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 10.54.7. 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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