The Stacks project

\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*}

Lemma 10.55.2. Let $S$ be a graded ring, which is finitely generated over $S_0$. Then for all sufficiently divisible $d$ the algebra $S^{(d)}$ is generated in degree $1$ over $S_0$.

Proof. Say $S$ is generated by $f_1, \ldots , f_ r \in S$ over $S_0$. After replacing $f_ i$ by their homogeneous parts, we may assume $f_ i$ is homogeneous of degree $d_ i > 0$. Then any element of $S_ n$ is a linear combination with coefficients in $S_0$ of monomials $f_1^{e_1} \ldots f_ r^{e_ r}$ with $\sum e_ i d_ i = n$. Let $m$ be a multiple of $\text{lcm}(d_ i)$. For any $N \geq r$ if

\[ \sum e_ i d_ i = N m \]

then for some $i$ we have $e_ i \geq m/d_ i$ by an elementary argument. Hence every monomial of degree $N m$ is a product of a monomial of degree $m$, namely $f_ i^{m/d_ i}$, and a monomial of degree $(N - 1)m$. It follows that any monomial of degree $nrm$ with $n \geq 2$ is a product of monomials of degree $rm$. Thus $S^{(rm)}$ is generated in degree $1$ over $S_0$. $\square$

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