Lemma 10.164.1 (033E)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.164: Descending properties
Lemma 10.164.1 (cite)

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Lemma 10.164.1. Let $R \to S$ be a ring map. Assume that

$R \to S$ is faithfully flat, and
$S$ is Noetherian.

Then $R$ is Noetherian.

Proof. Let $I_0 \subset I_1 \subset I_2 \subset \ldots$ be a growing sequence of ideals of $R$ . By assumption we have $I_ nS = I_{n +1}S = I_{n + 2}S = \ldots$ for some $n$ . Since $R \to S$ is flat we have $I_ kS = I_ k \otimes _ R S$ . Hence, as $R \to S$ is faithfully flat we see that $I_ nS = I_{n +1}S = I_{n + 2}S = \ldots$ implies that $I_ n = I_{n +1} = I_{n + 2} = \ldots$ as desired. $\square$

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