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1. Grothendieck of Abelian categories Roughly speaking, an is an such that finite exists, the kernel and the cokernel of a morphism exist, and the coimage of a morphism is isomorphic to its image (first isomorphism theorem holds). For example, if A is a noetherian , the category of finitely generated (left) A-modules is an abelian category. A morphism f : A → B in an abelian category A is a monomorphism if ker f = 0 and a epimorphism if coker f = 0. We say that the a sequence f g A −−−−→ B −−−−→ C is exact at B if ker g = Im f. A sequence of morphisms

· · · → Ai−1 → Ai → Ai+1 → · · · is said to be exact if it is exact at all Ai. The Grothendieck group K(A) of an abelian category A is an generated by the set {[A]} of symbols [A] of objects of A subject to the relations [A] = [A0] + [A00] whenever 0 → A0 → A → A00 → 0 is exact. Lemma 1.1. Let A be an abelian category and K(A) be its Grothendieck group. Then (1) [0] = 0, and (2) if A =∼ B, [A] = [B], and (3) [A ⊕ B] = [A] + [B]. Proof. To prove (1), we consider the 0 → 0 → 0 → 0 → 0. To prove (2), we consider the exact sequence 0 → 0 → A → B → 0. To prove (3), we consider 0 → A → A ⊕ B → B → 0.  The Grothendick group K(A) of A can be constructed as follows. Let F (A) be the generated by the isomorphism classes of objects of A and R(A) be the subgroup generated by [A] − [A0] − [A00] whenever 0 → A0 → A → A00 → 0 is exact. The F (A)/R(A) is the Grothendick group K(A). Theorem 1.1. Let A and B be an abelian category A. Then [A] = [B] in K(A) if and only if there exist short exact sequence 0 → C0 → C → C00 → 0 and 0 → D0 → D → D00 → 0 such that A ⊕ C ⊕ C00 ⊕ D is isomorphic to B ⊕ C ⊕ D ⊕ D00. An exact F : A → B on abelian categories is an additive functor such that for any exact sequence 0 → A0 → A → A00 → 0, the sequence 0 → F (A0) → F (A) → F (A00) → 0 is also exact.

Lemma 1.2. Let F : A → B be an exact functor of abelian categories. Then F induces a F∗ : K(A) → K(B). Proof. Let us define a map F : F (A) → F (B) by F ([A]) = [F (A)] and extend it additively to all elements of F (A). Since F is additive, F ([A] − [A0] − [A00]) = F ([A]) − F ([A0]) − F ([A00]) = [F (A)] − [F (A0)] − [F (A00)]. Since F is exact, the above identity implies that F (R(A)) ⊂ R(B). This shows that the map F∗ : K(A) → K(B) sending [A] + R(A) → [F (A)] + R(B) is well-defined. This map F∗ can be extend to a group homomorphism.  1 2

Let us abuse the use of the notation [A] : the image of [A] in K(A) will also be denoted by [A].

Lemma 1.3. Let F : A → B be an exact equivalence of categories. Then F∗ : K(A) → K(B) is an isomorphism of abelian groups.

Proof. Let G : B → A be the inverse of F. Then G ◦ F = 1A and F ◦ G = 1B. One can check that G∗ : K(B) → K(A) is the inverse of F∗ : K(A) → K(B).  Definition 1.1. Let A be a noetherian scheme and Coh(X) be the category of coherent sheaves on X. The category Coh(X) is an abelian category. We define the K0-group of the scheme X to be the Grothendieck group of Coh(X): K0(X) = K(Coh(X)). Let A be a commutative noetherian ring and X = Spec A be its corresponding noetherian affine scheme. We know that the global section functor Γ: Coh(X) → Mod(A) sending F → Γ(F ) gives an exact equivalence of abelian categories whose inverse is given by M 7→ M.f Lemma 1.3 implies that: Corollary 1.1. Let A be a noetherian ring and X = Spec A be its corresponding affine acheme. Then we have a group isomorphism: K0(X) =∼ K(Mod(A)). Now, let us define the Grothendieck group of a .

Let C be a triangulated category. The Grothendieck group K(C) of C is the abelian group generated by isomorphism classes of objects of C subject to the relation [A] = [A0] + [A00] whenever we have a distinguish triangle A0 → A → A00 → A0[1]. Theorem 1.2. Let A be an abelian category and Db(A) be its bounded derived category. Then the natural map K(A) → K(Db(A)) is an isomorphism of abelian groups.

Proof. This will be discussed later.  Let X be a noetherian scheme and Db(X) be the bounded derived category of coherent sheaves on X. Theorem 1.1 implies: Corollary 1.2. We have a natural isomorphism if abelian groups: K0(X) = K(Db(X)). Notice that furthermore, if X is regular, K0(X) =∼ K0(X). In this case, we obtain K0(X) =∼ K(Db(X)).