Although my number theory is really weak, I'm trying to understand the notion of etale coverings in this context. I think this could provide a very interesting point of view.

Let $U$ be an open subscheme of $\textrm{Spec} \ \mathbf{Z}$. The complement of $U$ is a divisor $D$ on $\textrm{Spec} \ \mathbf{Z}$.

Q. Can we classify the etale coverings of $U$? (Say of a given degree...)

This is what I (think I) know.

Given a finite etale morphism $\pi:V\longrightarrow U$, the normalization of $X$ in the function field $L$ of $V$ is $\textrm{Spec} \ O_L$. This is a nontrivial fact, of course.

So, in view of the previous remark, can we also say what $V$ itself should be?

I'm guessing this has to do with the etale fundamental group $\pi_1(U)$. The latter is, I believe, finitely generated. I think that has to do with the fact that there are only finitely many ramified covers of a certain degree, right?

Of course, we can complicate things by replacing $\textrm{Spec} \ \mathbf{Z}$ by $\textrm{Spec} \ O_K$.

Example. Take $U= \textrm{Spec} \ \mathbf{Z} -\{ (2)\}$ and let $V\rightarrow U$ be an finite etale morphism. Suppose that $V$ is connected and that let $K$ be its function field. The normalization of $\textrm{Spec} \ \mathbf{Z}$ in $K$ is of course $O_K$. The extension $\mathbf{Z}\subset O_K$ is unramified outside $(2)$ and (possibly) ramified at $(2)$. Can one give a description of $V$ here?

EDIT. I just realized one can also ask themselves a similar question for $\mathbf{P}^1_{\mathbf{C}}$. Or even better, for any Riemann surface $X$.

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    $\begingroup$ V is just O_K[1/2], right? $\endgroup$ – Kevin Buzzard Apr 28 '10 at 18:14
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    $\begingroup$ In general any open subset of the spectrum of a Dedekind domain is necessarily affine, and finite étale covers of affine schemes (being finite) are necessarily affine. You can work out this kind of question by thinking of a few things : since you're concerned with bases $S = Spec \mathcal{O}_K$, you know that all finite śtale covers $X\to S$ are going to be $Spec$ of some finite flat $O_K$-algebra... Then go back to the definition of unramified (i.e. calculate the stalk map for $Spec(B)\to Spec(A)$, it's just a the localization of $A\to B$ at a prime of $B$). $\endgroup$ – xuros Apr 28 '10 at 18:38
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    $\begingroup$ @Giovanni: for general Dedekind domain $A$, why should every open subset of the spectrum be affine? Why is the complement of the closed point corresponding to a maximal ideal $m$ affine (obvious if $m$ torsion in class group, but more generally...)? A good example: for complement of origin in an elliptic curve, removing a rational non-torsion point yields an open affine (via Riemann-Roch) but not basic open affine (= inverting some nonzero element). This is an embarrassingly elementary (but "useless") question about general Dedekind domains. I checked with two colleagues, as stumped as me. $\endgroup$ – BCnrd Apr 28 '10 at 21:00
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    $\begingroup$ @Giovanni: false alarm. By Serre's cohom. criterion, $U$ is affine. Consider extension $E$ of $O_U$ by nonzero coherent ideal $I$, so $E$ is rank-2 vector bundle on $U$. Extend $E$ to coherent $E'$ on $X = {\rm{Spec}}(A)$, and wlog kill torsion so $E'$ is vector bundle. Doing local calculation at $m$, can "intersect" $I$ with $E'$ to get saturated line bundle $L$ in $E'$ extending $I$ in $E$, so $N = E'/L$ is also line bundle. (I am using Dedekindness of $X$ all over the place.) Then on affine $X$ this exact sequence splits; restrict back to $U$. QED Huh. Is there an elementary proof? $\endgroup$ – BCnrd Apr 28 '10 at 21:33
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    $\begingroup$ @BCnrd: Any nonempty open U is the complement of a finite set S of closed points. Let A[1/S] be the set of x in Frac(A) whose valuation at each prime outside S is nonnegative. Then Spec A[1/S] --> Spec A is an open immersion with image U since this can be checked on a covering by basic open affines on which the primes in S become principal. $\endgroup$ – Bjorn Poonen Apr 29 '10 at 23:30

As Kevin points out, $V$ is indeed $\mathcal{O}_K[\frac{1}{2}]$ in your example. Your link to the fundamental group is also correct. $\pi_1(U)$ is the Galois group of the maximal extension of $\mathbb{Q}$ unramified outside 2 (since you can restrict attention to the connected etale $\mathbb{Q}$-algebras)*. More generally, in your original question, these are replaced by $\mathcal{O}_L$ with the primes above the support of $D$ inverted, and the Galois group of the maximal extension of $\mathbb{Q}$ unramified outside the support of $D$. These groups can get pretty horrendous, and so number-theorists tend to (at least in my view) study the more well-behaved but still very mysterious maximal pro-$p$-quotients of these groups. Now these pro-$p$ groups are not only finitely generated by work of Shafarevich ("Extensions with prescribed ramification points"), they are $d$-generated, where $d$ is the cardinality of the support of $D$ (for "tame" and not silly $D$). More impressively, the relation rank is also calculated/bounded (in this case, it's also $d$!!!), frequently leading to conclusions about when these groups are finite or infinite.

The "more complicated" starting point of $\mathcal{O}_K$ is in a sense not actually much more complicated. You're still asking for the Galois group of the maximal extension of $K$ unramified outside a finite set of primes, and Shafarevich's work gives formulas for the relation rank and generator rank for the pro-$p$-quotients.

The answer to your overall classification question is thus pretty difficult, and consume entire subdisciplines of number theory. As an example, Christian Maire has constructed number fields with a trivial class group but infinite unramified extensions -- you'd have to have a complete understanding of when this could happen before you could hope to prescribe all unramified extensions of a given degree, with or without certain primes inverted. There are certain cases where this can be done via, e.g., root discriminant bounds, but the story is far from being complete at this point.

As in Lars's answer, the situation is much much better understood for $\mathbb{P}_\mathbb{C}^1$ and Riemann surfaces than it is for number fields.

*: Actually, you have to be a little careful about 2-extensions. Etaleness doesn't pick up on whether or not infinite primes ramify. Everything's fine if you start with an totally imaginary field.

  • $\begingroup$ Thanks. Should be fixed up now -- anything dumb left is more likely ignorance than a typo. $\endgroup$ – Cam McLeman Apr 28 '10 at 21:21
  • $\begingroup$ This is great. Thanks. How is the situation on the 2-dimensional analogue? That is, what if we replace Spec $\mathbf{Z}$ by $\mathbf{P}^1_{\mathbf{Z}}$? Can one say something about the maximal pro-p quotients of the fundamental group? I think this is where things will get really interesting. For now, I will just focus on the "easy" 1-dimensional case, though. $\endgroup$ – Ariyan Javanpeykar Apr 29 '10 at 21:25
  • $\begingroup$ I'm surprised that nobody has pointed out the nice article by Mazur on the étale cohomology of number fields : numdam.org/numdam-bin/fitem?id=ASENS_1973_4_6_4_521_0 $\endgroup$ – Chandan Singh Dalawat Sep 5 '10 at 3:02

As to your addendum regarding the fundamental group of Riemann surfaces, the situation is as follows: If $X$ is a smooth projective algebraic curve of genus $g$ over an algebraically closed field $K$ of characteristic $0$, and if $U\subset X$ is obtained by removing $r$ distinct closed points, then $\pi_1(U)$ is the profinite completion of the surface group \[\left<a_1,\ldots, a_g, b_1,\ldots, b_g, c_1,\ldots, c_r | [a_1,b_1]\cdot\ldots\cdot[a_g,b_g]c_1\cdot\ldots\cdot c_r = 1\right>.\] Phrased more traditionally in terms of Galois theory of fields, if $K$ is the function field of $X$, then this group is precisely the Galois group of the maximal algebraic extension of $K$ which is unramified with respect to all valuations of $K$ except those corresponding to the $r$ points that were removed.

More concretely: The finite coverings of $U$ arise by taking finite extensions of $K$, unramified except possibly at the removed points, and taking the normalization of $U$ in $L$ (that is, $U$ is an affine curve given by a ring $A$ contained in $K$, the covering will be the spectrum of the normal closure of $A$ in $L$).

Edit: If $K$ is not algebraically closed and $U$ is geometrically connected, then there is a short exact sequence \[1\rightarrow \pi_1(U\times_K \bar{K}) \rightarrow \pi_1(U)\rightarrow Gal(\bar{K}/K)\rightarrow 1\] (one has to pick compatible base points), and if $X$ has a $K$-rational point, then this sequence splits, so the structure is "known", as far as the Galois group of the base is known.


Here is a result concerning Riemann surfaces which might be relevant (cf. your EDIT).

Fix a Riemann surface $X$, a discrete closed subset $D\subset X$ and put $X_0 =X\setminus D$ . Let $\mathcal RevRam (X;D)$ be the category with objects finite ramified coverings of $X$ (= proper non constant morphisms to $X$), étale over $X_0$. And let $\mathcal RevEt(X_0)$ be the category with objects finite étale coverings of $X_0$.

Theorem The restriction functor $\mathcal Revram (X;D) \to \mathcal Rev(X_0)$ is an equivalence of categories.

This is similar to Lars's interesting answer but the difference is that there is no assumption of algebraicity on the Riemann surfaces involved (for example $X$ could be an arbitrary open subset of $\mathbb C$)

  • $\begingroup$ 1. Shouldn't the objects of RevRam(X,D) be finite $\textbf{possibly}$ ramified coverings of $X$? 2. I'm guessing a morphism in RevRam(X,D) from $(Y,\pi)$ to $(Y^\prime,\pi^\prime)$ is a commutative diagram and similarly for Rev$(X_0)$. 3. How does the restriction functor act on morphisms? Given $(Y,\pi)$ in RevRam(X,D), I send this to $(\pi^{-1}(X_0),\pi|_{\pi^{-1}(X_0)}$, right? I don't see how I get a morphism in Rev$(X_0)$, though. Unless I demand something extra for morphisms in RevRam(X,D). I think this is really interesting. If true, can't one show eq. of cat. for sm. proj. varieties? $\endgroup$ – Ariyan Javanpeykar Apr 29 '10 at 21:37
  • $\begingroup$ Yes, dear Ariyan, you're right: "ramified" here means "POSSIBLY ramified". (This is abuse of language.) And a morphism $F:(Y,\pi) \to (Y', \pi ') $ in the category $\mathcal RevRam (X;D)$ gets sent to its restriction $F _0 : Y_0=\pi^{-1} (X_0) \to Y'_0=(\pi ')^{-1} (X_0)$, a morphism in the category $\mathcal Rev (X_0)$. Amusingly, I had asked myself in how much detail I should post my answer. I opted for terseness for the sake of fluidity of style, but maybe fluidity turned into sloppiness... $\endgroup$ – Georges Elencwajg May 1 '10 at 11:19
  • $\begingroup$ As to your last question: yes, there is a vast generalization of the above to higher dimensional varieties, due to Grauert-Remmert. You can read about it in SGA1 in the very interesting Exposé 12 by Mrs. M.Raynaud (not to be confused with Mr. Michel Raynaud ! ). $\endgroup$ – Georges Elencwajg May 1 '10 at 11:53

@Ariyan and BCnrd: Cf. Exercise 1.9, Section 4.1 of Q. Liu's wonderfully written book, "Algebraic Geometry and Arithmetic Curves". @Ariyan: Exercise: re-write the one-dimensional (i.e. classical) idele class group as a certain complement of $Spec\mathcal{O}_K$...cf. the intro to "Global class field theory" by Kato-Saito.


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