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 10.119.9. Let $R$ be a domain. Assume $R$ has the ascending chain condition for principal ideals. Then the same property holds for a polynomial ring over $R$.

Proof. Consider an ascending chain $(f_1) \subset (f_2) \subset (f_3) \subset \ldots $ of principal ideals in $R[x]$. Since $f_{n + 1}$ divides $f_ n$ we see that the degrees decrease in the sequence. Thus $f_ n$ has fixed degree $d \geq 0$ for all $n \gg 0$. Let $a_ n$ be the leading coefficient of $f_ n$. The condition $f_ n \in (f_{n + 1})$ implies that $a_{n + 1}$ divides $a_ n$ for all $n$. By our assumption on $R$ we see that $a_{n + 1}$ and $a_ n$ are associates for all $n$ large enough (Lemma 10.119.2). Thus for large $n$ we see that $f_ n = u f_{n + 1}$ where $u \in R$ (for reasons of degree) is a unit (as $a_ n$ and $a_{n + 1}$ are associates). $\square$


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