6

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.

stupid_question_bot
  • 1,007

1 Answers1

4

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.

Sempliner
  • 2,414
  • 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? – stupid_question_bot Apr 25 '20 at 00:24
  • 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. – Sempliner Apr 25 '20 at 00:45
  • if you only remove the node in the special fiber of $X$, then $X$ is also not $R$-smooth – stupid_question_bot Apr 25 '20 at 01:19
  • 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. – Sempliner Apr 25 '20 at 01:30
  • Why are you saying that in mixed characteristic the statement about connectivity is true? – stupid_question_bot Apr 25 '20 at 01:42
  • Because in any characteristic if you have an snc complement over a dvr the tame quotient of the fundamental group of the special fiber is isomorphic to the tame quotient of the fundamental group of the snc complement (this is in SGA 1 just below the discussion of XIII Corollary 2.8), in particular the map on fundamental groups is surjective if the snc complement is in mixed characteristic (which is equivalent to the fact that the pullback of a connected finite etale cover remains connected). I mention all of this in my answer stupid_question_bot. – Sempliner Apr 25 '20 at 01:53
  • Also I have updated my answer with a counterexample in equicharacteristic p. – Sempliner Apr 25 '20 at 02:00
  • In your Artin-Schreier counterexample, I don't think the special fiber is disconnected - it's only non-irreducible. – stupid_question_bot Apr 26 '20 at 23:53
  • If a scheme is etale over a smooth scheme it is smooth over the base, and thus if it were irreducible but not disconnected it could not be smooth, as there would be at least a node where two components intersect. – Sempliner Apr 27 '20 at 03:03
  • In this case we can be more specific: it demonstrably splits because it admits an explicit section. Namely one can take $\mathbb{F}_p[x] \to \mathbb{F}_p[x, y]/(y^p - y) \to \mathbb{F}_p[x]$ where the first map is the natural one and the second map is the map which sends $y$ to $0$ (one can actually send $y$ to literally any element of $\mathbb{F}_p$). Any etale cover which admits a section splits so... qed – Sempliner Apr 27 '20 at 03:06
  • Ah, that's a great point. Of course $y^p-y$ is also the product of two comaximal ideals, so there you can also see the splitting explicitly. – stupid_question_bot Apr 27 '20 at 19:39
  • In fact $p$ of them! But yes. – Sempliner Apr 27 '20 at 19:41
  • so at first your Artin-Schreier example would seem to imply that "tameness" is in some sense essential to the proof, hence your use of Grothendieck's result on tame fundamental groups. However, in mixed characteristic (say, working over $\mathbb{Z}_p$), tameness of the restriction to special fibers is implied to the base scheme being an NCD-complement (this is a version of Abhyankar's lemma), so I'm curious if in mixed characteristic you can get a proof without appealing to grothendieck's work on the tame fundamental group. – stupid_question_bot Apr 27 '20 at 19:55
  • Okay, so for a proper fppf morphism $X\rightarrow S$, the function on $S$ counting the number of connected components in the geometric fibers in $X$ is lower-semicontinuous. The argument is an easy consequence of Stein factorization (https://stacks.math.columbia.edu/tag/0bui)... – stupid_question_bot Apr 29 '20 at 04:36
  • I wonder how this meshes with your Artin Schreier example. That seems to imply that in any flat ``compactification'' of your curve, all the components of the special fiber have to intersect above infinity. E.g., what if you take the normalization of $\mathbb{P}^1_{\mathbb{F}_p[[t]]}$ inside the function field of your curve? Is it obvious to you this must happen? (ie, by means other than using stein factorization) – stupid_question_bot Apr 29 '20 at 04:37
  • Actually I guess the same stacks project lemma applied to the preimage above infinity in the hypothetical finite flat extension of your Artin-Schreier cover would imply that specializing can only decrease the cardinality of the finite geometric fiber. Hence above infinity in the special fiber there can only be one point, since there was only 1 point above infinity in the generic fiber. Interesting... – stupid_question_bot Apr 29 '20 at 05:33
  • Yeah the statement about Stein factorization is what I mentioned in the first paragraph of my answer. Your reasoning is totally correct, but I can offer an alternate galois theoretic interpretation: because P^1 is simply connected you know the cover must ramify at infinity, and because the extension is galois of degree p you know the ramification subgroup associated to infinity must be the whole galois group, whence it must be totally ramified (thus all the components meet). – Sempliner Apr 29 '20 at 06:28
  • I haven’t read that part of SGA 1 in a while but I assume it doesn’t use much more than Abhyankar’s lemma and perhaps come basic facts about cohomology. It could be a good thing to read if you want to develop your own proof of this fact about tame fundamental groups. – Sempliner Apr 29 '20 at 06:30