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

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

2. $S$ 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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