Lemma 10.55.8. Let $(R, \mathfrak m)$ be a local ring. Every finite projective $R$-module is finite free. The map $\text{rank}_ R : K_0(R) \to \mathbf{Z}$ defined by $[M] \to \text{rank}_ R(M)$ is well defined and an isomorphism.
Proof. Let $P$ be a finite projective $R$-module. Choose elements $x_1, \ldots , x_ n \in P$ which map to a basis of $P/\mathfrak m P$. By Nakayama's Lemma 10.20.1 these elements generate $P$. The corresponding surjection $u : R^{\oplus n} \to P$ has a splitting as $P$ is projective. Hence $R^{\oplus n} = P \oplus Q$ with $Q = \mathop{\mathrm{Ker}}(u)$. It follows that $Q/\mathfrak m Q = 0$, hence $Q$ is zero by Nakayama's lemma. In this way we see that every 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 $R$-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 $K_0(R) = \mathbf{Z}$. $\square$
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