Define an arithmetic scheme $X$ to be a separated, integral scheme, flat and finite type over $\mathbb{Z}$. I am interested in obtaining examples of finite étale covers of arithmetic schemes. I am mainly interested in the case of arithmetic surfaces, but I would enjoy examples of finite étale covers of arithmetic schemes of any dimension $>1$.

Let us restrict to schemes proper over $\mathbb{Z}$ to rule out the possibility of forming a finite integral extension and then restricting to an open subscheme on which the morphism is étale. Let us also restrict to geometrically connected (generic fiber?) to rule out the possibility of base ring extensions.

In the case where $X\to Spec(\mathbb{Z})$ is also smooth, the Katz-Lang finiteness theorem tells us that there will be at most finitely many abelian étale covers of $X$.

Abelian varieties can be a convenient way to create covers: if $A$ is an abelian variety, and $H$ a finite subgroup, then the dual isogeny to $A\to A/H$ is an étale cover of $A$. However, there are no abelian varieties over $Z$. I'm not sure whether this approach could be modified to produce an example.

It would be preferable to stick to regular or at least normal schemes, but that's not strictly necessary.

So, can anyone provide some nice examples of étale covers satisfying these conditions? If you would like to weaken some of the conditions to provide an interesting example (particularly, allowing nonproper or working over another $\mathcal{O}_K$), I guess I'll allow it.

To summarize, the optimal conditions on $X$ are regular, geometrically connected, proper over $\mathbb{Z}$.


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    $\begingroup$ I suppose you are familiar with the very interesting work in the opposite direction, giving interesting examples of simply connected arithmetic surfaces? Ihara's paper in Galois Groups over $\mathbb{Q}$ proves among other things that $\mathbb{P}_{\mathbb{Z}}^1$ minus $0$, $\infty$ and the $N$-th roots of unity is such an example, and then there is Bost's paper Potential theory and Lefschetz theorems for arithmetic surfaces. $\endgroup$ Commented Mar 3, 2016 at 4:18
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    $\begingroup$ In the discussion of abelian varieties the isogeny will not be etale at the primes dividing the order of $H$. So no examples arise in this way. (If you delete a single prime $p$ you do get examples. You could construct a smooth and proper curve over $\mathrm{Spec} \, \mathbb{Z} \setminus \{p\}$ and then pulling back multiplication by $[p^n]$ on the Jacobian gives infinitely many connected etale abelian covers. But this example is uninteresting as you said explicitly that you want your arithmetic surface to be complete.) $\endgroup$ Commented Mar 3, 2016 at 4:21
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    $\begingroup$ Note that, for all $d\geq 1$, a finite type scheme $X$ over $\mathbb Z$ has only finitely many finite etale covers $Y\to X$ of degree $d$. This is sometimes reformulated by saying that the fundamental group of an arithmetic scheme is small; see arxiv.org/abs/0803.2096 This statement is clearly a generalization of the classical finiteness theorem of Hermite-Minkowski $\endgroup$ Commented Mar 3, 2016 at 6:49
  • $\begingroup$ Also, there is work of Wiesend math.uchicago.edu/~drinfeld/Deligne's_conjecture_Manin_conf/… and Hofmann-Wiesend (see [HW] in previous link) on constructing finite etale covers of arithmetic schemes. $\endgroup$ Commented Mar 3, 2016 at 6:49
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    $\begingroup$ @VesselinDimitrov I only know a small amount about the opposite direction, and I'm not familiar with those references. I will check them out. Thanks for those as well as the remark. $\endgroup$ Commented Mar 4, 2016 at 0:04

1 Answer 1


You can, of course, use Bertini's theorem to make examples of a finite, flat, Galois extension with arbitrary finite Galois group $\Gamma$. Let $M$ be $\mathbb{Z}[\Gamma]$, the group ring of $\Gamma$ with coefficients in $\mathbb{Z}$. This is a finite, free $\mathbb{Z}$-module. Let $r>0$ be a positive integer. Then $M^{\oplus r}$ is also a finite, free $\mathbb{Z}$-module with a natural action of $\Gamma$. Form $S=\mathbb{Z}[M^{\oplus r}]$, the $\mathbb{Z}_{\geq 0}$-graded polynomial ring over $\mathbb{Z}$ such that the first graded piece $S_1$ equals $M^{\oplus r}$. Then $\mathbb{P}:=\text{Proj}(S)$ is a projective space over $\mathbb{Z}$ that has a natural action of $\Gamma$.

Denote by $U\subset\mathbb{P}$ the maximal open subscheme on which $\Gamma$ acts freely. Denote by $F$ the complement of $U$. The fibers of $F$ (over $\text{Spec}\ \mathbb{Z}$) have codimension at least $r$ inside the fibers of $\mathbb{P}$. Hence, assume $r$ is sufficiently large (say $r\geq 2$) so that the fibers of $F$ have codimension $\geq 2$. Denote by $q:\mathbb{P} \to Q$ the quotient of the action of $\Gamma$ on $\mathbb{P}$. This morphism is finite, and it is flat when restricted to $U$. The quotient $Q$ is a projective scheme, say $Q\subset \mathbb{P}^N_{\mathbb{Z}}$. By Bertini's Theorem, for sufficiently large integers $d$, there exist degree $d$ hypersurfaces in $\mathbb{P}^N_{\mathbb{Z}}$ whose intersection with $Q$ is a regular, $2$-dimensional scheme $X$ that is disjoint from the image of $F$. Define $\widetilde{X} = \mathbb{P}\times_Q X$. Then $\widetilde{X}\to X$ is a finite, flat, étale morphism that is Galois with Galois group $\Gamma$.

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    $\begingroup$ $\Gamma$ is allowed to be nonabelian? Then do we actually take $M$ and $S$ to be noncommutative rings or just polynomial rings with variables indexed by the elements of the group, or something else? Then $S$ is infinitely generated, by $M^{\otimes r}$? Or is $S_1 = \Gamma^{\otimes r}$? Sorry, I'm a little confused. $\endgroup$ Commented Mar 4, 2016 at 0:36
  • $\begingroup$ @PrimeRibeyeDeal. I am afraid that I have no idea what you are talking about. Could you please re-read what I wrote and then re-read what you wrote? $\endgroup$ Commented Mar 4, 2016 at 2:28
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    $\begingroup$ Sorry for being unclear. My first concern was that if we allow $\Gamma$ to be nonabelian, then the rings $M$ and $S$ would be noncommutative, and so $\text{Proj}(S)$ would not be defined. My second concern was that if we take $S = \mathbb{Z}[M^{\oplus r}] = \mathbb{Z}[(\mathbb{Z}[\Gamma])^{\oplus r}]$, then our graded ring would not be finitely generated. And I did mean to write $\oplus$ instead of $\otimes$. $\endgroup$ Commented Mar 4, 2016 at 17:19
  • $\begingroup$ The ring $S$ is the commutative polynomial ring over $\mathbb{Z}$ that is $\mathbb{Z}_{\geq 0}$-graded and whose first graded piece is $M^{\oplus r}$. Once again, please read what I wrote. $\endgroup$ Commented Mar 4, 2016 at 17:51
  • $\begingroup$ Ok, so we do take $S$ to be the polynomial ring with variables indexed by $r$ copies of the group. I got it now. Thanks! $\endgroup$ Commented Mar 4, 2016 at 18:06

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