
Lemma 10.54.7. Let $R$ 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. The $n$ generators of $P$ give a surjection $R^ n \to P$, and since $P$ is projective it follows that $R^ n \cong P \oplus Q$ for some projective module $Q$.

If $\mathfrak m \subset R$ is the maximal ideal, then $P/\mathfrak m$ and $Q/\mathfrak m$ are $R/\mathfrak m$-vector spaces, with $P/\mathfrak m \oplus Q/\mathfrak m \cong (R/\mathfrak m)^ n$. Say that $\dim P = p$, $\dim Q = q$, so $p + q = n$.

Choose elements $a_1, \ldots , a_ p$ in $P$ and $b_1, \ldots , b_ q$ in $Q$ lying above bases for $P/\mathfrak m$ and $Q/\mathfrak m$. The homomorphism $R^ n \to P \oplus Q \cong R^ n$ given by $(r_1, \ldots , r_ n) \mapsto r_1a_1 + \ldots + r_ pa_ p + r_{p + 1} b_1 + \ldots + r_ nb_ q$ is a matrix $A$ which is invertible over $R/\mathfrak m$. Let $B$ be a matrix over $R$ lying over the inverse of $A$ in $R/\mathfrak m$. $AB = I + M$, where $M$ is a matrix whose entries all lie in $\mathfrak m$. Thus $\det AB = 1 + x$, for $x \in \mathfrak m$, so $AB$ is invertible, so $A$ is invertible.

The homomorphism $R^ p \to P$ given by $(r_1, \ldots , r_ p) \mapsto r_1a_1 + \ldots + r_ pa_ p$ inherits injectivity and surjectivity from A. Hence, $P \cong R^ p$.

Next we show that the rank of a finite projective module over $R$ is well defined: if $P \cong R^\alpha \cong R^\beta$, then $\alpha = \beta$. This is immediate in the vector space case, and so it is true in the general module case as well, by dividing out the maximal ideal on both sides. If $0 \to R^\alpha \to R^\beta \to R^\gamma \to 0$ is exact, the sequence splits, so $R^\beta \cong R^\alpha \oplus R^\gamma$, so $\beta = \alpha + \gamma$.

So far we have seen that the map $\text{rank}_ R : K_0(R) \to \mathbf{Z}$ is a well-defined homomorphism. It is surjective because $\text{rank}_ R[R] = 1$. It is injective because the element of $K_0(R)$ with rank $\pm \alpha$ is uniquely $\pm [R^\alpha ]$. $\square$

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