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] = [A^{0}] + [A^{00}]
whenever 0 → A^{0} → A → A^{00}→ 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] − [A^{0}] − [A^{00}] when
0 → A^{0} → A → A^{00} → 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 → A_{1}⊕ A_{2} → A_{2} → 0, we obtain

[A_{1}⊕ A_{2}] = [A_{1}] + [A_{2}].

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 (A^{0}) + f (A^{00})

for any exact sequence 0 → A^{0} → A → A^{00} → 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 Vect_{F} 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 dim_{F} V = dim_{F}W. We define an additive function

ψ : ob VectF → Z^{≥0}, V 7→ dimFV.

For any exact sequences of K-vector spaces, 0 → V^{0}→ V → V^{00}→ 0, we have
dimFV = dimFV^{0}+ dimFV^{00}.

Hence ψ induces a group homomorphism

ψ : K(Vect_{F}) → 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 dim_{F}V = dim_{F} W for any
representative V of [V ] and W of [W ]. We find [V ] = [W ]. Hence [V ]−[W ] = 0 in K(Vect_{F}).

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

G ∼= Z^{n}⊕ 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/G_{t} is torsion free and thus free, and G/G_{t}∼= Z^{n}. Let us consider a map

ψ : ob Ab → Z≥0, G → rank(G/G_{t}).

If 0 → G^{0} → G → G^{00} → 0 is an exact sequence of finitely generated abelian groups, we
obtain an exact sequence of free modules

0 → G^{0}/G^{0}_{t}→ G/G_{t}→ G^{00}/G^{00}_{t} → 0
which implies that

rank(G/Gt) = rank(G^{0}/G^{0}_{t}) + rank(G^{00}/G^{00}_{t}).

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([Z^{n}] − [Z^{m}]) = 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 = A_{n}⊃ · · · ⊃ A_{1}⊃ A_{0} = 0

such that A_{i}/A_{i−1}∼= Zpi for some prime p_{i}. By induction^{1}, we can show that
[A] =

n

X

i=1

[A_{i}/A_{i−1}].

Hence [A] =Pn

i=1[Zpi]. This shows that elements of K(A) is generated by {[Zp] : p is a prime}.

Let

r_{p} : ob A → Z

to be r_{p}(A) = the number of A_{i}/A_{i−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

r_{p} : 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.

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 a_{k} = 0 since r_{p}(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⊃ A_{n−1}⊃ · · · ⊃ A_{0}= 0
of sub objects of A with each A_{i}/A_{i−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 : A^{0}→ A.