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Related question.

Let $A$ be a ring, and let $B$ be an $A$-algebra which is projective and finite as an $A$-module. Then the trace $B \rightarrow A$ can be defined. Let's say that $B$ is separable over $A$ if the trace induces an isomorphism of $A$-modules $B \rightarrow \operatorname{Hom}_A(B,A)$. For the purposes of this question, let's say that $B$ is an etale $A$-algebra if it is finite, projective, and separable.

Is it true that every etale $\mathbb Z$-algebra is of the form $\mathbb Z^n$? This is stated in these notes, example 1.1.13(ii).

It does not seem so easy to prove. If this is true, then combined with the much easier result, "if $L/K$ is an unramified extension of number fields, then $\mathcal O_L$ is an etale $\mathcal O_K$-algebra," we would obtain an algebraic proof that there are no nontrivial unramified extensions of $\mathbb Q$. I had thought that the only known proofs of this result relied on Minkowski's bound for the discriminant. Perhaps any proof that etale $\mathbb Z$-algebras are of the form $\mathbb Z^n$ must rely on this fact.

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    $\begingroup$ This fact is equivalent to Minkowski's theorem. I do not know of a different proof. $\endgroup$
    – Keerthi Madapusi
    Jul 23, 2018 at 18:38
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    $\begingroup$ A better term would be finite étale $A$-algebra, because there are also étale $A$-algebras that are not finite. $\endgroup$
    – R. van Dobben de Bruyn
    Jul 23, 2018 at 19:59
  • $\begingroup$ @R.vanDobbendeBruyn Do you know where to read about non-finite étale $\mathbb Z$-algebras? $\endgroup$
    – მამუკა ჯიბლაძე
    Jul 24, 2018 at 4:57
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    $\begingroup$ @მამუკაჯიბლაძე: I don't know a reference that treats the affine case exclusively; only the more general setting of étale morphisms of schemes. Examples include $\operatorname{Spec} \mathbb Z[1/n]$ and finite étale extensions thereof (e.g. $\mathbb Z[\zeta_n][1/n]$, the integral closure of $\mathbb Z[1/n]$ in the ray class field of modulus $n \cup \infty$), as well as products of such. $\endgroup$
    – R. van Dobben de Bruyn
    Jul 24, 2018 at 13:47
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    $\begingroup$ @მამუკაჯიბლაძე: admittedly, there might not be a reference that I truly like. Hartshorne leaves it at four exercises, and in EGA and the Stacks project it relies on a lot of theory already. A place to start could be Milne's notes on étale cohomology, or his book on the same subject (which is more technical). $\endgroup$
    – R. van Dobben de Bruyn
    Jul 24, 2018 at 14:12

1 Answer 1

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Let $X$ be a normal integral scheme. Then $\pi_1^\mathrm{et}(X)$ is isomorphic to $\mathrm{Gal}(K(X)^\mathrm{un}/K(X))$, where $K(X)^\mathrm{un}$ is the compositum of all finite extensions $F$ of $K(X)$ (in a fixed algebraic closure of $K(X)$) for which $X$ is unramified in $F$. Applying this to $X = \mathrm{Spec}(\mathbf{Z})$, it follows the result you are asking about is equivalent to Minkowski's theorem (that there are no nontrivial unramified extensions of $\mathbf{Q}$).

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  • $\begingroup$ By the way, the vanishing of $\pi_1^\mathrm{et}(\mathrm{Spec}(\mathbf{Z}))$ is just (by class field theory) the statement that $\mathbf{Z}$ is a PID. $\endgroup$
    – skd
    Jul 24, 2018 at 2:02
  • $\begingroup$ Class field theory, which can be proved algebraically, gives the vanishing of the abelianised fundamental group. Perhaps using non-abelian class field theory, one can handle the whole étale fundamental group? $\endgroup$
    – user19475
    Jul 24, 2018 at 3:53
  • $\begingroup$ @TKe Yes, you're right; sorry! I don't know anything about non-abelian class field theory, but I'd like to see a proof of this vanishing using that theory. $\endgroup$
    – skd
    Jul 24, 2018 at 4:30
  • $\begingroup$ Maybe one can edit the question accordingly? (I am not sure if I am allowed to change the question.) $\endgroup$
    – user19475
    Jul 24, 2018 at 4:34
  • $\begingroup$ @TKe I don't think the question's the problem, it was just my comment that was insufficient. The answer still works, but my comment only shows that class field theory implies that the abelianization of the etale fundamental group of Z vanishes. $\endgroup$
    – skd
    Jul 24, 2018 at 13:11

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