Galois Cohomology [Lecture notes] by Seyfi Türkelli PDF

By Seyfi Türkelli

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Example text

Thus, A ∈ CG . Here, we have a classical result, for the proof see [Sh,p 63]. 15. Let A be an abelian group. Then isomorphism classes of extensions of G by A are in 1-1 correspondence with the elements of H 2 (G; A). 16. Let p be a prime. Then, TFAE: i. cdp G ≤ 1 ii. Every extension of G by a finite abelian p−group splits iii. Every extension of G by a profinite p−group splits 29 Proof. 8. (iii) clearly p /A i /E /G / 0 be an extension in which implies (ii). So, assume (ii). Let 0 A is a profinite p−group.

Notice that H 1 (A0 ; Z/pZ) = 0 because one can find a nontrivial morphism ϕ : A0 → Z/pZ. ˜ < ∞. The Now, let’s set A˜ = ker(ϕi ). Then it is clear that A˜ is normal in E and (A0 : A) following short exact sequence 0 / A /A ˜ 0 / E/A ˜ / E/A0 O cHH HH HH ι0 ˜ ι HHH /0 G ˜ ˜ι). A contradiction. and our assumption allows us to define ˜ι which satisfies (A0 , ι0 ) ≺ (A, Let G = xα α be any group and Λ = {S G | (G : S) < ∞ and S contains almost ˆ = limΛ G/S. For all xα }. Then, one defines and denotes the profinite closure of G by G ←− a prime p, one considers the subset Λp of Λ containing the subgroups S with the property that (G : S) = pn for some n ∈ N.

By first part of inflation-restriction sequence 0 / H 1 (G/S; Z/pZ) inf / H 1 (G; Z/pZ) , one concludes that H 1 (G; Z/pZ) = 0 and gets a contradiction. The proposition above is one of the reasons to study the case dimension is one. Another reason is that there are natural examples of groups which have dimension 1, for instance ˆ = limZ/nZ. We will find an explicit criteria for such groups and we will prove that Z ←− ˆ = 1. cdZ An extension of G by an abelian group A for us is a short exact sequence 0→A→E→G→0 in which E is a topological group and the maps are continuous.

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Galois Cohomology [Lecture notes] by Seyfi Türkelli


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