Does every profinite group arise as the étale fundamental group of a connected scheme?

Equivalently, does every Galois category arise as the category of finite étale covers of a connected scheme?

Not every profinite group is an absolute galois group of a field (the only finite ones have order $1$ or $2$ by Artin-Schreier). Therefore we cannot restrict to spectra of fields. Perhaps one first has to check if every finite group arises as a fundamental group of a scheme. I don't even know enough examples to answer this question for cyclic groups. At least order $3$ is possible (see here, Remark 2).

If the answer turns out to be no, then I would like to know which profinite groups arise as fundamental groups.

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    $\begingroup$ What do you mean by "the only finite ones have order 1 or 2 by Artin-Schreier"? Sorry for such a basic question... $\endgroup$ Dec 1, 2012 at 9:38
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    $\begingroup$ A theorem of Artin-Schreier states: if $K$ is a field, $\bar K$ an algebraic closure of $K$ and ${\rm Gal}(\bar K/K)$ is a finite group, then it has order 2, $K$ has an ordering for which it is real closed, and $\bar K=K(\sqrt{-1})$. $\endgroup$
    – ACL
    Dec 1, 2012 at 13:31
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    $\begingroup$ That should be "nontrivial finite group." $\endgroup$ Dec 2, 2012 at 7:33

2 Answers 2


[Edit:] The answer should be positive, that is, every profinite group appears as the fundamental group of a scheme. Here is a sketch of proof.

First of all, I claim that for any finite group $G$ there exists a complex affine simply connected variety $X$ with a free action of $G$. Start from a faithful finite-dimensional representation $G \to \mathrm{GL}(V)$ of $G$, such that there is an open subset $U \subseteq V$ where the action of $G$ is free, and such that $V \smallsetminus U$ has codimension at least 2 in $V$. Then $X$ is obtained with an easy equivariant extension of Jouanolou's trick. This yields a $G$-covering $X \to X/G$; then $X/G$ is an affine variety, and its fundamental group is $G$.

Now take a profinite group $G = \projlim_{i\in I}G_i$; identify $G$ with the corresponding affine group scheme over $\mathbb C$, in the usual fashion. For each finite subset $J\subseteq I$ denote by $G_J$ the image of $G$ in $\prod_{j\in J}G_j$; clearly we have $G = \projlim_{J \subseteq I}G_J$.

For each $i$ take a complex affine simply connected variety $X_i$ with a free action of $G_i$. Consider the affine scheme $X := \prod_{i \in I}X_i$, with the resulting action of $G$, and the quotient $X/G$, which is the spectrum of the ring of invariants $\mathbb C[X]^G$. For each finite subset $J\subseteq I$ set $X^J := \prod_{j \in J}X_j$. The action of $G$ on $X^J$ factors through a free action of $G_J$. Furthermore, we have $\mathbb C[X]^G = \injlim_{J \subseteq I}\mathbb C[X^J]^{G_J}$, hence $X/G = \projlim X^J/G_J$.

Now, $X^J$ is simply connected, hence the fundamental group of $X^J/G_J$ is $G_J$. On the other hand, it is easy to see that the Galois category of $X$ is the inductive limit of the Galois categories of the $X^J$, so that its Galois group is precisely $\projlim G_J = G$.

[Edit2:] Let me clarify what I mean by the "equivariant Jouanolou trick". The theorem of Jouanolou says that if $U$ is a quasi-projective variety, there exists an locally trivial fibration in affine spaces $X\to U$, where $X$ is affine. What we need here is the statement that if $G$ is a finite group acting on $U$, then we can construct such a map $X \to U$ that is also $G$-equivariant. This is easy: start from a fibration in affine space $Y \to U$ with $Y$ affine, and take $X$ to be the fiber product over $U$ of all the $g^*Y$ for $g \in G$, with the obvious action of $G$.

  • $\begingroup$ Angelo, thank you very much. Could you explain what you mean precisely with "$X$ is obtained with an easy equivariant extension of Jouanolou's trick."? Also, that $X$ resembles somehow topologist's $BG$, right? $\endgroup$ Dec 1, 2012 at 19:00
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    $\begingroup$ Dear Martin, $X/G$ is an approximation of $BG$; by choosing the codimension of the complements of the $U_i$ to be higher and higher, and going to the limit in an appropriate sense, we can obtain an algebraic version of $BG$. This is an idea of Totaro, and is behind the Edidin and Graham construction of equivariant intersection theory. $\endgroup$
    – Angelo
    Dec 2, 2012 at 13:20
  • $\begingroup$ I have also heard "Jouanlou's trick" above in reference to work of Looijenga $\endgroup$ Apr 3, 2013 at 15:18
  • $\begingroup$ @Jason, could you give me some hint where I could find something like a Jouanolou trick (equivariantly or some analogue or anything like it) in Looijenga's work? $\endgroup$ Apr 5, 2013 at 13:05

Every finitely presented group occurs as the fundamental group of an irreducible complex variety (see below), so (at least) you get the completions of all those.

Simpson, Carlos. Local systems on proper algebraic $V$-manifolds. Pure Appl. Math. Q. 7 (2011), no. 4, Special Issue: In memory of Eckart Viehweg, 1675--1759. MR2918179


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