The Stacks project

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

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 $K_0(R) = \mathbf{Z}$. $\square$