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Source

This question came up in the discussion between Kevin Buzzard and Minhyong Kim in the comments to Smooth proper scheme over Z. It was 2 weeks ago, so I took the liberty of posting it as community wiki.

Question

Is there an example of smooth proper variety $X \to \mathop{\text{Spec}}\mathbb Z$ such that $\pi_1(X) \ne 0$?

About tags

We recently had other questions of the form "Example of ... with everywhere good reduction at $\mathbb Z$" (local-global, abelian varieties). I think it would be interesting to create a tag to group these. Thoughts?

Community
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Ilya Nikokoshev
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  • http://mathoverflow.net/questions/10569/smooth-proper-schemes-over-z-with-points-everywhere-locally – Anweshi Jan 07 '10 at 21:48
  • @Anweshi: I'm not sure what you mean, that seems like a different question to me. If that question somehow answers this one, I'll be happy to accept an answer! – Ilya Nikokoshev Jan 07 '10 at 21:51
  • The questions are related in that they both talk about smooth proper varieties over Z. But I don't see another relation. Minhyong and I had some dialogue about this via email last week but didn't come up with an example. – Kevin Buzzard Jan 07 '10 at 21:59
  • @Ilya. I didn't intend it as an answer. It was just a pointer that somebody else also posted an offshoot from the same question. I mean, you were speaking of creating tags and all. – Anweshi Jan 07 '10 at 22:46
  • Ah, I have this link as the "[counterexample to] local-global [principle for such maps]" in the brackets :) – Ilya Nikokoshev Jan 07 '10 at 22:56
  • If I wanted to start trying to produce an example, I guess I would take my insight from the other question and produce some smooth proper varieties Y over Z, and try to do so in a way that Y has a nontrivial group G of fixed-point-free (in the appropriate sense) automorphisms, and then take X = Y/G. – JSE Jan 08 '10 at 03:00
  • Basic observation: if G is fixed-point-free as in JSE's comment, so is any cyclic subgroup of G. So we may as well hunt for examples with just a single automorphism. – David E Speyer Jan 08 '10 at 11:58
  • I may as well record my failures. I thought about Kevin Buzzard's example. The automorphism group of a quadric is O(n). A finite order element of O(n) is diagonalizable. If it has an eigenvalue other then \pm 1, the corresponding eigenvector is a fixed point. Also, if 1 or -1 occur with multiplicity greater than 1, we get a fixed point. So quadrics don't work for n>2. – David E Speyer Jan 08 '10 at 12:02
  • I thought a little about Enriques surfaces but didn't get anywhere. I assume that Minhyong and Kevin would have also considered this, though. What progress did you make? – David E Speyer Jan 08 '10 at 12:03
  • Sorry for the delay David. Minhyong and I were prompted into conversation by Tyler Lawson. We just mused on whether the compactification of an M_{g,n} would work, but the problem is that these are stacks and furthermore that they tend to have trivial pi_1 anyway. – Kevin Buzzard Jan 09 '10 at 08:43
  • Is anyone still interested in this problem? Has anyone found a solution? I have been thinking a little bit about how one might prove that this is impossible. – Will Sawin May 08 '12 at 15:03

3 Answers3

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I think I have an argument that might work. The goal is to prove that this is impossible. There are some gaps in it.

Let $X$ be a connected smooth proper scheme over $\mathbb Z$. Clearly $\Gamma(X,\mathcal O_X)=\mathbb Z$. (If the ring had zero-divsors, it would indicate $X$ reducible, impossible, or $X$ non-reduced, thus ramified, impossible. If it were a ring of integers of a number field it would give ramification at some prime.) Since $H^1(X,\mathcal O_X)$ is the tangent space of the Picard scheme, and the Picard scheme is trivial, $H^1(X,\mathcal O_X)$ is trivial. (This probably requires smoothness of the Picard scheme. I'm not sure if that holds.) I need to assume that $H^2(X,\mathcal O_X)$ is torsion-free. (I would think that smoothness over a scheme should imply locally free higher pushforwards, which over an affine scheme implies torsion-free cohomology, but I don't know. This is true in characteristic 0 by Deligne, but we are obviously not in characteristic 0 here.)

We have the exact sequence $0\to \mathcal O_X \to \mathcal O_X \to \mathcal O_X/p\to 0$, with the first map multiplication by $p$. Taking cohomology and filling in what we know, we get

$ 0 \to \mathbb Z \to \mathbb Z \to H^0(X, \mathcal O_X/p) \to 0 \to 0 \to H^1(X,\mathcal O_X/p) \to H^2(X,\mathcal O_X)\to H^2(X,\mathcal O_X) $

which since those are also the cohomology groups of $X_P$, gives $\Gamma(X_p,\mathcal O_{X_p})=\mathbb F_p$, $H^1(X_p,\mathcal O_{X_p})=0$.

Now let $Y\to X$ be a cyclic etale cover of degree $p$. Artin-Schreier on $X$ gives $H^1_{et}(X_P,\mathbb Z/p)=\mathbb Z/p$. Thus there is a unique connected etale degree-$p$ cover of $X_p$, so it's the one you get by tensoring over $\mathbb F_p$ with $\mathbb F_{p^p}$. Since $\Gamma(Y_p,\mathcal O_{Y_p})=\mathbb F_p$, it is connected, and is not the result of tensoring anything with $\mathbb F_{p^p}$. This is a contradiction.

No cyclic etale covers of degree $p$ $\implies$ no cyclic etale covers $\implies$ no etale covers. (since ever group has a cyclic subgroup.)

Will Sawin
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  • For those dummies who do not feel good with schemes different than smooth complex manifolds. Can you comment ... I thought scheme over Z, means just Spec(Z), what are other examples ? and why Г(X,O_x) =Z) ? – Alexander Chervov May 12 '12 at 05:49
  • A scheme over $\mathbb Z$ is just a mildly more efficient way to say a scheme over $Spec \mathbb Z$. In fact, since $Spec \mathbb Z$ is a final object in the category of schemes, every scheme is a scheme over $Spec \mathbb Z$, including smooth complex manifolds. However, smooth complex manifolds cannot be smooth and proper over $Spec \mathbb Z$ because they are never finite-type. What you could do instead is take a complex manifold defined by some equation with integer coefficients inside $\mathbb P^n_{\mathbb C}$, and look at a the zero set of that equation in $\mathbb P^n_{\mathbb Z}$. – Will Sawin May 12 '12 at 06:21
  • This is the analogy by which number-theorist's "elliptic curves" are derived from the "elliptic curves", that are complex tori and were used to study elliptic functions. Such a scheme will certainly be proper over Spec $\mathbb Z$, but may or may not be smooth. – Will Sawin May 12 '12 at 06:24
  • I am not quite sure what the best way to compute $\Gamma(X,\mathcal O_X)$. It must be at least $\mathbb Z$, since the integers must be functions on any scheme, and must be of finite rank over $\mathbb Z$ since the morphism is proper. If it contained a nilpotent, a local ring somewhere would contain a nilpotent, and the scheme would not be smooth at that point. If it contained a zero-divisor, either the scheme would be disconnected or a local ring somewhere would have zero-divisors, also not smooth. And if it contains an algebraic integer, at some prime it contains a nilpotent. – Will Sawin May 12 '12 at 06:32
  • All rings but $\mathbb Z$ of finite rank over $\mathbb Z$ contain one of those. – Will Sawin May 12 '12 at 06:32
  • @Will Thank you. But still I do not quite understand the first sentence. "Let be a connected smooth proper scheme over . Clearly (X,O_x) =Z." I understand intuitively what you mean - but formally not. Is it definition of "connectedness" ? – Alexander Chervov May 12 '12 at 06:47
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    Regarding "may or may not be smooth" over $\mathbf{Z}$, it is worth clarifying that if you are dealing with curves of genus $>0$ (or abelian varieties of dimension $>0$), then it is never smooth over $\mathbf{Z}$ because it has bad reduction somewhere (Tate-Fontaine-Abrashkin). – Chandan Singh Dalawat May 12 '12 at 06:51
  • Indeed, otherwise the question would be a whole lot simpler to answer! – Will Sawin May 12 '12 at 07:07
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    I'm not clear on how the final implication in your argument works without knowing something strong about the etale fundamental group (e.g. that it is pro-solvable). – Tyler Lawson May 12 '12 at 12:30
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    Suppose $Y\to X$ is an etale galois cover, let C be a cyclic subgroup of the Galois group, then $Y\to Y^C$ is an abelian cover. I got the idea from David Speyer's comment. – Will Sawin May 12 '12 at 16:19
  • Ah, right, one could reduce to the cyclic case. Thanks. – Tyler Lawson May 12 '12 at 21:52
  • I don't think the Picard scheme needs to be smooth. It seems to me that Igusa's example of a non-reduced Picard scheme deforms smoothly to characteristic zero. The Picard scheme is generically smooth with an embedded curve at (2). It is generically positive dimensional, so it can't show up as a counterexample to this question, but I think it can be modified to be generically trivial ......... If you could make the Picard scheme vanish, it commutes with fibers, so $H^1(X_p; \mathcal O/p)$ vanishes, so the torsion in $H^2$ vanishes. – Ben Wieland May 13 '12 at 04:07
  • Isn't the Picard scheme of each fibre equal to the fibre of the Picard scheme if $X$ is proper and smooth over $\mathbf{Z}$? Doesn't this imply the smoothness of the Picard scheme? (This is how one proves that if a curve over a number field has good reduction at a place $v$, then its Jacobian has good reduction at $v$.) – Harry May 13 '12 at 07:00
  • @Ben: True. It remains to see if it can be modified that way. @Harry: I'm not sure how that implication would work. Do you have a reference? – Will Sawin May 13 '12 at 07:30
  • Isn't the Picard scheme of a smooth proper $\mathbf{F}_p$-scheme smooth? (Maybe I'm thinking too much about the case of a curve over $\mathbf{F}_p$.) – Harry May 13 '12 at 10:49
  • Smoothness at each point does not imply smooth family. – Will Sawin May 13 '12 at 15:30
  • Harry, no, in higher dimension, the Picard scheme have nilpotents. Igusa's example at 2 is to take the quotient by an involution of a product. One factor is an elliptic curve with action the inverse; the other factor should have free action so the quotient is smooth. – Ben Wieland May 13 '12 at 15:43
  • Igusa's example seems very difficult to extend to $\mathbb Z$ to me, since it's constructed from elliptic curves, and if those are non-smooth then it should also fail to be smooth. But of course that doesn't imply there's not some other example. – Will Sawin May 13 '12 at 19:09
  • Yes, Igusa's example is not going to work. However, your argument seems to be that scheme smooth and proper over $Z_p$ cannot have fundamental group $Z/p$, to which the Godeaux-Serre varieties are counter-examples. In fact, let me run your argument backwards: if a variety has geometric fundamental group $Z/p$, then by universal coefficients, the torsion of $H^2(X;Z)$ is $Z/p$. This shows up in coherent cohomology over $Z_p$ or $Z/p$, and thus the variety has non-reduced Picard scheme. The Picard scheme cannot be $Z/p$, but must be more like $\mu_p$. – Ben Wieland May 13 '12 at 22:41
  • My argument at several steps uses facts about $\mathbb Z$ that do not hold at $\mathbb Z_p$. For instance, the claim that the ring of global sections is $\mathbb Z$ is obviously false for $\mathbb Z_p$, which has unramified extensions. – Will Sawin May 14 '12 at 00:07
  • Just ran across this. As far as I understand it, the question you're trying to answer is whether or not there exist smooth proper $X$ over $\mathbb{Z}$ with $\pi_1(X)\not=0$. I agree this is interesting (and already hard enough), but a somewhat more interesting question is whether or not such $X$ exist with $\pi_1(X_{\overline{\mathbb{Q}}})\not=0$ (since specialization maps are very not injective). – Daniel Litt Sep 05 '17 at 03:12
  • @DanielLitt Well, if this question has a positive answer, then it's the more interesting one, right? Anyways, I think my question about Enriques surfaces is the first interesting special case of your question. – Will Sawin Sep 05 '17 at 05:15
  • Ok yeah, agreed! – Daniel Litt Sep 05 '17 at 12:25
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Here is a proof that if $X$ is smooth and proper over $\mathbb Z$ and of (relative) dimension $\leqslant 3$, then it is simply connected. The dimensional restriction is isolated to a particular step and I believe that theorem is conjectured to generalize to all dimensions.

Fontaine's letter to Messing proves that if $Y$ is smooth and proper over $\mathbb Z$, the Dolbeault cohomology $H^q(Y_{\mathbb Q};\Omega^p)$ vanishes off of the diagonal $p\ne q$ in low degree $p+q\leqslant 3$. I believe the low degree restriction is conjectured not to be necessary. By the Atiyah-Bott fixed-point formula, the Lefschetz number of an element of a finite group acting on a complex variety is the same as the Lefschetz number acting on its cohomology of the structure sheaf. Thus if $H^q(Y;\mathcal O)$ vanishes for $q>0$, Fontaine's theorem with $p=0$, then the Lefschetz number is $1$ and the action cannot be free. If $X$ were smooth and proper over $\mathbb Z$ with non-trivial pro-finite fundamental group*, then some finite cover $Y$ of $X$ would be canonical, thus defined over $\mathbb Z$ (eg, the composite of all covers of degree $\leqslant N$). Then $Y$ would be smooth and proper over $\mathbb Z$ with a free action by the finite covering group, a contradiction.

* If I recall correctly, there are varieties whose complex points have a nontrivial fundamental group, but that group has no finite quotients, and thus the étale fundamental group is trivial.

Ben Wieland
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  • Fontaine's result might extend as high as dimension $10$, but it probably fails in degree $11$ due to the Ramanujan Delta function, which appears in $H^0(\bar{M}_{1,11};\Omega^11)$. That's a stack and it's still simply connected, but the method of attack breaks down. See Kevin Buzzard and Dan Petersen here: http://mathoverflow.net/questions/97086/does-smooth-and-proper-over-mathbb-z-imply-rational – Ben Wieland May 16 '12 at 14:27
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This is wrong. But see my other answer arguing the opposite direction.

What's wrong is that Bertini's theorem fails over $\mathbb Z$. It works over infinite fields in single pencils. A version works over finite fields by allowing arbitrary degree and thus infinitely many choices. But high degree over $\mathbb Z$ is bad for smoothness. As Will points out, even in $P^2_{\mathbb Z}$, a high degree hypersurface is not smooth.

I think that Godeaux-Serre varieties exist integrally. Choose a prime $p$, let $G$ be the cyclic group of order $p$ and let $Z[G]$ be the group ring. Then the projectivization of the $n$-th power of the ring group $P(Z[G]^n)$ is an $pn-1$-dimensional variety with an action of $G$, generically free, with fixed set a disjoint union of $P^{n-1}$; 1 copy at $p$ and $p$ copies away from $p$. The quotient is not smooth, but a generic complete intersection of codimension $n$ misses the singular set and thus is smooth with fundamental group $G$.

I have never seen Godeaux-Serre varieties used in same characteristic, but when I looked up Igusa's example, I saw it asserted that not only does the construction work, they have non-reduced Picard scheme, evading Will's attack. But does this generic complete intersection argument work globally?

Ben Wieland
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  • What do you mean by "generic complete intersection"? What degree polynomials are you intersecting? Are they intended to be $G$-invariant polynomials? – Will Sawin May 13 '12 at 05:42
  • Yes, $G$-invariant polynomials; or, equivalently, work downstairs. The degree is a free parameter. The higher degree, the larger the space to choose from, avoiding any problematic space of fixed dimension. – Ben Wieland May 13 '12 at 05:59
  • Getting polynomials smooth over every $p$ might be tricky. It doesn't seem like the sort of thing that is impossible, but, for instance, you can't do it in dimension $2$ even with arbitrarily high degree. Do you think it would be possible to explicitly work out an example for some small $p$ and $n$, to show it exists? – Will Sawin May 13 '12 at 07:42
  • Oh, right, unbounded degree is bad for smoothness over $Z$. Since it doesn't work to produce a free action on a curve, it probably doesn't work to produce a free action on a simply connected space. More specifically, in the case $n=1$, avoiding the fixed point at $p$ requires an invariant monomial, thus minimum degree $p$, yielding nonnegative Kodaira dimension, making it implausible that it be smooth. – Ben Wieland May 15 '12 at 00:58
  • I was trying to construct a similar argument. I guessed the step "nonnegative Kodaira dimension => nonsmooth" purely on blind hope. Why is that true? (or implausibly untrue?) For n>1 all n monomials must have minimum degree p, else there are not enough monomials to avoid all the nonsingular points. You get the same computation for the Kodaira dimension nonnegative. – Will Sawin May 15 '12 at 02:41
  • If there's some good reason why negative Kodaira dimension is necessary for smoothness, Castlenuovo's theorem on the generic fiber would imply that all dimension 2 proper smooth surfaces over Spec Z are rational, thus simply connected, pushing the minimum possible dimension of a counterexample up to 3. – Will Sawin May 15 '12 at 03:31
  • I just meant that since there is a standard conjecture that smooth and proper over Z implies rational, that nonnegative Kodaira dimension would be a more surprising counter-example. Fontaine's letter to Messing probably already forces rationality in dimension 2. http://www.ams.org/mathscinet-getitem?mr=1274493 – Ben Wieland May 15 '12 at 17:36
  • The Atiyah-Bott fixed point formula plus the obvious (conjectural) generalization of Fontaine's result shows that a smooth proper variety over Z must have trivial (pro-finite) fundamental group. Fontaine's proven theorem covers the case of dimension 1-3. – Ben Wieland May 16 '12 at 02:02
  • Well then you should probably make that your answer to this question. – Will Sawin May 16 '12 at 02:31
  • For example, William Lang ('Classical Godeaux Surface in Characteristic p', Math. Ann. 256, 419-427 (1981)) and Rick Miranda ('Nonclassical Godeaux Surfaces in Characteristic Five', Proc. AMS 91, 9-11 (1984)) address this. I'm not sure whether their examples are defined over the integers as they choose generic hyperplane sections, but maybe a computer search can establish this by hand? Miranda's examples also show explicitly that the Picard scheme need not be smooth - it becomes non-reduced for p=5 (in fact ${\rm Pic}^0$ is isomorphic to the non-reduced scheme $\mu_5$) ... – Christian Liedtke May 16 '12 at 08:37
  • ... of course, these effects were known to Igusa and Serre already, but these examples are maybe more explicit – Christian Liedtke May 16 '12 at 08:38