By Grégory Berhuy
This publication is the 1st uncomplicated creation to Galois cohomology and its purposes. the 1st half is self contained and gives the fundamental result of the speculation, together with an in depth development of the Galois cohomology functor, in addition to an exposition of the overall conception of Galois descent. the full idea is inspired and illustrated utilizing the instance of the descent challenge of conjugacy periods of matrices. the second one a part of the e-book offers an perception of the way Galois cohomology can be valuable to resolve a few algebraic difficulties in numerous energetic examine subject matters, corresponding to inverse Galois idea, rationality questions or crucial size of algebraic teams. the writer assumes just a minimum history in algebra (Galois conception, tensor items of vectors areas and algebras).
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Let us prove now that the cohomology class of α does not depend on the choice of b. Let b ∈ B be another preimage of c under g. We then have g(b ) = c = g(b), so b = bf (a−1 ) = bf (a)−1 for some a ∈ A by assumption on g, and let α be the corresponding 1-cocycle. We then have f (ασ ) = f (a)b−1 σ·(bf (a−1 )) = f (a)f (ασ ) σ · f (a−1 ) = f (aασ σ·a−1 ), so by injectivity of f , this implies that α and α are cohomologous. This concludes the proof. Notice that a preimage under g of the base point of C Γ is the neutral element 1 ∈ B (since g is a morphism of pointed sets).
Pr ). Notice that E/Q is a ﬁnite Galois extension. Now assume that x ∈ ΩH . Since σp1 , . . , σpr ∈ H, and since they generate Gal(E/Q), we conclude that x ∈ Q by classical Galois theory. In order to get a Galois correspondence, we deﬁne a topology on the Galois group of Ω/k. 7. Let Ω/k be a Galois extension. 2 Galois theory 19 is a basis of open neighbourhoods of σ. We may now state the fundamental theorem of Galois theory. 8 (Fundamental theorem of Galois theory). Let Ω/k be a Galois extension.
Now, if ϕ is another extension of ι, then by the previous proposition, there exists ρ ∈ Gal(Ω/K) such that ϕ = ϕ ◦ ρ. Therefore, for every τ ∈ Gal(Ω /K ), we have τ ◦ ϕ = (τ ◦ ϕ) ◦ ρ−1 = ϕ ◦ (ϕ(τ ) ◦ ρ−1 ) = ϕ ◦ (ρ ◦ ϕ(τ ) ◦ ρ−1 ). We conclude as before. 4 The Galois group as a proﬁnite group Let Ω/k be a Galois extension. 4 shows in particular that an element σ ∈ Gal(Ω/k) is completely determined by its restrictions to ﬁnite Galois subextensions L/k of Ω/k. Intuitively, Gal(Ω/k) then should be completely determined by the ﬁnite groups Gal(L/k).
An Introduction to Galois Cohomology and its Applications by Grégory Berhuy