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*}

Proposition 10.57.7. Suppose that $S$ is a Noetherian graded ring and $M$ a finite graded $S$-module. Consider the function

\[ \mathbf{Z} \longrightarrow K_0'(S_0), \quad n \longmapsto [M_ n] \]

see Lemma 10.57.6. If $S_{+}$ is generated by elements of degree $1$, then this function is a numerical polynomial.

Proof. We prove this by induction on the minimal number of generators of $S_1$. If this number is $0$, then $M_ n = 0$ for all $n \gg 0$ and the result holds. To prove the induction step, let $x\in S_1$ be one of a minimal set of generators, such that the induction hypothesis applies to the graded ring $S/(x)$.

First we show the result holds if $x$ is nilpotent on $M$. This we do by induction on the minimal integer $r$ such that $x^ r M = 0$. If $r = 1$, then $M$ is a module over $S/xS$ and the result holds (by the other induction hypothesis). If $r > 1$, then we can find a short exact sequence $0 \to M' \to M \to M'' \to 0$ such that the integers $r', r''$ are strictly smaller than $r$. Thus we know the result for $M''$ and $M'$. Hence we get the result for $M$ because of the relation $ [M_ d] = [M'_ d] + [M''_ d] $ in $K_0'(S_0)$.

If $x$ is not nilpotent on $M$, let $M' \subset M$ be the largest submodule on which $x$ is nilpotent. Consider the exact sequence $0 \to M' \to M \to M/M' \to 0$ we see again it suffices to prove the result for $M/M'$. In other words we may assume that multiplication by $x$ is injective.

Let $\overline{M} = M/xM$. Note that the map $x : M \to M$ is not a map of graded $S$-modules, since it does not map $M_ d$ into $M_ d$. Namely, for each $d$ we have the following short exact sequence

\[ 0 \to M_ d \xrightarrow {x} M_{d + 1} \to \overline{M}_{d + 1} \to 0 \]

This proves that $[M_{d + 1}] - [M_ d] = [\overline{M}_{d + 1}]$. Hence we win by Lemma 10.57.5. $\square$


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