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Contents 1. Introduction 2 2. Preliminaries 4 2.1. Some results on ...

Contents 1. Introduction 2 2. Preliminaries 4 2.1. Some results on ...

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On the other hand, ∆ C and ε C are A-bilinear maps and in particular right A-modulemap so that we havei.e.∆ C ((a ⊗ c) b ′ ) = ( ∆ C (a ⊗ c) ) b ′ and ε C ((a ⊗ c) b ′ ) = ( ε C (a ⊗ c) ) b ′(228) ab ′ α ⊗ c α (1) ⊗ A 1 A ⊗ c α (2) = a ⊗ c (1) ⊗ A b ′ α ⊗ c α (2)and(229) ab ′ αε (c α ) = aε (c) b ′ .Then, for every c ∈ C and a ∈ A, we haveand[(A ⊗ ∆) ◦ ψ] (c ⊗ a) = (A ⊗ ∆) (a α ⊗ c α ) = a α ⊗ c α (1) ⊗ c α (2)(228)= 1 A ⊗ c (1) a α ⊗ c α (2) = ψ ( )c (1) ⊗ a α ⊗ cα(2)= (ψ ⊗ C) ( ) (c (1) ⊗ a α ⊗ c α (2) = [(ψ ⊗ C) ◦ (C ⊗ ψ)] c(1) ⊗ c (2) ⊗ a )= [(ψ ⊗ C) ◦ (C ⊗ ψ) ◦ (∆ ⊗ A)] (c ⊗ a)[r A ◦ (A ⊗ ε) ◦ ψ] (c ⊗ a) = [r A ◦ (A ⊗ ε)] (a α ⊗ c α ) = a α ε (c α )(229)= ε (c) a = [l A ◦ (ε ⊗ A)] (c ⊗ a) .3) Let M ∈ M C A (ψ) , that is ρC M is a right A-module map where A ⊗ C has a rightA-module structure given by(a ⊗ c) b ′ = aψ (c ⊗ b ′ ) = ab ′ α ⊗ c α .Since ρ C M is a right A-module map, then the comodule structure given by the compositeρ C M : M ρC M−→ M ⊗ C ≃ M ⊗ A A ⊗ C = M ⊗ A Cis a right A-module map and thus ( ( )M, ρM) C is a right C-comodule. C<strong>on</strong>versely, letM, ρCM be a right C-comodule, then we can c<strong>on</strong>sider215ρ C M : M ρC M−→ M ⊗ A C = M ⊗ A A ⊗ C ≃ M ⊗ Cas a right A-module map and thus we can see M as a (A, C, ψ)-entwined module.In fact, (226) just means that the map ρ C M is a right A-module map.□Theorem 9.31 ([SS, Lemma <str<strong>on</strong>g>1.</str<strong>on</strong>g>7]). Let C be a k-coalgebra and let A be a k-algebrasuch that (A, C, ψ) is an entwining structure. Then C = A ⊗ C is an A-coring.Assume that A C is flat (i.e. C is k-flat) and that ( A, m, ρA) C ∈ MCA (ψ). Let B =A coC . Then the following statements are equivalent:(a) the functor − ⊗ B A : Mod-B → M C A (ψ) is an equivalence of categories;(b) the can<strong>on</strong>ical map can : A ⊗ B A → A ⊗ C is an isomorphism (i.e. B ⊆ A isa C-Galois extensi<strong>on</strong>) and B A is faithfully flat.Proof. By Propositi<strong>on</strong> 9.30 we know that M C A = (Mod-A)C ≃ M C A (ψ) . By hypothesis( A, m, ρA) C ∈ MCA (ψ) and thus, by Lemma 9.23, C = A ⊗ C has a grouplikeelement, that is ρ C A (1 A) . Then we can apply Corollary 9.26 to c<strong>on</strong>clude. □

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