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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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