# Grothendieck Group of abelian Categories Let A be an abelian category

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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 = dimFW. We define an additive function

ψ : ob VectF → Z≥0, V 7→ dimFV.

For any exact sequences of K-vector spaces, 0 → V0→ V → V00→ 0, we have dimFV = dimFV0+ dimFV00.

Hence ψ induces a group homomorphism

ψ : K(VectF) → Z

defined by ψ([V ] − [W ]) = dimFV − dimFW. This is well-defined. In fact, this is a group isomorphism. To see this, ψ([V ] − [W ]) = 0, if and only if dimFV = 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([Fn] − [Fm]) = 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.

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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

G ∼= Zn⊕ Zd1 ⊕ · · · ⊕ Zdr. Here Zd= Z/dZ. Hence we obtain

[G] = n[Z] +

r

X

i=1

[Zdi].

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 part of G, then G/Gt is torsion free and thus free, and G/Gt∼= Zn. 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 → G0/G0t→ G/Gt→ G00/G00t → 0 which implies that

rank(G/Gt) = rank(G0/G0t) + rank(G00/G00t).

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 isomorphism; surjectivity can be proved by taking ψ([Zn] − [Zm]) = 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

such that Ai/Ai−1∼= Zpi for some prime pi. By induction1, we can show that [A] =

n

X

i=1

[Ai/Ai−1].

Hence [A] =Pn

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.

1We leave it to the readers as an exercise. You may consider the exact sequence 0 → Ai−1→ Ai→ Ai/Ai−1→ 0.

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COMPUTATION OF SOME EXAMPLES OF GROTHENDIECK GROUPS 3

SupposePr

i=1ai[Zpi] = 0 for distinct prime numbers p1, · · · , pr. Then rpk

r

X

i=1

ai[Zpi]

!

=

n

X

i=1

aiδik = ak.

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.

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