Lemma 29.15.6. 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 10.31.1. (Also: use the fact that the source of a quasi-compact morphism with quasi-compact target is quasi-compact.) $\square$

Comment #5043 by Taro konno on

I would like to propose that the statement of this lemma be changed to the following : Let $S$ be (locally) Noetherian scheme and $X$ be arbitrary scheme. If there exists a morphism (locally) of finite $f :X \longrightarrow S$ ,then $X$ is (locally) Noetherian.

Comment #5268 by on

@#5043: What would be the reason for that?

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