# Download Equivalence and Duality for Module Categories with Tilting by Robert R. Colby PDF

By Robert R. Colby

This ebook offers a unified method of a lot of the theories of equivalence and duality among different types of modules that has transpired during the last forty five years. extra lately, many authors (including the authors of this ebook) have investigated relationships among different types of modules over a couple of earrings which are caused through either covariant and contravariant representable functors, specifically, by means of tilting and cotilting theories. gathering and unifying the fundamental result of those investigations with leading edge and simply comprehensible proofs, the authors' supply an reduction to additional learn in this valuable subject in summary algebra.

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Additional resources for Equivalence and Duality for Module Categories with Tilting and Cotilting for Rings

Sample text

Proof. Since X generates Hom R (V, M) S , a straightforward argument shows µ (v f ) X → f ∈X f (v f ) defines an epimorphism V (X ) → TrV (M). To see that Hom R (V, µ) is epic, let g ∈ Hom R (V, TrV (M)) ⊆ Hom R (V, M). Then g = n (X ) denote the i=1 f i si where f i ∈ X and si ∈ S. Thus, letting ι f : V → V n (X ) ι fi si v, canonical injections and defining h ∈ Hom R (V, V ) via v → i=1 we obtain Hom R (V, µ)(h) = g. Now we are able to prove the following characterization of ∗-modules that is largely due to R.

3. If VR is a ∗-module with S = End(VR ), then H is exact on short exact sequences in Gen(VR ), and T is exact on short exact sequences in Cogen(VS∗ ). Proof. 2. 4. If S W R is a bimodule and Q R is injective, then for any N S , ExtiS (N , Hom R (W, Q)) ∼ = Hom R (ToriS (N , W ), Q) for all i ≥ 0. Proof. 1]. Now we have the necessary tools to give homological characterizations of Cogen(VS∗ ) for a ∗-module VR . 5. If VR is a ∗-module with S = End(VR ), then Cogen(VS∗ ) = Ker Tor1S ( , V ) = Ker Ext1S ( , V ∗ ).

Since VR is a quasi-progenerator, Gen(VR ) is closed under submodules. Thus, if X R ≤ V n , then X = TrV (X ), and if f : V → X, then there are s1 , . . , sn ∈ S such that f (v) = (s1 v, . . , sn v) for all v ∈ V. But then if b ∈ B, we have f (v)b = (s1 (vb), . . , sn (vb)) = f (vb) ∈ X. Thus X B = X and Hom B (V, X ) = Hom R (V, X ). In particular, (v1 , . . , vn )B = (v1 , . . , vn )R B = (v1 , . . , vn )R so that R is V -dense in B. Now if K ≤ VR and g : V → V /K is an Rhomomorphism, let g(v) = v + K .