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Let $\pi:X\rightarrow Y$ be a proper morphism to an irreducible variety and all fibers of $\pi$ are nonempty, irreducible, and of the same dimension.

Show $X$ must also be irreducible.

Thanks (Any hints would work too)

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  • $\begingroup$ I remember having seen this somewhere in EGA with a reference to Bourbaki's Topologie generale. Without properness assumption. But actually it's a good exercise in basic topology. $\endgroup$ – Martin Brandenburg Nov 24 '13 at 18:41
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    $\begingroup$ If you know that $\pi$ is open, then (independent of properness or dimension) this is a fact about topological spaces, proved as tag 004Z in the Stacks project. So, if you were to replace "proper" with "flat," then the result follows (because flat morphisms that are locally of finite presentation are open). I'm not sure how to do without flatness (if you imposed some additional conditions, e.g. $X$ equidimensional and $X$ and $Y$ regular, then flatness would hold). But presumably there is an easier way with the hypotheses you're given. $\endgroup$ – Keenan Kidwell Nov 24 '13 at 18:54
  • $\begingroup$ Related math.stackexchange.com/questions/242360/… $\endgroup$ – Ehsan M. Kermani Nov 24 '13 at 19:40
  • $\begingroup$ I see the proof for the case where $\pi$ is open. Here we know that $\pi$ is closed, but the analogous proof doesn't seem to work. $\endgroup$ – Gazerun Nov 25 '13 at 2:00
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This is exercise 11.4.C of Ravi Vakil's notes. I'll give a sketch of the solution.

First note that since $\pi$ is proper $X$ is finite type over a variety so it has finitely many irreducible components $Z_i$. Furthermore, $\pi$ is closed so $\pi(Z_i)$ is closed in $Y$ for each $i$ and we can conclude that some $\pi(Z_i) = Y$ by the irreducibility of $Y$. Call this component $Z_0$.

Let $X_y$ denote $\pi^{-1}(y)$. $X_y$ is irreducible and $X_y = \bigcup X_y \cap Z_i$ is a union of closed subsets so $X_y = X_y \cap Z_i \subset Z_i$ for some $i$. It then follows that if $x \in Z_i$ but not in $Z_j$ for $j \neq i$, then $X_{\pi(x)} \subset Z_i$.

By applying proposition 11.4.1 in Vakil's notes on the restriction of $\pi$ to $Z_i \to Y$, we see that $\dim X_y \cap Z_i \geq \dim Z_i - \dim \pi(Z_i)$ with equality on some open subset of $\pi(Z_i)$. However, for each $y$ there exists a $k$ such that $X_y \cap Z_k = X_y$ so in that case $\dim X_y = \dim Z_k - \dim \pi(Z_k)$ on some open subset. Since the fibers are equidimensional of dimension $d$, we can then conclude that $\dim Z_k - \dim \pi(Z_k) = d$ for every $k$. In particular, if $X_y$ intersects $Z_k$, then $\dim X_y \cap Z_k \geq d$ by the first inequality but $\dim X_y = d$ by assumption and $X_y \cap Z_k = X_y$. Therefore each connected component $Z_k$ is the union of fibers $X_y$ that intersect it. In particular, $Z_0$ is the union of fibers that intersect it, but every fiber intersects $Z_0$ so $Z_0$ is the union of all the fibers thus $Z_0 = X$ and $X$ is irreducible.

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  • $\begingroup$ Hi, I would like to ask why is it that when $X$ is finite type over a variety, it has finitely many irreducible components? Thanks! $\endgroup$ – enoughsaid05 Nov 3 '15 at 21:58
  • $\begingroup$ @enoughsaid05 $f : X \to Y$ being finite type over a variety $Y$ means two things: 1) for every affine $Spec B \subset Y$, and every affine $Spec A \subset f^{-1}(Spec B)$ has $A$ a finite type $B$-algebra by the natural map of pulling back (this is the meaning of locally of finite type). 2) $f$ is also quasi-compact: preimages of affines are quasi-compact... Now, Since $B$ is a variety, it is quasi-compact and finite type over a field $k$, hence a Noetherian scheme. As finitely generated rings over Noetherian rings are Noetherian (Hilbert Basis Theorem), each $A$ is Noetherian. $\endgroup$ – Lorenzo Najt Jan 3 '16 at 17:09
  • $\begingroup$ @enoughsaid05 It follows that $X$ is Noetherian - it is quasi-compact since it can be covered by finitely many of the $Spec A$ of the form 1), using quasi-compactness of the map $f$, plus quasi-compactness of $Y$. Then since affines are quasi-compact, and something which is covered by finitely many quasi-compacts is quasi-compact, the quasi-compactness of $X$ follows. (Noetherian schemes are locally Noetherian and also quasi-compact.) $\endgroup$ – Lorenzo Najt Jan 3 '16 at 17:10
  • $\begingroup$ @enoughsaid05 Now, it is a fact that Noetherian schemes have only finitely many irreducible components. This may be familiar a commutative algebra fact about the finiteness of primary decomposition of $(0)$ in the Noetherian ring case. This is because a Noetherian topological space (one that satisfies the descending chain condition on closed subsets) has this property, and Noetherian schemes have underlying topological spaces which are Noetherian spaces. $\endgroup$ – Lorenzo Najt Jan 3 '16 at 17:18
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    $\begingroup$ This is a great answer, but I think there's a slight mistake in the second sentence of the final paragraph. I think it should read "However, for each $k$, given any non-empty open subset of $\pi(Z_i)$, there exists a $y$ such that $X_y\cap Z_k = X_y$, so taking $y$ in the open subset where equality holds, $\rm{dim}X_y = \rm{dim}Z_k - \rm{dim}\pi(Z_k)$." I could be wrong, but I think we're trying to prove a result about an arbitrary component, rather than about an arbitrary point, and this replacement would make that work. $\endgroup$ – Tom Oldfield Jun 5 '16 at 2:39
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I just wanted to add that M. Mustaţă proves a slightly more general fact here - without assuming that the morphism is proper, but assuming $X$ is equidimensional, which is often useful: http://www-personal.umich.edu/~mmustata/Note1_09.pdf.

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