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

### 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?

• mathoverflow.net/questions/10569/… – Anweshi 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! – Ilya Nikokoshev 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. – Kevin Buzzard 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. – Anweshi Jan 7 '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 7 '10 at 22:56

## 3 Answers

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) ? – Alexander Chervov 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 $\mathbb P^n_{\mathbb Z}$. – Will Sawin 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). – Chandan Singh Dalawat 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). – Tyler Lawson 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. – Will Sawin 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/… – Ben Wieland 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? – Will Sawin 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. – Ben Wieland 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? – Will Sawin 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. – Ben Wieland 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. – Will Sawin May 15 '12 at 2:41