The Stacks project

Lemma 10.120.9. 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 10.120.2). 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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