The Stacks project

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.

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