Download An Introduction to Galois Cohomology and its Applications by Grégory Berhuy PDF

By Grégory Berhuy

This e-book is the 1st easy creation to Galois cohomology and its purposes. the 1st half is self contained and offers the elemental result of the speculation, together with a close development of the Galois cohomology functor, in addition to an exposition of the final conception of Galois descent. the total concept is influenced and illustrated utilizing the instance of the descent challenge of conjugacy sessions of matrices. the second one a part of the booklet provides an perception of ways Galois cohomology might be valuable to resolve a few algebraic difficulties in numerous energetic examine themes, reminiscent of inverse Galois idea, rationality questions or crucial size of algebraic teams. the writer assumes just a minimum historical past in algebra (Galois idea, tensor items of vectors areas and algebras).

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Moreover, for any finite Galois extension L/k, let XL = Gal(L/k), and for any L/k, L /k such that L ⊂ L , let πL,L be the group morphism defined by πL,L : Gal(L /k) −→ Gal(L/k) σ −→ σ|L . We obtain in this way a projective system of groups. 18. Let Ω/k be a Galois extension. Then we have an isomorphism of topological groups Gal(Ω/k) lim ←−Gal(L/k), L where L/k runs over all finite Galois subextensions of Ω/k. Proof. Let us consider the map Θ: Gal(Ω/k) −→ lim ←−Gal(L/k) L σ −→ (σ|L )L . This is clearly an abstract group morphism.

6. Let A, B, C be Γ-sets, and let A , B , C be Γ -sets. Assume that we have a commutative diagram with exact rows 1 /A 1  /A /B f h1  /B f /C g h2  /C g /1 h3 /1 satisfying the conditions explained at the beginning of the section. Let us denote by δ 0 and δ 0 the respective connecting maps. If ϕ : Γ −→ Γ is compatible with h1 , h2 and h3 , the diagram CΓ  C δ0 h3∗ Γ / H 1 (Γ, A) h1∗ δ 0  / H 1 (Γ , A ) is commutative. Proof. Let c ∈ C Γ , and let b ∈ B be any preimage of c under g. The cohomology class δ 0 (c) is represented by the cocycle α defined by the relations f (ασ ) = b−1 σ·b for all σ ∈ Γ.

Let K and K be two fields, let L/K be a field extension, and let ι : K −→ K be a ring morphism. Finally, let α ∈ L. For every extension ϕ : L −→ Kalg of ι, ϕ(α) is a root of ι(μα,K ). Proof. Write μα,K = X n + an−1 X n−1 + . . + a0 . Since ϕ is an extension of ι, we have ι(μα,K ) = X n + ϕ(an−1 )X n−1 + . . + ϕ(a0 ). Since ϕ is a ring morphism, we then get ι(μα,K )(ϕ(α)) = ϕ(μα,K (α)) = ϕ(0) = 0. Hence ϕ(α) is a root of ι(μα,K ) as claimed. 18. Let K, K be two fields, let ι : K −→ K be a ring morphism, and let α ∈ Kalg .

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