Let $R$ be a discrete valuation ring. Let $X\rightarrow\text{Spec }R$ be a smooth morphism with geometrically connected fibers of dimension 1. I'm happy to assume that $X$ is the complement of a normal crossings divisor inside a smooth projective $R$-curve.

Let $Y\rightarrow X$ be a finite etale map with $Y$ connected. Let's further assume the generic fiber of $Y$ is geometrically connected. Could $Y$ have a disconnected special fiber?

Surely not right? For some reason I'm blanking on how to argue this.

This came up when considering Galois closures. I want to say that if $Z\rightarrow X$ is a finite etale cover of curves over $R$ (both having geometrically connected special fibers), then the generic (resp. special) fiber of its Galois closure should be the Galois closure of its generic (resp. special) fiber.


1 Answer 1


Hi stupid_question_bot,

Unfortunately you seem to need some more assumptions for an easy proof, in particular properness would make this very easy, in general given a flat proper scheme with geometrically normal fibers one can show that the number of (geometric) components of the fibers is locally constant on the base, which would answer your question.

(EDIT: To be clear, the following is not a counter-example to the specific statement in the question, that comes later in this answer. I was just trying to point out that the proof would need some geometric input since it is false when the base is not normal.)

The counter example that I have in mind is as follows, take $\bar{X}$ to be the nodal cubic over $\mathbb{Z}_p$ ($\mathbb{P}^1$ glued together at two $\mathbb{Z}_p$ points: say $0, 1$ in a standard affine chart), let $\bar{Y} \to \bar{X}$ be a connected finite etale cover corresopnding to a nontrivial element of the geometric fundamental group of $\bar{X}$ (for definiteness, take the double cover given by two $\mathbb{P}^1$'s glued into a bigon and for safety let $p \neq 2$). Now let $X$ be the complement of the node in the special fiber, and let $Y$ be the pullback. Clearly while the generic fiber of $Y$ is connected the special fiber is not by inspection.

You can now complain: "oh but your $X$ is not an snc complement in a smooth scheme." In this case I was unable to say anything useful, except for that some results in SGA imply that this would be true if the cover $Y$ is tamely ramified over the snc divisor. Hope this example is helpful though, as it shows that the strong statement you made about connectivity of special fibers is not some total triviality.

EDIT: Update, bad news: there are even worse examples to be had here. Let $X$ be $\mathbb{A}^1_{\mathbb{F}_p[[t]]}$, then consider $Y$ the Artin-Schreier cover of $X$ given by the equation $Y^p - Y = x \cdot t$, then the special fiber of this etale cover splits but generically it defines a Galois Artin-Schreier cover.

  • $\begingroup$ Why would the generic fiber be connected in your example? It seems like $Y$ is just the complement of the two nodal sections in the bigon, so it should be a disjoint union of two $\mathbb{G}_m$'s? $\endgroup$ Apr 25, 2020 at 0:24
  • $\begingroup$ I am only removing the node in the special fiber in my example, the generic fiber is still a nodal cubic, to be even fancier you could remove a hyperplane section which passes through the residue disk of the node in the generic fiber but not through the generic fiber's node, and which hits the node in the special fiber, then neither fiber is proper and the desiderata are still all there. $\endgroup$
    – Sempliner
    Apr 25, 2020 at 0:45
  • $\begingroup$ if you only remove the node in the special fiber of $X$, then $X$ is also not $R$-smooth $\endgroup$ Apr 25, 2020 at 1:19
  • $\begingroup$ Yes, I mentioned this in my answer (see the third paragraph), this only partially relates to your question, in that it tells you what statements cannot possibly be true that would imply your statement. In particular in mixed characteristic the statement about connectivity of the special fiber is true, so I would not presume to contradict it. $\endgroup$
    – Sempliner
    Apr 25, 2020 at 1:30
  • $\begingroup$ Why are you saying that in mixed characteristic the statement about connectivity is true? $\endgroup$ Apr 25, 2020 at 1:42

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