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 properties1:
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
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
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