\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 10.56.2. Let $S$ be a $\mathbf{Z}$-graded ring containing a homogeneous invertible element of positive degree. Then the set $G \subset \mathop{\mathrm{Spec}}(S)$ of $\mathbf{Z}$-graded primes of $S$ (with induced topology) maps homeomorphically to $\mathop{\mathrm{Spec}}(S_0)$.

Proof. First we show that the map is a bijection by constructing an inverse. Let $f \in S_ d$, $d > 0$ be invertible in $S$. If $\mathfrak p_0$ is a prime of $S_0$, then $\mathfrak p_0S$ is a $\mathbf{Z}$-graded ideal of $S$ such that $\mathfrak p_0S \cap S_0 = \mathfrak p_0$. And if $ab \in \mathfrak p_0S$ with $a$, $b$ homogeneous, then $a^ db^ d/f^{\deg (a) + \deg (b)} \in \mathfrak p_0$. Thus either $a^ d/f^{\deg (a)} \in \mathfrak p_0$ or $b^ d/f^{\deg (b)} \in \mathfrak p_0$, in other words either $a^ d \in \mathfrak p_0S$ or $b^ d \in \mathfrak p_0S$. It follows that $\sqrt{\mathfrak p_0S}$ is a $\mathbf{Z}$-graded prime ideal of $S$ whose intersection with $S_0$ is $\mathfrak p_0$.

To show that the map is a homeomorphism we show that the image of $G \cap D(g)$ is open. If $g = \sum g_ i$ with $g_ i \in S_ i$, then by the above $G \cap D(g)$ maps onto the set $\bigcup D(g_ i^ d/f^ i)$ which is open. $\square$


Comments (5)

Comment #1260 by on

is missing from the conclusion. Some (homogeneous) localization at should be involved.

Comment #1263 by on

Nope, is not missing from the conclusion as the conclusion would be wrong if there was no such .

Comment #1272 by abcxyz on

Sorry, my mistake. Actually the way this result is formulated made me think so. In fact, the hypothesis is confusing. I'd rephrase it like this: Let be a -graded ring containing an invertible element of positive degree. Then the set..."

Comment #1273 by abcxyz on

Even better: "Let be a -graded ring containing a homogeneous invertible element of positive degree. Then the set..."

There are also:

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

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