
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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