
## 10.54 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 $K'_0(R)$,

2. for every short exact sequence $0 \to M' \to M \to M'' \to 0$ we have the relation $[M] = [M'] + [M'']$,

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

4. all relations in $K_0'(R)$ 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.

Lemma 10.54.1. If $R$ is an Artinian local ring then the length function defines a natural abelian group homomorphism $\text{length}_ R : K_0'(R) \to \mathbf{Z}$.

Proof. The length of any finite $R$-module is finite, because it is the quotient of $R^ n$ which has finite length by Lemma 10.52.6. And the length function is additive, see Lemma 10.51.3. $\square$

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

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

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

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

4. 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 10.54.2. 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 10.54.3. Let $k$ be a field. Then $K_0(k[x]) = K_0'(k[x]) = \mathbf{Z}$.

Since $R = k[x]$ is a principal ideal domain, any finite projective $R$-module is free. In a short exact sequence of modules

$0 \to M' \to M \to M'' \to 0$

we have $\text{rank}(M) = \text{rank}(M') + \text{rank}(M'')$, which gives $K_0(k[x]) = \mathbf{Z}$.

As for $K_0'$, the structure theorem for modules of a PID says that any finitely generated $R$-module is of the form $M = R^ r \times R/(d_1) \times \ldots \times R/(d_ k)$. Consider the short exact sequence

$0 \to (d_ i) \to R \to R/(d_ i) \to 0$

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 any torsion part “disappears” in $K_0'$. Again the rank of the free part determines that $K_0'(k[x]) = \mathbf{Z}$, and the canonical homomorphism from $K_0$ to $K_0'$ is an isomorphism.

Example 10.54.4. Let $k$ be a field. Let $R = \{ f \in k[x] \mid f(0) = f(1)\}$, compare Example 10.26.4. In this case $K_0(R) \cong k^* \oplus \mathbf{Z}$, but $K_0'(R) = \mathbf{Z}$.

Lemma 10.54.5. 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 10.54.6. Let $R$ be an Artinian local ring. The map $\text{length}_ R : K_0'(R) \to \mathbf{Z}$ of Lemma 10.54.1 is an isomorphism.

Proof. Omitted. $\square$

Lemma 10.54.7. Let $R$ 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. The $n$ generators of $P$ give a surjection $R^ n \to P$, and since $P$ is projective it follows that $R^ n \cong P \oplus Q$ for some projective module $Q$.

If $\mathfrak m \subset R$ is the maximal ideal, then $P/\mathfrak m$ and $Q/\mathfrak m$ are $R/\mathfrak m$-vector spaces, with $P/\mathfrak m \oplus Q/\mathfrak m \cong (R/\mathfrak m)^ n$. Say that $\dim P = p$, $\dim Q = q$, so $p + q = n$.

Choose elements $a_1, \ldots , a_ p$ in $P$ and $b_1, \ldots , b_ q$ in $Q$ lying above bases for $P/\mathfrak m$ and $Q/\mathfrak m$. The homomorphism $R^ n \to P \oplus Q \cong R^ n$ given by $(r_1, \ldots , r_ n) \mapsto r_1a_1 + \ldots + r_ pa_ p + r_{p + 1} b_1 + \ldots + r_ nb_ q$ is a matrix $A$ which is invertible over $R/\mathfrak m$. Let $B$ be a matrix over $R$ lying over the inverse of $A$ in $R/\mathfrak m$. $AB = I + M$, where $M$ is a matrix whose entries all lie in $\mathfrak m$. Thus $\det AB = 1 + x$, for $x \in \mathfrak m$, so $AB$ is invertible, so $A$ is invertible.

The homomorphism $R^ p \to P$ given by $(r_1, \ldots , r_ p) \mapsto r_1a_1 + \ldots + r_ pa_ p$ inherits injectivity and surjectivity from A. Hence, $P \cong R^ p$.

Next we show that the rank of a finite projective module over $R$ is well defined: if $P \cong R^\alpha \cong R^\beta$, then $\alpha = \beta$. This is immediate in the vector space case, and so it is true in the general module case as well, by dividing out the maximal ideal on both sides. If $0 \to R^\alpha \to R^\beta \to R^\gamma \to 0$ is exact, the sequence splits, so $R^\beta \cong R^\alpha \oplus R^\gamma$, so $\beta = \alpha + \gamma$.

So far we have seen that the map $\text{rank}_ R : K_0(R) \to \mathbf{Z}$ is a well-defined homomorphism. It is surjective because $\text{rank}_ R[R] = 1$. It is injective because the element of $K_0(R)$ with rank $\pm \alpha$ is uniquely $\pm [R^\alpha ]$. $\square$

Lemma 10.54.8. Let $R$ be a local Artinian ring. There is a commutative diagram

$\xymatrix{ K_0(R) \ar[rr] \ar[d]_{\text{rank}_ R} & & K_0'(R) \ar[d]^{\text{length}_ R} \\ \mathbf{Z} \ar[rr]^{\text{length}_ R(R)} & & \mathbf{Z} }$

where the vertical maps are isomorphisms by Lemmas 10.54.6 and 10.54.7.

Proof. By induction on the rank of $M$. Suppose $\left[M\right] \in K_0(R)$. Then $M$ is a finite projective $R$-module over a local ring, so M is free; $M \cong R^ n$ for some $n$. The claim is that $\text{rank} (M) \text{length}_ R (R) = \text{length}_ R(M)$, or equivalently that $n\text{length}_ R(R) = \text{length}_ R (R^ n)$ for all $n \geq 1$. When $n = 1$, this is clearly true. Suppose that $(n-1) \text{length}_ R(R) =\text{ length}_ R(R^{n-1})$. Then since there is a split short exact sequence

$0 \to R \to R^ n \to R^{n-1} \to 0$

by Lemma 10.51.3 we have

\begin{eqnarray*} \text{length}_ R(R^ n) & = & \text{length}_ R(R) + \text{length}_ R(R^{n-1}) \\ & = & \text{length}_ R(R) + (n-1) \text{length}_ R(R) \\ & = & n\text{length}_ R(R) \end{eqnarray*}

as desired. $\square$

Comment #674 by Keenan Kidwell on

In the second paragraph at the top of 00JC, "generators of for" should be "generators for," or "generators of."

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