The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

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.

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