The Stacks project

Proposition 10.58.7. 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 10.58.6. 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 10.58.5. $\square$


Comments (0)

There are also:

  • 5 comment(s) on Section 10.58: Noetherian graded rings

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 00K1. Beware of the difference between the letter 'O' and the digit '0'.