\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*}

The Stacks project

Lemma 10.56.9. Let $R$ be a ring. Let $S$ be a graded $R$-algebra. Let $f \in S_{+}$ be homogeneous. Assume that $S$ is of finite type over $R$. Then

  1. the ring $S_{(f)}$ is of finite type over $R$, and

  2. for any finite graded $S$-module $M$ the module $M_{(f)}$ is a finite $S_{(f)}$-module.

Proof. Choose $f_1, \ldots , f_ n \in S$ which generate $S$ as an $R$-algebra. We may assume that each $f_ i$ is homogeneous (by decomposing each $f_ i$ into its homogeneous components). An element of $S_{(f)}$ is a sum of the form

\[ \sum \nolimits _{e\deg (f) = \sum e_ i\deg (f_ i)} \lambda _{e_1 \ldots e_ n} f_1^{e_1} \ldots f_ n^{e_ n}/f^ e \]

with $\lambda _{e_1 \ldots e_ n} \in R$. Thus $S_{(f)}$ is generated as an $R$-algebra by the $f_1^{e_1} \ldots f_ n^{e_ n} /f^ e$ with the property that $e\deg (f) = \sum e_ i\deg (f_ i)$. If $e_ i \geq \deg (f)$ then we can write this as

\[ f_1^{e_1} \ldots f_ n^{e_ n}/f^ e = f_ i^{\deg (f)}/f^{\deg (f_ i)} \cdot f_1^{e_1} \ldots f_ i^{e_ i - \deg (f)} \ldots f_ n^{e_ n}/f^{e - \deg (f_ i)} \]

Thus we only need the elements $f_ i^{\deg (f)}/f^{\deg (f_ i)}$ as well as the elements $f_1^{e_1} \ldots f_ n^{e_ n} /f^ e$ with $e \deg (f) = \sum e_ i \deg (f_ i)$ and $e_ i < \deg (f)$. This is a finite list and we see that (1) is true.

To see (2) suppose that $M$ is generated by homogeneous elements $x_1, \ldots , x_ m$. Then arguing as above we find that $M_{(f)}$ is generated as an $S_{(f)}$-module by the finite list of elements of the form $f_1^{e_1} \ldots f_ n^{e_ n} x_ j /f^ e$ with $e \deg (f) = \sum e_ i \deg (f_ i) + \deg (x_ j)$ and $e_ i < \deg (f)$. $\square$


Comments (0)

There are also:

  • 6 comment(s) on Section 10.56: Proj of a graded ring

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 07Z2. Beware of the difference between the letter 'O' and the digit '0'.