Lemma 10.55.1. If $R$ is an Artinian local ring then the length function defines a natural abelian group homomorphism $\text{length}_ R : K'_0(R) \to \mathbf{Z}$.

## 10.55 K-groups

Let $R$ be a ring. We will introduce two abelian groups associated to $R$. The first of the two is denoted $K'_0(R)$ and has the following properties^{1}:

For every finite $R$-module $M$ there is given an element $[M]$ in $K'_0(R)$,

for every short exact sequence $0 \to M' \to M \to M'' \to 0$ of finite $R$-modules we have the relation $[M] = [M'] + [M'']$,

the group $K'_0(R)$ is generated by the elements $[M]$, and

all relations in $K'_0(R)$ among the generators $[M]$ are $\mathbf{Z}$-linear combinations of the relations coming from exact sequences as above.

The actual construction is a bit more annoying since one has to take care that the collection of all finitely generated $R$-modules is a proper class. However, this problem can be overcome by taking as set of generators of the group $K'_0(R)$ the elements $[R^ n/K]$ where $n$ ranges over all integers and $K$ ranges over all submodules $K \subset R^ n$. The generators for the subgroup of relations imposed on these elements will be the relations coming from short exact sequences whose terms are of the form $R^ n/K$. The element $[M]$ is defined by choosing $n$ and $K$ such that $M \cong R^ n/K$ and putting $[M] = [R^ n/K]$. Details left to the reader.

**Proof.**
The length of any finite $R$-module is finite, because it is the quotient of $R^ n$ which has finite length by Lemma 10.53.6. And the length function is additive, see Lemma 10.52.3.
$\square$

The second of the two is denoted $K_0(R)$ and has the following properties:

For every finite projective $R$-module $M$ there is given an element $[M]$ in $K_0(R)$,

for every short exact sequence $0 \to M' \to M \to M'' \to 0$ of finite projective $R$-modules we have the relation $[M] = [M'] + [M'']$,

the group $K_0(R)$ is generated by the elements $[M]$, and

all relations in $K_0(R)$ are $\mathbf{Z}$-linear combinations of the relations coming from exact sequences as above.

The construction of this group is done as above.

We note that there is an obvious map $K_0(R) \to K'_0(R)$ which is not an isomorphism in general.

Example 10.55.2. Note that if $R = k$ is a field then we clearly have $K_0(k) = K'_0(k) \cong \mathbf{Z}$ with the isomorphism given by the dimension function (which is also the length function).

Example 10.55.3. Let $R$ be a PID. We claim $K_0(R) = K'_0(R) = \mathbf{Z}$. Namely, any finite projective $R$-module is finite free. A finite free module has a well defined rank by Lemma 10.15.8. Given a short exact sequence of finite free modules

we have $\text{rank}(M) = \text{rank}(M') + \text{rank}(M'')$ because we have $M \cong M' \oplus M'$ in this case (for example we have a splitting by Lemma 10.5.2). We conclude $K_0(R) = \mathbf{Z}$.

The structure theorem for modules of a PID says that any finitely generated $R$-module is of the form $M = R^{\oplus r} \oplus R/(d_1) \oplus \ldots \oplus R/(d_ k)$. Consider the short exact sequence

Since the ideal $(d_ i)$ is isomorphic to $R$ as a module (it is free with generator $d_ i$), in $K'_0(R)$ we have $[(d_ i)] = [R]$. Then $[R/(d_ i)] = [(d_ i)]-[R] = 0$. From this it follows that a torsion module has zero class in $K'_0(R)$. Using the rank of the free part gives an identification $K'_0(R) = \mathbf{Z}$ and the canonical homomorphism from $K_0(R) \to K'_0(R)$ is an isomorphism.

Example 10.55.4. Let $k$ be a field. Then $K_0(k[x]) = K'_0(k[x]) = \mathbf{Z}$. This follows from Example 10.55.3 as $R = k[x]$ is a PID.

Example 10.55.5. Let $k$ be a field. Let $R = \{ f \in k[x] \mid f(0) = f(1)\} $, compare Example 10.27.4. In this case $K_0(R) \cong k^* \oplus \mathbf{Z}$, but $K'_0(R) = \mathbf{Z}$.

Lemma 10.55.6. Let $R = R_1 \times R_2$. Then $K_0(R) = K_0(R_1) \times K_0(R_2)$ and $K'_0(R) = K'_0(R_1) \times K'_0(R_2)$

**Proof.**
Omitted.
$\square$

Lemma 10.55.7. Let $R$ be an Artinian local ring. The map $\text{length}_ R : K'_0(R) \to \mathbf{Z}$ of Lemma 10.55.1 is an isomorphism.

**Proof.**
Omitted.
$\square$

Lemma 10.55.8. Let $(R, \mathfrak m)$ be a local ring. Every finite projective $R$-module is finite free. The map $\text{rank}_ R : K_0(R) \to \mathbf{Z}$ defined by $[M] \to \text{rank}_ R(M)$ is well defined and an isomorphism.

**Proof.**
Let $P$ be a finite projective $R$-module. Choose elements $x_1, \ldots , x_ n \in P$ which map to a basis of $P/\mathfrak m P$. By Nakayama's Lemma 10.20.1 these elements generate $P$. The corresponding surjection $u : R^{\oplus n} \to P$ has a splitting as $P$ is projective. Hence $R^{\oplus n} = P \oplus Q$ with $Q = \mathop{\mathrm{Ker}}(u)$. It follows that $Q/\mathfrak m Q = 0$, hence $Q$ is zero by Nakayama's lemma. In this way we see that every finite projective $R$-module is finite free. A finite free module has a well defined rank by Lemma 10.15.8. Given a short exact sequence of finite free $R$-modules

we have $\text{rank}(M) = \text{rank}(M') + \text{rank}(M'')$ because we have $M \cong M' \oplus M'$ in this case (for example we have a splitting by Lemma 10.5.2). We conclude $K_0(R) = \mathbf{Z}$. $\square$

Lemma 10.55.9. Let $R$ be a local Artinian ring. There is a commutative diagram

where the vertical maps are isomorphisms by Lemmas 10.55.7 and 10.55.8.

**Proof.**
Let $P$ be a finite projective $R$-module. We have to show that $\text{length}_ R(P) = \text{rank}_ R(P) \text{length}_ R(R)$. By Lemma 10.55.8 the module $P$ is finite free. So $P \cong R^{\oplus n}$ for some $n \geq 0$. Then $\text{rank}_ R(P) = n$ and $\text{length}_ R(R^{\oplus n}) = n \text{length}_ R(R)$ by additivity of lenghts (Lemma 10.52.3). Thus the result holds.
$\square$

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Comment #674 by Keenan Kidwell on