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.
Best Answer
[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 $G$.