By Grégory Berhuy

This ebook is the 1st easy advent to Galois cohomology and its purposes. the 1st half is self contained and gives the fundamental result of the idea, together with a close building of the Galois cohomology functor, in addition to an exposition of the overall idea of Galois descent. the complete thought is inspired and illustrated utilizing the instance of the descent challenge of conjugacy sessions of matrices. the second one a part of the booklet supplies an perception of ways Galois cohomology will be necessary to resolve a few algebraic difficulties in different lively learn issues, equivalent to inverse Galois idea, rationality questions or crucial measurement of algebraic teams. the writer assumes just a minimum heritage in algebra (Galois conception, tensor items of vectors areas and algebras).

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Additional resources for An Introduction to Galois Cohomology and its Applications

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Again, we only prove the claim for n = 1 in the case of Γ-groups. We have to show that α and β = f∗ (α) are cohomologous for every 1-cocycle α ∈ Z 1 (Γ, A). First notice that applying the cocyclicity relation to τ = σ −1 gives σ −1 ·ασ = ασ−1 −1 for σ ∈ Γ. Now we have βσ = ρ−1 ·αρσρ−1 = ρ−1 ·(αρ (ρ·ασρ−1 )) = (ρ−1 ·αρ )ασρ−1 , since Γ acts on A by group automorphisms. Hence we get βσ = (ρ−1 ·αρ )(ασ σ·αρ−1 ) = αρ−1 −1 ασ σ·αρ−1 . Setting a = αρ−1 −1 then shows that α and β are cohomologous. 22.

Now let σ : ks −→ kalg be a k-linear embedding. 2). Then σ|L : L −→ kalg is a k-embedding of L into kalg . Since L/k is a Galois extension, we have σ(x) ∈ L. In particular σ(x) is separable, since L/k is separable. Therefore, we have proved that σ(ks ) ⊂ ks . To prove the missing inclusion, let x ∈ ks and let L /k be a finite Galois extension containing x . Since L /k is a Galois extension, σ|L is a k-automorphism of L . Hence, there exists x ∈ L ⊂ ks such that σ|L (x) = x . Thus we have x = σ|L (x) = σ(x), and therefore, σ(ks ) = ks .

U ∈N Let Γ be a profinite group, and let A be a Γ-group (resp. a Γ-module if n ≥ 2). 25 that we have maps inf U,U : H n (Γ/U, AU ) −→ H n (Γ/U , AU ), for all U, U ∈ N , U ⊃ U and a map fU : H n (Γ/U, AU ) −→ H n (Γ, A). The following lemma follows from direct computations. 32. The sets H n (Γ/U, AU ) together with the maps inf U,U form a directed system of pointed sets (resp. of groups if A is abelian). Moreover, we have fU = fU ◦ inf U,U . We now come to the main result of this section. 33. Let Γ be a profinite group, and let A be a Γ-group.

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