Contents 1. Introduction 2 2. Preliminaries 4 2.1. Some results on ...
Share Embed Flag
Download ePaper

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

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

eprints.unife.it
from eprints.unife.it More from this publisher
12.07.2015 Views

182(208)(209)(210)(211)(h1 ) 0 ⊗ h2 ⊗ ( h 1) 1= (h 2 ) 1 ⊗ (h 2 ) 2 ⊗ S (h 1 )h 1 h 2 = ε H (h) 1 A(hl) 1 ⊗ (hl) 2 = l 1 h 1 ⊗ h 2 l 2a 0 (a 1 ) 1 ⊗ (a 1 ) 2 = 1 A ⊗ a.9.5. The Schauenburg situation is the particular case when T = A co(H) = k. Hencewe haveA = Mod-k,B = Mod-k where T = End eC (Σ) = A co(H) = k.A = − ⊗ k A : A = Mod-k −→ A = Mod-kC = − ⊗ k H : A = Mod-k −→ A = Mod-kB = − ⊗ k A = A : B = Mod-k −→ B = Mod-k where A = Hom A (A A , A A )Ψ = − ⊗ k ψ : AC = − ⊗ k H ⊗ k A −→ CA = − ⊗ k A ⊗ k H(ψ : H ⊗ k A → A ⊗ k H, h ⊗ k a ↦→ can C can−1C (1 A ⊗ k h) a )L = − ⊗ k A : Mod-k −→ Mod-A ∼ = A AW = Hom A (A, −) : Mod-A −→ Mod-k˜C = − ⊗ A A ⊗ k H ∼ = ⊗ k H : Mod-A −→ Mod-Al = eC ρ L : L = − ⊗ k A −→ ˜CL = − ⊗ k A ⊗ A A ⊗ k H ∼ = − ⊗ k A ⊗ k HLet us assume thatThen,k = commutative ringH = Hopf algebraA/k = faithfully flat H-Galois extension with ρ H A : A → A ⊗ HA co(H) = k1 Acan : A ⊗ A → A ⊗ H defined by setting can (a ⊗ b) = ab 0 ⊗ b 1γ : H → A ⊗ A defined by setting γ (h) := can −1 (1 A ⊗ h) = h 1 ⊗ h 2τ : A → A ⊗ A ⊗ Aa ↦→ a 0 ⊗ γ (a 1 ) = a 0 ⊗ a 1 1 ⊗ a 2 1is a pretorsor. We want to construct the comonads C and D as in Theorem 6.5where A = B = Mod-k and P = Q = − ⊗ A. First, let us consider ω l = − ⊗ k ̂ωland ω r = − ⊗ k ̂ω r : − ⊗ A ⊗ A → − ⊗ A ⊗ A ⊗ A wherêω l = ( σ A ⊗ A ⊗ A ) ◦ (A ⊗ τ) : A ⊗ A → A ⊗ A ⊗ Âω l (a ⊗ b) = ab 0 ⊗ b 1 1 ⊗ b 2 1̂ω r = (u A ⊗ A ⊗ A) : A ⊗ A → A ⊗ A ⊗ Âω r (a ⊗ b) = 1 A ⊗ a ⊗ b.

Let ω = ω l −ω r and ̂ω = ̂ω l −̂ω r . First of all we want to prove that for any k-moduleX we haveKer (ωX) = Ker (X ⊗ ̂ω) = X ⊗ Ker (̂ω) .Since A is faithfully flat over k we equivalently prove thatNote thatA ⊗ Ker (̂ω)Ker (X ⊗ ̂ω ⊗ A) = X ⊗ Ker (̂ω) ⊗ A. A ⊗ A ⊗ AA⊗ b ω lA⊗c ω r A ⊗ A ⊗ A ⊗ Awith respect to the map m A ⊗ A ⊗ A is a contractible equalizer so that alsoKer (̂ω) A ⊗ Abω lcω r A ⊗ A ⊗ Ais a contractible equalizer (see the dual case of [BW, Proposition 3.4 (c)]). Hence,by Proposition ong>2.ong>20, ( it is preserved by any functor. Since(C, i) = Equ Fun ω l = − ⊗ ̂ω) l , ω r = − ⊗ ̂ω r we have that C = − ⊗ Ĉ where{∑Ĉ = a i ⊗ b i | ∑ ( ) ( aib i) ⊗ ( b i) 1⊗ ( b i) 2= 1 0 1 1 A ⊗ ∑ }a i ⊗ b i .Note that γ is a fork for ̂ω l and ̂ω r , in fact( ̂ωl ◦ γ)(h) = ̂ω ( l h 1 ⊗ h 2) = h 1 h 2 0 ⊗ ( )h 2 1 ( )1 ⊗ h2 2 (206)1 = 1 A ⊗ h 1 ⊗ h 2and(̂ω r ◦ γ ) (h) = ̂ω ( r h 1 ⊗ h 2) = 1 A ⊗ h 1 ⊗ h 2 .) ( )Since(Ĉ, î = Equ ̂ωl , ̂ω r , by the universal property of the equalizer, there existsa unique functorial morphism ϕ : H → Ĉ such that î ◦ ϕ = γ. In our case thismeans that Imγ ⊆ C and henceϕ : H → Ĉh ↦→ h 1 ⊗ h 2 .We want to prove that ϕ is an isomorphism. We computeLet us setand let us computeand[(A ⊗ ϕ) ◦ can] (a ⊗ b) = (A ⊗ ϕ) (ab 0 ⊗ b 1 ) = ab 0 ⊗ b 1 1 ⊗ b 2 ong>1.ong>ψ : A ⊗ C → A ⊗ Aa ⊗ b ⊗ d ↦→ ab ⊗ d[ψ ◦ (A ⊗ ϕ) ◦ can] (a ⊗ b) = ψ ( ab 0 ⊗ b 1 1 ⊗ b 2 1)= ab0 b 1 1 ⊗ b 2 1[(A ⊗ ϕ) ◦ can ◦ ψ] (a ⊗ b ⊗ d) = [(A ⊗ ϕ) ◦ can] (ab ⊗ d)(211)= a ⊗ b183

Let ω = ω l −ω r and ̂ω = ̂ω l −̂ω r . First of all we want to prove that for any k-moduleX we haveKer (ωX) = Ker (X ⊗ ̂ω) = X ⊗ Ker (̂ω) .Since A is faithfully flat over k we equivalently prove thatNote thatA ⊗ Ker (̂ω)Ker (X ⊗ ̂ω ⊗ A) = X ⊗ Ker (̂ω) ⊗ A. A ⊗ A ⊗ AA⊗ b ω lA⊗c ω r A ⊗ A ⊗ A ⊗ Awith respect to the map m A ⊗ A ⊗ A is a c<strong>on</strong>tractible equalizer so that alsoKer (̂ω) A ⊗ Abω lcω r A ⊗ A ⊗ Ais a c<strong>on</strong>tractible equalizer (see the dual case of [BW, Propositi<strong>on</strong> 3.4 (c)]). Hence,by Propositi<strong>on</strong> <str<strong>on</strong>g>2.</str<strong>on</strong>g>20, ( it is preserved by any functor. Since(C, i) = Equ Fun ω l = − ⊗ ̂ω) l , ω r = − ⊗ ̂ω r we have that C = − ⊗ Ĉ where{∑Ĉ = a i ⊗ b i | ∑ ( ) ( aib i) ⊗ ( b i) 1⊗ ( b i) 2= 1 0 1 1 A ⊗ ∑ }a i ⊗ b i .Note that γ is a fork for ̂ω l and ̂ω r , in fact( ̂ωl ◦ γ)(h) = ̂ω ( l h 1 ⊗ h 2) = h 1 h 2 0 ⊗ ( )h 2 1 ( )1 ⊗ h2 2 (206)1 = 1 A ⊗ h 1 ⊗ h 2and(̂ω r ◦ γ ) (h) = ̂ω ( r h 1 ⊗ h 2) = 1 A ⊗ h 1 ⊗ h 2 .) ( )Since(Ĉ, î = Equ ̂ωl , ̂ω r , by the universal property of the equalizer, there existsa unique functorial morphism ϕ : H → Ĉ such that î ◦ ϕ = γ. In our case thismeans that Imγ ⊆ C and henceϕ : H → Ĉh ↦→ h 1 ⊗ h 2 .We want to prove that ϕ is an isomorphism. We computeLet us setand let us computeand[(A ⊗ ϕ) ◦ can] (a ⊗ b) = (A ⊗ ϕ) (ab 0 ⊗ b 1 ) = ab 0 ⊗ b 1 1 ⊗ b 2 <str<strong>on</strong>g>1.</str<strong>on</strong>g>ψ : A ⊗ C → A ⊗ Aa ⊗ b ⊗ d ↦→ ab ⊗ d[ψ ◦ (A ⊗ ϕ) ◦ can] (a ⊗ b) = ψ ( ab 0 ⊗ b 1 1 ⊗ b 2 1)= ab0 b 1 1 ⊗ b 2 1[(A ⊗ ϕ) ◦ can ◦ ψ] (a ⊗ b ⊗ d) = [(A ⊗ ϕ) ◦ can] (ab ⊗ d)(211)= a ⊗ b183

logo

products

Resources

Company

Terms of service
Privacy policy
Cookie policy
Cookie settings
Imprint
Change language
Made with love in Switzerland
© 2024 Yumpu.com all rights reserved

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!