Lemma 28.14.6. Let $f : X \to S$ be a morphism. If $S$ is (locally) Noetherian and $f$ (locally) of finite type then $X$ is (locally) Noetherian.

Proof. This follows immediately from the fact that a ring of finite type over a Noetherian ring is Noetherian, see Algebra, Lemma 10.30.1. (Also: use the fact that the source of a quasi-compact morphism with quasi-compact target is quasi-compact.) $\square$

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