\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 10.56.7. Let $S$ be a graded ring. Let $\mathfrak p \subset S$ be a prime. Let $\mathfrak q$ be the homogeneous ideal of $S$ generated by the homogeneous elements of $\mathfrak p$. Then $\mathfrak q$ is a prime ideal of $S$.

Proof. Suppose $f, g \in S$ are such that $fg \in \mathfrak q$. Let $f_ d$ (resp. $g_ e$) be the homogeneous part of $f$ (resp. $g$) of degree $d$ (resp. $e$). Assume $d, e$ are maxima such that $f_ d \not= 0$ and $g_ e \not= 0$. By assumption we can write $fg = \sum a_ i f_ i$ with $f_ i \in \mathfrak p$ homogeneous. Say $\deg (f_ i) = d_ i$. Then $f_ d g_ e = \sum a_ i' f_ i$ with $a_ i'$ to homogeneous par of degree $d + e - d_ i$ of $a_ i$ (or $0$ if $d + e -d_ i < 0$). Hence $f_ d \in \mathfrak p$ or $g_ e \in \mathfrak p$. Hence $f_ d \in \mathfrak q$ or $g_ e \in \mathfrak q$. In the first case replace $f$ by $f - f_ d$, in the second case replace $g$ by $g - g_ e$. Then still $fg \in \mathfrak q$ but the discrete invariant $d + e$ has been decreased. Thus we may continue in this fashion until either $f$ or $g$ is zero. This clearly shows that $fg \in \mathfrak q$ implies either $f \in \mathfrak q$ or $g \in \mathfrak q$ as desired. $\square$


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