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 $n$. By faithful flatness, extending and contracting gives the same ideal, meaning that $I = R \cap IS$ for each ideal $I$ in $R$ (Lemma 10.82.11). So $I_ n = I_{n+1} = I_{n+2} = \ldots $ as desired. $\square$
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