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Let $X \longrightarrow Y$ be a flat and finitely presented morphism where $X$ is a scheme and $Y$ is a Noetherian (or locally Noetherian) algebraic stack, then will this imply $X$ is also Noetherian (or locally Noetherian)? I am using here the definition of a Noetherian stack as in Stacks Project which states that $Y$ is a Noetherian algebraic stack if for any affine scheme $U$ and any smooth morphism $U \xrightarrow{smooth} Y$, $U$ is Noetherian.

If not, can someone give me an explicit example?

P.S: I have posted this in Math Stack Exchange but didn't get a reply and so I have reposted it here.

Thanks in advance!

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  • $\begingroup$ For a ring $R$, being Noetherian can be checked after faithful base change, $R\to S$, in the following sense. For a set $\{J_\alpha\}_{\alpha\in A}$ of ideals of $R$, if the associated set of ideals$\{J_\alpha S\}_{\alpha \in A}$ has a maximal element $J_\beta S$, then also $J_\beta$ is a maximal element of the original set. Indeed, suppose that $J_\beta \subset J_\alpha$ is a proper inclusion, i.e., $J_\alpha/J_\beta$ is nonzero. Since $S$ is faithful, also $J_\alpha S/J_\beta S$ is nonzero. $\endgroup$
    – Jason Starr
    Commented May 12, 2017 at 13:27
  • $\begingroup$ The answer is yes even if $X\to Y$ is not flat and $X$ is not a scheme. This follows from stacks.math.columbia.edu/tag/04YF (1) $\Leftrightarrow$ (2): X is locally noetherian iff there exists a smooth surjective map $U\to X$ where U is a locally noetherian scheme. $\endgroup$
    – Marc Hoyois
    Commented May 12, 2017 at 13:34
  • $\begingroup$ For an algebraic stack, there is a smooth, faithfully flat morphism $U\to Y$ from an affine scheme $U$ to the stack $Y$. Thus, also $X\times_Y U \to X$ is smooth and faithfully flat. By my previous comment, it suffices to check that $X\times_Y U$ is Noetherian. $\endgroup$
    – Jason Starr
    Commented May 12, 2017 at 13:34
  • $\begingroup$ @Marc Hoyois: I don't see how is that true without any condition! For any non Noetherian ring $A$, consider the morphism $Spec(A) \longrightarrow Spec(k)$ where $k$ is a field. $Spec(k)$ is certainly Noetherian but $Spec(A)$ is not. $\endgroup$
    – Sam
    Commented May 13, 2017 at 11:27
  • $\begingroup$ I guess I must mention this simple example was pointed out to me in stackexchange. $\endgroup$
    – Sam
    Commented May 13, 2017 at 11:29

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