INFINITE GALOIS THEORY

Frederick Michael Butler

A THESIS

in

Mathematics

Presented to the Faculties of the University of Pennsylvania in PartialFulfillment of the Requirements for the Degree of Master of Arts

2001

Supervisor of Thesis

Graduate Group Chairperson

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1 Introduction

Nowhere else in the realm of basic abstract algebra does one see such an elegantinteraction of topics as in the subject of Galois theory. It brings the subject offield extensions of finite degree together with the subject of finite groups, givinga bijective correspondence between intermediate fields of a Galois extension andsubgroups of the Galois group of this extension.

In this paper we will briefly discuss this correspondence in the case of a finiteextension of fields before moving on to an extensive discussion of the correspondencein the infinite case. This discussion will begin with basic definitions and theoremsinvolving projective families and projective limits. We will then move on to profinitegroups, which are the projective limit of a specific kind of projective family, and seehow they relate to Galois groups of infinite Galois extensions. Next we discuss theKrull topology, a topology that the Galois group of an infinite Galois extensionpossesses, and see its use in extending the classical fundamental theorem of Galoistheory to infinite extensions. We are then finally prepared to state and prove thefundamental theorem extended to the infinite case. At this point in the paper weembark upon a brief discussion of absolute Galois groups and we state (withoutproof) a theorem of Artin and Schreier, which demonstrates that there are manysituations in which this infinite theory applies. We finish the paper by consideringmany examples of infinite Galois extensions, applying all that we have discussed inthe rest of the paper.

An attempt was made to be a thorough as possible in development of the theory,omitting proofs only when they are technical in nature or too long or difficult for thisparticular paper. This paper only covers the basic theory, however, and numerousreferences are given for further reading.

2 A Brief Review of Classical Galois Theory

In this section we will recall the basic definitions and theorems that compriseclassical Galois theory, then observe via an example that the fundamental theoremcannot carry over to the case of infinite extensions without change. For an extensivediscussion of classical Galois theory see [1] Chapters 13 and 14, or [5] Chapters Vand VI. Let us begin with some definitions.

We shall say that an algebraic field extension K/F is separable if for every α ∈ K,the minimal polynomial mα,F (x) of α over F has no repeated roots in a splittingfield of mα,F (x). An element α ∈ K whose minimal polynomial mα,F (x) over Fhas no repeated roots will also be called separable. The extension K/F is normal ifevery irreducible polynomial f(x) ∈ F [x] which has one root in K splits into linear

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factors over K. The following theorem provides several useful equivalent definitionsof a normal extension.

Theorem 2.1. Let K/F be an algebraic field extension, and let F denote the alge-braic closure of F (so K ⊆ F ). Then the following are equivalent:

(1) K/F is a normal extension;(2) K is the splitting field of a family of polynomials {fi(x)}i∈I , where I is an

arbitrary index set and each fi(x) ∈ F [x];(3) Every embedding σ : K ↪→ F which fixes F induces an automorphism of K.

A proof of Theorem 2.1 can be found in [5], Chapter V §3. Note that the theoremdoes not make the assumption that the extension K/F is finite, just algebraic. Thuswe will be able to make use Theorem 1.1 when we are considering the case of infinitealgebraic extensions as well. Now we can define a Galois extension K/F as one whichis normal and separable. The Galois group of the field extension K/F is the groupof automorphisms of K which fix F , and is denoted Gal(K/F ). If H ≤ Gal(K/F )for some Galois field extension K/F , we call the subfield K ⊇ E ⊇ F of elements ofK fixed by H the fixed field of H, and we denote it F(H). The next theorem givesseveral equivalent definitions of a Galois field extension.

Theorem 2.2. Let K/F be an algebraic field extension. Then the following areequivalent:

(1) K/F is a Galois extension;(2) F(Gal(K/F )) = F ;If it is also the case that K/F is a finite extension, then (1) and (2) are equivalent

to the following:(3) |Gal(K/F )| = [K : F ].

A proof of Theorem 2.2 is in [5] Chapter VI §1. Note that again we made noassumption that the extension is finite in (1) and (2) of Theorem 2.2. We state onemore result which will prove useful in later sections of this paper.

Theorem 2.3. (Isomorphism Extension Theorem) Let σ : E → E ′ be a fieldisomorphism, S = {fi(x)}i∈I a set of polynomials in E[x], and let S ′ = {σ(fi(x))} ⊆E ′[x] be the image of S under σ. Let K be the splitting field for S over E, K ′ thesplitting field of S ′ over E ′. Then there is an isomorphism τ : K → K ′ such thatτ |E = σ. Furthermore, if α ∈ K and β is a root of σ(mα,E(x)) then τ can be chosensuch that τ(α) = β.

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For a proof of Theorem 2.3 see [7] Chapter 1 §3. Let us quickly note the situationin which we will use Theorem 2.3 in this paper. We will have E = E ′ and K = K ′,and E is a Galois extension of another field F , σ ∈ Gal(E/F ). Then we will have Kis a Galois extension of F with K ⊇ E ⊇ F (hence K/E will be Galois as well), andwe will seek a τ ∈ Gal(K/F ) which extends σ to all of K. Theorem 2.3 guaranteesthat such a τ exists. We are now ready to state the main result of classical finiteGalois theory.

Theorem 2.4. (Fundamental Theorem of Classical Galois Theory) Let K/Fbe a finite Galois extension, and let G = Gal(K/F ). Then there is a bijectionbetween the set of fields K ⊇ E ⊇ F and the set of groups 1 ≤ H ≤ G. The mapsare E 7→ Gal(K/E) and H 7→ F(H). We also have the following:

(1) If E1, E2 ⊆ K correspond to H1, H2 ≤ G respectively, then E1 ⊆ E2 iffH2 ≤ H1;

(2) If E1, E2 ⊆ K correspond to H1, H2 ≤ G respectively, then E1 ∩ E2 corre-sponds to 〈H1, H2〉 and E1E2 corresponds to H1 ∩H2.

If E = F(H) where H ≤ G then we also have:(3) [K : E] = |H| and [E : F ] = |G : H|;(4) K/E is always Galois and Gal(K/E) = H;(5) E is Galois over F iff H / G, in which case Gal(E/F ) ' G/H.

Theorem 2.4 is proved in [1] Chapter 14 §2. One sees that Theorem 2.4 provesuseful in many situations, since it allows us to find out information about the inter-mediate fields of a Galois extension from the subgroups of the Galois group of theextension, and vice versa. This translation of a given problem into a related one isa pervasive technique in mathematics.

Because the fundamental theorem often proves useful for just the reason statedabove, we would like to extend it to the case of infinite Galois extensions. Luckilyour definition of a Galois extension carries over without change from the finite caseto the case of infinite algebraic extensions. Thus we will agree to say that an infinitealgebraic field extension is Galois if it is normal and separable. Unfortunately wequickly learn that the fundamental theorem as it stands does not hold for infinitealgebraic extensions. We can see this fact in the example that follows.

Example 2.5. Let S = {√p ∈ N | p prime}, K = Q(S), and consider the fieldextension K/Q. K is the splitting field of {x2−p | p ∈ N, p prime} so K/Q is normalby Theorem 2.1, and each polynomial x2 − p is separable hence K/Q is separable.Thus by our definition, K/Q is an infinite Galois extension. Any σ ∈ Gal(K/Q)must send

√p to a root of its minimal polynomial over Q (which is x2−p), hence to√

p or −√p, so we see that Gal(K/Q) is an infinite elementary abelian two group.That is, Gal(K/Q) ' ∏∞

i=1 Z/2Z. However Gal(K/Q) can also be thought of as aninfinite dimensional vector space V over the finite field F = Z/2Z. We see that the

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dual vector space V ∗ = {φ : V → F | φ is a linear transformation} is uncountable,so {ker φ | φ ∈ V ∗} is also uncountable. We see this second fact because for anyv ∈ V and φ ∈ V ∗, φ(v) = 0 or φ(v) = 1, hence ker φ tells us what the value of φis on all v ∈ V ; namely, φ(v) = 0 if v ∈ ker φ and φ(v) = 1 if v ∈ V − ker φ. Sincefor each φ, V/ ker φ ' F2, we see that each ker φ ≤ G is a subgroup of index 2, andthere are uncountably many such subgroups. However, there are only countablymany quadratic extensions of Q ({Q(

√q) | q ∈ Q, q is not a perfect square} is a

complete list), hence there is no longer a bijection between intermediate fields andsubgroups of the Galois group. This paper will explain what sort of correspondenceexists for infinite Galois extensions.

3 Projective Families and Limits

In this section we begin to lay the foundation of infinite Galois theory with theideas of a projective family and a projective limit. We will then see that the Galoisgroup of an infinite Galois extension is actually the projective limit of a specificprojective family. Let us begin with a basic definition.

Definition 3.1. Let A be an index set, {Ga}a∈A a family of groups. Assume thatA is partially ordered (by ≤) and that for all a, b ∈ A, there exists c ∈ A such thata ≤ c and b ≤ c. Assume for all a, b ∈ A with a ≤ b, we have a group homomorphismφb,a : Gb → Ga, consistent in that if a ≤ b ≤ c, then φb,a◦φc,b = φc,a. Also φa,a = 1Ga

(that is, the identity in Ga) for all a ∈ A. We say that {Ga, φb,a} is a projectivefamily of groups.

Example 3.2. Let A = N where m ≤ n iff m|n,Gn = Z/nZ, and φn,m : Z/nZ →Z/mZ is the trivial map if m 6= n, and φn,n = idGn .

Example 3.3. Again A = N and m ≤ n iff m|n,Gn = Z/nZ, and φn,m : Z/nZ →Z/mZ is the natural map a + nZ 7→ a + mZ.

Example 3.4. Let G be any group, A the family of all normal subgroups of G offinite index. For N ∈ A, let GN = G/N . Say N1 ≤ N2 iff N2 ⊆ N1; we see that ifN1, N2 ∈ A then N1 ∩N2 ∈ A, N1 ∩N2 ⊆ N1, N1 ∩N2 ⊆ N2, so N1 ∩N2 ≥ N1 andN1 ∩N2 ≥ N2. Finally let φN2,N1 : GN2 → GN1 be the natural map gN2 7→ gN1.

Note that while Examples 3.2 and 3.3 consist of the same index set, partial order,and groups, the maps are different. Thus it is important to specify the maps as wellas the other information about a projective family. Also note Example 3.4 well asit will become a central focus in our discussion of Galois theory.

One familiar with category theory (see [1] Appendix II for a brief overview) isacquainted with the idea of a universal object, which we seek to define next for aprojective family.

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Definition 3.5. A projective limit of the projective family {Ga, φb,a} is a group Gtogether with a collection of homomorphisms φa : G → Ga such that the followinghold:

(1) If a ≤ b, then φb,a ◦ φb = φa;(2) Given any group H and a collection of homomorphisms ρa : H → Ga satis-

fying (1) above, there exists a unique group homomorphism χ : H → G such thatφa ◦ χ = ρa for all a ∈ A.

A projective limit of {Ga, φb,a} is denoted lim←−a∈AGa or sometimes lim←−Ga, where

the set A and the maps φb,a are understood from context. For the following exam-ples the corresponding projective families are those in Examples 3.2, 3.3, and 3.4respectively.

Example 3.6. Here lim←−Z/nZ =∏

n∈N Z/nZ, where ψm :∏

n∈N Z/nZ → Z/mZ isjust the natural projection map.

Example 3.7. In this case lim←−Z/nZ =∏

p∈N, p prime Zp, the direct product of the

p-adic integers over all primes p. This group is often denoted Z; we will see muchmore of it later.

Example 3.8. Here since we did not specify G, we cannot really describe lim←−GN

except to say it is called the profinite completion of G. We see that Example 3.7 isa specific example of a profinite completion.

One of course hopes that such objects as projective limits of groups exist, andare at least up to isomorphism unique; the following proposition establishes both ofthese facts.

Proposition 3.9. For a projective family {Ga, φb,a} of groups, a projective limit{G, φa} exists; and if {H, ρa} is another projective limit, then the map χ : H → Gas in Definition 3.5 (2) is an isomorphism.

Proof. Define G ≤ ∏a∈A Ga by letting (ga)a∈A ∈ G iff for all a ≤ b one has φb,a(gb) =

ga. Clearly G 6= ∅; it contains (1a)a∈A. Let φa : G → Ga be the projection (that is,the ordinary projection map πa :

∏a∈A Ga → Ga restricted to the set G), which is

clearly a homomorphism. Then we see that φb,a ◦ φb = φa. This proves condition(1) of Definition 3.5.

Now suppose {H, ρa} also satisfies for a ≤ b, that φb,a ◦ ρb = ρa, where theρa are also group homomorphisms. Define a map χ : H → ∏

a∈A Ga by settingχ(h) = (ρa(h))a∈A. Then we clearly have πa(χ(h)) = πa((ρa(h))a∈A) = ρa(h), soπa ◦ χ = ρa for all a ∈ A. Also χ is clearly a homomorphism since each of the ρa

are. If a ≤ b then πa ◦ χ = ρa = φb,a ◦ ρb = φb,a ◦ (πb ◦ χ), so for any h ∈ H, χ(h)satisfies πa(χ(h)) = φb,a(πb(χ(h))). This implies that χ(h) ∈ G by our definitionof G, so χ : H → G. Since φa = πa|G, we see also that χ satisfies φa ◦ χ = ρa.

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Finally, if χ′ : H → G is another homomorphism satisfying φa ◦ χ′ = ρa for allh ∈ H and a ∈ A, then the ath entry of χ′(h) must be ρa(h) for all h ∈ H anda ∈ A, which forces χ(h) = χ′(h) for all h ∈ H. Hence χ = χ′, so χ is unique.This proves condition (2) of Definition 3.5, so G is a projective limit of the givenprojective family.

Next if {G, ψa} and {H, ρa} are both projective limits, we know that there existsa unique map χ1 : H → G such that ψa ◦ χ1 = ρa, and there exists a unique mapχ2 : G → H such that ψa = ρa ◦ χ2 for all a ∈ A. Thus ρa = ρa ◦ χ2 ◦ χ1, andψa = ψa ◦ χ1 ◦ χ2 for all a ∈ A. The universal property implies that there is onlyone map σ : G → G such that ψa = ψa ◦ σ, and only one map τ : H → H suchthat ρa = ρa ◦ τ . However, 1G satisfies ψa = ψa ◦ 1G, and 1H satisfies ρa = ρa ◦ 1H .Thus we have σ = 1G and τ = 1H , so we have χ1 ◦ χ2 = 1G and χ2 ◦ χ1 = 1H , andχ1 : H → G is an isomorphism with χ−1

1 = χ2.

Now that the notion of a projective limit has been established, the Galois groupof an infinite algebraic extension can be explored. From this point on in the paper,let us fix the following notation. Let K/F be an infinite Galois extension, G =Gal(K/F ), N = {H | H = Gal(K/E) for some E ∈ I} where I = {E | K ⊇ E ⊇F, E/F finite Galois}. If H ≤ G, we denote the fixed field of H by F(H) as we doin the finite case, and for σ ∈ G we will use F(σ) as an abbreviation for F(〈σ〉).

Lemma 3.10. Let K/F be an infinite Galois extension, G = Gal(K/F ), and letH ∈ N , so H = Gal(K/E) for some E ∈ I. Then H / G, and Gal(E/F ) ' G/H.

Proof. Since E/F is a normal extension (because it is Galois), the map θ : G →Gal(E/F ) given by θ(σ) = σ|E is a well defined map to Gal(E/F ) (see Theorem 2.1,equivalence (3)). Given τ ∈ Gal(E/F ), τ can be extended to a τ ′ ∈ Gal(K/F ) (soτ ′|E = τ) by Theorem 2.3; thus θ is onto. Finally, ker(θ) = Gal(K/E) = H, so H/Gand by the first isomorphism theorem for groups we have Gal(E/F ) ' G/H.

Now let us discuss the partial ordering that can be placed on I. We shall say thatE1 ≤ E2 for two fields E1, E2 ∈ I if E1 ⊆ E2. We see that {Gal(E/F ), φE2,E1}E∈Iforms a projective family with maps φE2,E1 : Gal(E2/F ) → Gal(E1/F ) for E1 ≤ E2

defined by φE2,E1(σ) = σ|E1 . φE2,E1 is a well defined homomorphism by Theorem2.1, and clearly satisfies the compatibility condition.

In light of Lemma 3.10 we see that we can define another projective family asfollows. Define a partial ordering of N by declaring H1 ≤ H2 if H2 ⊆ H1. In thiscase our projective family will be {G/H, φH2,H1}H∈N where φH2,H1 : G/H2 → G/H1

is given by φH2,H1(σH2) = σH1. We know that for H ∈ N , H /G so G/H is a group,and if H1 ≤ H2 (so H2 ⊆ H1) the map φH2,H1 is a well defined homomorphism, alsoclearly compatible.

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However, if H ∈ N then H = Gal(K/E) for some E ∈ I, and by Lemma3.10 we know that then Gal(E/F ) ' G/H. It is also clear that if H1, H2 ∈ N ,hence H1 = Gal(K/E1) and H2 = Gal(K/E2) for E1, E2 ∈ I, then H1 ≤ H2 inN iff E1 ≤ E2 in I. Finally, in this situation we see that the maps φH2,H1 andφE2,E1 correspond exactly, thus the projective families {Gal(E/F ), φE2,E1}E∈I and{G/H, φH2,H1}H∈N are the same. We will make use of this fact, using the descriptionthat is most convenient for our purposes.

We note here that by considering the projective family {G/H, φH2,H1}H∈N , weclearly have a collection of homomorphisms φH : G → G/H which are compatiblewith the maps φH2,H1 . Namely, φH(σ) = σH. The universal property of the projec-tive limit implies that there exists a unique homomorphism χ′ : G → lim←−H∈N G/H;

since lim←−H∈N G/H ' lim←−E∈I Gal(E/F ), we also have a homomorphism χ : G →lim←−E∈I Gal(E/F ). We shall see momentarily that this homomorphism is in fact anisomorphism. We need a lemma first.

Lemma 3.11. Let α1, α2, . . . , αn ∈ K. Then there is an E ∈ I such thatα1, α2, . . . , αn ∈ E.

Proof. Let K ⊇ E ⊇ F be the splitting field of {mαi,F (x)}ni=1 over F . Clearly E is

normal over F, and since K ⊇ E and K/F is separable, E/F is separable, so E/Fis Galois. Finally, [E : F ] ≤ ∏n

i=1 deg(mαi,F (x))! < ∞, so E ∈ I.

Theorem 3.12. Let K/F be an infinite algebraic extension, with notation the sameas above. Then the homomorphism χ : G → lim←−E∈I Gal(E/F ) is an isomorphism.

Proof. First note that for each Gal(E/F ), E/F is a finite Galois extension. Hencethe uniqueness of χ implies that that χ is the map σ 7→ (σ|E)E∈I since this map isa compatible homomorphism. χ is injective, because χ(σ) = (1E)E∈I implies thatσ|E = 1E for every finite Galois extension E/F . However, since K =

⋃E∈I E, one

has that σ = 1K . Finally, χ is onto. Let (σE)E∈I ∈ lim←−E∈I Gal(E/F ). Define σas follows. Let α ∈ K, so there exists some E ∈ I such that α ∈ E by Lemma3.11. Then let σ(α) = σE(α) for such an E as above. In this way one defines σ onall of K, and thus obtains σ ∈ G. Since E1 ≤ E2 iff Gal(E1/F ) ≤ Gal(E2/F ), ifα ∈ E1 ≤ E2 then σE1(α) = σE2(α), so σ is well defined. Then χ(σ) = (σE)E∈I , soχ is onto and hence χ is an isomorphism.

We see that the Galois group of an infinite algebraic extension K/F is in asense “pasted” together from all of the Galois groups of the Galois extensions ofF contained in K in a way that agrees on the overlap. Theorem 3.12 tells usformally that a Galois group of an infinite algebraic extension is a projective limitof a projective family of finite groups. This important notion will be considered inthe next section.

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4 Profinite Groups

In this section we explore profinite groups. We shall see that a profinite grouppossesses a natural topology under which it forms a topological group (that is,multiplication and inversion are continuous maps), and we will look at the propertiesof such a group as a topological space.

Definition 4.1. A group G is a profinite group if G = lim←−Ga for a projective family{Ga, φb,a}, where each Ga is a finite group.

We see from Theorem 3.12 that Galois groups of infinite algebraic extensions arein fact profinite groups. They will of course provide the most important examplefor us.

Now let us consider the natural topology on a profinite group. Let G = lim←−Ga

be a profinite group, so G is the projective limit of the projective family {Ga, φb,a}of finite groups. Give each Ga the discrete topology, then give

∏a∈A Ga the product

topology, and finally give lim←−Ga ⊆ ∏a∈A Ga the subspace topology. The next

proposition provides us with some evidence that this is the “right” topology to givea profinite group.

Proposition 4.2. Under the topology described above, a profinite group G is atopological group. That is, the operations p : G×G → G such that p(g, h) = gh andi : G → G such that i(g) = g−1 are continuous in this topology.

Proof. The topology we are giving to G = lim←−Ga is none other than the producttopology by definition (where each Ga has the discrete topology), which is generatedby the sub-basis

⋃a∈A{π−1

a ({ga}) | ga ∈ Ga}. Thus it suffices to check that theinverse image of one of these sub-basic open sets under the maps p and i is open,which we do now. Let π−1

a ({ga}) be such an open set. Then it is easy to checkthat p−1(π−1

a ({ga})) =⋃

h∈Gaπ−1

a ({gah})× π−1a ({h−1}) which is clearly open in the

product topology on G × G. Similarly, it is easy to see that i−1(π−1a ({ga})) =

π−1a ({g−1

a }), also clearly open in the profinite group topology on G. Thus G formsa topological group under this topology.

Now we need one more definition, which might look strangely familiar at first.

Definition 4.3. Let A be an index set, {Xa}a∈A a family of topological spaces.Assume that A is partially ordered (by ≤) and for all a, b ∈ A with a ≤ b thereexists a continuous map φb,a : Xb → Xa consistent in that if a ≤ b ≤ c thenφb,a ◦ φc,b = φc,a, and φa,a = 1Xa for all a ∈ A. We say {Xa, φb,a} is a projectivefamily of topological spaces. A projective limit of such a family, denoted lim←−Xa is

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a topological space X together with a collection of continuous maps φa : X → Xa

such that:(1) If a ≤ b then φb,a ◦ φb = φa;(2) Given another topological space Y and collection of maps ρa : Y → Xa

satisfying (1) above, there exists a unique continuous map χ : Y → X such thatφa ◦ χ = ρa for all a ∈ A.

We note that in our definition of the profinite group topology, each Ga is giventhe discrete topology, which makes the maps φb,a : Gb → Ga continuous (as any mapf : X → Y where X has the discrete topology is continuous). Also, the maps φa :G → Ga (G and φa defined analogously as in Proposition 3.9) are simply projectionmaps πa :

∏a∈A Ga → Ga restricted to G, hence are also certainly continuous.

Thus we see that we are dealing with projective families and projective limits oftopological groups, which can be thought of either in the context of Definitions 3.1and 3.5 as simply groups, or in the context of Definition 4.3. By thinking of thesegroups in the second sense as purely topological spaces, we shall see that a profinitegroup under this natural topology has some perhaps surprising characteristics. Inorder to prove this fact we need the following lemma. The proof is fairly technical;it can be found in [13] Chapter 1 §1.

Lemma 4.4. Let {Ga, φb,a}a∈A be an projective family of Hausdorff topologicalspaces. Then G = lim←−Ga is closed in

∏a∈A Ga.

Theorem 4.5. Let G be a profinite group, and equip G with the topology describedabove. Then as a topological group, G is compact, Hausdorff, and totally discon-nected.

Proof. Suppose G = lim←−Ga for some projective family {Ga, φb,a}a∈A. Each Ga is afinite group with the discrete topology, hence each Ga is compact, Hausdorff, andtotally disconnected under this topology. It is an elementary fact from topology thata product of Hausdorff spaces is Hausdorff and a product of totally disconnectedspaces is totally disconnected, where the product of the spaces is given the producttopology. Thus

∏a∈A Ga is Hausdorff and totally disconnected. It is also an ele-

mentary fact from topology that a subspace of a Hausdorff space is Hausdorff and asubspace of a totally disconnected space is totally disconnected, where the subspaceis given the subspace topology. Thus lim←−Ga ⊂

∏a∈A Ga is Hausdorff and totally

disconnected. Finally, since each Ga is compact, the Tychanoff theorem (proven in[8] Chapter 5 §1) tells us that

∏a∈A Ga is compact. Then Lemma 4.4 shows that

lim←−Ga is closed in∏

a∈A Ga, hence lim←−Ga is compact being a closed subspace of acompact space.

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An immediate corollary of Theorem 4.5 is the following.

Corollary 4.6. Let G = Gal(K/F ), K/F an infinite Galois extension. Then withthe profinite group topology G is compact, Hausdorff, and totally disconnected.

While Theorem 4.5 may seem strange, it is not completely without intuition.The simplicity of the proof demonstrates this fact. The next theorem however, theproof of which can be found in [10] Chapter 1 §1, is more surprising.

Theorem 4.7. Let G be a compact, Hausdorff, totally disconnected topologicalgroup. Then G is a profinite group.

Thus we see that profinite groups and compact, Hausdorff, totally disconnectedtopological groups are in fact identical objects. There is one final result of thissection in the following theorem, whose proof can be found in [13] Chapter 3 §3.

Theorem 4.8. Let G be a profinite group. Then G is the Galois group of some fieldextension.

Thus if we collect all the results of this section we see that Galois group, profi-nite group, and compact, Hausdorff, totally disconnected topological group are allequivalent concepts. There is another way to define a topology on the Galois groupof Galois extension, which we will discuss at length in the next section. We shallsee that this topology turns out to be useful in our quest to extend the classicalfundamental theorem to the infinite case.

5 The Krull Topology

In this section we define the Krull topology on the Galois group of an infiniteGalois extension, and derive some of the properties of this topology. While at firstthe Krull topology may seem to be a more workable definition of a topology on aGalois group, we shall see that it is really nothing more than the profinite grouptopology of the previous section. Consider the following lemmas, which will beuseful later.

Lemma 5.1. Let G = Gal(K/F ) for some Galois extension K/F , and let N be asdefined in the previous section. Then

⋂N∈N N = {1}, and for all σ ∈ G we have

that⋂

N∈N σN = {σ}.

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Proof. Let τ ∈ ⋂N∈N N and let α ∈ K. Lemma 3.11 implies that there is an E ∈ I

with α ∈ E. Let H = Gal(K/E) ∈ N . Then τ ∈ H because τ ∈ ⋂N∈N N , but

then τ fixes E, so τ(α) = α. Thus τ fixes every α ∈ K, and therefore τ = 1 so⋂N∈N N = {1}.If τ ∈ ⋂

N∈N σN , then τσ−1 ∈ ⋂N∈N N , so τσ−1 = 1, and hence τ = σ.

Lemma 5.2. If N1, N2 ∈ N , then N1 ∩N2 ∈ N .

Proof. Let N1 = Gal(K/E1) and N2 = Gal(K/E2) for E1, E2 ∈ I. Because E1 andE2 are finite Galois over F, so is E1E2, so E1E2 ∈ I. However, Gal(K/E1E2) =N1 ∩ N2, because σ ∈ N1 ∩ N2 iff σ|E1 = 1E1 and σ|E2 = 1E2 iff E1, E2 ⊆ F(σ) iffE1E2 ⊆ F(σ) iff σ ∈ Gal(K/E1E2). Hence N1 ∩N2 = Gal(K/E1E2) ∈ N .

Now we are ready to take a major step toward our original goal with the followinglemma.

Lemma 5.3. Let K/F be an infinite Galois extension, G = Gal(K/F ). ThenB = {σH | σ ∈ G, H ∈ N} forms a basis for a topology on G.

Proof. Each open set is a union of cosets σH hence an arbitrary union of open sets isalso a union of such cosets, so in this topology an arbitrary union of open sets is open.G is open because G = Gal(K/F ), and F/F is a finite Galois extension of degree1. The main thing to check is that open sets are closed under finite intersections.It suffices to check this for two elements of the basis, which we do now. If τ1H1

and τ2H2 are two basis elements, let τ ∈ τ1H1 ∩ τ2H2. Then τH1 = τ1H1 andτH2 = τ2H2, so τ1H1 ∩ τ2H2 = τH1 ∩ τH2 = τ(H1 ∩H2). Lemma 5.2 implies thatH1 ∩ H2 ∈ N , hence τ(H1 ∩ H2) is open. Finally for some H ∈ N with H 6= G,choose τ1, τ2 ∈ G such that τ1H 6= τ2H (which we can do, otherwise H = G). Thenτ1H ∩ τ2H = ∅, and ∅ is open, so B is indeed the basis for a topology on G.

In light of Lemma 5.3 we can define the following.

Definition 5.4. Let K/F be an infinite Galois extension, G = Gal(K/F ). TheKrull topology on G is the topology with basis all cosets σH, where σ ∈ G, H =Gal(K/E), and E/F is a finite Galois extension.

Immediately we see something very interesting about the Krull topology. IfH ∈ N then H = Gal(K/E) and E/F is a finite Galois extension. By Lemma 3.10we know that Gal(E/F ) ' Gal(K/F )/H, so |G : H| is also finite. Thus there existsσ1, σ2, . . . σn−1 such that G = H∪σ1H∪· · ·∪σn−1H, so G−H is also a union of opensets. Therefore H is both an open set and a closed set. Thus the Krull topologyhas a basis of sets which are both closed and open, sometimes called clopen sets.

13

At this point we recall that a Galois group of an infinite algebraic extension isalso a profinite group, hence it possesses a natural topology as such a group. Wehave already seen that for G = Gal(K/F ) where K/F is infinite Galois and I asabove, the map χ : G → lim←−E∈I Gal(E/F ) given by χ(σ) = (σ|E)E∈I is a groupisomorphism. In fact, more is true.

Proposition 5.5. Keeping the notation as above, giving G the Krull topology andlim←−E∈I Gal(E/F ) the profinite group topology, the map χ : G → lim←−E∈I Gal(E/F )is a homeomorphism of topological spaces.

Proof. We already know that χ is a group isomorphism, so χ is bijective. Theopen sets in G are generated by the basis {σH | σ ∈ G,H ∈ N} and by thesub-basis

⋃E∈I{π−1

E ({σ}) | σ ∈ Gal(E/F )} in lim←−E∈I Gal(E/F ), where we use πE

to denote the ordinary projection map restricted to lim←−E∈I Gal(E/F ). First let

us check that χ is continuous. χ−1(π−1E ({σ})) = {τ ∈ G | τ |E = σ} = {τ ∈

G | τ is an extension of σ to K} =⋃

τ∈G τGal(K/E) where the union is taken overall such τ which extend σ, and which is clearly open in G by definition of the Krulltopology.

Now let us check that χ−1 is continuous, which is equivalent to checking thatχ is an open map. Let σH be a basic open set for the Krull topology on G, soσ ∈ G and H = Gal(K/E) for some E ∈ I. Then χ(σH) = {(στL)L∈I | τL|E =1L∩E} = {(τL)L∈I | σ−1τL|E = 1L∩E} = {(τL)L∈I | τL|E = σ|L∩E} = π−1

E ({σ|E})which is also open in lim←−E∈I Gal(E/F ). Thus χ : G → lim←−E∈I Gal(E/F ) is ahomeomorphism.

Proposition 5.5 trivially yields the following corollaries.

Corollary 5.6. Equipped with the Krull topology, the Galois group G of an infinitealgebraic extension forms a topological group. That is, the maps p : G × G → Gsuch that p(g, h) = gh and i : G → G such that i(g) = g−1 are continuous under theKrull topology.

Corollary 5.7. Equipped with the Krull topology, the Galois group of an infinitealgebraic extension is compact, Hausdorff, and totally disconnected.

Now we finally have all the tools necessary to discuss what sort of fundamentaltheorem of Galois theory exists for the infinite algebraic case. Let us discuss it now.

14

6 The Fundamental Theorem

In this section we will state and prove the fundamental theorem for infinite Galoistheory, and look at a familiar example to which this theorem applies. We are onthe brink of proving this generalized fundamental theorem. We will require just onemore lemma, which we state and prove now.

Lemma 6.1. Let K/F be a Galois extension, G = Gal(K/F ), H ≤ G and letH ′ = Gal(K/L) where L = F(H). Then H ′ = H where H denotes the closure of Hin the Krull topology on G.

Proof. Since every element of H fixes L by definition of L, we have H ≤ H ′. Nowtake σ ∈ G−H ′. Then there is an α ∈ L with σ(α) 6= α. Choose E ∈ I with α ∈ E(which we know we can do by Lemma 3.11) and let N = Gal(K/E). Then for anyτ ∈ N , τ(α) = α, so στ(α) = σ(α) 6= α. Hence σN is an open neighborhood of σdisjoint from H ′, so G−H ′ is open and hence H

′is closed.

Finally, we want to show that H ′ ⊆ H. Then we will have H ⊆ H ′ ⊆ H and H ′

is closed, so H ′ = H. Let σ ∈ H ′ and again choose any N ∈ N , N = Gal(K/E)with E ∈ I. Let H0 = {ρ|E | ρ ∈ H} ≤ Gal(E/F ), where Gal(E/F ) is finite.Since the fixed field of H0 is F(H) ∩ E, that is, L ∩ E, the classical fundamentaltheorem shows that H0 = Gal(E/E ∩ L). Since σ ∈ H ′, σ|L = 1L, so σ|E ∈ H0.Thus there is a ρ ∈ H with ρ|E = σ|E, and thus σ−1ρ ∈ Gal(K/E) = N soρ ∈ σN ∩H. Thus for every σ ∈ H ′ and every basic open neighborhood σN of σ,we have (σN∩H ′)−{σ} 6= ∅, so σ ∈ H. Therefore we have H ′ ⊆ H, so H ′ = H.

Now, finally, we are ready to state and prove our generalized fundamental theo-rem, valid for infinite Galois extensions.

Theorem 6.2. (Fundamental Theorem of Infinite Galois Theory) Let K bea Galois extension of F , and let G = Gal(K/F ).

(1) With the Krull topology on G the maps E 7→ Gal(K/E) and H 7→ F(H)give an inclusion-reversing correspondence between intermediate fields K ⊇ E ⊇ Fand closed subgroups H ≤ G.

(2) If E corresponds to H then the following are equivalent:(a) |G : H| < ∞;(b) [E : F ] < ∞;(c) H is open.

(3) If the conditions in (2) are satisfied, then |G : H| = [E : F ].(4) For any closed subgroup H ≤ G where H = Gal(K/E), we have H / G

iff E/F is Galois. If this is the case then there exists a group isomorphism θ :Gal(E/F ) → G/H.

15

Proof. (1) If K ⊇ E ⊇ F (not necessarily E ∈ I) then K/E is normal and separable,hence Galois. Thus E is the fixed field of Gal(K/E) by Theorem 2.2. If H ≤ G,then Lemma 6.1 shows that Gal(K/F(H)) = H. Thus we have that H = Gal(K/E)for some K ⊇ E ⊇ F iff H is closed, so the maps L 7→ Gal(K/L) and H 7→ F(H)give the desired correspondence between intermediate fields and closed subgroups.

(2) (|G : H| < ∞⇒ H is open) Let K ⊇ E ⊇ F , H = Gal(K/E), and suppose|G : H| < ∞. Then G −H is a finite union of closed cosets (because H is closed),so H is open.

(H is open ⇒ [E : F ] < ∞) If H = Gal(K/E) is open then H contains somebasic open neighborhood of 1, so N ⊆ H for some N ∈ N . If L = F(N) then E ⊆ L;L ∈ I, so [L : F ] < ∞ and [L : F ] = [L : E][E : F ] < ∞ implies [E : F ] < ∞.

([E : F ] < ∞ ⇒ |G : H| < ∞) Finally, if [E : F ] < ∞ then choose L ∈ I withE ⊆ L (which can always be done by Lemma 3.11), and let N = Gal(K/L). ThenN ≤ H since E ⊆ L, so |G : H| ≤ |G : N | < ∞.

(3) If H = Gal(K/E) is open, then by definition E/F is a finite Galois extension.By Lemma 3.10 we know that Gal(E/F ) ' G/H, and by Theorem 2.2 equivalence(3) we know that |Gal(E/F )| = [E : F ]. Thus |G : H| = [E : F ].

(4) Suppose H / G is closed in G, so H = Gal(K/E). Let α ∈ E, and let f(x)be the minimal polynomial of α over F . If β ∈ K is another root of f then there isa σ ∈ G with σ(α) = β. If τ ∈ H then τ(β) = σ−1(στσ−1(α)) = σ−1(α) = β sinceστσ−1 ∈ H. Thus β is in the fixed field of H which is E, so f splits over E. ThusE/F is normal, E/F is separable since K/F is, so E/F is Galois.

Conversely if E/F is Galois, then the map θ : G → Gal(E/F ) such that θ(σ) =σ|E is well defined (by Theorem 2.1 because in particular E/F is a normal extension),and ker θ = Gal(K/E) = H / G. θ is also surjective by Theorem 2.3 so G/H 'Gal(E/F ) and we are finished.

So it is indeed possible to extend the fundamental theorem of classical Galoistheory to infinite algebraic extensions, and the results are quite stunning in theirsimplicity. The alert reader will note that this new theorem does indeed extend ourold theorem, as the following example illustrates.

Example 6.3. If K/F is a Galois extension and [K : F ] < ∞, then the Krulltopology on Gal(K/F ) is discrete. This is because K/F is a finite Galois extension,hence Gal(K/K) = {1} is open. Thus every subgroup of Gal(K/F ) is closed,so we obtain our original bijective correspondence between intermediate fields andsubgroups.

We will look at a number of examples of extensions which do not reduce to thefinite case in a little while. First, we will convince ourselves that such examplesexist and are plentiful in number, which we do in the next section.

16

7 Absolute Galois Groups

In this section we will explore the notions of ordered fields and real closed fieldsen route to stating a result concerning the existence of infinite Galois extensions.This discussion leads to the notion of an absolute Galois group, a useful type ofGalois group for an infinite Galois extension.

Now that we have defined these Galois groups and established a fundamentaltheorem for them, an appropriate question to pose is whether this new theorem is atall useful. It could be the case that infinite Galois extensions are rarely encountered,so the study of their Galois groups would be in many ways fruitless. We will statea result due to Artin and Schreier, which tells us that infinite algebraic extensionsare actually quite common. First, a few definitions.

Definition 7.1. A field F can be ordered if there exists a set P ⊂ F (the set ofpositive elements) such that P is closed under addition and multiplication, and Fis the disjoint union of the three sets P , {0}, and −P = {−p | p ∈ P}.

Of course we call such a field F ordered because we can define a linear orderingon it by declaring for a, b ∈ F that a < b iff (b − a) ∈ P . We can also deducemore about an ordered field F . It is clear that if a < b (as defined above) thena + c < b + c for every c ∈ F and ap < bp for every p ∈ P , so a2 = (−a)2 > 0 forevery a ∈ F×. This implies that if

∑ni=1 a2

i = 0 where ai ∈ F , then each ai = 0. Inparticular, if each ai = 1 then we obtain that

∑ni=1 1 6= 0 for all values of n. Thus

we see that any ordered field must have characteristic 0. Examples of well knownordered fields are R and Q.

Definition 7.2. A field F is real closed if F is ordered (with positive element setP ), every x ∈ P has a square root in F , and every polynomial f(x) ∈ F [x] of odddegree has a root in F .

Examples of real closed fields are R and Q ∩ R. We can basically think of realclosed fields as being “almost” algebraically closed. The following theorem makesthis statement precise.

Theorem 7.3. A field F is real closed iff√−1 /∈ F (that is, the polynomial x2 + 1

has no roots in F ) and K = F (√−1) is algebraically closed.

A proof of Theorem 7.3 can be found in [3] Chapter 11 §2. We now state theresult by Artin and Schreier which lends some validation to the study of this infiniteGalois theory.

17

Theorem 7.4. Let K be an algebraically closed field, F ( K a proper subfield suchthat [K : F ] < ∞. Then F is real closed, and K = F (

√−1).

The proof of Theorem 7.4 can be found in [3] Chapter 11 §7. Let us discuss fora moment the importance that Theorem 7.4 has for us. If we pick F in the theoremto be any field and let K = F , we see that if [F : F ] = n < ∞, then F is a realclosed field, F = F (

√−1), and hence n = 2. Thus if F is not a real closed field,then it must be the case that [F : F ] = ∞. Of course, there are many fields whichfall into the category of non real closed: Q, Fp, and C(x) to name a few. We seethat this Galois theory for infinite algebraic extensions is indeed quite useful, sincethe majority of fields which we encounter are not real closed.

There is another wrinkle in the situation. We know from elementary field theorythat every algebraic extension of either a field of characteristic 0 or a finite field isseparable. Thus when our field F is in one of these categories, the field extensionF /F is separable, clearly normal since F contains all roots of every polynomialf(x) ∈ F [x], and hence is Galois. However, there are cases when F /F is not aseparable extension as we shall see in the following example, so F /F cannot beGalois.

Example 7.5. Let F = F2(t), where t is transcendental. Then√

t ∈ F because√

tis a root of the polynomial x2 − t over F , but x2 − t = (x −√t)2 is not separable.Hence F /F is not a separable extension, so it cannot be Galois.

Definition 7.6. Let K/F be a field extension. The separable closure of F in K,denoted Fsep, is {x ∈ K | x is separable over F}. When Fsep is written withoutreference to a particular extension field K of F , we will mean the separable closureof F in F .

Definition 7.7. Let K/F be a field extension. An element α ∈ K is inseparableover F if the minimal polynomial mα,F (x) of α over F has a repeated root. We sayα is purely inseparable over F if mα,F (x) has only one root, namely α. The extensionK/F is also called purely inseparable if every α ∈ K is purely inseparable as above.

Several remarks need to be made about these definitions, most of which areelementary facts from classical field and Galois theory. First of all, it is fairly routineto show that Fsep forms a field, so we will not prove it here. Second we note thatwe only encounter field extensions which are not separable when we are consideringfields of characteristic p > 0, so of course any purely inseparable extension mustbe of a field of prime characteristic. It is also not hard to show that if K/F is anextension of fields of characteristic p > 0 and α ∈ K is inseparable over F , thenthere exists a minimal n ∈ N, n > 0 and a polynomial f(x) ∈ F [x] which is separableand irreducible over F such that mα,F (x) = f(xpn

). For more details see [7] ChapterI §4. These remarks will be useful in proving the following propositions.

18

Proposition 7.8. Let K/F be algebraic, and let Fsep denote the separable closureof F in K. Then the extension K/Fsep is purely inseparable.

Proof. If α ∈ Fsep, then mα,Fsep(x) = x − α is clearly purely inseparable. Thus letα ∈ K − Fsep and consider mα,Fsep(x). From the above remarks there exists n > 0and f(x) ∈ Fsep[x] separable and irreducible such that mα,Fsep(x) = f(xpn

). Leta = αpn

. Then f(a) = 0 and since f is irreducible we have f(x) = ma,Fsep(x). Alsosince f is separable we have a ∈ Fsep, so clearly mα,Fsep(x) = xpn − a = (x − α)pn

(here we are using that the characteristic is p > 0) and hence α is purely inseparableover Fsep.

Proposition 7.9. Let F be a field, F its algebraic closure, and Fsep its separableclosure in F . Then Fsep/F is a Galois extension, and Gal(F /F ) ' Gal(Fsep/F ).

Proof. Let α ∈ Fsep. Then the minimal polynomial mα,F (x) of α over F has norepeated roots, so if β is also a root of mα,F (x) then β ∈ Fsep as well. Thusmα,F (x) splits over Fsep, so Fsep/F is a normal extension. Therefore Fsep/F isGalois. Now the map θ : Gal(F /F ) → Gal(Fsep/F ) given by σ 7→ σ|Fsep is awell-defined group homomorphism, and ker θ = Gal(F /Fsep). For any α ∈ F andany τ ∈ Gal(F /Fsep), τ(α) must be a root of mα,Fsep(x). However by Proposition7.8 we know that α is purely inseparable over Fsep, so mα,Fsep(x) has only oneroot, namely α. Thus τ(α) = α for all α ∈ F , so ker θ = {1}. Finally we haveGal(F /F ) ' Gal(Fsep/F ).

Definition 7.10. The group G = Gal(Fsep/F ) is the absolute Galois group of thefield F .

More about absolute Galois groups can be found in [7] chapter IV §18. In thenext section we will look at several examples of Galois groups of infinite algebraicextensions, many of which will be absolute Galois groups of various fields.

8 Examples

In this section we are finally able to enjoy the fruits of our labors. We willconsider infinite Galois extensions of many different types of fields, and computeabsolute Galois groups whenever possible. In order to get warmed up, let us firstlook at a familiar example and apply what we have learned in this paper.

19

Example 8.1. Let S = {√p | p ∈ N and p is prime} and K = Q(S). From theintroduction it is known that K/Q is Galois, and Gal(K/Q) ' ∏∞

i=1 Z/2Z. Let usconvince ourselves that this is really the projective limit over Gal(E/F ), E/F finiteGalois extensions. We clearly have that K is generated by the set

⋃p primeQ(

√p).

Any intermediate field E, K ⊇ E ⊇ Q can be written as E = Q(√

p1, . . . ,√

pn) fordistinct primes p1, . . . , pn ∈ N. Thus if we order these intermediate fields by ⊆ andE1 ≤ E2, then E1 = Q(

√p1, . . . ,

√pn) and E2 = Q(

√p1, . . . ,

√pn,√

q1, . . . ,√

qm)for distinct primes p1, . . . , pn, q1, . . . , qm ∈ N. Then Gal(E1/Q) is a quotient ofGal(E2/Q), via the map φ : Gal(E2/Q) → Gal(E1/Q) given by φ(σ) = σ|E1 . Fromthese facts it is easy to see that Gal(K/Q) = lim←−p prime

Gal(Q(√

p)) ' ∏∞i=1 Z/2Z

as before. Note that the fact that Gal(K/Q) has uncountably many subgroups ofindex 2, while Q only has countably many quadratic extensions, does not contradictthe fundamental theorem, as “most” of these subgroups of index 2 are not closed inthe Krull topology.

Example 8.2. Consider the absolute Galois group of Fp for any prime p. It isknown from classical Galois theory that any algebraic extension of Fp is separable,hence in particular Fp is a separable extension of Fp. Thus the absolute Galoisgroup of Fp is actually Gal(Fp/Fp). It is also known from classical Galois theorythat for any n ∈ N, Fp has a unique Galois extension of degree n (which we call Fpn)with Galois group Gal(Fpn/Fp) ' Z/nZ, and also that Fpm ⊆ Fpn iff m|n. Thuswe let A = N, say m ≤ n iff m|n, and let φn,m : Gal(Fpn/Fp) → Gal(Fpm/Fp) berestriction. That is, φn,m : Z/nZ→ Z/mZ sends a mod n to a mod m, and hence

Gal(Fp/Fp) = lim←−Z/nZ =∏

p∈N, p is prime Zp, also known as Z.

The next example makes use of the Kronecker-Weber Theorem, the proof ofwhich is vastly outside the scope of this paper. A proof can be found in [6] ChapterX §3. Recall that an extension of the rational numbers of the form Q(ζn) for somenth root of unity ζn is called a cyclotomic extension of Q.

Theorem 8.3. Let K/Q be a finite extension. Then K/Q is abelian iff K is con-tained in a cyclotomic extension of Q.

Example 8.4. Consider the extension Qab/Q, where Qab denotes the maximalextension of Q which is abelian. The Kronecker-Weber Theorem implies that Qab =Q({ζn|n ∈ N}). Clearly Q({ζn|n ∈ N}) is the splitting field of {xn − 1|n ∈ N} overQ, a collection of separable polynomials. Hence Qab/Q is Galois. For any σ ∈ G =Gal(Qab/Q), once the value of σ(ζn) is known for all n ∈ N then σ will be completelydetermined on all of Qab. We know that σ must send a particular ζn to a root of itsminimal polynomial; that is, σ(ζn) = ζk

n for some k ∈ N such that (k, n) = 1. Forfixed n ∈ N, we know from classical Galois theory that Gal(Q(ζn)/Q) ' (Z/nZ)×,

20

the multiplicative group of units of Z/nZ. We also see that if m|n, say n = md, thenζm = ζd

n. Thus if σ(ζn) = ζkn, then σ(ζm) = σ(ζd

n) = σ(ζn)d = ζkdn = (ζd

n)k = ζkm. Now

if we think of σ as an element of lim←−n∈NGal(Q(ζn)/Q) ' lim←−n∈N(Z/nZ)× (written

as σ = (σn)n∈N) we see by the above argument that for m|n we must have σn ≡ σm

mod m. If we let Aj = (Z/jZ)×, partially ordered by m ≤ n iff m|n, then we havethe maps φn,m : (Z/nZ)× → (Z/mZ)×, where φn,m(a mod n) = a mod m. ThusGal(Qab/Q) ' lim←−n∈N(Z/nZ)×, which is isomorphic to the multiplicative group of

units Z× of Z.

It is important to point out that while the infinite Galois extension Qab/Q isunderstood completely, this is not the case for the extension Q/Qab. These aretwo very important extensions, since any understanding of them would facilitateunderstanding of the Galois extension Q/Q. In turn, any understanding of theGalois group Gal(Q/Q) would be indispensable in solving the following problem,which is the topic of much current research.

Inverse Problem of Galois Theory. Given a finite group G, does there exist afinite Galois extension K of Q such that Gal(K/Q) ' G?

More can be read about the Inverse Galois Problem in [12]. Note that a moregeneralized version of the Inverse Galois Problem can be posed with Q replacedby any field K. Now let us return to the extension Q/Qab. There is currently aconjecture as to what Gal(Q/Qab) is as a profinite group. To understand it, we firstneed the concept of a free profinite group.

Definition 8.5. Let X be a set, G the abstract free group on X. Let G = lim←−G/Hwhere the projective limit is taken over all H / G such that |G : H| < ∞ (that is,G/H is a finite group) and H contains all but a finite number of elements in thegenerating set X. Then G is called the free profinite group on X.

For a fairly complete basic discussion of free profinite groups, see [13] Chapter5. We are now ready to state the conjecture mentioned earlier, which is due toShafarevich. More can be read about it in [5] Chapter VI §14, and more extensivereferences are given there.

Conjecture 8.6. Consider the Galois extension Q/Qab. Then Gal(Q/Qab) is iso-morphic to a free profinite group on countably many generators.

Let us pause for a moment and reflect upon the significance of Conjecture 8.6.It is known from elementary group theory that any group which is generated byn elements is a quotient of the free group on n generators. This fact comes fromthe universal property of the free group; see [1] Chapter 6 §3. If we let G be theabstract free group on countably many generators, then the free profinite groupG on countably many generators is the projective limit over all finitely generated

21

quotients of G. Thus if G ' Gal(Q/Qab) then in particular for any finite groupH the projection map πH : G ³ H is a continuous map from Gal(Q/Qab) onto H(since H is one of the groups over which the projective limit is taken). Thus ker πH =π−1

H ({1}) is open in G, so ker πH = Gal(Q/E) for some finite Galois extension E

of Qab. Then we obtain that H ' G/ ker πH ' G/Gal(Q/E) ' Gal(E/Qab), hencethe Inverse Galois Problem would have an affirmative answer over Qab.

Example 8.7. Now let us consider an example which may seem less natural at first,but which finds its quite natural motivation in the realm of algebraic geometry. Letthe base field be F = C(x), define S = { n

√x | n ∈ N}, and then let K = F (S). Again

we have K/F is Galois because K is the splitting field of {yn − x | n ∈ N} over F ,hence normal, and an extension of fields of characteristic 0, hence separable. For eachfinite extension K ⊇ E ⊇ F where E is of the form E = F ( n

√x), Gal(E/F ) ' Z/nZ.

This fact comes from Kummer theory of finite Galois extensions. (See [5] Chapter VI§8.) Also if m|n, say n = md, and σ ∈ Gal(K/F ) is such that σ( n

√x) = ζk

nn√

x, then

σ( m√

x) = σ( n√

xd) = σ( n

√x)d = ζkd

nn√

xd

= ζkm

m√

x. Thus we see that Gal(K/F ) 'lim←−Z/nZ, where the indexing set N is partially ordered via m ≤ n iff m|n, and form ≤ n we define φn,m : Z/nZ → Z/mZ by φn,m(a mod n) = a mod m. Hence

again the Galois group is isomorphic to Z as in Example 8.2.

Example 8.7 also has an important geometric interpretation, which we brieflydiscuss now. The field C(x) is the function field of the complex projective lineP1(C) (from now on we will write P1 in place of P1(C)), also known as the Riemannsphere. Any algebraic extension of the field C(x) (for example, one of the formC(x)[ n

√x] as above) corresponds to a cover of P1 in the algebraic geometric sense;

we shall refer to such coverings as branched coverings. Each extension of the formC(x)[ n

√x] ' C(x)[y]/(yn − x) ' C(y) corresponds to an n-fold branched covering

of P1 by P1 via the map y 7→ yn, branched only at 0 and ∞. This branchedcovering is closely related to a covering that has no branch points, which we shallcall just a covering. That is, the map fn : P1 − {0, ∞} → P1 − {0, ∞} foreach n ∈ N is a covering map (where fn(z) = zn for z ∈ P1 − {0, ∞}), andP1−{0, ∞} is an n-sheeted covering space of P1−{0, ∞} under the map fn. Thebase space of this covering is also P1−{0, ∞}, which has as its fundamental groupπ1(P1 − {0, ∞}) = Z. See [12] Chapter 4 §1 for more on the terminology in thisparagraph. Note that each of the finite Galois groups occurring in Example 8.7 is aquotient of π1(P1 − {0, ∞}), and the profinite completion of π1(P1 − {0, ∞}) is Zas in Example 8.7. These fact are not accidental, as we shall see now.

Given a covering f : E → B, it is easy to show that the set {g : E →E | g is a homeomorphism, f ◦g = f} forms a group under composition of functions,called the group of deck transformations of the covering f : E → B and writtenDeck(f). When E is a connected topological space and Deck(f) acts transitively onsome (and hence each) fiber f−1(b) for b ∈ B, we say that f : E → B is a Galoiscovering. The next proposition ties together these definitions with Example 8.7 and

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the comments in the the previous paragraph. It is proven in [12] Chapter 4 §1, alongwith more details concerning the definitions in this paragraph.

Proposition 8.8. Let f : E → B be a Galois covering and G = Deck(f). Let e ∈ Eand b = f(e). Then there is a unique surjective homomorphism Φe : π1(B, b) ³ Gsuch that Φe([γ]) maps the endpoint of the lift beginning at the point b of γ (oftendenoted [γ]b) to the point b.

Thus we see from Proposition 8.8 that any group G which occurs as the Galoisgroup of a covering f : E → B must be a quotient of the fundamental group ofthe base space. This geometric point of view is often used in the current researchinvolving Galois theory. The following theorem, which is proved in [12] Chapter 4§2, is very important in that respect.

Theorem 8.9. (Riemann Existence Theorem) Let R = (G,P, (Cp)p∈P ) be aramification type, where G is a finite group, P is a finite set of points in P1, foreach p ∈ P , Cp is a conjugacy class of G, and |P | = r. Then there exists a finiteGalois covering of P1−P of ramification type R iff there exist generators g1, . . . , gr

of G such that g1 . . . gr = 1 and gi ∈ Cpifor 1 ≤ i ≤ r.

For more details on ramification type, see [12] Chapter 2 §2. An almost imme-diate corollary of Theorem 8.9 is the following.

Corollary 8.10. Given any finite group G, there exists a finite Galois extension Kof C(x) such that G ' Gal(K/C(x))

Corollary 8.10 is proven in the following way. The group generated by r elementsg1, . . . , gr subject to only the relation that g1 . . . gr = 1 is the same as the free groupon r− 1 generators. Thus given a finite group G, we choose r sufficiently large suchthat G can be generated by r− 1 elements, say h1, . . . , hr−1. Choose distinct pointsp1, . . . , pr ∈ P1. Now if we denote the conjugacy class in G of g ∈ G by CG(g),then Theorem 8.9 implies that there exists a finite Galois cover of ramificationtype (G,P, (Cp)p∈P ) where P = {p1, . . . , pr}, Cpi

= CG(gi) for 1 ≤ i ≤ r − 1and Cpr = CG((g1 . . . gr−1)

−1). This cover corresponds to a Galois field extensionK/C(x) with Galois group G. Thus we see from Theorem 8.10 that the InverseGalois Problem has an affirmative answer over the field C(x). Now on to a newexample.

Example 8.11. Again let the base field F = C(x), and let K = C(x) be thealgebraic closure of F . Then Gal(C(x)/C(x)) is the free profinite group on the set

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C. In fact, this result also holds if C is replaced by an arbitrary algebraically closedfield K. Neither of these facts is obvious, and more can be found about them in [13]Chapter 5 §1.

Before we consider the next example, let us establish a few facts.

Definition 8.12. Let F be a field. The ring of formal Laurent series over F ,denoted F ((x)), is the collection of all sums of the form

∑∞n=−m anxn for some

m ∈ N, where each ai ∈ F .

Addition and multiplication are defined in the ring of formal Laurent series asnatural extensions of addition and multiplication of polynomials in the field F (x)of rational functions over F . The 0 in this ring is the series 0 + 0x + 0x2 + . . . andthe 1 is the series 1 + 0x + 0x2 + . . . . It is clear that under these operations, F ((x))forms a ring. We see that a stronger fact is actually true.

Proposition 8.13. For any field F , F ((x)) is also a field.

Proposition 8.13 is proved in [3] Chapter 7 §10. Now that we know that F ((x))is a field, we can consider its absolute Galois group. In the case when F is analgebraically closed field of characteristic 0, the field F ((x)) has the following simpledescription.

Proposition 8.14. Let F be an algebraically closed field of characteristic 0. ThenF ((x)) = [F ((x))]({ n

√x | n ∈ N}).

Proposition 8.14 is another fact the proof of which is outside the scope of thispaper. See [9] Chapter IV §2 for more details and a proof. Now, finally, we havethe proper knowledge to discuss the next example.

Example 8.15. Consider the field C((x)). Since C((x)) has characteristic 0, weknow that its absolute Galois group is really just Gal(C((x))/C((x))). Proposition8.14 shows that C((x)) = [C((x))]({ n

√x | n ∈ N}). Since each n

√x has minimal

polynomial yn−x over C((x)), and each yn−x has roots n√

x, ζnn√

x, . . . ζn−1n

n√

x, wesee very similarly as in Examples 8.2 and 8.7 that Gal(C((x))/C((x))) ' lim←−Z/nZ,

again also known as Z.

There is a theorem which is the analogue of Proposition 8.14 in the case whenthe field F is algebraically closed of characteristic p > 0. It makes use of the conceptof a generalized power series, which is an expression of the form

∑i∈Q ait

i whereai ∈ F and the support of the series (that is, the set of all i such that ai 6= 0)is a well-ordered subset of Q. The set of generalized power series over a field F ,

24

under the natural operations of addition and multiplication, forms a ring. Thetheorem analogous to Proposition 8.14 states that a particular subset of the ring ofgeneralized power series over an algebraically closed field F of characteristic p > 0is the algebraic closure of F ((x)); see [4] for details.

The next example requires some basic knowledge of the p-adic numbers, anextensive discussion of which can be found in [2].

Example 8.16. The structure of Galois extensions of Qp is discussed in detail in[9], and all of the assertions in this example are proved there in Chapter IV §1 and§2. Such extensions are constructed as a tower of three extensions K ⊇ L ⊇ E ⊇Qp in the following manner. The (unramified) extension E/Qp has Galois groupGal(E/Qp) ' Z/nZ for some n ∈ N. The (tamely ramified) extension L/E hasGalois group Gal(L/E) ' Z/mZ for m such that (m, p) = 1. Finally, the (wildlyramified) extension K/L has Galois group Gal(K/L) ' P where P is a group oforder pk for some k ∈ N. The Galois group of the extension K/E is a semi-directproduct; that is, Gal(K/E) ' P o Z/mZ subject to the constraints above. Thischaracterization tells us that not every group can be realized as a Galois group overQp. For example, the alternating group A5 is not a Galois group over Qp for anyp, reasoned as follows. If it were, then it would have to be of the form discussedabove. However, the group Gal(E/Qp) is cyclic, which would imply that A5 hasa normal subgroup H such that A5/H is cyclic. But A5 is simple, so the onlysubgroup with this property is H = A5, which implies that E = Qp. Then sinceGal(K/E) = Gal(K/Qp), A5 must have a normal subgroup N which is a p-groupsuch that A5/N is cyclic of order relatively prime to p. The only normal subgroupsof A5 are A5 and {1}, and only {1} is a p-group. Clearly A5/{1} ' A5 is not cyclic,so A5 is not a Galois group over Qp. However, it is the case that every group of theform P oZ/mZ as above occurs as a Galois group over some finite cyclic extensionE of Qp. From this fact we see that many non-abelian groups occur as Galois groupsover Qp. Hence Qp is an example of a field over which there are many groups thatdo not occur as Galois groups of extensions, but there are also many groups whichdo occur. The structure of the absolute Galois group of Qp is fairly complicated. Inparticular, many non-abelian groups are quotients of Gal(Qp/Qp), while A5 is not.

9 Conclusion

In this paper we began with the basic ideas from classical Galois theory, andbuilt upon them to apply the ideas to infinite Galois field extensions. We saw thatthe classical fundamental theorem of Galois theory could not be extended withoutchange to the case of infinite Galois extensions. In fact we had to borrow from thesubject of topology in order to do so. It is my hope that the example section providedthe reader with ample evidence that Infinite Galois theory has many applications.

25

This theory, while useful in determining information about field extensions, alsoturns up often in many other branches of mathematics.

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References

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[2] Fernando Q. Gouvea. p-adic Numbers. Springer-Verlag, New York, 1997.

[3] Nathan Jacobson. Basic Algebra II. W.H. Freeman and Co., New York, 1989.

[4] Kiran S. Kedlaya. The Algebraic Closure of the Power Series Field in PositiveCharacteristic. on the web at http://front.math.ucdavis.edu/author/Kedlaya-K*, December 1999. manuscript, to appear in Proceedings of the AMS.

[5] Serge Lang. Algebra. Addison-Wesley Publishing Co., Melino Park CA, 1984.

[6] Serge Lang. Algebraic Number Theory. Springer-Verlag, New York, 1986.

[7] Patrick Morandi. Field and Galois Theory. Springer-Verlag, New York, 1996.

[8] James R. Munkres. Topology: A First Course. Prentice-Hall, Englewood CliffsNJ, 1974.

[9] Jean Pierre Serre. Local Fields. Springer-Verlag, New York, 1979.

[10] Jean Pierre Serre. Galois Cohomology. Springer-Verlag, New York, 1997.

[11] Stephen S. Shatz. Profinite Groups, Arithmetic, and Geometry. PrincetonUniversity Press, Princeton NJ, 1972.

[12] Helmut Volklein. Groups as Galois Groups: An Introduction. Cambridge Uni-versity Press, Cambridge UK, 1996.

[13] John S. Wilson. Profinite Groups. Oxford University Press, Oxford UK, 1998.