By Grégory Berhuy
This booklet is the 1st basic advent to Galois cohomology and its functions. the 1st half is self contained and offers the elemental result of the idea, together with a close building of the Galois cohomology functor, in addition to an exposition of the overall thought of Galois descent. the complete thought is stimulated and illustrated utilizing the instance of the descent challenge of conjugacy sessions of matrices. the second one a part of the publication offers an perception of ways Galois cohomology should be valuable to resolve a few algebraic difficulties in different energetic examine issues, corresponding to inverse Galois idea, rationality questions or crucial size of algebraic teams. the writer assumes just a minimum historical past in algebra (Galois conception, tensor items of vectors areas and algebras).
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Extra resources for An Introduction to Galois Cohomology and its Applications
14. The set H 1 (Γ, A) is a pointed set, and any abstract group G may be considered as a pointed set, whose base point is the neutral element. 15. If A is a Γ-module, the set Z 1 (Γ, A) is an abelian group for the pointwise multiplication of functions. This operation is compatible with the equivalence relation, hence it induces an abelian group structure on H 1 (Γ, A). We would like now to deﬁne higher cohomology groups. Let A be a Γ-module (written additively here) and n ≥ 0. We set C 0 (Γ, A) = A, and if n ≥ 1, we denote by C n (Γ, A) the set of all continuous maps from Γn to A.
3 and the previous result that A lim −→ A . 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.
Since surjectivity is clear by deﬁnition of f , we just need to check that f is injective. Assume that f (xi /∼ ) = f (xj /∼ ) for some i, j ∈ I, xi ∈ Xi , xj ∈ Xj . Let k ∈ I, k ≥ i, j. By assumption on the fi ’s, we have f (xi /∼ ) = fi (xi ) = fk (ιik (xi )), and similarly f (xj /∼ ) = fk (ιjk (xj )). Now using the equality f (xi /∼ ) = f (xj /∼ ) and the injectivity of fk , we get ιik (xi ) = ιjk (xj ), and thus xi /∼ = xj /∼ . This concludes the proof. 31. Let Γ be a proﬁnite group, and let A be a Γ-set.
An Introduction to Galois Cohomology and its Applications by Grégory Berhuy