The Stacks project

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 properties1:

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

  2. for every short exact sequence $0 \to M' \to M \to M'' \to 0$ of finite $R$-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 $K'_0(R)$ 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.

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

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:

  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 10.55.2. 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

\[ 0 \to M' \to M \to M'' \to 0 \]

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

\[ 0 \to (d_ i) \to R \to R/(d_ i) \to 0 \]

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 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

\[ 0 \to M' \to M \to M'' \to 0 \]

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

\[ \xymatrix{ K_0(R) \ar[rr] \ar[d]_{\text{rank}_ R} & & K'_0(R) \ar[d]^{\text{length}_ R} \\ \mathbf{Z} \ar[rr]^{\text{length}_ R(R)} & & \mathbf{Z} } \]

where the vertical maps are isomorphisms by Lemmas 10.55.7 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 lengths (Lemma 10.52.3). Thus the result holds. $\square$

[1] The definition makes sense for any ring but is rarely used unless $R$ is Noetherian.

Comments (1)

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."


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00JC. Beware of the difference between the letter 'O' and the digit '0'.