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.
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.