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Does it follows from conductor's property in Dedekind domains? Where I can find a reference?

Konstantin

David Roberts
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Konstantin
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    What do you mean by simply connected? – Will Chen Sep 12 '18 at 04:16
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    If you're asking if it has trivial étale fundamental group, this just follows from the fact that $\mathbb Q$ has no unramified extensions. – Kevin Casto Sep 12 '18 at 04:31
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    By a theorem of Minkowski, there are no unramified extensions of $\mathbb{Q}$. – Felipe Voloch Sep 12 '18 at 04:32
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    Thanks a lot! Simply connected means trivial etale covers only. – Konstantin Sep 12 '18 at 06:03
  • See also https://mathoverflow.net/questions/26491/is-there-a-ring-of-integers-except-for-z-such-that-every-extension-of-it-is-ram –  Sep 13 '18 at 09:26
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    It sounds awkward to phrase the question as a question about the "space" $Spec(Z)$. This seems to be a question about the scheme, not just the underlying topological space, isn't it? – YCor Sep 13 '18 at 16:23
  • @YCor: the topological space $\operatorname{Spec} \mathbb Z$ is contractible: you can contract everything to the generic point. But I suspected this was probably not the question... – R. van Dobben de Bruyn Sep 13 '18 at 17:39
  • @R.vanDobbendeBruyn Yes but this does not technically mean that the question (about trivial étale $\pi_1$) is not only about the underlying space. – YCor Sep 13 '18 at 18:15
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    That typo was killing me. I had to fix it. – David Roberts Sep 14 '18 at 06:14

1 Answers1

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Here's a nice way to see that $\mathrm{Spec}(\mathbb{Z})$ is étale simply connected. Since $\mathrm{Spec}(\mathbb{Z})$ is a normal scheme every connected finite étale cover of $\mathrm{Spec}(\mathbb{Z})$ is of the form $\mathrm{Spec}(\mathcal{O}_K)$ for some number field $K$, as $\mathcal{O}_K$ is the normalization of $\mathbb{Z}$ in $K$ over $\mathbb{Q}$. Now suppose that $K$ has degree $n=r_1+2r_2$ where $r_1$ and $r_2$ are the numbers of real and complex embeddings respectively of $K$, and let $\Delta_K$ be the discriminant of $K$. Then by the Minkowski bound (see Neukirch) every class in the ideal class group of $K$ contains an integral ideal of norm at most the Minkowski constant $M_K=\sqrt{\vert\Delta_K\vert}(\frac{4}{\pi})^{r_2}\frac{n!}{n^n}$. But by definition an integral ideal of $\mathcal{O}_K$ has norm at least $1$ so that $1\leq M_K$ and hence $\sqrt{\vert\Delta_K\vert}\geq(\frac{\pi}{4})^{r_2}\frac{n^n}{n!}\geq(\frac{\pi}{4})^{n/2}\frac{n^n}{n!}$; in particular if $K$ is a nontrivial extension of $\mathbb{Q}$ we have $\vert\Delta_K\vert>1$, and since a prime $p\in\mathbb{Z}$ ramifies in $K$ precisely if $p$ divides $\Delta_K$ it follows that the finite étale cover $\mathrm{Spec}(\mathcal{O}_K)\rightarrow\mathrm{Spec}(\mathbb{Z})$ is somewhere unramified unless it is trivial, which implies $\widehat{\pi}_1(\mathrm{Spec}(\mathbb{Z}))=0$.

In general for $\mathcal{O}_K$ the ring of integers of a number field $K$ the maximal Abelian quotient $\widehat{\pi}^\mathrm{ab}_1(\mathrm{Spec}(\mathcal{O}_K))$ is isomorphic to the narrow class group $\mathrm{Cl}^+_K=I_K/P^+_K\simeq\mathrm{Gal}(H^+_K/K)$ where where $H^+_K$ is the narrow class field of $K$ (the maximal Abelian extension of $K$ which is unramified outside the finite places), and $I_K/P^+_K$ is the group of fractional ideals of $\mathcal{O}_K$ modulo the group of principal fractional ideals $(\alpha)=\alpha\mathcal{O}_K$ whose generator $\alpha\in K$ is totally positive for each real embedding.

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