# Non-simply-connected smooth proper scheme over Z?

### 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 $\pi_1(X) \ne 0$?

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?

• mathoverflow.net/questions/10569/… Jan 7 '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! Jan 7 '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. Jan 7 '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. Jan 7 '10 at 22:46
• Ah, I have this link as the "[counterexample to] local-global [principle for such maps]" in the brackets :) Jan 7 '10 at 22:56

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.)

• 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) ? May 12 '12 at 5:49
• 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 $\mathbb P^n_{\mathbb Z}$. May 12 '12 at 6:21
• 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). May 12 '12 at 6:51
• 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). May 12 '12 at 12:30
• 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. May 12 '12 at 16:19

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.

• 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: mathoverflow.net/questions/97086/… May 16 '12 at 14:27

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?

• What do you mean by "generic complete intersection"? What degree polynomials are you intersecting? Are they intended to be $G$-invariant polynomials? May 13 '12 at 5:42
• 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. May 13 '12 at 5:59
• 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? May 13 '12 at 7:42
• 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. May 15 '12 at 0:58
• 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. May 15 '12 at 2:41