TITLE: The etale fundamental group of a field
QUESTION [7 upvotes]: Background and motivation:
I am teaching the "covering space" section in an introductory algebraic topology course.  I thought that, in the last five minutes of my last lecture, I might briefly sketch how to compute the "fundamental group of a field," primarily as a way of illustrating the analogy between Galois theory and covering space theory, but also because this is the sort of thing that makes me eager to learn more about a subject when I am on the receiving end of a lecture.  
Unfortunately, as yet, I myself know little more about the etale fundamental group than the definition and a bit of motivation (although I am certainly planning to learn more).  In particular, I realized that there is an obvious question I don't know how to answer.  When defining the fundamental group in an algebraic setting, we define it as an inverse limit because there is, in general, no natural analogue of a universal cover.  However, if we are looking at Spec k, there is an obvious candidate, namely the Spec of the algebraic closure of k.
Question:
Let $k$ be a field, with algebraic closure $K$.  How does the etale fundamental group of Spec $k$ compare to the automorphism group of $K$ over $k$?  Assuming they are different, what (in very general terms) are the reasons for working with one rather than the other?

REPLY [9 votes]: Also for the étale fundamental group there is in fact always some universal
cover. However, in the abstract way that Grothendieck formulated the theory of
coverings a universal cover would only exist as a pro-object. It then depends on
the precise situation (and the precise definitions) whether this pro-object can
be realised as an actual covering. Consider the case of a field $k$, the
question is whether 
$$\mathrm{Spec}(\overline{k}) \rightarrow \mathrm{Spec}(k)$$
where $k\subseteq\overline k$ is a separable closure, is considered an étale map
or not (depending on your preferences only finite separable $k$-algebras may
give you étale covers).
Here is, I think, an even more instructive example. Consider $\mathbb C^\ast$ as
the algebraic variety $\mathrm{Spec}(\mathbb C[\mathbb Z])$, where the group
algebra $\mathbb C[\mathbb Z]$ is nothing but the Laurent polynomial ring
$\mathbb C[t,t^{-1}]$. A topological universal cover is given by
$\exp\colon\mathbb C \rightarrow \mathbb C^\ast $ whereas an étale universal cover
is given by the pro-covering 
$$\{\mathrm{Spec}(\mathbb C[1/n\mathbb Z])\rightarrow
\mathrm{Spec}(\mathbb C[\mathbb Z])\}$$ 
(with transition maps induced by inclusions $1/n\mathbb Z\subseteq 1/mn\mathbb Z$). If one accepts such maps as étale covers one may take the direct limit of the rings involved and get an étale universal cover 
$$\mathrm{Spec}(\mathbb C[\mathbb Q]) \rightarrow\mathrm{Spec}(\mathbb{C}[\mathbb Z]).$$ 
This algebraic universal cover is not (analytically) isomorphic to $\exp\colon\mathbb C \rightarrow \mathbb C^\ast $; the fibres of the algebraic
cover can be identified with the profinite completion of $\mathbb Z$ while the
fibres of the analytic/topological universal have fibres $\mathbb Z$
itself. There is a map of covers 
$$\mathbb C^\ast \rightarrow \mathrm{Spec}(\mathbb C[\mathbb Q])$$ 
which on fibres correspond to the inclusion of $\mathbb Z$ into its profinite completion.