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Background. I am trying to solve Exercise 11.4.C in Vakil's Foundations of Algebraic Geometry (November 18, 2017 draft)

(Exercise 11.4.C) Suppose $\pi: X \to Y$ is a proper morphism to an irreducible variety, and all the fibers of π are nonempty, and irreducible of the same dimension. Show that X is irreducible.

using Theorem 11.4.1:

(Theorem 11.4.1) Suppose $\pi: X \to Y$ is a morphism of irreducible $k$-varieties, with $\dim X = m$ and $\dim Y = n$. Then there exists a nonempty open subset $U \subset Y$ such that for all $q \in U$, the fiber over $q$ has pure dimension $m − n$, or is empty.

Confusion. I find an answer at https://math.stackexchange.com/questions/579527/ but the usage of Theorem 11.4.1 in the accepted answer seems unrigorous to me. The reason is that in the answer, the author applies Theorem 11.4.1 to the restriction of $\pi: X\to Y$ to $Z_i \to \pi(Z_i)$, where $Z_i$ is an irreducible component of $X$ and $\pi(Z_i)$ is the scheme-theoretic image of $Z_i$; the assumption of Theorem 11.4.1 is that the morphism is between irreducible $k$-varieties (i.e. integral $k$-scheme of finite type over $k$), but $Z_i$ and $\pi(Z_i)$ need not be quasi-compact. (If we read the proof of Theorem 11.4.1, we see that the second reduction made in Vakil's proof of Theorem 11.4.1 does not work without the quasi-compactness assumption, as $Z_i$ need not be covered by finitely many affine opens.)

Questions. I would like to ask: does Theorem 11.4.1 hold if we replace the phrase "irreducible $k$-varieties" with the phrase "integral $k$-schemes locally of finite type"? And is the use of Theorem 11.4.1 in the cited solution to Exercise 11.4.C actually fine? Thanks in advance.

WLOG
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    If $Y$ is an irreducible variety and $\pi:X\to Y$ is proper, then $X$ is finite type over a field (proper = finite type + separated + universally closed, composition of finite type morphisms is finite type) and therefore both $Z$ and $\pi(Z)$ are quasi-compact for every $Z\subset X$. – KReiser Jul 12 '22 at 01:06
  • @KReiser Just to be sure, is the logic that $X$ is finite type over a field and hence a Noetherian scheme, and every subset of a Noetherian topological space is quasi-compact? I think I understand. Thank you! – WLOG Jul 12 '22 at 01:16
  • Yes. If you like, I can post this as an answer, but it wouldn't contain any discussion for the other part of your question. (I think the other part is perhaps less relevant given my comment, but it's up to you if that's enough.) – KReiser Jul 12 '22 at 01:25
  • @KReiser Yes, you have answered my main question. So, please post your idea as an answer, and I will accept it when I see it. – WLOG Jul 12 '22 at 01:34

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If $Y$ is an irreducible variety and $\pi:X\to Y$ is proper, then $X$ is finite type over a field (proper = finite type + separated + universally closed, composition of finite type morphisms is finite type). Therefore $X$ and $Y$ are both noetherian and every subset of $X$ or $Y$ is quasi-compact.

KReiser
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