1. Grothendieck Group of Abelian Categories Let a Be an Abelian Category

COMPUTATION OF SOME EXAMPLES OF GROTHENDIECK GROUPS

1. Grothendieck Group of abelian Categories Let A be an abelian category. The Grothendieck group of A is an abelian group generated by the isomorphism classes [A] of objects A of A, subject to the relations [A] = [A0] + [A00] whenever 0 → A0 → A → A00 → 0 is an exact sequence in A. This group can be constructed as follows. Let F (A) be the free abelian group generated by isomorphism classes of objects [A] of A and R(A) be the subgroup of F (A) generated by elements [A] − [A0] − [A00] when 0 → A0 → A → A00 → 0 is exact. The quotient group F (A)/R(A) is the Grothendieck group K(A). The image of [A] in K(A) is still denoted by [A]. Using the exact sequence 0 → A1 → A1 ⊕ A2 → A2 → 0, we obtain

[A1 ⊕ A2] = [A1] + [A2]. Given an abelian category A, we denote ob A the class of objects of A. Let G be an abelian group. A function f : ob A → G is said to be additive if f(A) = f(A0) + f(A00) for any exact sequence 0 → A0 → A → A00 → 0. One can verify that an additive function f : ob A → G induces a group homomorphism f : K(A) → G. Let us compute some examples.

Proposition 1.1. Let VectF be the category of finite dimensional vector spaces over a field ∼ F. Then K(VectF ) = Z. Proof. Two finite dimensional F -vectors spaces are isomorphic if and only if they have the same dimension. In other words, let V and W be finite dimensional F -vector spaces. Then [V ] = [W ] if and only if dimF V = dimF W. We define an additive function

ψ : ob VectF → Z≥0,V 7→ dimF V. For any exact sequences of K-vector spaces, 0 → V 0 → V → V 00 → 0, we have 0 00 dimF V = dimF V + dimF V . Hence ψ induces a group homomorphism

ψ : K(VectF ) → Z defined by ψ([V ] − [W ]) = dimF V − dimF W. This is well-defined. In fact, this is a group isomorphism. To see this, ψ([V ] − [W ]) = 0, if and only if dimF V = dimF W for any representative V of [V ] and W of [W ]. We find [V ] = [W ]. Hence [V ]−[W ] = 0 in K(VectF ). We prove that ψ is a monomorphism. To show that it is surjective, we simply use the fact that ψ([F n] − [F m]) = n − m for any n, m ≥ 0. This completes the proof of our assertion. Proposition 1.2. Let Ab be the category of finitely generated abelian groups. Then ∼ K(Ab) = Z. 1 2 COMPUTATION OF SOME EXAMPLES OF GROTHENDIECK GROUPS

Proof. If G is a ﬁnitely generated abelian group, then there exists n ≥ 0 and d1, ··· , dr such that ∼ n G = Z ⊕ Zd1 ⊕ · · · ⊕ Zdr . Here Zd = Z/dZ. Hence we obtain r X [G] = n[Z] + [Zdi ]. i=1 Let f : Z → Z be the homomorphism deﬁned by f(x) = dx. Then Im f = dZ and ker f = 0. These give us a short exact sequence

0 → Z → Z → Zd → 0, This shows that [Z] = [Z] + [Zd], and thus [Zd] = 0. Notice that if we denote Gt the torsion ∼ n part of G, then G/Gt is torsion free and thus free, and G/Gt = Z . Let us consider a map ψ : ob Ab → Z≥0,G → rank(G/Gt). If 0 → G0 → G → G00 → 0 is an exact sequence of finitely generated abelian groups, we obtain an exact sequence of free modules 0 0 00 00 0 → G /Gt → G/Gt → G /Gt → 0 which implies that 0 0 00 00 rank(G/Gt) = rank(G /Gt) + rank(G /Gt ). In other words, ψ is an additive function. We obtain a group homomorphism ψ : K(Ab) → Z. Notice that if ψ(G) = 0, then G is torsion and thus [G] = 0. Hence ψ is in fact a group n m isomorphism; surjectivity can be proved by taking ψ([Z ] − [Z ]) = n − m. This also shows that K(Ab) is the free abelian group generated by [Z]. Proposition 1.3. Let A be the category of finite abelian groups. Then K(A) is the free abelian group generated by {[Zp]: p is a prime}. Proof. Any finite abelian group A has a composition series

A = An ⊃ · · · ⊃ A1 ⊃ A0 = 0 ∼ 1 such that Ai/Ai−1 = Zpi for some prime pi. By induction , we can show that n X [A] = [Ai/Ai−1]. i=1 Pn Hence [A] = i=1[Zpi ]. This shows that elements of K(A) is generated by {[Zp]: p is a prime}. Let rp : ob A → Z to be rp(A) = the number of Ai/Ai−1 isomorphic to Zp. This is a well-deﬁned function (independent of choice of composition series) by the Jordan Holder theorem. Then rp induces a well-deﬁned group homomorphism

rp : K(A) → Z by rp([Zq]) = δpq for any primes p, q. Here we use rp for its induced map. Claim the set {[Zp]: p is a prime} is Z-linearly independent.

1 We leave it to the readers as an exercise. You may consider the exact sequence 0 → Ai−1 → Ai → Ai/Ai−1 → 0. COMPUTATION OF SOME EXAMPLES OF GROTHENDIECK GROUPS 3

Pr Suppose i=1 ai[Zpi ] = 0 for distinct prime numbers p1, ··· , pr. Then r ! n X X rpk ai[Zpi ] = aiδik = ak. i=1 i=1

We obtain that ak = 0 since rp(0) = 0 for all prime p. This proves the linear independence of {[Zp]: p is a prime}. We complete the proof of our assertion. An object A in an abelian category A is simple if A has no proper subobject 2. We say that A has ﬁnite length if there exists a composition series

A = An ⊃ An−1 ⊃ · · · ⊃ A0 = 0 of sub objects of A with each Ai/Ai−1 simple. Theorem 1.1. Let A be an abelian category such that every object of A has ﬁnite length. Then the Grothendick group K(A) is a free abelian group with basis {[S]: S is simple}.

Proof. The proof is similar to the proof of Proposition 1.3.

2A subobject of an object A in an abelian category is a monomorphism j : A0 → A.