By Grégory Berhuy

This e-book is the 1st simple creation to Galois cohomology and its functions. the 1st half is self contained and offers the fundamental result of the speculation, together with a close building of the Galois cohomology functor, in addition to an exposition of the overall thought of Galois descent. the entire concept is inspired and illustrated utilizing the instance of the descent challenge of conjugacy sessions of matrices. the second one a part of the e-book supplies an perception of the way Galois cohomology might be precious to resolve a few algebraic difficulties in different energetic learn themes, comparable to inverse Galois concept, rationality questions or crucial measurement 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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**Example text**

Since T is ﬁnite, the set (i) Ut ∩ U0 = i,t Ut t is an open neighbourhood of 1 in Γ. Since Γ is a proﬁnite group, there 32 Cohomology of proﬁnite groups exists a normal open subgroup U of Γ contained in U0 . Now let s = (σ1 , . . , σn ) ∈ Γn . By choice of T , there exists t ∈ T such that s ∈ Ut , (i) so for i = 1, . . , n we may write σi = τi ui for some ui ∈ Ut . We then (i) (i) get σ1 U ×· · ·×σn U ⊂ Ut , since U ⊂ Ut for all i and Ut is a subgroup. Hence for all s ∈ σ1 U × · · · × σn U , we have αs = αt ∈ AUt , since α is constant on Ut .

N is well-deﬁned, and satisﬁes the required conditions. (3) ⇒ (1) Assume that we have an open normal subgroup U and a map α(U ) : (Γ/U )n −→ AU satisfying (3). Let V be an open subset of A. We have to prove that α−1 (V ) is open in Γ. Since α−1 (V ) = α−1 ({v}), v∈V it is enough to show that α−1 ({v}) is open for every v ∈ V . If v does not lie in the image of α, this is obvious, so we can assume that α−1 ({v}) is not empty. Let s = (σ1 , . . , σn ) ∈ α−1 ({v}). ,σn = αs = v. Thus α−1 ({v}), contains an open neighbourhood of s.

In particular, for any ﬁnite Galois subextension L/k of Ω/k, we have Gal(Ω/k)/Gal(Ω/L) Gal(L/k). All these results and their proofs may be found in [42], Chapter IV. See also [26], Chapter 4. 3 Morphisms of Galois extensions To continue this section on Galois theory, we would like to have a closer look at morphisms of Galois extensions. 9. Let K and K be two ﬁelds, let Ω/K two Galois extensions (not necessarily ﬁnite), and let ι : a ring morphism. Assume that there exist two extensions of ι,i = 1, 2.