The Stacks project

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

This proves that $[M_{d + 1}] - [M_ d] = [\overline{M}_{d + 1}]$. Hence we win by Lemma 10.58.5. $\square$