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*}

Lemma 10.54.8. 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 10.54.6 and 10.54.7.

Proof. By induction on the rank of $M$. Suppose $\left[M\right] \in K_0(R)$. Then $M$ is a finite projective $R$-module over a local ring, so M is free; $M \cong R^ n$ for some $n$. The claim is that $\text{rank} (M) \text{length}_ R (R) = \text{length}_ R(M)$, or equivalently that $n\text{length}_ R(R) = \text{length}_ R (R^ n)$ for all $n \geq 1$. When $n = 1$, this is clearly true. Suppose that $(n-1) \text{length}_ R(R) =\text{ length}_ R(R^{n-1})$. Then since there is a split short exact sequence

\[ 0 \to R \to R^ n \to R^{n-1} \to 0 \]

by Lemma 10.51.3 we have

\begin{eqnarray*} \text{length}_ R(R^ n) & = & \text{length}_ R(R) + \text{length}_ R(R^{n-1}) \\ & = & \text{length}_ R(R) + (n-1) \text{length}_ R(R) \\ & = & n\text{length}_ R(R) \end{eqnarray*}

as desired. $\square$


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