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

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

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.

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