By Mikhail Borovoi

During this quantity, a brand new functor $H^2_{ab}(K,G)$ of abelian Galois cohomology is brought from the class of attached reductive teams $G$ over a box $K$ of attribute $0$ to the class of abelian teams. The abelian Galois cohomology and the abelianization map$ab^1:H^1(K,G) \rightarrow H^2_{ab}(K,G)$ are used to provide a functorial, nearly particular description of the standard Galois cohomology set $H^1(K,G)$ whilst $K$ is a host box.

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**Example text**

1) ab1 v 1 H 1 (Kv , G) −−−− → Hab (Kv , G) it is clear that ab1 takes X(G) into X1ab (G). Write temporarily abX for the restriction of ab1G to X(G). We prove the injectivity of abX . 12 the map 1 ab1G × loc∞ : H 1 (K, G) → Hab (K, G) × Π H 1 (Kv , G) ∞ is injective. Since loc∞ (X(G)) = 1, we conclude that the restriction abX of ab1G to X(G) is injective. 1 We prove the surjectivity of abX . Let h ∈ X1ab (G) ⊂ Hab (K, G). Then 1 loc∞ (h) = 1 ∈ Π Hab (Kv , G). Hence the element ∞ 1 h × 1 ∈ Hab (K, G) × Π H 1 (Kv , G) ∞ 1 lies in the fiber product over Π Hab (Kv , G).

8 that the image of Hi (K, M, K i × ¯ H (K, M, A ) is isomorphic to coriv : ⊕Tvi (M ) → T (M ). ker v∈V Let ∆ be the image of Γ in Aut M . Let K be the corresponding Galois extension (so that Gal(K /K) = ∆), and for v ∈ V let ∆v be a decomposition group of v (defined up to conjugation). ABELIAN GALOIS COHOMOLOGY OF REDUCTIVE GROUPS 33 Let v ∈ V. If v ∈ Vf and ∆v = ∆v (up to conjugation), then im coriv ⊂ im coriv . ) Let v ∈ V∞ ; then ∆v is cyclic. By Chebotorev’s density theorem there are infinitely many places v ∈ Vf such that ∆v is conjugate to ∆v .

1) we reduce the assertion to the well known (cf. 25) case of a torsion free module M . The lemma is proved. 6. For any h ∈ Hi (K, M, K i × ¯ v ) is zero for v ∈ / S. 5 that for any ξ ∈ Hi (K, M, A ¯ × ). This exists a finite set S ⊂ V such that ξ comes from H1 (K, M, ⊕(Kv ⊗ K) S K implies the corollary. 7. We want to describe the map ¯ → Hi (K, M, C) ¯ × ) = Hi (K, M, (K ¯ ⊗ Kv )× ) → Hi (K, M, A) ¯ Hi (Kv , M, K v for i = 1, 2. 8. 4. Proof: We consider the case i = 1; the case i = 2 can be treated similarly.