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$\DeclareMathOperator\Spec{Spec}$Does there exist a scheme $X/{\operatorname{Spec}(\mathbb{Z})}$ such that $\pi_1^\text{ét}(X)=\smash{\hat{\mathbb{Z}}}^2$?

It's well known that $\pi_1^\text{ét}(\Spec(\mathbb{F}_p))=\hat{\mathbb{Z}}$, for $p$ a prime. However, the scheme $$X_{p,q}:=\Spec(\mathbb{F}_p)\times_{\Spec(\mathbb{Z})}\Spec(\mathbb{F}_q)$$ is empty if $p\neq q$, and isomorphic to $\Spec(\mathbb{F}_p)$ if $p=q$, so neither work. The idea was to use the natural map, $$\pi_1^\text{ét}(X\times_{S}Y)\rightarrow \pi_1^\text{ét}(X)\times_{\pi_1^\text{ét}(S)} \pi_1^\text{ét}(Y)$$ somehow. So does such a scheme exist?

Edit:

As @Will Chen pointed out in the comments 1 2, $X=$ an elliptic curve or $\mathbb{G}_m^2$ work. However, that's not really the question I meant to ask. I'd like the map to be an embedding$$X\hookrightarrow\Spec(\mathbb{Z}).$$ The reason is $\Spec(\mathbb{Z})$ has cohomological dimension $3$ (up to $2$-torsion). And $$\Spec(\mathbb{F}_p)\hookrightarrow\Spec(\mathbb{Z})$$ is an embedding, and $\Spec(\mathbb{F}_p)$ has cohomological dimension 1. So also, $X$ with $\operatorname{cd}(X)=2$.

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    $\begingroup$ Of course, we could take the product of F_p and F_q, which just translates to disjoint union of the Spec but that’s a silly example and I’m not sure if you want that. $\endgroup$
    – Cranium Clamp
    Commented Sep 10 at 20:50
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    $\begingroup$ @CraniumClamp I don't think one normally defines $\pi_1$ for a disconnected space as the products of the $\pi_1$s of the components. Usually it's just taken to be undefined. $\endgroup$
    – Will Sawin
    Commented Sep 10 at 20:55
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    $\begingroup$ Do you have any requirements on the map $X\rightarrow\text{Spec }\mathbb{Z}$? If not, the fundamental group of an elliptic curve over $\mathbb{C}$ is $\hat{\mathbb{Z}}^2$. $\endgroup$
    – Will Chen
    Commented Sep 10 at 21:01
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    $\begingroup$ You can also take a product of $\mathbb{G}_m$'s over $\mathbb{C}$. $\endgroup$
    – Will Chen
    Commented Sep 10 at 21:09
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    $\begingroup$ @RJAcuña that condition doesn’t leave you many options. Connected closed subschemes are just F_p, and open subschemes all include into Z[1/p], whose fundamental group I’m pretty sure is nonabelian… $\endgroup$
    – Will Chen
    Commented Sep 10 at 22:45

1 Answer 1

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As stated, the answer is no. (There is no connected subscheme of $\text{Spec }\mathbb{Z}$ with etale fundamental group $\widehat{\mathbb{Z}}^2$)

A closed subscheme of $\mathbb{Z}$ is a union of closed points, and the only connected closed subscheme is $\text{Spec }\mathbb{F}_p$ for some $p$, which has $\pi_1 = \widehat{\mathbb{Z}}$.

An open subscheme of $\mathbb{Z}$ has the form $\text{Spec }\mathbb{Z}[1/n]$. If $n$ is a unit, then by Minkowski's theorem $\text{Spec }\mathbb{Z}$ has trivial fundamental group. If $n$ is not a unit, it is divisible by a prime $p$, and the inclusion $\text{Spec }\mathbb{Z}[1/n]\hookrightarrow\text{Spec }\mathbb{Z}[1/p]$ induces a surjection on fundamental groups (this follows from the observation that a connected finite etale cover of the target remains connected when pulled back to the source). On the other hand, $\text{Spec }\mathbb{Z}[1/p]$ has nonabelian fundamental group: The number field $\mathbb{Q}(\zeta_p,p^{1/p})$ has nonabelian Galois group $\mathbb{Z}/p\rtimes(\mathbb{Z}/p)^\times$, and its ring of integers is finite etale over $\mathbb{Z}[1/p]$. Thus, $\text{Spec }\mathbb{Z}[1/n]$ also has nonabelian fundamental group.

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  • $\begingroup$ Given that you mentioned “as stated” what would be the legit question that could have interesting answers? $\endgroup$
    – RJ Acuña
    Commented Sep 11 at 14:57
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    $\begingroup$ @RJAcuña It's a legit question. It's just that there aren't that many subschemes of Spec $\mathbb{Z}$, so the answer is not that interesting. You might be able to ask a better question by enlarging the category of possible answers so that the answer is tautologically "yes", and then asking for the smallest/nicest subcategory that contains your desired object. My answer shows that the desired object is not a subscheme, but maybe if you relax the requirements a bit it is a sub(something). $\endgroup$
    – Will Chen
    Commented Sep 11 at 18:54
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    $\begingroup$ (this is kind of like the approach to moduli problems. Even if the moduli-thing isn't a scheme, you shouldn't stop there. Instead, first note that the moduli-thing is tautologically a category. Then you show maybe that it's a stack, maybe even a Deligne-Mumford stack, maybe even smooth, maybe even an algebraic space, maybe even a scheme! At each stage you can also identify the obstructions to the desired thing lying in the smaller category. This way you can get a much better understanding of the situation.) $\endgroup$
    – Will Chen
    Commented Sep 11 at 18:58

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