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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?

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They're the same. The algebraic closure is more or less by definition the inverse limit of all finite extensions of k, so the absolute Galois group is more or less by definition the inverse limit of the Galois groups of all finite extensions of k. –  Qiaochu Yuan Apr 18 '10 at 3:28
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Er, if k is perfect, anyway. –  Qiaochu Yuan Apr 18 '10 at 3:29
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And the pretentious but absolutely essential caveat about choice of basepoint carries over on the algebraic side to a pretentious but absolutely essential caveat about choice of algebraic closure! –  Hunter Brooks Apr 18 '10 at 3:29
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@Qiaochu: A separable closure is a direct limit, since it is a union of separable algebraic extensions of the field. The Galois correspondence is contravariant. –  S. Carnahan Apr 18 '10 at 4:04
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If you read French, you should look up Douady, Régine; Douady, Adrien Algèbre et théories galoisiennes. 1 & 2. CEDIC, Paris, 1979. 222 pp. ISBN: 2-7124-0709-1 The first sentence from the review in MR by C. Small is These two volumes provide a rich, dense, unrelentingly Bourbakian approach to both algebraic Galois theory (finite and infinite) and the analogous topological theory (coverings of Riemann surfaces). –  Chandan Singh Dalawat Apr 18 '10 at 4:20
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3 Answers

up vote 8 down vote accepted

Essentially by definition, the etale fundamental group of a field $k$ is the group of $k$-automorphisms of a separable algebraic closure $k^{\operatorname{sep}}$ over $k$. If $k$ is perfect, then this is (well-defined up to an inner automorphism) $\operatorname{Aut}(\overline{k}/k)$. On the other hand, any $k$-automorphism of $k^{\operatorname{sep}}$ extends uniquely to an automophism of the algebraic closure $\overline{k}$. Therefore in every case the etale fundamental group of $k$ is isomoprhic to $\operatorname{Aut}(\overline{k}/k)$.

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The étale fundamental group is defined using a reconstruction procedure that applies with minimal alteration to topological covering spaces. Tyler Lawson explained the correspondence in his answer to a previous question. For the case of fields, covering spaces correspond to spectra of finite étale $k$-algebras (which are isomorphic to finite products of finite separable extensions of $k$), and connected covering spaces correspond to spectra of finite separable extensions of $k$.

The category whose objects are finite étale $k$-algebras and whose morphisms are $k$-algebra maps is anti-equivalent to the category whose objects are finite sets equipped with continuous actions of a certain profinite group and whose morphisms are equivariant maps of sets. This group (the fundamental group) is uniquely defined up to isomorphism by this property (and is actually defined as the automorphism group of a certain functor). The fundamental group acts continuously on any separable closure of $k$ by field automorphisms, and in fact makes up all of its $k$-algebra automorphisms. You can check this by using the fact that each element of a separable closure lies in some finite separable extension of $k$, together with the compatibility afforded by the anti-equivalence of categories. If you pass to spectra, you get a covariant equivalence, and the maps of spectra of fields are the analogues of covering maps.

In the topological world, the reconstruction from the category of covering spaces of your space gives you a fundamental group that is usually not profinite, but if you were to restrict your view to covering maps with finite fibers, then the profinite completion of that group would appear.

It's not clear to me that this quantity of material can be reasonably covered in 5 minutes, but since you're giving a motivating discussion, I don't see a problem with using the Clark étale fundamental group.

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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.

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I strongly agree with Torsten's point. In "reasonable circumstances" in algebraic geometry (just as in "reasonable circumstances" in topology) there is a universal cover, if you just say things in the right way. Kirsten Wickelgren and I attempted to describe this in front.math.ucdavis.edu/0902.3464 (although opinions on what we did may vary). –  Ravi Vakil Apr 18 '10 at 22:19
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