mathrm{Gal}(L/K), a,b∈(P(L)∩K)/P(K)a, b \in (\mathscr{P}(L) \cap K)/\mathscr{P}(K), we have [στ,a)=[σ,a)+[τ,a)[\sigma \tau, a) = [\sigma, a) + [\tau, a), [σ,a+b)=[σ,a)+[a,b)[\sigma, a + b) = [\sigma, a) + [a, b)).

For the right-hand side of (15) is an element of Fp\mathbf{F}_p which depends only on the residue class of P(x)mod  P(K)\mathscr{P}(x) \mod \mathscr{P}(K); this proves the first assertion; the second may be verified without difficulty.

For every Galois extension LL of KK let us write

aL:(P(L)∩K)/P(K)→Hom(Gal(L/K),Fp) a_L : (\mathscr{P}(L) \cap K)/\mathscr{P}(K) \to \mathrm{Hom}(\mathrm{Gal}(L/K), \mathbf{F}_p)
aL′:Gal(L/K)→Hom((P(L)∩K)/P(K),Fp) a_L' : \mathrm{Gal}(L/K) \to \mathrm{Hom}((\mathscr{P}(L) \cap K)/\mathscr{P}(K), \mathbf{F}_p)

for the homomorphisms obtained from the above Z-bilinear mapping (V, p. 87).

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

Cited by

Nothing.

Cites

Nothing.

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