I'm sure this must be covered somewhere, but all the references I have only treat this in very special cases (mostly when working over fields).

Suppose $f : X\rightarrow S$ is smooth of finite presentation with geometrically connected fibers of dimension 1, and let $g_i : S\rightarrow X$ be finitely many sections. Let $X^\circ := X - \sqcup_i g_i(S)$ and let $p^\circ : Y^\circ\rightarrow X^\circ$ be finite etale.

I would like a procedure to find a branched cover of $X$ which agrees with $p^\circ$ over $X^\circ$, ideally a construction that commutes with (arbitrary) base change.

One way is to consider the normalization $X'$ of $X$ in $Y^\circ$ (though this only commutes with smooth base change), so we get a commutative diagram, which I don't know how to draw here, but the point is that the following two compositions are equal:

$$Y^\circ\stackrel{p^\circ}{\rightarrow} X^\circ\hookrightarrow X$$ and $$Y^\circ\stackrel{h}{\rightarrow} X'\stackrel{\nu}{\rightarrow} X$$

Under the mild assumption that $S$ is Nagata, we know that $\nu$ is finite (http://stacks.math.columbia.edu/tag/03GR), so lets assume $S$ is Nagata.

Further, by Zariski's main theorem (http://stacks.math.columbia.edu/tag/02LQ) , we know that $h$ is an open immersion.

My questions are:

  1. Intuitively, what does this relative normalization do? I really don't have any intuition for how relative normalization will behave over the $g_i$'s... I guess some more specific questions are:
  2. If it's possible to extend $p^\circ : Y^\circ\rightarrow X^\circ$ to an etale cover of $X$, will the normalization $\nu$ be etale? (ie, does the normalization "prefer" "unramified completions" of the cover?)
  3. When will $X'$ be smooth over $S$?
  4. Is this the "best" way to recover a branched cover from its etale locus? Unfortunately, as Jason Starr pointed out, it does not commute with general base change. Is there a way to do this that does commute with base change?

References would be appreciated.

EDIT: Perhaps I shouldn't have used "reconstruct" in the title, since that implies that there exists a unique/canonical/best candidate for an extension of $p^\circ$ to a finite map over $X$. Basically I'm asking if there are decent alternatives to the normalization, and also for some intuition about how the normalization behaves.

  • $\begingroup$ One thing to bear in mind: the formation of $X'$ is not compatible with base change. If $S$ is $\text{Spec}\ \mathbb{C}[t]$, if $X$ is $\text{Spec}\ \mathbb{C}[t,x]$, if $X'$ is $\text{Spec}\ \mathbb{C}[t,x,y]/\langle t - (y-x)(y+x) \rangle$, and if $X^o$, $Y^o$ are the corresponding etale loci, then the fiber over $t=0$ of the normalization $X'$ of $X$ in $Y^o$ is different from the normalization of the fiber over $t=0$ of $X$ in the fiber over $t=0$ of $Y^o$. $\endgroup$ Sep 14, 2015 at 17:06
  • $\begingroup$ I see that you want the complement of the $X^o$ to be a union of sections. So base change by $\mathbb{C}[t] \mapsto \mathbb{C}[s]$, $t\mapsto - s^2$. Then the complement of $X^o$ is the union of the sections $g_+$ and $g_{- }$ with $g_+^*s = +x$ and $g_{- }^*s = -x$. $\endgroup$ Sep 14, 2015 at 17:10
  • $\begingroup$ @JasonStarr Thanks for pointing that out. I guess it only commutes with smooth base change. $\endgroup$
    – Will Chen
    Sep 14, 2015 at 17:19
  • $\begingroup$ The question is, in what category are you looking for a "functorial" way of reconstructing the normalization? Since normalization is not compatible with arbitrary base change, you should not expect a functor that extends, for instance, to the scheme of an Artin ring. So that obstructs Question 3. You mention smooth morphisms to $X$. If you work with the larger category of normal schemes together with a dominant morphism to $X$, it is straightforward to give a functor that is represented by $X'$: the functor of rational transformations to $Y^o$ over $X$. $\endgroup$ Sep 15, 2015 at 11:54
  • $\begingroup$ @JasonStarr I'm not necessarily asking to reconstruct the normalization. I suppose I shouldn't have used the word "reconstruct" since that assumes there is a unique/canonical/best candidate for a finite map $p$ extending $p^\circ$. My basic two questions are - is there a functor that is at least as good as normalization at "extending $p^\circ$?", and failing that, I was hoping for someone to explain the intuition behind how normalization behaves above the possibly ramified locus. $\endgroup$
    – Will Chen
    Sep 16, 2015 at 0:28

1 Answer 1


2) Yes, if the cover is finite etale. (Maybe we need to be over a Noetherian base as well?) All the functions on the cover are going to be integral over the base and contained in the field of fractions of the open subset. Furthermore, there will be no additional functions of this type in the field of fractions of the cover (check this etale-locally, where it becomes obvious.)

3) In the case of wild ramification, where the $g_i$ don't intersect, smoothness implies some additional conditions on the ramification that you can see by looking at transverse curves. If your etale cover extends smoothly, then the extension by zero of the pushforward of the constant sheaf from that etale cover has no vanishing cycles, which implies (Deligne's semicontinuity theorem) that the Swan conductor is constant.

Also when it's smooth the standard argument that gives you the structure of the Galois group of local fields applies and says something about the structure of the cover - you can split it into an unramified part, a tame part, and a wild part. When it's not smooth, this argument fails, and the result can fail as well.

The conditions I describe on monodromy are invariant under inseparable base change, because it preserves the etale topology. But smoothness is not invariant under inseparable base change. Take a finite etale cover $Y^\circ \to X^\circ$ that extends to $Y \to X$, with $Y$ a family of nontrivial smooth projective curves over $S$. Take some inseparable morphism $S \to T$ which $X$ descends down but $Y$ does not (Maybe $X$ is $\mathbb P^1$, $Y$ is an elliptic curve with nonconstant $j$ invariant, and $T$ is the ring generated by the $p$th power of the $j$ invariant). Finite etale covers always descend, so $Y^{\circ}$ descends, but it no longer has a smooth compactification. (Smooth compactifications of curves are unique, so we also know there isn't some other compactification that does the job.)

There may be analogues of the first two phenomena in characteristic zero when $g_i$ intersect.

4) The normalization is optimal in a pretty clear sense: Given any finite cover that extends your cover, all the function son it are integral, and all lie in the ring of functions on the open subset that is your original cover, so they are all functions on the normalization. Hence any other branched cover $Y$ that extends $X^{circ}$ has a map $X' \to Y$.

When you look at an individual fiber, so this is a map of curves, note that in a birational map of proper curves, the target always has worse singularities (by some reasonable measurement) than the source. So in this sense, it's optimal.

Using this, we can show that no base change-invariant construction exists. Simply consider families which contain the same etale cover, but which have arbitrarily bad singularities. If you consider the cover $y^2 = (x - t)(x-2t)(x-3t) \dots (x-nt)$, at $t=0$ you get the singularity $y^2 =x^n$, which gets worse and worse as $n$ goes to $\infty$. However on the open set where $x\neq t, 2t, 3t, \dots, nt$, restricted to the fiber $t=0$, the singularity only depends on $n$ mod $2$.

1) Intuitively, to me, the relative normalization makes each fiber as smooth as it can be considering the nearby fibers. So if the nearby fibers are really quite ramified (in number of ramification points or in Swan conductor), but a single fiber is not so ramified, then that difference will show up in a singularity of the special fiber. I don't know how to make this precise.


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