The Inverse Galois Problem: The Rigidity Method

Amin SaiedDepartment of Mathematics,Imperial College London,SW7 2AZ, UK

CID:00508639

June 24, 2011

Supervisor: Professor Martin Liebeck

Abstract

The Inverse Galois Problem asks which finite groups occur as Galois groups of extensions of Q,and is still an open problem. This project explores rigidity, a powerful method used to showthat a given group G occurs as a Galois group over Q. It presents the relevant theory required tounderstand this approach, including Riemann’s Existence Theorem and Hilbert’s IrreducibilityTheorem, and applies them to obtain specific conditions on the group G in question, which, ifsatisfied say that G occurs as a Galois group over the rationals.

These ideas are then applied to a number of groups. Amongst other results it is explicitlyshown that Sn, An and PSL2(q) for q 6≡ ±1 mod 24 occur as Galois groups over Q. Theproject culminates in a final example, highlighting the power of the rigidity method, as theMonster group is realised as a Galois group over Q.

Contents

1 Introduction 21.1 Preliminary Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 A Motivating Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Riemann’s Existence Theorem 52.1 The Fundamental group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Galois Coverings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.3 Coverings of the Punctured Sphere . . . . . . . . . . . . . . . . . . . . . . . . . . 8

3 Hilbertian Fields 133.1 Regular Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133.2 Hilbertian Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143.3 Galois Groups Over Hilbertian Fields . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.3.1 Sn as a Galois group over k . . . . . . . . . . . . . . . . . . . . . . . . . . 20

4 Rigidity 214.1 Laurent Series Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214.2 Finite Extensions of Λ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224.3 Branch Points . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244.4 Rigid Ramification Types . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264.5 Structure Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

5 The Rigidity Criteria 335.1 Descent of the Base Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 335.2 Rigidity Criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

6 Applications of Rigidity 416.1 Sn and An as Galois groups over Q . . . . . . . . . . . . . . . . . . . . . . . . . . 416.2 PSL2(q) for q 6≡ ±1 mod 24 as a Galois group over Q . . . . . . . . . . . . . . . 436.3 M12 as Galois group over Q . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 466.4 The Ree Groups 2G2(q

2) as Galois groups over Qab . . . . . . . . . . . . . . . . . 476.5 Realising the Monster over Q . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

7 Conclusion 51

References 52

A GAP Code 53

B Characters for M12 55

1

Chapter 1

Introduction

By associating a certain group to a given polynomial, Evariste Galois was able to prove thatthere is no general solution by radicals to a polynomial equation of degree 5 or higher. Hisapproach provided not only a beautiful answer to this classical problem, but also offered a greatinsight into why it is possible to solve polynomial equations of degree less than 5 in general.

Galois’ idea was to associate to a polynomial p(x) a certain group Gal(p(x)) which permutesthe roots of p(x). In general, given a polynomial with rational coefficients, the Galois group isdefined on a certain field extension E/Q, where E is the splitting field of p(x). It is here thata natural question arises.

Inverse Galois Problem: Which groups can be realised as Galois groups over Q?

The problem was first approached by Hilbert, in 1892, and remains unsolved today. This projectdevelops the most successful approach to the question thus far, namely the concept of rigidity.Introduced by John Thompson in 1984 in his breakthrough paper [11], this method has alreadybeen used to realise numerous families of groups including, and perhaps most notably, all but2 of the 26 sporadic groups.

A striking feature of the material is the vast diversity of topics it draws from, including grouptheory, field theory, Galois theory, algebraic topology, Riemann surface theory and number the-ory. This project develops the relevant theorems from these areas and pulls them together toform a strict group theoretic condition, rigidity, which if satisfied in a group G, along with someother conditions, will guarantee a positive solution to the Inverse Galois Problem for G.

Some background material is set up and some basic results from Galois theory are stated.Proofs, where not provided, and a more thorough description of the background can be foundin [2] and [10].

1.1 Preliminary Results

A field extension E/K is algebraic if every element of E is algebraic over F . An algebraicfield extension is said to be normal if every irreducible polynomial f(x) ∈ K[x] which has aroot in L, has all its roots in L. An algebraic field extension E/K is separable if every x ∈ Lhas a separable minimal polynomial over K, that is, the minimal polynomial has distinct roots.A Galois extension is an algebraic field extension E/K which is normal and separable. Thedegree of an extension E/K is the dimension of E as a vector space over K and is denoted

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|E : K|. An extension is said to be finite if it has finite degree. Moreover for nested extensionsE/K, K/F , the degree of E/F obeys the tower law |E : F | = |E : K||K : F | as in [10].

Definition 1.1.1. Let E/K be a finite Galois extension. Then the Galois group is defined as

Gal(E/K) := {σ ∈ Aut(E) : σ(x) = x ∀ x ∈ K}

This next basic result is fundamental to Galois theory and will be used throughout, oftenwithout reference.

Lemma 1.1.2. Let f(x) ∈ K[x] and let E/K be a Galois extension. If α ∈ E is a root of f(x)and σ ∈ Gal(E/K), then σ(α) ∈ E is a root of f(x).

Proof. Let f(x) = anxn + · · · + a0 with ai ∈ K. Now f(α) = 0 so anα

n + · · · + a0 = 0. Nowapplying σ ∈ Gal(E/K) gives

σ(anαn + · · ·+ a0) = anσ(α)n + · · ·+ an = 0

because σ fixes K pointwise. Hence f(σ(α)) = 0

Theorem 1.1.3. (The Primitive Element Theorem)Let E/K be a finite Galois extension. Then ∃ α ∈ E such that E = K[α].

I now quote the main theorem of Galois theory, from [2].

Theorem 1.1.4. Let E/K be a finite Galois extension with Galois group G = Gal(E/K). Foran intermediate field F such that K ⊆ F ⊆ E let λ(F ) be the subgroup of G leaving F fixed.Then λ is a one-to-one map of the set of all such intermediate fields F onto the set of subgroupsof G with the following properties:

1. λ(F ) = Gal(E/F )

2. F = EGalE/F = Eλ(F ) where EH denotes the fixed field of H ≤ G in E.

3. For H ≤ G, λ(EH) = H

4. |E : F | = |λ(F )|, |F : K| = |G : λF )|

5. F is a normal extension of K if and only if λ(F ) is a normal subgroup of G. When thisis the case

Gal(F/K) = Gal(E/K)/Gal(E/F )

This correspondence between subgroups of Gal(E/K) and intermediate fields F betweenE and K is referred to as the Galois correspondence, and it too will be used throughout thisproject. I now quote another classic result from [10].

Theorem 1.1.5. (Artin’s Theorem) If G is a finite group of automorphisms of a field E andif the fixed field of G in E is K then E/K is a finite Galois extension with Gal(E/K) = G

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1.2 A Motivating Example

How does one find a group as a Galois group? In the true nature of the original idea of Galoistheory, one might think to look at polynomials f(x) ∈ Q[x]. To find the Galois group of a poly-nomial one must find its splitting field Ef , which is the smallest field extension of Q containingthe roots of f [10]. Lets consider the dihedral group of the square D8. Can it be realised as aGalois group over Q?

Consider the irreducible polynomial

f(x) = x4 − 6x2 + 2 ∈ Q[x]

In general the splitting field Ef of an irreducible polynomial of degree n can have degree upto n! over Q, and thus the Galois group is a subgroup of Sn [10]. Therefore in our case thesplitting field Ef has degree dividing 24 (by Lagrange). This is why it was a good idea to lookat a polynomial of degree 4, as D8 is the unique subgroup of S4 of order 8. Proceed by findingthe roots of f(x) by first setting y = x2 and finding the roots of y2 − 6y + 2, which are 3±

√7.

An extension of degree 2 has been created, namely Q(√

7). Now find the roots of f(x) to be

±α :=

√3 +√

7 ± β :=

√3−√

7

Hence the splitting field of f is Ef = Q(α, β). Notice that αβ =√

9− 7 =√

2 /∈ Q(α). Now

β ∈ Q(α) ⇐⇒ αβ ∈ Q(α). Therefore β /∈ Q(α). Each equation x2 = 3+√

7 and x2 =√

3−√

7produces an extension of degree 2, and so we have that Ef is a degree 8 extension of Q. Therefore|Gal(f)| = |Gal(Ef/Q)| = 8, and since D8 is the unique subgroup of S4 of order 8 we have that

Gal(x4 − 6x2 + 2) ∼= D8

So by a clever choice of polynomial one was able to realise D8 as a Galois group over Q. Thisapproach essentially tries to construct a field extension which yields the desired group as itsGalois group. However, this will become increasing more difficult as the group in question be-comes more complicated. For instance, how would one use this technique to realise the Monstergroup. The mind boggles.

This motivates the need to develop a more powerful technique. This will be rigidity. Rigiditytakes the opposite approach, that is, rather than creating a field extension which yields thedesired group G, rigidity is a condition on the group itself, which if satisfied, along with someother conditions, ensures the existence of a field extension over Q with Galois group G. Afundamental concept in this vein is Riemann’s Existence Theorem.

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Chapter 2

Riemann’s Existence Theorem

Riemann’s Existence Theorem is one of the foundations on which rigidity is built. Eventually itwill ensure the existence of a field extension of C(x), the complex function field, whose Galoisgroup is G, where G satisfies some specific conditions. It is based on coverings of the puncturedRiemann sphere, and this chapter begins by setting up the relevant background in algebraictopology (for a more thorough development see [3]). It goes on to explore Galois coverings andends with the proof of a topological version of the theorem.

2.1 The Fundamental group

Recall some basic notions from algebraic topology. Let S be a topological space. A path fromp to q is a continuous map γ : I → S with γ(0) = p, γ(1) = q. C(S, p, q) denotes the set of allpaths in S from p to q. Two paths γ0, γ1 ∈ C(S, p, q) are homotopic if ∃ a continuous mapΓ : I × I → S with Γ(0, t) = γ0(t),Γ(1, t) = γ1(t) for all t ∈ I and Γ(s, 0) = p,Γ(s, 1) = q for alls ∈ I. Homotopy of paths is an equivalence relation, and we denote the equivalence class of γby [γ]. The set of homotopy classes of closed paths (i.e. with same start point and end point)based at p in S is denoted π1(S, p).

Consider two paths g, h : I → S such that g(1) = h(0). Define the product of these twopaths by

hg : I → S t 7→{

g(2t) t ∈ [0, 12 ]h(2t− 1) t ∈ [12 , 1]

The path γ has a natural inverse under this product γinv : I → S defined by γinv(t) = γ(1− t),which traverses γ backwards. Moreover, this product induces a product of homotopy classes.We are now able to define the fundamental group.

Definition 2.1.1. The set π1(S, p) of homotopy classes based at p in S together with the aboveproduct is a group, called the fundamental group of S based at p.

The groups π1(S, p) and π1(S, q) are isomorphic if ∃ a path γ ∈ C(S, p, q). Then an isomor-phism is given by [a] 7→ [γaγinv]..

A (topological) manifold is topological space that is second countable, Hausdorff and locallyEuclidean. From now on we consider S, a manifold. A surjective map f : R → S is calleda covering if ∀ p ∈ S ∃ an open neighbourhood U such that f−1(U) is a disjoint union ofopen sets in R, each of which is mapped homeomorphically onto U by f . Such a U is called

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an admissible neighbourhood. It follows from this definition that R, the covering space, is lo-cally Euclidean. Indeed each admissible neighbourhood U is locally Eulclidean, and f−1(U) ishomeomorphic to U . Let f : R → S be a covering, and γ a path in S. Then γ is a lift of γ iff ◦ γ = γ. Recall the homotopy lifting property from [3],

Theorem 2.1.2. Let f : R → S be a covering. For each a ∈ f−1(p) ∃! lift of γ to R withstarting point a. Moreover, if two paths in S are homotopic, then their lifts are homotopic ifthey have the same start point.

Define the fiber of p as f−1(p). The fundamental group π1(S, p) acts naturally on the fiberf−1(p) thusly: [γ] ∈ π1(S, p) sends a ∈ f−1(p) to b, the endpoint of the unique lift of γ startingat a. Suppose S is connected. Then this action is transitive if and only if R is connected. Nowrecall from [3] two results on coverings

Lemma 2.1.3. Let f : R→ S be a covering and suppose S is connected. Let R1 be a connectedcomponent of R. Then the restriction of f to R1 is a covering. Moreover, if f−1(p) ⊂ R1 thenR1 = R is connected.

Lemma 2.1.4. Let f : R→ S and h : T → S be two coverings with R and T connected. Let p= f(a)=h(b) with a ∈ R and b ∈ T . Suppose [γ]a = a ⇐⇒ [γ]b = b. Then ∃ a homeomorphismα : R→ T with h ◦ α = f and α(a) = b.

Corollary 2.1.5. If π1(S, p) is trivial then each covering f : R → S with R connected is ahomeomorphism.

Proof. Let h = Id : S → S. Then f : R→ S and h : S → S are two coverings and R and S areconnected by hypothesis. Now since π1(S, p) is trivial we can use 2.1.4 to get a homeomorphismα : R→ S. But h ◦ α = f i.e. α = f so indeed f is a homeomorphism.

For a covering f : R → S the isomorphisms α : R → R are called deck transformations,and they form a group under composition, denoted Deck(f) or Deck(R/S) if f is understood[3]. The group Deck(f) acts naturally on the fiber f−1(p) thusly: for a ∈ f−1(p), α ∈ Deck(f)then α(a) ∈ f−1(p).

For a covering f : R → S, a path γ from p to q and a ∈ f−1(p) denote by γa the endpoint ofthe lift of γ with initial point a. This gives us a bijection between the fibers f−1(p) and f−1(q),and moreover, this bijection commutes with the action of Deck(f), that is α(γa) =γ α(a) [12].For γ a loop at p we have [γ] ∈ π1(S, p). As a result we have that Deck(f) commutes with theaction of π1(S, p)

Lemma 2.1.6. Let R be connected. If α ∈ Deck(f) fixes a point a ∈ R then α = Id. Moreover,if G is a subgroup of Deck(f) that acts transitively on a fiber f−1(p) then G = Deck(f).

Proof. Let γ be a path from a to b, some b ∈ R. Then γ = f ◦ γ, and γ is the lift of γ withinitial point a and end point γa = b. Now α ∈ Deck(f) fixes a by assumption, that is α(a) = a.But then α(b) = α(γa) =γ α(a) =γ a = b. Now b was arbitrary, so α = Id.

Consider again a ∈ f−1(p). For each α ∈ Deck(f) there is β ∈ G with β(α(b)) = b Butthen by above βα = Id⇒ α = β−1 ∈ G.

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2.2 Galois Coverings

If f : R→ S is a covering then the cardinality of f−1(p) is locally constant over S. Hence if S isconnected then this cardinality is constant on all S. The cardinality is then called the numberof sheets or the degree of f [3].

Definition 2.2.1. A covering f : R → S is called a Galois covering if R is connected andDeck(f) acts transitively on each fiber f−1(p), p ∈ S.

Lemma 2.2.2. Let f : R → S be a Galois covering and let H := Deck(f). The number ofsheets of f is equal to the order of H.

Proof. Exhibit a bijection from H to a general fiber. Let a ∈ f−1(p). Consider map θa : H →f−1(p) by α 7→ α(a).θa injective: α, β ∈ H. Suppose α(a) = β(a) then β−1α(a) = a. By 2.1.6 then β−1α = Id ⇒β = α therefore θa is injective.θa surjective: f is a Galois covering, so acts transitively on the each fiber. Therefore θa is clearlysurjective.

Intuitively, if the degree of f is n then each fiber has n components, and H := Deck(f)permutes them transitively among themselves.

Proposition 2.2.3. Let f : R → S be a Galois covering. For a ∈ R, p ∈ S there is a uniquesurjective homomorphism

Φa : π1(S, p)→ Deck(f)

[γ] 7→ Φa([γ])

where Φa([γ]) : [γ]a 7→ a. Here [γ]a is the endpoint of the lift of γ starting at a.

Proof. π1(S, P ) acts transitively on f−1(p) so the map Ψa : π1(S, p)→ f−1(p) by [γ] 7→ [γ]a issurjective. But composition of surjections is surjective, so θ−1a ◦Ψa : π1(S, p)→ H is surjective(as θa is a bijection). Now let Φa : [γ] 7→ ((θ−1a ◦ Ψa)[γ])−1. Since π1(S, p) commutes withthe action of H, it is trivial but tedious to check that Φa is indeed a homomorphism with

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the required property. Suppose that Φ′a is another surjective homomorphism with the sameproperties. That is Φ′a([γ]) : [γ]a 7→ a. But then

Φa([γ])−1 ◦ Φ′a([γ]) : a 7→ a

thus by 2.1.6, Φa([γ])−1 ◦ Φ′a([γ]) = Id ⇒ Φa([γ]) = Φ′a([γ]). This shows that indeed Φa isunique.

Example: The Punctured Disc

Let K(r) := {z ∈ C : 0 < |z| < r} be a disc of radius r minus the origin. It suffices to considerK = K(1) as all K(r) are homeomorphic to K. Let H := {z ∈ C : Re(z) < 0} the open left halfplane. Let i denote the fixed choice of

√−1 in C. Consider the map f∞ : H → K by z 7→ ez.

Now ez = ez′ ⇐⇒ z′ = λm(z) := z + 2πmi for m ∈ Z.

Let z ∈ H and let V = Bε(z), 0 < ε ≤ 1, the open disc of radius ε about z contained inH. Then V ∩ λm(V ) = ∅ for m ∈ Z. Define U := f∞(V ), open in K, and for each m ∈ Z themap f∞ restricted to λm(V ) is a homeomorphism onto U , thus U is an admissable neighbour-hood for q = f∞(z) and thus f∞ : H→ K is a covering.

Now by construction each λm is a deck transformation of f∞. Moreover it is clear that thegroup G = {λm : m ∈ Z} acts transitively on the fiber f−1∞ (p) hence f∞ is a Galois coveringand Deck(f) = G (by 2.1.6).

Now consider the associated map Φa : π1(K, p) → Deck(f∞). I claim it is an isomorphism.Indeed, suppose [g] ∈ ker(Φa). Lift g to a loop at a. But π1(H, a) = {1} is trivial [3]. Hencethe lift of g is homotopic to a constant path and thus g is homotopic to a constant path at pi.e. [g] = 1. So Φa is injective. It is by construction surjective and thus indeed is an isomorphism.

Define a path h : I → H by h(t) = a + 2πit. Let h = f∞ ◦ h, then h(t) = pe2πit. Now[h]a = h(1) = a+ 2πi. Thus Φa([h]) : a+ 2πi 7→ a⇒ Φa([h]) : z 7→ z − 2πi. Therefore π1(K, p)is infinite cyclic and is generated by [h].

2.3 Coverings of the Punctured Sphere

Let P1 := C∪{∞}. P1 is a topological space. We can view it as R2 ∪{ a point} and so see thatit is homeomorphic to S2 [3]. Let P be a finite subset of P1. It is our goal to show that finiteGalois coverings of P1\P correspond to finite Galois extensions of C(x).

Define the open disc D(p, r) of radius r about p by

D(p, r) =

{{z ∈ C : |z − p| < r} p ∈ C{z ∈ C : |z| > 1

r} ∪ {∞} p =∞

Let f : R → P1\P be a finite Galois covering. Let D = D(p, r) for some p ∈ P such that Ddoesn’t contain any other elements of P . Define D∗ = D \ p. Then D∗ ⊂ P1\P . We can nowdefine

κp : D∗ → K(r)

z 7→{z − p p 6=∞

1z p =∞

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Let E be a connected component of f−1(D∗), and define fE = κp ◦ f |E . This is a coveringf |E : E → K(r). Indeed the covering f restricts to a covering f−1(D∗)→ D∗. So compositionwith the homeomorphism κp gives a covering f−1(D∗) → K(r) which restricts to a coveringE → K(r) by 2.1.3. E is called a circular component of level r over p. Let 0 < r < r. Thereis a bijection between cirular components E of level r and circular components E of level r overp. Indeed, there is precisely one E ⊂ E, and f

Eis the restriction of fE to E. This allows us to

define an equivalence relation on the set of circular components over p: E ∼ E′ if E ⊂ E′ orE′ ⊂ E. The equivalence classes of ∼ are called the ideal points of R over p.

Lemma 2.3.1. Deck(f) permutes the components E of f−1(D∗) transitively. If HE is thestabiliser of E in H := Deck(f) then restriction to E gives an isomorphism HE → Deck(fE).

Proof. H acts on f−1(D∗) and hence permutes the components E. Let h ∈ H. If h maps onepoint of E to E′, some other component, then h maps all of E to E′, since components arepairwise disjoint. So if h maps some point in E to some other point in E then h ∈ HE i.e.stabilises E. Let q ∈ D∗ and consider the set FE = f−1(q) ∩ E, a fiber of the f |E . Sincef is a Galois covering, H acts transitively on the fiber f−1(q) any two points of FE can bemapped into each other by an h ∈ H ,and so this h must be in HE . So we have that HE actstransitively on FE . We have a homomorphism HE → Deck(fE),, which is injective by 2.1.6 andwhose image is a subgroup of Deck(fE) acting transitively on FE . Thus by 2.1.6, again, eachsubgroup is all of Deck(fE) and thus restriction gives the required isomorphism.

As a result, and by the example of the punctured disc, HE is cyclic. Let hE be the distin-guished generator of HE .

Let p∗ ∈ D∗, p = κp(p∗) and λ(t) = κ−1p (pe2πit), a loop in D∗ based at p∗. Let b ∈ R and

define q0 = f(b). Further, let δ be a path in P1\P joining q0 to p∗. Lift δ to δ via f with initialpoint b. Then δ has endpoint b∗ in some component E. Let λ be the lift of λ via f with initialpoint hE(b∗). Then λ is the lift of the path pe2πit via fE with initial point hE(b∗). Thus theendpoint of λ is b∗ [12].

Let h ∈ H := Deck(f) and E′ = h(E) another connected component. Then hhEh−1 = hE′

where hE is the distinguished generator of HE . Therefore the stabilisers hE form a conjugacyclass Cp of Deck(f) which depends only on p and f . Indeed h ◦ λ is the lift of λ via f with

initial point h(hE(b∗)) ∈ h(E) = E′. Now h ◦ λ has endpoint h(b∗) and hE′ maps the endpointof h ◦ λ to the initial point. So hE′(h(b∗)) = h(hE(b∗)) ⇒ h−1E h−1hE′h(b∗) = b∗ therefore by

2.1.6 is identity i.e. hhEh−1 = hE′ . If E ⊂ E then h

E= hE so indeed the class doesn’t depend

on the radius of the disc [12].

Define γ = δinvλδ be a closed path in P1\P based at q0, and the map Φb : π1(P1\P, q0) → Hsends [γ] to an element in Cp. Indeed the path (hE ◦ δ) is the lift of δ starting at hE(b). Thisends at b therefore Φb([γ]) = hE ∈ Cp.

Definition 2.3.2. Let Cp be the conjugacy class of stabilisers of the components E of f−1(D∗).Then, given f , Cp depends only on p. Let e be the common order of the elements of Cp.

Then e is equal to the degree of the covering fE : E → K(r) for any component E, and inparticular, Cp = {1} if and only if fE is a homeomorphism [12].

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Ideal Points

Let f : R → P1\P . The ideal points are the equivalence classes of ∼ defined on the circularcomponents by E ∼ E′ if E ⊂ E′ or E′ ⊂ E. So the number of ideal is equal to the numberof componants E of f−1(D∗). Now D∗ = D \ p, so the circular components E are homeo-morphic to a disc minus the center, the ideal points can be thought of as the missing centers.Since Deck(f) permutes the components transitively 2.3.1 the number of ideal points over p is≤ |Deck(f)|. Infact it is intuitively clear that the the number of ideal points is less than orequal to the number of sheets of the covering.

Let f : R → P1\P be a finite Galois covering. Let R be the disjoint union of R and allideal points over all p ∈ P . Define a topology on R by V ⊂ R is open if V ∩R is open in R andthe following holds: for each ideal point π ∈ V there is an E ∈ π with E ⊂ V . This makes Rinto a connected compact Hausdorff space [12]. Note that if π is an ideal point then for eachE ∈ π the set E ∪ {π} is an open neighbourhood of π. Moreover, each neighbourhood of πcontains such a set.

Now we can extend f : R → P1\P to a continuous surjective map f : R → P1 by settingf(π) = p where π is an ideal point over p. Further, each α ∈ Deck(f) extends uniquely to ahomeomorphism α : R→ R by f ◦ α = f .

Lemma 2.3.3. Let X be a topological space with open subsets Xj, j ∈ J , such that X =⋃j

Xj.

Then each path δ : I → X is homotopic to a product of finitely many paths δν with each δν insome Xj.

Proof. Consider the preimages of Xj under δ. They clearly cover I. I is a closed bounded subsetof R so by Hiene-Borel is compact, thus there exists a finite subcover. Choose δ−1(X1), . . . , δ

−1(Xs)with s minimal. Order the Iν such that inf(Iν) < inf(Iν+1). Then Iν ∩ Iν+1 6= ∅ by minimal-ity of s. Let tν be in this overlap. Then we have 0 = t0 < t1 < · · · < ts−1 < ts = 1 andδ([tν , tν+1]) lies entirely within Xj for some j. Thus after a suitable re-parametrisation theresult is obtained.

Let p1, . . . , pn ∈ C distinct and set S = C\{p1, . . . , pn}. It is possible to choose a pointsuch that the straight line between this point and the pi hit no other pj . Let q0 ∈ S be such apoint. Can write pi = q0 + ρie

iνi . Relabel the pi so that the vi are increasing. Here of courseνi represents the angle the line from q0 to pi makes with the real line. Now divide the complexplane up, into connected components Si, with rays, M1, . . . ,Mn emanating from q0 such thateach pi lies in Si. Let Di be a disc about pi entirely contained within Si. Define paths γistarting from q0, travelling along the straight line from q0 to pi until it hits the boundary ∂Di,traversing the boundary once clockwise, and travelling back to q0 along this same line.

10

Theorem 2.3.4. [γ1], . . . , [γn] generate π1(S, q0).

Proof. Enlarge Si to S′i = {z ∈ C : dist(a, Si) < ε} such that S′i ∩ Dj 6= ∅ for i 6= j. Thenthe S′i form an open cover of C and thus the S′i\pi form an open cover of S. Now consider ageneral loop in S based at q0. By 2.3.3 γ can be decomposed into a product of paths δν withν = 1, . . . , s such that each δν is a path in Tν := S′iν\{piν}. Let κν be the straight line path fromq0 to δν lying in Tν ∩ Tν−1. Now the path ων = κinvν+1δνκν is a loop at p0 (note, let κs+1 be theconstant path at q0). Now γ is homotopic to the product of the ων . Also, Tν is homeomorphicto the punctured disc K which has infinite cyclic fundamental group as calculated earlier, andis generated by the γi with i = iν . So each ων is homotopic to some γmi ,m ∈ Z. But thenγ is a product of these ων , so is homotopic to a product of γmiν , so indeed the γi generate thefundamental group.

I now quote a result from [12]

Theorem 2.3.5. Let S = C\{p1, . . . , pn}. Let G be a group with generators g1, . . . , gn. Then∃ a Galois covering f : R→ S, an isomorphism θ : Deck(f)→ G and a point b ∈ f−1(q0) suchthat the composition of θ with the surjection Φb : πq(S, q0) → Deck(f) from 2.2.3 maps [γi] togi for i = 1, . . . , n. Now G ∼= Deck(f).If G is finite then f : R → P1\{p1, . . . , pn,∞} is a finite Galois covering, and the associatedconjugacy classes Ci = Cpi and C∞ of G as in 2.3.1 are such that gi ∈ Ci and (g1 · · · gn)−1 ∈ C∞for i = 1, . . . , n.

The proof of this theorem is omitted due its length. It can be found in [12]. As a sketch,consider the line from q0 to pi. Let Li be the continuation of this line emanating from pi. LetQ = C\{L1, . . . , Ln}. Then, roughly speaking, R is obtained by glueing sheets Q× {g}, whichare indexed by g ∈ G, along the rays Li × g. We see that G acts naturally on the indices, andthis gives the permutation of the sheets required for the identification with Deck(f).

Lemma 2.3.6. The fundamental group π1(S, q0) where S = C\{p1, . . . , pn} is freely generatedby [γ1], . . . , [γn].

Proof. Let G be the free group generated by g1, . . . , gn. Then by 2.3.5 there exists a homomor-phism π1(S, q0)→ G sending [γi] to gi. But there is also a homomorphism from G→ π1(S, q0)sending gi to [γi]. Composition both ways is identity as it fixes either all gi which generate G, or

11

all [γi] which generate π1(S, q0). Hence G is isomorphic to π1(S, q0) and under this isomorphismgi ↔ [γi] i.e. π1(S, q0) is generated by [γ1], . . . , [γn].

It follows that since S2 minus a point is homeomorphic to C, that S2\{p1, . . . , pn+1} ishomeomorphic to S = C\{p′1, . . . , p′n} and thus is free of rank n.

The spaces S2 and S2\{p} both have trivial fundamental group, the latter is seen by its asso-ciation with R2 or C as mentioned above [3].

Our next goal is to use Theorem 2.3.4 and 2.3.5 to prove a topological version of Riemann’sexistence theorem. First, consider a finite Galois covering f : R → P1\P , where P is a finitesubset of P1. Let H = Deck(f) and as in definition 2.3.2, let Cp be the class of H associatedto p. Consider the points giving rise to non-trivial conjugacy classes P ′ := {p ∈ P : Cp 6= {1}}.Thus we have a triple (H,P ′, (Cp)p∈P ′). We say two such triples (H,P,C), (H ′, P ′,C′) areequivalent if P = P ′ and there is an isomorphism from H to H ′ mapping Cp to C ′p.This definesan equivalence relation on such triples.

Definition 2.3.7. The (topological) ramification type of the covering f is the equivalenceclass of the triple (H,P ′, (Cp)p∈P ′) denoted [H,P ′, (Cp)p∈P ′ ].

The ramification type arises from a well defined triple depending only on f : R → P1\P .Indeed, H = Deck(f) and P ′ ⊂ P such that Cp 6= {1} ⇐⇒ p ∈ P ′. Given such a Galoiscovering another is obtained from a change of coordinates in the following way. Consider ahomeomorphism,

g : P1 −→ P1

sending z ∈ C to z− p0 if p0 ∈ C, and ∞ 7→ ∞. Or sending z ∈ C\{0} to 1z ,∞ 7→ 0 and 0 7→ ∞

Then f := g ◦ f : R → P1\g(P ) is another finite Galois covering with Deck(f) = Deck(f).Hence the ramification type of f is [H, g(P ′), (Cg−1p)p∈g(P ′)].

Theorem 2.3.8. (Topological) Riemann’s Existence TheoremLet T = [G,P, (Kp)p∈P ] be a ramification type. Let r = |P | and label the elements of P asp1, . . . , pr. Then there exists a finite Galois covering of P1\P of ramification type T if and onlyif there exist generators g1, . . . , gr of G with g1 · · · gr = 1 and gi ∈ Kpi.

Proof. We can ensure, possibly by a change of coordinates, that ∞ ∈ P , and can label theelements such that pr =∞.Firstly, suppose f : R → P1\P is a finite Galois covering of ramification type T . Chooseq0 ∈ P1\P and paths γ1, . . . , γr−1 as for Theorem 2.3.4. Let γr = (γ1 · γr−1)inv. Consider thefiber f−1(q0), and fix some point b in this fiber. Then Φb : π1(P1\P, q0) → G is a surjectivehomomorphism, as in 2.2.3. Let gi := Φb([γi]). Then by Theorem 2.3.4 the gi generate G andby Theorem 2.3.5, gi ∈ Cpi = Ci. Finally, γ1 · · · γr = 1 by the definition of γr. Thus g1 · · · gr = 1as Φb is a homomorphism.Conversely, suppose ∃ generators g1, . . . , gr of G with g1 · · · gr = 1 and gi ∈ Kpi . Apply Theorem2.3.5 which gives the existence of a finite Galois covering f : R → P1\P and an isomorphismbetween Deck(f) and G such that gi lies in the class of Cpi associated to pi for i = 1, . . . , r− 1.gr = (g1 · · · gr−1)−1 ∈ C∞ = Cpr . Hence Deck(f) ∼= G, and Cp = Kp ∀ p ∈ P , so indeed f hasramification type T as needed.

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Chapter 3

Hilbertian Fields

It was Hilbert who first systematically approached the Inverse Galois Problem, and his ideasstill form a crucial role in the functioning of the rigidity method. His main idea allows a Galoisgroup over Q(x), the rational function field, to be realised as a Galois group over Q. Thischapter explores the theory of so called hilbertian fields and concludes with the statement ofHilbert’s Irreducibility Thorem.

3.1 Regular Extensions

Given a FG-extension k′/k and x = (x1, . . . , xm) with the xi’s algebraically independent, whatcan we say about k′(x)/k(x)?

Lemma 3.1.1. Given FG-extension k′/k and x as above, then k′(x)/k(x) is a FG-extensionand Gal(k′(x)/k(x)) ∼= Gal(k′/k).

Proof. Extend action of G on k′ to k′(x) by g ∈ G then g(xi) = xi for i = 1, . . .m. G clearlyfixes k(x) and hence k′(x)/k(x) is a FG-extension with Galois group G.

Remark 3.1.2. By the Galois correspondence we see that any intermediate field between k′(x)and k(x) is of the form k′′(x) and [k′′(x) : k(x)] = [k′′ : k]

Definition 3.1.3. L is said to be regular over k if k is algebraically closed in L. Say L/k isregular or, sometimes, geometric.

Lemma 3.1.4. Let k be an algebraic closure of k. If f(x, y) ∈ k(x)[y] is an irreducible polyno-

mial over k(x), and if K = k(x)[y]/

(f) is the corresponding field extension of k(x), then

K is regular over k ⇐⇒ f is irreducible over k(x)

Proof.(⇐) Let k be the algebraic closure of k in K. In other words, K is regular over k iff k = k.

Let f an irreducible poly over k(x) and let α be a root of f .

We have a homomorphism

ϕ : k(x)[y] −→ k(x)(α)

h(y) 7→ h(α)

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where k(x)(α) ∼= k(x)[y]/ker(ϕ) = k(x)[y]

/(f) = K

Then α satisfies a polynomial f(y) ∈ k(x)[y] of degree [K : k(x)] and f divides f . Indeed

[K : k(x)] =[K : k(x)]

[k(x) : k(x)]≤ [K : k(x)] (by tower law)

Hence deg(f) ≤ deg(f)

Therefore if k 6= k then f not irreducible over k(x)[y] and hence not irreducible over k(x)[y].

(⇒) Assume K is regular over k, (then k = k). Let k′ be a FG-extension of k and defineK ′ to be the composite of K and k′(x) in an algebraic closure of k(x).

Then by 3.1.1 k′(x)/k(x) is a FG-extension. And by Remark 3.1.2. K ∩ k′(x) is of the formk′′(x) for k′′ an intermediate field between k′ and k.

Now k′′ ⊂ k = k hence k′′ = k i.e. K ∩ k′(x) = k(x). Since k′(x)/k(x) is Galois, it fol-lows that [K ′ : k′(x)] = [K : k(x)]

3.2 Hilbertian Fields

Definition 3.2.1. A field k is called hilbertian if for each irreducible polynomial f(x, y) ∈k[x, y] with deg(f) ≥ 1 there are infinitely many b ∈ k such that f(b, y) ∈ k[y] is irreducible.

The polynomial fb(y) = f(b, y) is called a specialisation.

Remark 3.2.2. Let k be a field. Let b be algebraic over k. That is, b satisfies

anbn + an−1b

n−1 + · · ·+ a0 = 0

Define a monic polynomial g(y) = yn+(a0an−1n +a1a

n−2n y+ · · · an−1yn−1) such that g(ban) = 0.

Lemma 3.2.3. Let k be a field, and let f(x, y) ∈ k[x, y] be separable when viewed over as apolynomial in one variable f(y) ∈ k(x)[y]. Then there are only finitely many b ∈ k such thatf(b, y) that are not separable.

Proof. By remark we may assume that f is monic in y. Viewed as a polynomial in one variabley f ∈ k(x)[y] is separable, therefore the discriminant Disc(f) is a non-zero element of k(x). SayDisc(f) = h(x). Then for b ∈ k fb(y) has discriminant h(b) ∈ k. This is zero if and only if b isa root of h, which only happens finitely often. Therefore the specialisation fb is not separableonly finitely often.

Let D be an integral domain, and let F be its field of fractions. D is called a factorialdomain if there exists P ⊂ A such that 0 /∈ P and any b ∈ D can be written uniquely as a

product b = u∏p∈P

pn(p) with u a unit of D, and this product is finite (i.e. n(p) = 0 for all but

finitely many p). The best known examples of factorial domains are Z and k[x] the polynomialring, both of which are PIDs. It is known that all PIDs are factorial domains [8]. In fact, apolynomial ring in n variables is known to be factorial [5].

Let f(y) ∈ D[y] be a polynomial over the factorial domain D such that deg(f) ≥ 1. f(y)is irreducible in D[y] if and only if it is irreducible in F [y], and f(y) is called primitive if thegcd of all nonzero coefficients is 1. For all g(y) ∈ F [y] ∃! d ∈ F such that dg(y) is primitive [5].

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Lemma 3.2.4. Let f(x1, . . . , xs) ∈ k[x1, . . . , xs] be in a polynomial in s > 1 variables. Then f isirreducible if and only if when viewed as a polynomial in one variable fs(xs) ∈ k[x1, . . . , xs−1][xs],fs is irreducible and primitive.

Proof. Clearly if f(x1, . . . , xs) is irreducible then fs(xs) is irreducible and primitive. Con-versely, suppose fs(xs) is irreducible and primitive. If f = gh can view this as fs = gh forg, h ∈ k[x1, . . . , xs] then, WLOG, g ∈ k[x1, . . . , xs−1] else contradicts irreducibility of fs overk[x1, . . . , xs−1][xs]. f primitive⇒ g a unit in k[x1, . . . , xs−1], which is a factorial domain. Hencef is irreducible when viewed as a polynomial in s variables.

Consider the finite Galois extension L/k of degree n with G = Gal(L/k). Let κ ⊆ k suchthat k is the field of fractions of κ i.e. k = (κ∗)−1κ. Let a generate L over k such that thereexists f(y) ∈ κ[y] monic, satisfying a with deg(f) = n.

Lemma 3.2.5. Suppose there exists A ⊂ L such that a ∈ A, A is invariant under G andthe field generated by κ and A, κ[A] = κ[a]. Then there exists w ∈ κ such that for each ringhomomorphism φ : κ→ k′, k′ a field, φ(w) 6= 0 and φ extends to a map

Φ : κ[a]→ L′

where L′/k′ is a finite Galois extension with Galois group G′ = Gal(L′/k′). L′ is generatedover k′ by a′ := Φ(a). Let f ′ be the polynomial obtained by applying φ to all the coefficients off . Then f ′(a′) = 0. Furthermore, if f ′ is irreducible then there exists a unique isomorphismG→ G′ sending g 7→ g′ such that

Φ(g(x)) = g′(Φ(x)) (3.1)

for all g ∈ G, x ∈ κ(a).

Proof. L/k is a Galois extension ⇒ f(y) ∈ κ[y] is separable i.e. has no repeated roots. Hencethe discriminant Disc(f) 6= 0 ∈ κ. Now φ(Disc(f)) = Disc(f ′). Consider φ : κ→ k′ such thatDisc(f ′) 6= 0 i.e. such that f ′(y) is separable. Extend φ : κ → k′ to a map ψ : κ[y] → k′[y]where ψ(y) = y. This map induces a homomorphism

Ψ : κ[y]/fκ[y] → k′[y]

/f ′k′[y]

since f ′ is obtained by applying φ to all the coefficients of f , so ψ(f(y)) = f ′(y) by definition.Here fκ[y] denotes the ideal generated by f in κ[y].

There is a natural isomorphism ζ : κ[y]/fκ[y] → κ[a] by h 7→ h(a). Now define

η := Ψ ◦ ζ−1 : κ[a]→ κ′[y]/

(f ′)

Here we are taking the quotient by (f ′), the ideal generated by f ′ in κ′[y]. A larger field isobtained by taking the quotient of κ′[y] by (g′), for g′ an irreducible factor of f ′. This field is

naturally a field extension of κ′[y]/

(f ′) , and so composing with η there is an extension of φ to

the map

κ[a]→ κ′[y]/

(f ′) → κ′[y]/

(g′) = L′

and this is Φ. Need to show that L′ is a Galois extension of k′. Now G permutes the conju-gates a1, . . . , an of a amongst themselves, and since by hypotheses A is invariant under G, allconjugates of a lie in A ⊂ κ[A] = κ[a]. Write

f(y) = (y − a1) · · · (y − an)

15

then applying φ to this get

f ′(y) = (y − a′1) · · · (y − a′n)

where a′i = φ(ai). Now f ′(a′) = 0, so a′1, . . . , a′n are the conjugates of a′, and so the conjugates

of a′ all lie in L′ over k′, and so L′/k′ is a normal extension. But f ′ is separable, so thereforeL′/k′ is separable i.e. L′/k′ is indeed a finite Galois extension.

Finally, suppose f ′ is irreducible. It is separable, i.e. has distinct roots i.e. a′1, . . . , a′n are

distinct. There is a unique g′i ∈ G′ mapping a′ to a′i by properties of Galois groups. Similarlythere is a unique gi ∈ G mapping a to ai. Hence there is a bijection G → G′ sending gi → g′i.Need to show this is an isomorphism satisfying (3.1). Pick x ∈ κ[a]. That is, there is polynomialh(y) ∈ κ[y] such that x = h(a). Let h′(y) = Φ(h(y)) ∈ κ′[y]. Then

g′i(Φ(x)) = g′i(Φ(h(a))) = g′i(h′(a′)) = h′(a′i) = Φ(h(ai)) = Φ(gi(h(a))) = Φ(gi(x))

This shows (3.1). Now if g1, g2 ∈ G then (g1g2)′(a′) = (g1g2)

′(Φ(a)) = Φ((g1g2)a) = g′1(Φ(g2(a))) =g′1g′2(Φ(a′)). Therefore the map G→ G′ sending gi 7→ g′i is a homomorphism, and it is a bijec-

tion, so it is an isomorphism, as required.

Remark 3.2.6. The condition that κ[A] = κ[a] in Lemma 3.2.5 can be removed. Indeed, x ∈ Acan be written x =

∑n−1i=1 bia

i with bi ∈ k. Since k is the field of fractions of κ, can choose c ∈ κsuch that cbi ∈ κ. Recall that Disc(f) is a non-zero element of κ, as f(y) ∈ κ[y] is separable.Define d = cDisc(f) and let κ = κ[d−1]. Then cbi ∈ κ ⇒ cDisc(f)bi ∈ κ ⇒ dbi ∈ κ ⇒ bi ∈ κ.Therefore x ∈ κ[a] for all x ∈ A. So κ[A] = κ[a]. This is the case in Lemma 3.2.5.

This next result forms the basis of the key theorem in this chapter, which in turn is a keyconcept in the theory of rigidity when applied to the Inverse Galois Problem.

Lemma 3.2.7. Let L/k(x) be a finite Galois extension. Then there is a monic polynomialf(x, y) ∈ k[x, y] and a generator a of L over k(x) such that f(x, a) = 0. Moreover, for allbut finitely many b ∈ k, if the specialised polynomial fb(y) = f(b, y) is irreducible in k[y] then

k[y]/

(fb) is Galois over k with Galois group isomorphic to Gal(L/k(x)).

Proof. a generates L over k(x), so a satisfies a polynomial f(y) ∈ k(x)[y]. Let a1, . . . , an bethe conjugates of a over k(x). Can view f as a polynomial in two variables f ∈ k[x, y], and byRemark 3.2.2 we may assume f is monic in y. Write

f(y) = (y − a1) · · · (y − an) (3.2)

Now a is amongst the conjugates, hence f(a) = f(x, a) = 0.

For the last part let b ∈ k. There is a natural homomorphism

φb : k[x]→ k

h(x) 7→ h(b)

Now by Lemma 3.2.5 to φb : k[x]→ k. Obtain f ′(y) by applying φb to the coefficients of f(y).Then by definition of φb, f

′(y) = f(b, y) = fb(y). Lemma 3.2.5 says that if there exists w ∈ k[x]

such that φb(w) 6= 0 then the field k[y]/

(fb) is Galois over k with Galois group Gal(L/k(x)).

Now φb(w) = w(b) is zero only when b is a root of w, i.e. only finitely often. This completesthe proof.

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Lemma 3.2.8. Let L/k(x) be a finite Galois extension, and let l be a finite extension of k suchthat l ⊂ L. Then if p(x, y) ∈ l[x, y] is irreducible as a polynomial in one variable y over l(x)such that the roots of p are contained in L, then for all but finitely many b ∈ k, if fb(y) ∈ k[y]is irreducible then pb(y) = p(b, y) ∈ l[y] is irreducible, where f is as in Lemma 3.2.7

Proof. Write p(x, y) = p0(x)∏ri=1(y − bi) with bi ∈ L, p0(x) ∈ l[x]. Let bi ∈ A ⊂ L containing

a generator of L over k, say a (this is the A from Lemma 3.2.5). Recall Φ : κ[A] → L′. Letb′i = Φ(bi). Let A contain a generator of l over k. Then Φ is defined on l too. Now assumingfb is irreducible, then Φ maps l isomorphically to a subfield of L′. Then applying Φ to p(x, y)gives p(b, y) = p0(b)

∏ri=1(y − b′i). p(x, y) is irreducible, hence separable as we are assuming k

has characteristic 0. By Lemma 3.2.5 Gal(L/l(x)) is mapped onto Gal(L′/l). By properties ofGalois groups, Gal(L/l(x)) acts transitively on the bi, and so Gal(L′/l) acts transitively on theb′i. By Lemma 3.2.3 there are only finitely many b ∈ k for which p(b, y) is not separable, andthere are finitely many b such that p0(b) = 0, namely the roots of p0. So for all b ∈ k excludingthese finitely many, the polynomial pb(y) is separable, and Gal(L′/l) acts transitively on theroots. Therefore pb is irreducible.

Lemma 3.2.9. Let L/k(x) be a finite Galois extension. There are finitely many polynomialspi(x, y) ∈ k[x, y] which are irreducible when viewed as polynomials in one variable y over k(x)and such that for all but finitely many b ∈ k, if none of the pi(b, y) has a root in k(x) then fb(y)is irreducible in k[y], where f is as in Lemma 3.2.7.

Proof. By Lemma 3.2.7, f(x, y) ∈ k[x, y] is irreducible when viewed as a polynomial in onevariable y over k(x), and f is of the form f(y) = (y−a1) · · · (y−an) where ai are the conjugatesof a over k(x), where a is a generator of L over k(x), as in (3.2). Hence L/k(x) has degree n.Let I ⊂ {1, . . . , n}. Now

∏i∈I(y − ai) /∈ k(x)[y], as f(y) is irreducible over k(x). Hence it has

some coefficient cI /∈ k(x). Recall from Lemma 3.2.5 the map Φ : κ[A] → L′ where the set Acontained a and its conjugates. Let pI be the an irreducible polynomial satisfying cI over k(x).Now suppose that fb is not irreducible. Then there exists an I of the form described above suchthat

∏i∈I(y−ai) lies in k[y]. Now d = Φ(cI) is a coefficient of this polynomial, therefore d ∈ k.

But now applying Φ to pI(x, ci) = 0 gives pi(b, d) = 0.

Proposition 3.2.10. The field k is hilbertian if and only if, given a finite extension l/k, andirreducible polynomails pi(x, y) ∈ l(x)[y] for i = 1, . . . ,m, there are infinitely many b ∈ k suchthat pi(b, y) are irreducible in l[y] for i = 1, . . . ,m.

Proof.(⇒) Let p1(x, y), . . . , pm(x, y) ∈ l(x)[y]. Form a set A of all the roots of these m polynomials inan algebraically closed field containing l(x). Let L/l(x) be a finite extension such that A ⊂ Land L is Galois over k(x). Now apply Lemma 3.2.8.(⇐) Clear.

Proposition 3.2.11. The field k is hilbertian if and only if, for any irreducible pi(x, y) ∈ k(x)[y]for i = 1, . . . ,m, there are infinitely many b ∈ k such that none of the specialised polynomialspi(b, y) ∈ k[y] has a root in k

Proof. (⇒) Follows from the above proposition and Lemma 3.2.9. (⇐) is again clear.

Propositions 3.2.10 and 3.2.11 are essentially equivalent definitions for hilbertian fields, andI will use them interchangeably. In this set up, hilbertian fields k have a particularly usefulproperty, that if a finite group G occurs as a Galois group over k(x1, . . . , xn), then G also occursas a Galois group over k. I first need to establish some basic properties of hilbertian fields. The

17

properties established so far apply to polynomials in two variables, but this next remark on theKronecker specialisation, provides a bridge to polynomials in s > 1 variables.

Remark 3.2.12. Let f(x1, . . . , xs) be a polynomial in s > 1 variables over a hilberitan field k.The Kronecker specialisation of f is defined by first letting d = df be an integer greater thenany power of any variable xi occurring in f , and then by specialising thusly

Sdf(x, y) = f(x, y, yd, yd2, . . . , yd

s−2) (3.3)

a polynomial in two variables x and y. Now, write Sdf(x, y) = g(x)∏i gi(x, y) with gi irre-

ducible over k[x, y] and g ∈ k[x]. k is hilbertian i.e. there are infinitely many b ∈ k such thatgi(b, y) remains irreducible over k[y].

Let Sdf(x, y) = g(x)H1(x, y)H2(x, y). Then there exists unique polynomials h1(x1, . . . , xs),h2(x1, . . . , xs) such that Sdh1 = gH1 and Sdh2 = H2 and the highest power of x2, . . . , xs occur-ring is less then d [12].

Lemma 3.2.13. Let k be a hilbertian field with f(x1, . . . , xs) ∈ k[x1, . . . , xs] irreducible over k.Then there are infinitely many b ∈ k such that f(b, x2, . . . , xs) remains irreducible over k.

Proof. Let Sdf(x, y) = g(x)∏i gi(x, y) be the Kronecker specialisation of f . Suppose for a con-

tradiction that f(b, x2, . . . , xs) = h1(x2, . . . , xs)h2(x2, . . . , xs) is reducible, and consider theirKronecker specialisations Sdh1(y) and Sdh2(y). Let b be such that the gi(b, y) remain ir-reducible over k[y] from Remark 3.2.12, and also such that b is not a root of g(x). ThenSdf(b, y) = Sdh1(y)Sdh2(y) = g(b)

∏i gi(b, y). So Sdh1(y), Sdh2(y) are products of the irre-

ducible factors gi(b, y). Let H1(x, y), H2(x, y) be the product of the corresponding gi(x, y).Hence write Sdf(x, y) = g(x)H1(x, y)H2(x, y).

There exists unique polynomials h1(x1, . . . , xs), h2(x1, . . . , xs) such that Sdh1 = gH1 and Sdh2 =H2 as in Remark 3.2.12. Let f = h1h2. If the highest power of f occuring is less than d, thenf = f . But f is irreducible by hypothesis, hence this is a contradiction. Therefore the highestpower of f occurring is greater than or equal to d. Hence f contains a monomial

f0(x1)xi22 · · ·x

iss (3.4)

such that ij ≥ d some j ∈ [2, . . . , s]. Now Sdf(b, y) = Sdh1(y)Sdh2(y) and Sdf(x, y) =

g(x)H1(x, y)H2(x, y) = Sdh1Sdh2. Therefore, hi(b, x2, . . . , xs) is a scalar multiple of hi(x2, . . . , xs),for i = 1, 2 ⇒ f(b, x2, . . . , xs) is a scalar multiple of f(b, x1, . . . , xs). Hence the monomial mustvanish upon evaluation at b ∈ k else contradicting that d is greater then any power of anyvariable xi in f . This means that f0(b) = 0 from (3.4). But this clearly happens only finitelyoften. Hence it is the case that there are infinitely many b ∈ k such that f(b, x2, · · · , xs) remainsirreducible over k.

The following corollary shows that it is possible to specialise numerous variables in anirreducible polynomial f(x1, . . . , xs) ∈ k[x1, . . . , xs] in a hilbertian field k and still have anirreducible polynomial.

Corollary 3.2.14. Let k be a hilbertian field and f(x1, . . . , xs) an irreducible polynomial ins > 1 variables. Then for any p(x1, . . . , xs−1) ∈ k[x1, . . . , xs−1] there are b1, . . . , bs−1 ∈ k suchthat p(b1, . . . , bs−1) 6= 0 and f(b1, . . . , bs−1, xs) is irreducible.

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Proof. By induction on s. Let s = 2, and suppose f(x1, x2) irreducible. Then by Lemma 3.2.13there exists infinitely many b ∈ k such that f(b, x2) is irreducible over k. Hence statement holds.Fix s > 2 and suppose statement true for s − 1. View p(x1, . . . , xs−1) ∈ k[x1, . . . , xs−1] as apolynomial p1(x2, . . . , xs−1) ∈ k(x1)[x2, . . . , xs−1], with some coefficients polynomials in x1, saycj(x1). Then by Lemma 3.2.13 there exists b1 ∈ k such that fb1(x2, . . . , x2) := f(b1, x2, . . . , xs)is irreducible and cj(b1) 6= 0 for some j. Then pb1(x2, . . . , xs−1) := p(b1, x2, . . . , xs−1) is non-zeroas some of its coefficients cj(b1) are non-zero. But now by the inductive hypothesis on pb1 a poly-nomial in s−2 variables, we have that there exists b2, . . . , bs−1 ∈ k such that pb1(b2, . . . , bs−1) 6= 0and fb1(b2, . . . , bs) is irreducible. Therefore b1, . . . , bs−1 ∈ k works, and the inductive step isproved.

Lemma 3.2.15. Any finitely generated extension of a hilbertian field is hilbertian.

Proof. Let l be a finite extension of k. Then for f ∈ l[x, y] an irreducible polynomial, then thereare infinitely many b ∈ k and hence in l such that f(b, y) is irreducible in l[y], by definition(see Proposition 3.2.10. Hence l is hilbertian. Now consider a purely transcendental extensionK = k(x1, . . . , xm). Let R = k[x1, . . . , xm], and f(x, y) ∈ K[x, y] irreducible in y. Then f isirreducible in y over K(x). Then WLOG f ∈ R[x, y], so by Lemma 3.2.4 f is irreducible asa polynomial in x1, . . . , xm, x, y. Hence by Lemma 3.2.13 there are infinitely many b ∈ k suchthat f(x1, . . . , xm, b, y). Therefore, when viewed as a polynomial in one variable y over R, andhence over K, f(x1, . . . , xm, b, y) = f(y) ∈ R[y] ⊂ K[y] is irreducible. Therefore K is hilbertian.

Let K = k(a1, . . . , an) be a finitely generated extension of k. If x1, . . . , xm is a maximalalgebraically independent subset of a1, . . . , an over k, then K is finite over the purely transcen-dental extension k(x1, . . . , xm). Indeed K is algebraic over k(x1, . . . , xm) and hence finite overk(x1, . . . , xm). Hence any finitely generated extension K of k is a finite extension of k or a finiteextension of a purely transcendental extension of k. Either way K is hilbertian.

3.3 Galois Groups Over Hilbertian Fields

This section provides results that will prove invaluable later on, and forms one of the corearguments when using rigidity to realise groups as Galois groups over Q.

Theorem 3.3.1. Let k be a hilbertian field and G a finite group. If G occurs as a Galois groupover k(x1, . . . , xm) then G occurs as a Galois group over k.

Proof. By induction on m. For m = 1 use Lemma 3.2.7. Let m > 1. Now k is hilbertian byhypothesis, therefore k(x1, . . . , xm−1) is a finitely generated extension of k and so by Lemma3.2.15 is hilbertian. Then k(x1, . . . , xm) = k(x1, . . . , xm−1)(xm) so we reduce to the case m = 1and use Lemma 3.2.7 again.

Definition 3.3.2. Let G be a finite group. Then G occurs regularly over k if there is afinite Galois extension L of k(x1, . . . , xm), regular over k with G = Gal(L/k(x1, . . . , xm)).

Theorem 3.3.3. If G occurs regularly over k then G occurs regularly over every finitely gen-erated extension l of k. Thus G is a Galois group over l if l is hilbertian.

Proof. G occurs regularly over k, say G = Gal(K/k(x1, . . . , xm)), and K is regular over k bydefinition. Then there is irreducible f ∈ k(x1, . . . , xm)[y] such that K is of the form

K = k(x1, . . . , xm)[y]/

(f)

19

with f(x1, . . . , xm, y) ∈ k(x1, . . . , xm)[y] irreducible over l(x1, . . . , xm) by Lemma 3.1.4. Con-

struct a field extension of l(x1, . . . , xm) thusly, L = l(x1, . . . , xm)[y]/

(f) . Now the roots of f

over l(x1, . . . , xm) are contained in K already. Hence L/l(x1, . . . , xm) is a Galois extension, andwe have Gal(L/l(x1, . . . , xm)) = Gal(K/k(x1, . . . , xm)). Finally, if l is hilbertian then G occursas a Galois group over l by Theorem 3.3.1.

3.3.1 Sn as a Galois group over k

Consider the polynomial f(y) ∈ k(x1, . . . , xn)[y] defined by

f(y) = yn + x1yn−1 + · · ·+ xn

Let a1, . . . , an be the roots of f(y). Then Sn acts naturally on the roots in the following way,let σ ∈ Sn then

σ : (a1, . . . , an) 7→ (aσ(1), . . . , aσ(n))

This action extends to an automorphism σ : k(a1, . . . , an) 7→ k(a1, . . . , an). In this way wehave constructed an action of Sn on k(a1, . . . , an). Consider the fixed field F of this actionin k(a1, . . . , an). Then F contains k(x1, . . . , xn), and further |k(a1, . . . , an) : F| = |Sn| = n!.But since a1, . . . , an are the roots of a polynomial of degree n over k(x1, . . . , xn) it follows that|k(a1, . . . , an)/k(x1, . . . , xn)| ≤ n!. Therefore F = k(x1, . . . , xn) and so by Artin’s theoremGal(k(a1, . . . , an)/k(x1, . . . , xn)) = Sn. Therefore Sn occurs regularly over k. Hence if k isHilbertian then Sn occurs as the Galois group over k.

Given that the Inverse Galois Problem asks about finding groups as Galois groups of extensionsof Q, the importance of this next result speaks for itself. I quote from [12] without proof

Theorem 3.3.4. (Hilbert’s Irreducibility Theorem)The field Q is hilbertian.

As a corollary to this we see that Sn occurs as a Galois group over Q. In general, if we canrealise a group over the rational function field Q(x) then Hilbert’s irreducibilty theorem coupledwith Theorem 3.3.1 says that G occurs as a Galois group over Q. Now Riemann’s ExistenceTheorem will give a way of realising groups over the complex function field C(x). This motivatessome discussion on decent of the base field, in other words, if G occurs as a Galois group overC(x), when can we say that G occurs as a Galois group over Q(x)? To answer this questionone first needs to introduce rigidity.

As a digression, the chapter concludes with a nice little application of Hilbert’s Irreducibil-ity Theorem. Consider the polynomial f(x, y) = y2 − g(x). Theorem 3.3.4 says that if g(b) is aperfect square in Q for all b ∈ Q then g is the square of a polynomial [14].

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Chapter 4

Rigidity

This chapter investigates the idea of branch points and how they are used in the developmentof an algebraic ramification type. This will finally allow the algebraic version of Riemann’sExistence Theorem to be stated. It is at this point that rigidity will be defined, and a firstglimpse at the role it plays in the Inverse Galois Problem will be observed. The chapter endswith an important result that allows a character theoretic approach to be applied. This is usedin many applications of rigidity in conjunction with [1].

4.1 Laurent Series Fields

Let k be a field.

Define Λ the set of all sequences (ai)i∈Z with ai ∈ k such that ∃ N ∈ Z with ai = 0 forsome i < N .

Then Λ is a field with addition (ai)i∈Z+(bi)i∈Z = (ai+bi)i∈Z and multiplication (ai)i∈Z·(bj)j∈Z =(cn)n∈Z where cn =

∑i+j=n aibj [12]

k is a subfield of Λ. Indeed, can embed k in Λ by a ∈ k 7→ (ai)i∈Z with a0 = a and ai = 0otherwise.

Define a sequence t by (ai) with a1 = 1 and ai = 0 else.

Define k[t] the ring generated by k and t. Then k[t] is a subring of Λ, it is the polynomialring in one variable over k.

As a purely formal notation write,

(ai) =∑

aiti

The field operations of addition and multiplication defined correspond to formal addition andmultiplication of Laurent Series and thus Λ is called the field of formal Laurent series overk and denoted by k((t)).

The ring of formal power series over k consisting of all∑ait

i is denoted k[[t]].

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Polynomials over k[[t]]

Have a ring homomorphism

ϕ : k[[t]] −→ k

(ai) 7→ a0

Consider a polynomial F (y) ∈ k[[t]][y]. By applying ϕ to all it’s coefficients we obtain Fϕ(y) ∈k[y]

Lemma 4.1.1. Let F (y) ∈ k[[t]][y] monic. If Fϕ(y) ∈ k[y] factors as

Fϕ = g · h

with g, h ∈ k[y] and g, h coprime, then F factors as

F = G ·H

with G,H ∈ k[[t]][y] monic and Gϕ = g, Hϕ = h.

4.2 Finite Extensions of Λ

For each e ∈ Z \{0}, let Ze−1 = {i/e : i ∈ Z}. Then Ze−1 forms an additive group isomorphicto Z.

Z a subgroup of Ze−1 of index e

Define Λe the set of sequences (aj)j∈Ze−1 with aj ∈ k. As before, Λe is a field. In fact there isan isomorphism

Λe ∼= Λ

(aj)j∈Ze−1 7→ (bi)i∈Z

with bi = ai/e. Under this isomorphism the element t ∈ Λ τ ∈ Λe where

τ = (aj)j∈Ze−1 where a1/e = 1, aj = 0 else.

We have Λ a subfield of Λe consisting of all sequences (aj)j∈Ze−1 with aj ∈ k : aj = 0 ∀ j /∈ Z.

Lemma 4.2.1. Suppose k contains a primitive eth root of unity ζe. Then Λe is Galois over Λof degree e and it has cyclic Galois group. Furthermore Λe = Λ(τ) with τ e = t.

Proof. Consider the automorphism of Λe,

ω :∑

i∈Z biτi →

∑i∈Z(biζ

ie)τ

i

It fixes elements of the form∑

i∈Z biτi with bi = 0 unless ζie = 1 =⇒ bi = 0 unless e divides i.

That is, the fixed field of ω is Λ.

Therefore, ω an automorphism of Λe with fixed field Λ so Λe is Galois over Λ with Ga-lois group 〈ω〉. Further, we have that ω has order e (since ζe is an eth root of unity) hence[Λe : Λ] = |Gal(Λe/Λ)| = |〈ω〉| = e.

Finally, NTS that no non-trivial element of G = Gal(Λe/Λ) fixes τ . That is, no non-trivialelement of 〈ω〉 fixes τ . Indeed ω(τ) = ζeτ , therefore ωr(τ) = ζre τ = τ =⇒ e|r =⇒ ωr is trivialelement of G. Thus Λe = Λ(τ).

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Now I’ll prove a useful result following [12].

Lemma 4.2.2. Let k be an algebraically closed field with characteristic 0. Suppose F (y) ∈k[[t]][y] monic, non-constant polynomial. Then F has a root in Λe for some e ∈ N.

Proof. Suppose F of minimal degree and has no root in any Λe, for a contradiction. Thenn = degF ≥ 2. Have

F (y) = yn + λn−1yn−1 + · · ·+ λ1y + λ0 (λr ∈ k[[t]])

Can eliminate yn−1 term by replacing F with

F (y) = F (y − λn−1n

)

to assume F has zero yn−1 coefficient.

Claim If Fϕ(y) 6= yn then F factors as F = GH with G,H ∈ k[[t]][y] monic, non-constantpolynomials.

Indeed, since k is algebraically closed the polynomial Fϕ ∈ k[y] factors as a product of moniclinear polynomials. If these factors are not all equal then can write Fϕ = g · h as in 4.1.1 andthus the claim follows. However, if all the linear factors are the same, i.e. if Fϕ(y) = (y − a)n

with a ∈ k, then the yn−1 coefficient is −na hence a = 0 (as char k = 0) and Fϕ(y) = yn

contradicting the assumption.

So if Fϕ(y) 6= yn then F factors as above, contradicting the minimality of F . Therefore Fϕ = yn

which is to say that ϕ(λr) = 0 ∀ r =⇒ all λr have zero constant term.

Now, it cannot be the case that all terms in all coefficients λr are zero as then would haveλr = 0 ∀ r and hence F (y) = yn which has zero as a root contradicting our hypothesis. Con-sider only those λr 6= 0. Define mr the lowest power of t occurring with non-zero coefficient inλr, that is

λr = a · tmr+ higher terms

with a ∈ k\{0}. Define

u := infr

{mr

n− r

}Then u ∈ Q. Write u = d/e with d, e ∈ Z>0. Now consider the polynomial

F ∗(y) = τ−dnF (τdy) = yn +n−2∑r=0

λrτd(r−n)yr ∈ Λe[y]

The yr coefficient is a Laurent series in τ of the form

λrτd(r−n) = a · tmrτd(r−n) + higher terms

= a · τEr + higher terms

where Er = e(n− r)( mrn−r − u). We have that Er = 0 for at least one r. Hence the coefficientsof F ∗ are power series in τ , and for at least one r, this power series has nonzero constant term.Thus F ∗ϕ 6= yn, so by the above claim F ∗ = GH factors over k[[t]]. Then degH < n = degF

hence has a root in Λe(τ1/e′) by minimality of n. Thus F ∗ has a root in Λ(τ1/e

′) = Λee′ . Thus

F also has a root in Λee′ .

23

Theorem 4.2.3. Let k an algebraically closed field, characteristic 0. Let ∆ be a field extensionof Λ = k((t)) of finite degree e. Then ∆ = Λ(δ) with δe = t.

Proof. Let γ be a primitive element for the finite extension ∆/Λ, that is write ∆ = Λ(γ). Thenγ satisfies an irreducible polynomial F (y) ∈ Λ[y]. WLOG we may assume that F (y) is a monic,non-constant polynomial. Then by 4.2.2, F has a root γ1 in some Λe1 ∴ ∆ ⊂ Λe1 .

Now, Gal(Λe1/Λ) is cyclic order e1, hence for each e diciding e1 there exists an intermediatefield between Λ and Λe1 of degree e over Λ. Therefore ∆ = Λe = Λ(δ) where δe = t.

Remark 4.2.4. The theorem says that for every finite extension ∆/Λ, it is the case that∆ ∼= Λe, for some e. But this is just what was proved in 4.2.2, that each polynomial over Λ hasa root in some Λe.

4.3 Branch Points

Definition 4.3.1. A compatible system (ζe)e∈N consists of primative eth roots of unity suchthat whenever e = e1 · e2 then ζe2e = ζe1

Set Pk1 = k ∪ {∞}.

For p ∈ P1k define an isomorphism

vp : k(x)→ k(t)

a 7→ a ∀ a ∈ k

x 7→{t+ p p 6=∞1/t p =∞

Let L/k(x) be a finite Galois extension. Set G = Gal(L/k(x)), and let p ∈ P1k.

Definition 4.3.2. Let γ be a primitive element of L/k(x). Then γ satisfies an irreduciblepolynomial F (y) ∈ k(x)[y]. Let Fvp be the polynomial obtained by applying vp to the coefficientsof F (y).

Lemma 4.3.3. We can extend vp to an isomorphism v : L → Lv, where Lv is a subfield ofsome FG-extension ∆ of Λ. Then the group Gal(∆/Λ) leaves Lv invariant.

Proof. Let H be an irreducible factor of Fvp , and define

∆ := Λ[y]/

(H)

a finite extension of Λ. Now H has a root γ1 in ∆ and therefore so does Fvp . Therefore we canextend vp to an isomorphism v from L = k(x)[y] to Lv = k(t)[γ1] ⊂ ∆.

Now Lv is generated by the roots of Fvp , and since Gal(∆/Λ) permutes these roots thenGal(∆/Λ) leaves Lv invariant.

Define gv := v−1 · ω · v, where ω is the generator of Gal(∆/Λ). Then gv ∈ G.

Lemma 4.3.4. If ∆ is another FG-extension of Λ with subfield Lv, and v : L → Lv is anisomorphism extending vp, then gv and gv lie in the same conjugacy class of G.

24

Proof. We may assume that ∆ and ∆ both lie in a FG-extension ∆0 of Λ. Then Lv and Lv areboth subfields of ∆0 generated over k(t) by roots of Fvp in ∆0. Hence Lv = Lv.

Set h = v−1v. Then h ∈ G. Indeed, v : L → Lv = Lv and v−1 : Lv → L, isomorphisms.Hence h = v−1 · v : L→ L and thus h ∈ G

Let ω0 be generator of Gal(∆0/Λ). Restriction to ∆ (resp. ∆) gives the generator of Gal(∆/Λ)(resp. Gal(∆/Λ)). We have

gv = v−1 · ω0 · v = h−1v−1 · ω0 · vh = h−1 · gv · h

Therefore gv and gv lie in the same conjugacy class, say Cp.

Definition 4.3.5. Call the conjugacy class containing gv and gv in the proof of 4.3.4 the classof G associated with p. Denote Cp. Once L is fixed, this class depends only on p.

Definition 4.3.6. Let eL,p be the (common) order of elements of Cp. Call this the ramificationindex of L at p. If L and p are clear from context then drop the subscript and just write e.

Lemma 4.3.7. All irreducible factors H of Fvp in Λ[y] have degree e. Furthermore, we cantake ∆ as in 4.3.3 to be ∆ = Λe.

Proof. Can restrict Gal(∆/Λ) → Gal(Lv/k(t)). This is an injective homomorphism. Indeed ifwe write ∆ = Λ(γ1) we have Gal(Λ(γ1)/Λ)→ Gal(k(t)[γ1]/k(t)) which has trivial kernel.

The degree [∆/Λ] =order of ω =order of ω|Lv .

Now gv = v−1ωv hence order of ωLv = order of gv = e.

Finally, by Theorem 4.2.3, since [∆/Λ] = e, we have ∆ ∼= Λe.

Definition 4.3.8. Let L/k(x) be a FG-extension, and let p ∈ P1k. Then p is a branch point

of L/k(x) if eL,p > 1.

Remark 4.3.9. That is the same as saying that the class of G associated to p is non-trivial.

Lemma 4.3.10. Let L/k(x) and L′/k(x) be FG-extensions such that L′ ⊂ L. Let ρ : G→ G′ byrestriction, where G = Gal(L/k(x)), G′ = Gal(L′/k(x)). Then ρ maps the class of G associatedwith p to the class of G′ associated with p. That is, ρ(Cp) = C ′p.

Proof. Let v : L→ ∆ as in Lemma 4.3.3. Then restrict to L′ to get v′ = v|L′ , an isomorphismfrom L′ to a subset of ∆, extending vp.

Therefore gv′ = (v′)−1ωv′ = v−1ωv|L′ = (gv)|L′ . Therefore restriction to L′ does indeed mapCp to C ′p, and the lemma is proved.

Theorem 4.3.11. Branch cycle argumentLet L/k(x) and L′/k(x) be FG-extensions of degree n. Let α ∈ Aut(k) and let m ∈ Z such thatα−1(ζn) = ζmn . Suppose α extends to an isomorphism λ : L → L′ fixing x. Let λ∗ denote theinduced group isomorphism from G to G′ (in the usual notation) mapping g 7→ λgλ−1. Then,

C ′α(p) = λ∗(Cp)m

25

Proof. Fix p ∈ P1k. Then e = eL,p divides n = |G| by Lagrange, say n = e · s. Then by

compatibility of ζi’s we have

ζsn = ζe ⇒ α−1(ζe) = α−1(ζsn) = ζmsn = ζme (4.1)

Extend α to an automorphism of Λe = k((τ)) sending Σbiτi 7→ Σα(bi)τ

i. Call this automor-phism α. As before let ω be the generator of G = Gal(Λe/Λ) with ω(τ) = ζeτ . Then we have,using (4.1)

α−1ωα(τ) = α−1ω(τ) = α−1(ζeτ) = ζme τ = ωm(τ)

By 4.3.7 can extend vp to an isomorphism v from L to a subfield of Λe. Then define v′ := α·v·λ−1,an isomorphism extending vα(p) from L′ to a subfield of Λe. Indeed it fixes all K ∈ k since λ isan extension of α and hence λ−1(K) = α−1(K) and α sends K 7→ α(K).

Consider,

v′(x) = αvλ−1(x)

= αv(x) (since λ is identity on x)

=

{α(t+ p) p 6=∞α(1/t) p =∞

=

{t+ α(p) p 6=∞

1/t p =∞

Now gv′ ∈ C ′α(p) as before is

gv′ = (v′)−1ωv′ = λv−1α−1 · ω · αvλ−1

= λv−1 · ωm · vλ−1

= λ · gmv λ−1

= λ∗(gv)m

proving the theorem.

4.4 Rigid Ramification Types

Let k be an algebraically closed subset of C and let P1 = C ∪ {∞}

Consider the triple (G,P,C) where G is a group, P is a finite subset of P1 and C = (Cp)p∈P .Define an equivalence relation on this triple by

(G,P,C) ∼ (G′, P ′,C′)

iff ∃ an isomorphism from G→ G′ which sends Cp 7→ C ′p for each p ∈ P and P = P ′. (This isclearly an equivalence relation!)

Definition 4.4.1. The set of equivalence classes of (G,P,C) is called a (algebraic) ramifica-tion type and denoted T = [G,P,C].

Remark 4.4.2. For a FG-extension L/k(x) and a set P ⊂ P1 we have canonical choice for atriple as above; we take G = Gal(L/k(x)) and C the set of classes of G associated with p, thebranch points of L/k(x) over P1. Therefore, the Inverse Galois Problem can be reformulated inthis language asking if one can find an extension of Q of type T = [G,P,Cp] for any group G.

26

Recall the topological version of Riemann’s Existence Theorem 2.3.8. There is an algebraicanalogue in which the topological ramification type and the algebraic ramification type areidentified using Riemann surface theory. Under this identification there is an algebraic versionof Riemann’s Existence Theorem [12].

Theorem 4.4.3. Riemann’s Existence TheoremLet T = [G,P,C] be a type and say P = {p1, . . . , pr}. Then there exists an FG-extension ofC(x) of type T iff ∃ generators σ1, . . . , σr of G with σ1 · · ·σr = 1 and σi ∈ Cpi for i = 1, . . . , r.

So if one can find a generating set of a finite group G satisfying these properties thenRiemann’s Existence Theorem guarantees the existence of an extension of C(x) of the desiredtype. I now introduce the notion of rigidity as an extra condition on this generating set. As afirst application, I will show that rigidity ensures a unique extension of desired type in Theorem4.4.3, but as we will see in later chapters rigidity will be crucial in guaranteeing an extensionof Q of the desired type, which is of course the aim.

Definition 4.4.4. Let (C1, . . . , Cr) be a tuple of conjugacy classes in group G. We say it is arigid tuple if

1. There exist generators σ1, . . . , σr of G with σ1 · · ·σr = 1 and σi ∈ Ci for i = 1, . . . , r.

2. If σ′1, . . . , σ′r is another set of generators with that property then there exists a unique

g ∈ G such that gσig−1 = σ′i.

The uniqueness of g is equivalent to saying G has trivial center. Indeed if 1 6= z ∈ Z(G) andg ∈ G as in definition, then gσig

−1 = σ′1 ⇒ (zg)σi(zg)−1 = z(gσig−1)z−1 = zσiz

−1 = σ′i, sincez commutes with σ′i. But then g is not unique, which is a contradiction.

Definition 4.4.5. If instead of a unique element g ∈ G such that gσig−1 = σ′i, there exists a

general automorphism γ ∈ Aut(G) such that γ(σi) = σ′i, then we say (C1, . . . , Cr) is a weaklyrigid tuple.

Since γ ∈ Aut(G) is defined on (σ1, . . . , σr), and these generate G, then γ is uniquelydefined on all G. Indeed, let h ∈ G. Then since {σi} generate G can write h = σ

ni1i1· · ·σnikik

,with ij ∈ [1, . . . , r] and nij ∈ Z, so

γ(h) = γ(σni1i1· · ·σnikik

) = γ(σi1)ni1 · · · γ(σik)nik

Lemma 4.4.6. Let (C1, . . . , Cr) be a rigid tuple of conjugacy classes, and let γ ∈ Aut(G) suchthat γ fixes each Ci. Then γ ∈ Inn(G) is an inner automorphism, i.e. acts by conjugation.

Proof. Define σ′i = γ(σi). Then σ′i ∈ Ci since γ fixes each Ci, and σ′1 · · ·σ′r = γ(σ1) · · · γ(σr) =γ(σ1 · · ·σr) = γ(1) = 1. Let h ∈ G, then ∃ h′ ∈ G such that h = γ(h′) since γ is an automor-phism. Now, h′ = σ

ni1i1· · ·σnikik

, since G is generated by the σi, so we have

h = γ(h′) = γ(σni1i1· · ·σnikik

) = (σ′i1)ni1 · · · (σ′ik)nik

So the σ′ generate G too. Now by rigidity of (C1, . . . , Cr) we have that ∃ g ∈ G such thatgσig

−1 = σ′i. But by remarks made above, since γ is unique, for x ∈ G, γ(x) = gxg−1. That is,γ is an inner automorphism.

Definition 4.4.7. A type T = [G,P,C] is called rigid if the classes Ci form a rigid tuple.

Theorem 4.4.8. For each rigid type there is a unique FG-extension of C(x) of this type up toC(x) isomorphism.

27

Proof. Existence follows from RET. For uniqueness, suppose L1/C(x) and L2/C(x) are bothof the same type, and suppose that type is rigid. WLOG let L1 and L2 lie in a common FG-extension L of C(x). Let G = Gal(L/C(x)), Gi = Gal(Li/C(x)) for i = 1, 2. Then define

ρi : G → Gi by restriction, and for p ∈ P, denote by Cp (resp. C(i)p ) the class of G (resp. Gi)

associated with p. Then ρi(Cp) = C(i)p by 4.3.10. Let p1, . . . , pr be the brach points of L.

By RET there exist generators σ1, . . . , σr of G with the properties that σ1 · · ·σr = 1 andσj ∈ Cpj . Then ρi(σ1), . . . , ρi(σr) generate Gi. It is clear that they also satisfy the same prop-erties.

L1 and L2 are of the same type, so by definition there exists an isomorphism ε : G2 → G1

that maps C(2)p 7→ C

(1)p .

Consider ε(ρ2(σ1)), . . . , ε(ρ2(σr)). They clearly generate G1. We have

ε(ρ2(σ1)) · · · ε(ρ2(σr)) = ε(ρ2(σ1) · · · ρ2(σr))= ε(ρ2(σ1 · · ·σr))= ε(ρ2(1)) = 1

Also, σi ∈ Cpi ⇒ ρ2(σi) ∈ C(2)pi ⇒ ε(ρ2(σi)) ∈ C(1)

pi .

So ε(ρ2(σ1)), . . . , ε(ρ2(σr)) satisfy the same properties as ρ1(σ1), . . . , ρ1(σr). So by definition4.4.4 (of rigidity) ∃ g ∈ G1 : g · ε(ρ2(σi)) · g−1 = ρ1(σi) for i = 1, . . . , r. Define,

δ : G1 → G1

σ 7→ σg = g · σ · g−1

action by conjugation. So we have δ(ε(ρ2(σi))) = ρ1(σi) Then we have an isomorphism γ :=δ · ε : G2 → G1 such that

γ(ρ2(σi)) = δ · ε(ρ2(σi))= ρ1(σi)

That is ρ1 = γ · ρ2, so ρ1 and ρ2 have the same kernel. The fixed field in L of this kernel equalsL1 and L2, therefore L1 = L2.

Remark 4.4.9. We could actually prove a stronger theorem by replacing the condition ofrigidity with weakly rigid. The proof goes the same way except δ is just a general automorphism.

4.5 Structure Constants

It can in general be difficult to find rigid tuples by hand. In this section I present a formula thattranslates the condition of uniqueness of generators up to conjugacy (i.e. Definition 4.4.4(2))into a condition on the characters of the conjugacy classes. This can be checked using thecharacter tables of groups in [1] or using a computer (e.g. using GAP). I conclude the chapterby using the aforementioned formula to find rigid tuples in SL2(8) and S4.

A class vector C is just a collection of conjugacy classes {C1, . . . , Cr}. Write σ = {σ1, . . . , σr}.So to say σ1, . . . , σr generate G is the same as saying 〈σ〉 = G. Lets redefine a rigid tuple as in[7]. Firstly,

28

Definition 4.5.1. Denote the set of all generating s-systems by Σs(G) i.e. Σs(G) := {σ ∈Gs : 〈σ〉 = G, σ1 · · ·σs = 1}

Then define,Σ(C) := {σ ∈ Σs(G) : σi ∈ Ci}

So Σ(C) consists of generating s-systems that satisfy the first part of the definition of rigidityDefinition 4.4.4(1). The second part of this definition is that if another such system σ′ existsthen there exists an element g such that the action by conjugation by g on σ gives σ′. Thisaction is an inner automorphism, and so motivates the following definition,

l(C) :=∣∣∣Σ(C)

/Inn(G)

∣∣∣Definition 4.5.2. A class vector C is called rigid if l(C) = 1

Now consider the set of not necessarily generating s-systems

Σ(C) := {σ ∈ Gs : σi ∈ Ci, σ1 · · ·σr = 1}

G acts by component-wise conjugation, i.e. (σ1, . . . , σr)→ (gσ1, . . . ,g σr), so similarly quotient

out by Inn(G) and define,

n(C) :=∣∣∣Σ(C)

/Inn(G)

∣∣∣Definition 4.5.3. n(C) is the normalised structure constant of C.

Remark 4.5.4. Σ(C) is an enlargement of Σ(C), so l(C) ≤ n(C) with equality only whenΣ(C) = Σ(C).

Let

Qn := Σ(C)/Inn(G)

Proposition 4.5.5. If C is a class vector of conjugacy classes Ci of a finite group G then

n(C) =∑

[σ]∈Qn

Z(G)

CG(〈σ〉)

Proof. Let G act on Σ(C) by conjugation (as above). The class equation gives,

|Σ(C)| =∑

[σ]∈Qn

|G : CG(〈σ〉)| (4.2)

Now,

n(C) =|Σ(C)||Inn(G)|

=∑

[σ]∈Qn

|G : CG(〈σ〉)||Inn(G)|

(by 4.2)

=∑

[σ]∈Qn

|G|/|Inn(G)||CG(〈σ〉)|

Now G/Z(G) ∼= Inn(G)⇒ |G|/|Inn(G)| = |Z(G)| ⇒ result.1

1This follows from the first isomorphism theorem, let h : G → Aut(G) by g 7→ πg then Im(h) = Inn(G) andKer(h) = Z(G).

29

Theorem 4.5.6. Let C a class vector in a finite group G, and let s ≥ 2. Then

n(C) = |Z(G)|∑

χ∈Irr(G)

|G|s−2

χ(1)s−2

s∏i=1

χ(σi)

|CG(σi)|(σi ∈ Ci)

Proof. χ ∈ Irr(G). Let R : G → GLn(C) be corresponding representation. Schur’s lemma ⇒for each σ ∈ G ∃ ω(σ) ∈ C such that

1

|G|∑ρ∈G

R(σρ) = ω(σ)In (ω(σ) =χ(σ)

χ(1))

It is clear then that for σ, τ ∈ G

1

|G|∑ρ∈G

R(σρτ) = ω(σ)In ·R(τ) =χ(σ)

χ(1)R(τ)

Furthermore, for ρ = (ρ1, ρ2) ∈ G2, σ1, σ2, τ ∈ G we have,

1

|G|2∑ρ∈G2

R(σρ11 σρ22 τ) =

χ(σ1)

χ(1)· χ(σ2)

χ(1)·R(τ)

By induction we can extend this, for ρ = (ρ1, . . . , ρs), to,

1

|G|s∑ρ∈Gs

R(σρ11 , . . . , σρss τ) =

χ(σ1) · · ·χ(σs)

χ(1)sR(τ)

Set τ = 1 and take traces,

1

|G|s∑ρ∈Gs

χ(σρ11 , . . . , σρss ) =

χ(σ1) · · ·χ(σs)

χ(1)sχ(1) =

χ(σ1) · · ·χ(σs)

χ(1)s−1

Multiply both sides by χ(1)|G|s−1. LHS becomes(χ(1)|G|s−1

) 1

|G|s∑ρ∈Gs

χ(σρ11 · · ·σρss ) =

1

|G|∑ρ∈Gs

χ(1)χ(σρ11 · · ·σρss )

Hence equation becomes,

1

|G|∑ρ∈Gs

χ(1)χ(σρ11 · · ·σρss ) = |G|s−1χ(σ1) · · ·χ(σs)

χ(1)s−2(4.3)

Define

m(C) :=∑ρ∈Gs

1

|G|∑

χ∈Irr(G)

χ(1)χ(σρ11 · · ·σρss )

Summing over both sides of (4.3) over χ ∈ Irr(G) gives

m(C) = |G|s−1∑

χ∈Irr(G)

χ(σ1) · · ·χ(σs)

χ(1)s−2

Now

n(C) =1

|Inn(G)||{σ ∈ C : σ1 · · ·σs = 1}|

30

may be written as

n(C) =m(C)

|Inn(G)|

s∏i=1

1

|CG(σi)|

And so, recalling that|G|

|Irr(G)|= |Z(G)| =⇒ 1

|Irr(G)|=|Z(G)||G|

we get,

n(C) =|Z(G)||G|

m(C)

s∏i=1

1

|CG(σi)|

=|Z(G)||G|

|G|s−1∑

χ∈Irr(G)

χ(σ1) · · ·χ(σs)

χ(1)s−2

= |Z(G)|∑

χ∈Irr(G)

|G|s−2

χ(1)s−2

s∏i=1

χ(σi)

|CG(σi)|

And the theorem is proved.

This next result is key and is used frequently in applications of rigidity [7].

Corollary 4.5.7. A class vector is C is rigid if the following hold:

1. G = 〈σ〉 with σi ∈ Ci : σ1 · · ·σs = 1

2.∑

χ∈Irr(G)

χ(σ1) · · ·χ(σs)

χ(1)s−2=|CG(σ1)| · · · |CG(σs)||G|s−2|Z(G)|

Proof. (2.) ⇒ n(C) = 1. Recall from remark 4.5.4 that l(C) ≤ n(C). Well (1.) ⇒ l(C) ≥ 1.Therefore, together we get that l(C) = 1 and so C is a rigid class vector.

Examples

1. The Special Linear Group, SL2(8)Using GAP to compute the character table of SL2(8) (see Appendix A, Figure A.1) let’sverify that the class vector C = (9a, 9b, 9c) is rigid; firstly lets check condition (2.):

LHS =

(1 · 1 · 1

1+

1 · 1 · 17

+ABC

7+BCA

7+CAB

7+−1 · −1 · −1

8

)= (1 +

1

7+

1

7+

1

7+

1

7− 1

8) =

81

56

Now to calculate the sizes of the centralisers, the size of the conjugacy classes have beenfound, again in Appendix A, Figure A.1, together with the formula |Cl(σ)| = |G : CG(σ)|.Hence

RHS =9 · 9 · 9504 · 1

=729

504=

81

56

and thus condition (2.) is satisfied. Now lets verify condition (1.):

Assume for a contradiction that the condition is false, that is, σ ∈ C : σ1σ2σ3 = 1 ⇒〈σ〉 ( G. Only maximal subgroups of G with order divisible by 9 are isomorphic to D18

[7] and hence 〈σ〉 = Z9 ⇒ σ2 either σ21 or σ71 and σ3 either σ41 or σ51.

31

But σ1σ2σ3 = σ1σi1σ

j1 6= 1 for i=2 or 7, j=4 or 5 ⇒⇐ Contradiction.

And so we have verified that C is a rigid class vector in SL2(8). Now we have a rigid typewith G = SL2(8) and thus there exists a unique FG-extension of C(x) of this type uptoC(x)-isomorphism.

2. The Symmetric Group, S4Let σ1 = (34), σ2 = (123), σ3 = (3421). Lets verify that C = (C(2), C(3), C(4)) forms arigid class vector.

First show that 〈σ〉 = S4. One calculates

σσ21 = (1 4) (1 4)σ2 = (2 4) (2 4)σ3 = (1 2)

(1 2)σ3 = (1 3) (1 2)σ2 = (2 3) (2 3)σ3 = (1 4)

and hence by conjugation one sees that all transpositions are in 〈σ〉 and therefore 〈σ〉 = S4.Moreover σ1σ2σ3 = 1 by inspection. It remains to verify the second condition using thecharacter table of S4 which I generate using GAP, and using the formula Corollary 4.5.7∑

χ∈Irr(G)

χ(σ1) · · ·χ(σs)

χ(1)s−2=|CG(σ1)| · · · |CG(σs)||G|s−2|Z(G)|

See Appendix A, Figure A.2 for the character table of S4, which we can use to calculate,

LHS =∑

χ∈Irr(G)

χ(σ1)χ(σ2)χ(σ3)

χ(1)3−2=

(1 · 1 · 1

1+

(−1) · 1 · (−1)

1

)= 2

and for the size of the conjugacy classes, which can be used to calculate,

|CG((3 4))| = 4, |CG((1 2 3))| = 3, |CG((3 4 2 1))| = 4.

Therefore,

RHS =|CG((3 4))||CG((1 2 3))||CG((3 4 2 1))|

|S4||Z(S4)|=

4 · 3 · 424 · 1

= 2

and thus Corollary 4.5.7 gives that C is a rigid class vector in Sn

32

Chapter 5

The Rigidity Criteria

Now we have defined rigidity, and found some rigid class vectors. Riemann’s Existence Theo-rem then guarantees the existence of an extension over C(x) of the desired type, but this hasn’ttold us anything about extensions of Q. In this chapter the importance of rigidity in our con-text of trying to find groups as Galois groups over Q is revealed in the Rational Rigidity Criteria.

Riemann’s Existence Theorem and Hilbert’s Irreducibility Theorem are combined to give acondition on G, a finite group, that if satisfied gives a positive solution to the Inverse GaloisProblem for G. If G forms a rigid type, Riemann’s Existence Theorem gives an extension overC(x) of that type. In this chapter a condition, rational rigidity, is shown to ensure an extensionof Q(x) of that same type. Hilbert’s Irreducibility Theorem provides the final step by concludingthat, since Q is hilbertian, G is realisable over Q.

5.1 Descent of the Base Field

Definition 5.1.1. L/k(x) is defined over κ if there exists subfield Lκ ⊆ L, Galois over κ(x)and regular over κ and such that [L : k(x)] = [Lκ : κ(x)]

Suppose L is defined over κ. Consider the Galois extension Lκ/κ(x). Let θ be a primitiveelement of the extension, that is Lκ = κ(x)(θ). That is to say, θ satisfies an irreducible poly-nomial F (y) ∈ κ(x)[y] of degree n = [L : k(x)]. By Lemma 3.1.4, since Lκ is regular over κby definition, F (y) remains irreducible over κ′(x)[y], where κ′ is an intermediate field betweenκ and k. Now the roots of F (y) are all contained in Lκ′ , so Lκ′ is Galois over κ′(x). Theirreducibility of F (y) over κ′(x) implies that Lκ is a degree n extension over κ′(x), and it isgeometric too by Lemma 3.1.4. Therefore L is defined over κ′, for any intermediate field κ′

between κ and k. This proves the following result:

Lemma 5.1.2. If L is defined over κ and θ is a primitive element of the extension Lκ/κ(x)then L = k(x)(θ)

Proof. Simply notice that k can be seen as an intermediate field κ′, thus L is defined over k.Hence k(x)(θ) is an extension of degree n, but it is a subfield of L, therefore L = k(x)(θ) asrequired.

So we have that if L is defined over κ then it is defined over all intermediate fields κ ⊆ κ′ ⊆ k,and the field Lκ′ was generated by the roots of the minimal polynomial F (y) of the primitiveelement θ for the extension Lκ/κ(x). But by Lemma 5.1.2 L = k(x)(θ) and so the Galoisgroup Gal(L/k(x)) acts naturally to permute these roots. The following result is immediatelyobtained

33

Lemma 5.1.3. Gal(L/k(x)) ∼= Gal(Lκ/κ(x)) via restriction to Lκ.

And so one sees the utility of so called fields of definition in solving the Inverse Galoisproblem as stated, to find groups as Galois groups over Q, given that Riemann’s ExistenceTheorem only guarantees extensions of C(x) of desired type.

Lemma 5.1.4. Let L be defined over κ and suppose that either Gal(L/k(x)) has trivial centeror that κ is algebraically closed. Then Lκ is the unique subfield of L, Galois over κ(x) of degreen and regular over κ.

Proof. Suppose that Lκ is another subfield of L satisfying the properties of Definition 5.1.1.Consider the subfield K of L generated by Lκ and Lκ. Since these are both Galois over κ(x)then so is K . Let κ′ be the algebraic closure of κ in K. Then κ′ and thus κ′(x) are invariantunder Gal(K/κ(x)), so κ′(x) is Galois over κ(x). Let θ be a primitive element for Lκ/κ(x) asin Lemma 5.1.2. Then, Lκ ⊆ K ⇒ |K : κ(x)| ≥ |κ′(x)(θ) : κ′(x)| = n. Let γ be a primitiveelement for K/κ′(x). By Lemma 3.1.4 the minimal polynomial for γ remains irreducible overk(x). Therefore |K : κ′(x)| ≤ n = |L/k(x)|. Therefore |K/κ′(x)| = n. That is K = κ′(x)(θ).

If κ is algebraically closed then κ′ = κ. So K = κ′(x)(θ) = κ(x)(θ) = Lκ by definition ofθ. So indeed Lκ is unique.

Now suppose Gal(L/k(x)) has trivial center. We have that κ′(x)∩Lκ = κ′(x)∩ κ(x)(θ) = κ(x)since κ is algebraically closed in Lκ by definition. The same holds for Lκ by assumption thatLκ satisfies the same properties as Lκ. Hence by the Galois correspondence

Gal(K/κ(x)) = Gal(K/Lκ)×Gal(K/κ′(x))

Gal(K/κ(x)) = Gal(K/Lκ)×Gal(K/κ′(x))

Moreover, by Lemma 5.1.3, Gal(K/κ′(x)) ∼= Gal(L/k(x)), which by assumption has trivialcenter. Therefore Gal(K/Lκ) = G(K/Lκ) = CGal(K/κ(x))(Gal(K/κ

′(x))). Therefore Lκ = Lκand indeed Lκ is unique.

Consider the finite Galois extension L/k(x) with primitive element θ satisfying the irre-ducible polynomial F (y) ∈ k(x)[y]. Let {θi}ni=1 be the roots of F . There exists polynomialsfi(y) ∈ k(x)[y] such that θi = fi(θ). The coefficients of F and the fi lie in k(x) and so lie inκ′(x) for a finitely generated extension κ′ of κ. Let Lκ′ = κ′(x)(θ), Galois over κ′(x) and regularover κ′. Hence L is always defined over a finitely generated extension κ′ of κ.

Lemma 5.1.5. Suppose M/κ(x) is a finite Galois extension which is regular over κ. Then thereexists a finite Galois extension L/k(x), defined over κ with Lκ ∼= M , where Lκ as in Definition5.1.1.

Proof. Let γ be a primitive element for the extension M/κ(x). Define L := k(x)(γ), a finiteGalois extension of k(x) of the same degree as M/κ(x). Now by the same argument as inLemma 5.1.2 in reverse, get that L is defined over κ with Lκ = M .

Have P1 = C ∪ {∞}. Recall that to a finite Galois extension L/k(x) we associate a triple(G,P,C) where G = Gal(L/k(x)), P is the set of branch points and Cp is the class of Gassociated to p ∈ P . It is from this association that the notion of ramification type arises. Nowhow does the ramification type of L/k(x) relate to that of Lκ/κ(x) when L is defined over κ?If κ is algebraically closed the following theorem answers this,

34

Theorem 5.1.6.

1. Let L be defined over κ. Then L/k(x) is naturally of the same ramification type at Lκ/κ(x)if κ is algebraically closed.

2. If K is an algebraically closed subfield of C then for each weakly rigid type T there is atmost one extension of K(x) of this type.

Proof.

1. Let θ be a primitive element for Lκ/κ(x). Then by Lemma 5.1.2, since L is defined overκ, L = k(x)(θ). Let p ∈ κ ∪ {∞} and define vp : k(x) → k(t) as when defining branchpoints. Similrly, this extends to v : L → ∆, where ∆ is a finite Galois extension ofΛ = k((t)). Now v(κ(x)) = κ(t) and hence v(Lκ) = v(κ(x)(θ)) = κ(t)(v(θ)). Thus, byrestricting to Lκ one obtains v : Lκ → ∆κ where ∆κ := κ((t))(v(θ)). Hence restrictionyields an isomorphism between Gal(∆/Λ) and Gal(∆κ/κ((t))), mapping the class Cp of

Gal(L/k(x)) associated with p to the class Cp of Gal(Lκ/κ(x)) associated with p, for allp ∈ κ∪{∞}. So the question now is, where are the branch points of L/k(x) and Lκ/κ(x)?θ satisfies monic irreducible polynomial F (y) ∈ κ(x)[y]. The discriminant D(x) ∈ κ(x) isnon-zero because F is irreducible and hence separable. Let p ∈ κ with D(p) 6= 0. ViewF (y) ∈ κ(x)[y] as F (x, y) ∈ κ[x, y]. Then (vpF )(y) = F (t + p, y). So (vpF ) is a monicpolynomial in y with coefficients in κ[t]. By Lemma 4.1.1 then vpF factors as a product oflinear factors in Λ[y] if F (p, y) is separable, which is the case when D(p) 6= 0. And theseare precisely the points with eL,p = 1 by 4.3.7. Therefore, branch points, which are thepoints where eL,p 6= 1 are roots of D(x), and so are algebraic over κ. But κ is algebraicallyclosed by assumption, so the branch points of L/k(x) lie in κ∪{∞}, and clearly so do thebranch points of Lκ/κ(x). Thus the restriction isomorphism yields an equivalence of thetriple (Gal(L/k(x)), P, Cp) with (Gal(Lκ/κ(x)), P, Cp) in other words both extensions areof the same ramification type.

2. Suppose for a contradiction that M and N are different finite Galois extensions of K(x)of weakly rigid ramification type T . By Lemma 5.1.5 there is a finite Galois extensionL of C(x) defined over K such that LK is isomorphic to M , and similarly there existsfinite Galois extension L′/C(x) such that L′K is isomorphic to N . Then by (1) L and L′

are of the same type, T . By Remark 4.4.9 then L and L′ are C(x)-isomorphic. Now byLemma 5.1.4, since K is algebraically closed in C, we have that LK ⊂ L is unique, andsimilarly, L′K ⊂ L′ is unique. But L isomorphic to L′ implies that LK is isomorphic toL′K i.e. M = N , which is a contradiction.

α-Isomorphisms

Consider α ∈ Aut(k). We can extend this naturally to an automorphism of k(x) fixing x.Suppose we have two finite Galois extenions of the same base field i.e.

L/k(x) L′/k(x)

and suppose there is a field isomorphism λ : L → L′ which acts in the same way on the basefields of these extensions, as automorphisms of k(x), that is λ|k(x) = α. Such a field isomorphism

35

is called an α-isomorphism. By their very nature, each α-automorphism λ induces a groupisomorphism

λ∗ : Gal(L/k(x))→ Gal(L′/k(x))

g 7→ λgλ−1

It is plain to see that this is indeed an isomorphism. It is a homomorphism since λ∗(g)λ∗(h) =λgλ−1λhλ−1 = λghλ−1 = λ∗(gh). Now λ : L → L′ is by definition a field isomorphism, fromwhich λ∗ is an isomorphism follows.

Let α ∈ Aut(k) and L/k(x) a finite Galois extension. Construct an α-isomorphism from α

as follows: Take L of the form k(x)[y]/

(F ) with F (y) ∈ k(x)[y] an irreducible polynomial.

Extend α to an automorphism of k(x, y) fixing x and y, denote this by α. Now α induces a ringisomorphism λ,

λ : k(x)[y]/

(F ) → k(x)[y]/

(αF )

Moreover λ|k(x) = α, so indeed λ is an α-isomorphism.

Lemma 5.1.7. Let L be defined over κ and α ∈ Aut(k). If α|κ = Id then there exists anα-isomorphism λ : L→ L with the induced map λ∗ = Id on Gal(L/k(x)).

Proof. L is defined over κ so by 5.1.2, if θ is a primitive element of Lκ/κ(x) then L = k(x)(θ).That is, we have

L = k(x)[y]/

(F ) Lκ = κ(x)[y]/

(F )

where F (y) ∈ κ(x)[y] is minimal polynomial of θ. Now construct an α-isomorphism as describedabove, where λ : L→ L′ and

L′ = k(x)[y]/

(αF )

but by assumption α|κ = Id, so α(F ) = F . Therefore L′ = L. Finally, for g ∈ Gal(L/k(x)),have λ∗(g)(θ) = λgλ−1(θ). But λ fixes Lκ, and g(θ) ∈ Lκ. So

λg(λ−1θ) = λ(gθ) = g(θ)

so indeed λ∗ = Id.

Let G a group with N C G. If K ≤ G such that G = KN,K ∩ N = {1} and K C G thenG = K ×N . This is called a normal compliment [9].

Lemma 5.1.8. Let G be a group with N,K C G forming a normal compliment. Then G is adirect product G = K ×N .

Proof. Consider [k, n] = k−1n−1kn

k−1n−1k ∈ N since N CG, so [k, n] ∈ Nn−1kn ∈ K since K CG, so [k, n] ∈ K

So [k, n] ∈ K ∩ N = {1}. This holds for all k ∈ K,n ∈ N so have [K,N ] = 1. Now,since G = KN all g ∈ G can be written as g = kn for some k ∈ K,n ∈ N . I claim thisexpression is unique. Indeed suppose g = k1n1 = k2n2 for k1, k2 ∈ K,n1, n2 ∈ N . Then

36

k−12 k1 = n2n−11 ∈ K ∩N = {1}. So k−12 k1 = 1⇒ k1 = k2 and n2n

−11 = 1⇒ n1 = n2 so indeed

the expression is unique. Define a map

ϕ : G→ K ×Ng = kn 7→ (k, n)

Claim ϕ is an isomorphism:g1, g2 ∈ G then write g1 = k1n1, g2 = k2n2. Then g1g2 = k1n1k2n2 = k1k2n1n2 since [K,N ] = 1.So ϕ(g1)ϕ(g2) = (k1, n1)(k2, n2) = (k1k2, n1n2) = ϕ(g1g2), so ϕ is a homomorphism. Since everyg ∈ G is of the form kn for k ∈ K,n ∈ N ϕ is onto, hence is an isomorphism. This completesthe proof.

Now we consider the descent from κ to κ, where κ is the algebraic closure of κ in C. Asa technicality note that the group Gal(κ/κ) is naturally a profinite group. For my purposes Iview it as an abstract group, but it has the property that every for finite Galois extension κ′ ofκ inside κ the restriction map Gal(κ/κ)→ Gal(κ′/κ) is a surjection [12].

Lemma 5.1.9. Let k = κ and let L/k(x) be a finite Galois extension whose Galois groupGak(L/k(x)) has trivial center. Then L is defined over κ if and only if for each α ∈ Gal(κ/κ) ∃an α-isomorphism λ : L→ L with λ∗ = Id.

Proof. Suppose L is defined over κ. α ∈ Gal(κ/κ) ⇒ α|κ = Id by definition of Galois group.Then by Lemma 5.1.7 there does indeed exists an α-isomorphism λ : L → L which induces atrivial map on the Galois group.Conversely, suppose for each α ∈ Gal(κ/κ) ∃ α-isomorphism λ : L → L with λ∗ = Id. L isalways defined over a finitely generated extension κ1 of κ. Set L1 := Lκ1 . Let α1 ∈ Gal(κ1/κ).Since restriction Gal(κ/κ) is surjective we can extend α1 to an element α ∈ Gal(κ/κ), which byassumption has an α-isomorphism λ : L→ L associated to it, such that λ∗ = Id. Gal(L/k(x))has trivial center by hypothesis. Consider λ(L1). This is a subgroup of L, Galois over κ1(x)and regular over κ1. But Lemma 5.1.4 says that if L1 is the unique group with these properties.Thus L1 = λ(L1). So λ1 := λ|L1 : L1 → L1. That is λ1 ∈ Aut(L1). But λ1 is an α1-isomorphism for α1 ∈ Gal(κ1/κ) ∼= Gal(κ1(x)/κ(x)). Therefore (λ1)|κ(x) = α1 which fixes

κ(x) and so λ1 ∈ Gal(L1/κ(x)). Now, λ∗ = Id so λ1gλ−11 = g for all g ∈ Gal(L/k(x)). But

Gal(L/k(x)) ∼= Gal(L1/κ1(x)) by 5.1.3. So for g1 ∈ Gal(L1/κ1(x)),

λ1g1 = g1λ1

Hence λ1 ∈ CGal(L1/κ(x))(Gal(L1/κ1(x))) =: C. This holds for each α ∈ Gal(κ1/κ), so C

surjects onto Gal(L1/κ(x))/Gal(L1/κ1(x)) := H/G, that is H = GC. Moreover G ∩ C lies in

the center of Gal(L1/κ1(x)) ∼= Gal(L/k(x)) which has trivial center, so G ∩C = {1}. This is anormal compliment and so, by Lemma 5.1.8, we get that H = G× C, or in full,

Gal(L1/κ(x)) = Gal(L1, κ1(x))× C

Let Lκ be the fixed field of C in L1. Then Gal(Lκ/κ(x)) = H/C = G ∼= Gal(L/k(x)). Now, Lκis the fixed field of C by definition and κ1(x) is the fixed field of Gal(L1/κ1(x)) by definitionof Galois group. Hence their intersection is the fixed field of CGal(L1/κ1(x)) = Gal(L1/κ(x))which is κ(x). But Lκ ⊂ L1 and L1 is regular over κ1 therefore Lκ is regular over κ, showingthat L is defined over κ, as required.

37

5.2 Rigidity Criteria

A standard definition in group theory is that of a rational conjugacy class; we say C, a conjugacyclass in a group G is called rational if Cm = C for all m ∈ Z coprime to the order of G [9].Now we extend this definition to ramification types T ,

Definition 5.2.1. A ramification type T = [G,P, (Cp)p∈P ] is κ-rational if P ⊂ κ ∪ {∞} andfor each p ∈ P and α ∈ Gal(κ/κ), α(p) ∈ P and Cα(p) = Cmp where m ∈ Z is such that

α−1(ζn) = ζmn for ζn an nth root of unity in κ where n = |G|.

Notice the similarity between the condition in this definition and the condition in the BranchCycle Argument, Theorem 4.3.11. The significance of this similarity is revealed in the nexttheorem,

Theorem 5.2.2. Let k = κ and let L/k(x) be an FG-extension with ramification type T

1. If L is defined over κ then T defines a κ-rational ramification type.

2. If the ramification type T is rigid and κ-rational, then L is defined over κ.

Proof.

1. L is defined over κ. Let α ∈ Gal(κ/κ). Then α|κ(x) = Id, so by Lemma 5.1.7 there is anα-isomorphism λ : L → L with λ∗ = Id. Now let m ∈ Z such that α−1(ζn) = ζmn . Thenby the Theorem 4.3.11 (Branch Cycle Argument), since α extends to an α-isomorphismλ : L → L with λ(x) = x, the class Cα(p) = λ∗(Cp)

m = Cmp . Hence T is a κ-rationalramification type.

2. T is rigid so G = Gal(L/k(x)) has trivial center. Firstly, fix some α ∈ Gal(κ/κ). Foreach α ∈ Aut(k) there is an α-isomorphism λ : L → L′ where L′ is some finite Galoisextension of k(x). Let G′ = Gal(L′/k(x)). For each p ∈ P1 = κ ∪ {∞} let Cp be theclass of G associated with p, and let C ′p be the class of G′ associated with p. Now T isκ-rational by hypothesis, therefore α(p) ∈ P , say α(p) = q, and so Cα(p) = Cq = Cmp . Soby Theorem 4.3.11, C ′q = λ∗(Cmp ) = λ∗(Cq). Now λ∗ is an isomorphism from G → G′,and it takes the class Cq to C ′q. Therefore (G,P,Cp) ∼ (G′, P, C ′p), and so L/k(x) andL′/k(x) have the same ramification type T . Now k = κ is an algebraically closed subfieldof C by definition, T is a rigid type, so by Lemma 5.1.6 there is at most one finite Galoisextension of this type up to k(x)-isomorphism. So L and L′ are k(x)-isomorphic. Let thisisomorphism be µ : L′ → L ,and define a = µλ : L → L. Then a|k(x) = µλ|k(x) = Idso a is an α-isomorphism. Recall from Lemma 5.1.9 that if for each α ∈ Gal(κ/κ) thereexists and α-isomorphism λ : L → L with λ∗ = Id then L is defined over κ. Thus anα-isomorphism satisfying this property will complete the proof.

C ′q = λ∗(Cq) ⇒ a∗(Cq) = µ∗λ∗(Cq) = µ∗(C ′q) = Cq as µ : L′ → L is a k(x)-isomorphism.Therefore a∗ fixes all the classes Cq. By Lemma 4.4.6 then, a is an inner automorphism.Given h ∈ G, say a∗(h) = ghg−1 with g ∈ G. g ∈ G induces g∗ : G → G by h 7→ ghg−1.Now define b = g−1a. Then b∗ = (g∗)−1a∗ = Id. Moreover b|k(x) = g−1|k(x)a|k(x). Buta is an α-isomorphism so is identity on k(x), and g ∈ G = Gal(L/k(x)) so fixes k(x)by definition of Galois group. Therefore b is an α-isomorphism with b∗ = Id, and thiscompletes the proof.

38

A usefull observation is that T = [G,P, (Cp)p∈P ] is κ-rational if P ⊂ κ∪{∞} and each classCp is rational in the usual sense. Indeed, if α ∈ Gal(κ/κ) and p ∈ P ⊂ K ∪ {∞} then α(p) = pand so by rationality of classes Cmp = Cp = Cα(p).

Theorem 5.2.3. Given a rigid and κ-rational ramification type T = [G,P,C] there exists aunique finite Galois extension L/C(x) of this type, defined over a purely transcendental extensionκ(t1, . . . , ts) of κ and G occurs regularly over κ.

Proof. T is rigid, so by 4.4.8, there exists a unique finite Galois extension of C(x) of this type,say L/C(x). L is defined over a finitely generated extension of κ, say κ1 = κ(a1, . . . , ar). Ift1, . . . ts is a collection of elements among a1, . . . , ar maximal with respect to being algebraicallyindependent, then κ1 is finitely generated and algebraic over κ0 = κ(t1, . . . , ts), so κ1 is finiteover the purely transcendental extension κ0 of κ. Let k = κ0 = κ1. Then by Lemma 5.1.2 Lis defined over k, and Lκ/κ(x) is of the same type as L/k(x) (Theorem 5.1.6), and this type isrigid and κ-rational by hypothesis.

Given an element g ∈ Gal(κ0/κ0), since κ ⊂ κ0, can restrict g to g ∈ Gal(κ/κ). ThereforeT is κ-rational ⇒ T is κ0-rational. Hence Lκ has rigid and κ0-rational type. Hence by Theo-rem 5.2.2 Lκ is defined over κ0. Let M be the subfield of Lκ, regular of κ0, Galois over κ0(x),from the definition 5.1.1. Then M is also a subfield of L with the same properties⇒ L is definedover κ0. Now by Lemma 5.1.3

G = Gal(L/C(x)) ∼= Gal(M/κ0(x)) = Gal(M/κ(t1, . . . , ts, x))

M is regular over κ0 by definition, and κ0 is regular over κ, therefore M is regular over κ. Thatis, H occurs regularly over κ.

Recall that the Inverse Galois Problem seeks to realise groups as Galois groups of extensionsof Q. With that in mind, consider the special case of the theorem, with κ = Q. This gives acriterion for realising groups over Q.

Theorem 5.2.4. (Rational Rigidity Criterion)Let G be a group with a rigid tuple (C1, . . . , Cr) such that Ci are rational conjugacy classes.Then G occurs regularly over Q.

Proof. This is an immediate corollary of Theorem 5.2.3 with P ⊂ Q ∪ {∞}.

In more complicated examples it is often difficult to find rigid tuples that are rational. Oftenone uses a field that provides a less strict κ-rationality condition. The best and most commonlyused example of this is the field Qab, generated by all roots of unity. Let α ∈ Gal(Q/Qab).Then α(ζn) = ζn. Thus if P ⊂ Qab ∪ {∞} the α(p) = p and so Cα(p) = Cp trivially satisfies the

definition of Qab-rationality. Hence,

Corollary 5.2.5. (Qab Rigidity Criterion)Let G be a group with rigid tuple (C1, . . . , Cr). Then G occurs regularly over Qab

Recall that the definition of rational classes was extended to ramification types T in 5.2.1.Now generalise to define a κ-rational class vector C.

Definition 5.2.6. Let C = (C1, . . . , Cr) be a class vector in G, a finite group. Then C isκ-rational if Cm1 , . . . , C

mr is a permutation of C1, . . . , Cr for each m ∈ Z such that there exists

α ∈ Gal(κ/κ) with α−1(ζn) = ζmn where n = |G|

39

It would be preferable to have a general condition that applied to class vectors, rather thanspecifically on ramification types as in Theorem 5.2.3. Hence

Theorem 5.2.7. (The General Rigidity Criterion)Let κ ⊂ C, and let G be a finite group. If C is a rigid and κ-rational class vector then G occursregularly over κ.

Proof. Need to show that a κ-rational class vector ensures a κ-rational ramification type. LetZ = Gal(κ(ζn)/κ). If α ∈ Z then α−1(ζn) = ζmn . Hence define an action of Z on the classvector C by α(C) = Cm. Let Z1 be the stabiliser of C1 in C. Consider the quotient Z

/Z1 =

{α1Z1, . . . , αsZ1}. Now α1(C1), . . . , αs(C1) are distinct. Moreover, by κ-rationality they are asubset of C. WLOG say αi(C1) = Ci (possibly relabelling some of the classes). Now Z1 ≤ Z =Gal(κ(ζn)/κ), so Z1 acts on κ(ζn). Consider the fixed field κ1 of this action. By properties ofGalois groups, since Z1 ≤ Z, and Z fixes κ, then κ1/κ is an extension. Let p1 be a primitiveelement for the extension. Define pi = αi(p1). Then Z acts on p1, . . . , ps. Now the stabiliserof p1 is Z1, which is the stabiliser of C1 under Z. Hence there is an isomorphism pi 7→ Ci.Further, Cα(p) = α(Cp) = Cmp , as is required for a κ-rational ramification type. However,the set {p1, . . . , ps} needs to be extended to have size r = |C|. Do so by induction. LetP = {p1, . . . , pr}, and define T = [G,P, (Cp)p∈P ]. Then this is a κ-rational ramification type.Now by Theorem 5.2.3, G occurs regularly over κ.

Recall Hilbert’s Irreducibility Theorem, which said that Q is a hilbertian field. Therefore,if G occurs regularly over Q, then G occurs as a Galois group over Q. This gives the mostappropriate form of a rigidity criterion in the context of the Inverse Galois Problem over Q,this is the Rigidity criterion over Q.

Theorem 5.2.8. (Rigidity Criterion over Q)Let G be a finite group with rigid and rational class vector C. Then G occurs regularly over Q.Therefore G is a Galois group over Q, and the Inverse Galois Problem has positive solution forG.

40

Chapter 6

Applications of Rigidity

In this chapter, the results obtained to this point are applied to show that the Inverse GaloisProblem has a positive solution in these examples. Rigidity is applied to the symmetric andalternating groups on n letter. Although Sn has already been shown to have a positive solution tothe Inverse Galois Problem as an application of Hilbert’s Irreducibility Theorem, it is presentedagain as a nice example of how rigidity can be used. Then, by imposing a congruence conditionon p, PSL2(p) is realised as a Galois group over Q. The value of the character theoreticformula is then highlighted with several examples. It is particularly useful when approachingthe sporadic groups, as their character tables are known. To demonstrate this M12 and theMonster are realised. As was alluded to earlier, it is often difficult to ensusre rationality of theclass vector, and as a result, often one has to settle for a realisation over a different field. Anexample of such a case is presented as the Ree groups in characteristic 3 are realised over Qab.

6.1 Sn and An as Galois groups over Q

In Sn the cycle shape determines the conjugacy class. Now conjugation by a given cycle onan element of the same cycle shape does not change the cycle shape, therefore the conjugacyclasses of Sn are rational. It remains to find a rigid class vector. Then by Theorem 5.2.8 Snwill occur as a Galois group over Q. I will construct such a rigid tuple.

Let a = (1 · · ·n− 1) and b = (n− 1 n n− 2 · · · 1) be an (n− 1)-cycle and an n-cycle. Now letC(i) denote the conjugacy class of i-cycles in Sn. Then a ∈ C(n−1), b ∈ C(n).

Claim 1: A subgroup of Sn containing a and b is transitive.Pf: Is clear. a takes x 7→ x + 1 for x ∈ [1, . . . , n − 2], b takes x 7→ x − 1 for x ∈ [2, . . . , n − 2]and takes n− 1 7→ n. So by repeated application of a and/or b one can take any x ∈ [1, . . . , n]to any other. Thus H is indeed transitive.

Claim 2: A transitive subgroup of Sn, with n ≥ 2 which contains an (n − 1)-cycle and atransposition, is the whole of Sn.Pf: Let H be said subgroup. Without loss of generality we may assume the (n − 1)-cycle isa = (1 · · ·n− 1) (possibly after relabelling the elements). Let (i j) be the transposition. SinceH is transitive on [1, . . . , n] by hypothesis, there exists h ∈ H such that h(j) = n. Moreover,the element (i j)h = h(i j)h−1 = (hi hj) = (x n) ∈ H by closure. Now am(x n)a−m =(am(x) am(n)). Nut a fixes n, so (am(x) am(n)) = (x + m n) where m ∈ [1, . . . , n] and wherex + m cycles through (1 · · ·n − 1). Therefore by repeated application of a to (x n) we obtain(1 n), (2 n), . . . , (n − 1, n) ∈ H. Finally, (i n)(j n)(i n) = (i j) ∈ H. Therefore H contains all

41

transposition and thus H = Sn.

Let C = (C(2), C(n−1), C(n)) and let σ1 = (n − 1 n) ∈ C(2), σ2 = (1 · · ·n − 1) ∈ C(n−1), σ3 =(n − 1 n n − 2 · · · 1) ∈ C(n). Then the subgroup generated by σ = (σ1, σ2, σ3) is the whole ofSn by the two claims. Moreover,

σ1σ2σ3 = (n− 1 n)(1 · · ·n− 1)(n n− 1 n− 2 · · · 1) = (n− 1 n)(n− 1 n) = 1

Therefore σ ∈ C generates Sn and satisfies σ1σ2σ3 = 1. It remains to show that if σ′ ∈ C isanother triple with the same properties then there exists τ ∈ Sn such that τσiτ = σ′i.

Claim 3: Let (C1, C2, C3) be a class vector in H with generators g1, g2, g3 such that gi ∈ Ci andg1g2g3 = 1. If H has trivial center and for each g′2 ∈ C2 with (g1g

′2)−1 ∈ C3 and H = 〈g1, g′2〉

there is h ∈ H with hg1h−1 = g1 and hg′2h

−1 = g2, then (C1, C2, C3) is a rigid class vector.Pf: Clear from definition of rigidity.

Now let σ′1 ∈ C(1). If σ2σ′1 ∈ C(n) then σ′1, σ2 generate Sn by Claim 1 and Claim 2. This

σ′1 is the g′2 in Claim 3. Well if σ2σ′2 is an n-cycle then σ′2 must be of the form σ′2 = (i n). Now

σ2(i n)σ−12 = (σ2(i) σ2(n)) = (i+ 1 n)

as in the proof of claim 2. Therefore repeated conjugation by σ2 will map (i n) to (n−1 n) = σ1.That is to say that σ′1 conjugates to σ1. Now it is clear tht σ2 commutes with σ2 and to whateverthe power of σ2 that conjugates σ′1 to σ1 will conjugate σ2 to itself. Thus the condition in Claim3 is satisfied, and hence the class vector is rigid.

Therefore the class vector C = (C(2), C(n−1), C(n)) is a rigid and rational class vector in Sn, andhence by Theorem 5.2.8, Sn occurs as a Galois group over Q.

I now appeal to a result of Serre to realise An over Q. For the proof I follow [12].

Lemma 6.1.1. Let G be a group with rigid and rational class vector C = (C1, C2, C3). Then ifH ≤ G such that |G : H| = 2 then H occurs regularly over Q.

Proof. First construct an appropriate ramification type T . Let P = {p1, p2, p3} where pi ∈ Qare distinct. Then T = [G,P,C] is a rigid and rational ramification type. Now Theorem 5.2.3says there exists a finite extension M/C(x) of type T , moreover G occurs regularly over Q.That is, M/C(x) is defined over a purely transcendental extension κ0 = Q(t1, . . . , ts), so bydefinition there exists subfield M0 := Mκ0 regular over κ0, Galois over C(x) and such that|M0 : κ0(x)| = |M/C(x)|. And by Lemma 5.1.3 Gal(M0/κ0(x)) = Gal(M/C(x)) = G. Fix suchan isomorphism. Now a subgroup H ≤ G of index 2 corresponds to a quadratic extension N0

of κ0(x) inside M0 by the Galois correspondence. Say N0 = κ0(x)(√f) with f ∈ κ0[x] a non-

constant polynomial with no multiple irreducible factors and hence no multiple roots over C.Let N = C(x)(

√f). Then N ⊂M is a quadratic extension of C(x) by Lemma 3.1.4. By Lemma

4.3.7 the ramification index e of points from N/C(x) and M/C(x) are the same. Therefore thebranch points of N/C(x) are branch points of M/C(x). Now the branch points of M/C(x) arep1, p2, p3, by construction of the type T . Now we may take f of the form

∏(x − pi), where

pi is a branch point of N/C(x), and ∞ is a branch point if and only if the number of branchpoints is odd [12]. That is to say that there is always an even number of branch points forquadratic extensions. Assume without loss of generality that p1, p2 are the branch points ofN/C(x). Then f(x) = b(x− p1)(x− p2) with b ∈ κ0 Thus

N0 = κ0(√f) = κ0

(√f/

(x− p2))

(6.1)

42

Let z2 = c(x−p1)(x−p2) . Then (6.1)= κ0(x)(z). Finally, notice that x ∈ κ0(z) since z2 is a linear

function of x. Therefore N0 = κ0(x)(z) = κ0(z) is a rational function field. Therefore H =Gal(M0/N0) = Gal(M0/κ0(z)) and so H occurs regularly over κ0 and therefore also occursregulary over Q by Theorem 5.2.3.

We are now able to realise An as a Galois group over Q as a corollary to this, and giventhat |Sn : An| = 2.

6.2 PSL2(q) for q 6≡ ±1 mod 24 as a Galois group over Q

A New Rigidity Criterion

For this example it is convenient to give a new rigidity criterion that applies to groups modulotheir center. Let C = (C1, . . . , Cr) be a class vector in a group H. Recall, we say C is weaklyrigid in H if ∃ generators σ1, . . . , σr of H with σ1 · · ·σr = 1 and σi ∈ Ci such that if σ′1, . . . , σ

′r

is another generating set with the same property then ∃ γ ∈ Aut(H) such that γ(σi) = σ′i.

Definition 6.2.1. We say C is quasi-rigid in H if it is weakly rigid and the automorphismin the definition is inner i.e. γ ∈ Inn(H).

For a rigid class vector we require the existence of a unique element h ∈ H such thathσih

−1 = σ′i. This uniqueness is equivalent to saying that H has trivial center. For a weakly-rigid class vector, since γ is defined on a generating set, then it must be unique. Thus in therigid case any automorphism of H fixing each Ci is inner. Therefore C is quasi-rigid and hastrivial center ⇐⇒ C is rigid.

Lemma 6.2.2. Let k = κ and let L/k(x) be an FG-extension with Galois group G. Let L′

be the fixed field of Z(G). If the ramification type of L is quasi-rigid and κ-rational then L′ isdefined over κ.

Proof. As in the proof of Theorem 5.2.2(2), for each α ∈ Gal(κ/κ) we get an α-automorphismλ of L with λ∗ = Id. Let θ generate L over k(x). Let θ1, . . . , θt be the conjugates of θ overκ(x). Then θi ∈ L since L is Galois over κ(x). Let L1 := κ(x)(θ1, . . . , θt). Then L1 is anFG-extension of κ(x). Let κ1 be the algebraic closure of κ in L1. Then L is defined over κ1as in Lemma 5.1.4, and G1 = Gal(L1/κ1(x)) ∼= G. Again, as in proof of Lemma 5.1.4, G1 isnormal in H := Gal(L1/κ(x)). Let C = CH(G1), then H = G1C. Now LC1 is Galois over κ(x)with Galois group

H /C = (G1C) /C ∼= G1

/(G1 ∩ C) = G1

/Z(G1)

where here LC1 denotes the fixed field of C in L1. We have that κ(x) = κ1(x) ∩ LC1 followsfrom H = G1C. Thus LC1 is regular over κ. Furthermore, from Lemma 5.1.3, we have LC1 ⊂LZ(G1)1 ⊂ LZ(G) = L′. Thus L′ is defined over κ with L′κ = LC1 .

If we now replace Theorem 5.2.2 with this lemma in the proof of the Theorem 5.2.7 we get

Theorem 6.2.3. If C is a quasi-rigid and κ-rational class vector in H then H/Z(H) occurs

regularly over κ.

43

The Groups PSL2(q) as Galois Groups

We consider only p odd and firstly show that PSL2(q) occurs regularly over Qab for powers ofodd primes. Then, by imposing a condition on the prime p we show PSL2(p) occurs regularlyover Q.

Let q be an odd prime.

Lemma 6.2.4. Let U1 =

(1 10 1

). Let U ∈ SL2(q) of trace 2, not upper triangular. Then

∃ A ∈ SL2(q) such that A ∈ CSL2(q)(U1) and A conjugates U into a matrix of the form

(1 0c 1

)

Proof. Let q odd prime. Consider U =

(a bc d

)∈ SL2(q) not upper triancular, so c 6= 0.

Take A of the form A =

(1 m0 1

), which clearly centralises U1. Then

AUA−1 =

(a+mc ∗

c ∗

)

With m = c−1(1 − a) get AUA−1 =

(1 ∗c ∗

). But, trace(U) = 2 ⇒ trace(AUA−1) = 2 and

AUA−1 ∈ SL2(q)⇒ det(AUA−1) = 1, thus

AUA−1 =

(1 0c 1

)

Lemma 6.2.5. There are precisely two conjugacy classes of non-identity elements of SL2(q)with trace 2, say C1, C2, where C1 is the class containing U1.

Proof. Non-identity elements in SL2(q) of trace 2 are of the form

(1 u0 1

),

(1 0v 1

)where

u, v 6= 0 in Fq. I claim these matrices are in C1 if and only if u or −v are square in Fq,respectively. Indeed, let U ∈ SL2(q) of trace 2. By Lemma 6.2.4, if U is not upper triangular

then it is conjugate to some

(1 0c 1

). We have

(0 1−1 0

)(1 0v 1

)(0 1−1 0

)−1=

(1 −v0 1

)(6.2)

So we have that U is conjugate to some

(1 u0 1

)(note: if U is upper triangular then this is

trivial). Need to see when a matrix of this form is conjugate to U1. Suppose B ∈ SL2(q) issuch that

BU1B−1 =

(1 u0 1

)U1 has eigenvalue 1 with corresponding eigenspace E1 = Sp{(1, 0)}. Then B must be uppertriangular as it fixes this eigenspace. Combining this with the fact that det(B) = 1 we see that

44

B is of the form

(w ∗0 w−1

). Now

(w ∗0 w−1

)U1

(w ∗0 w−1

)−1=

(1 w2

0 1

)Hence by 6.2 we have proved the claim, and the lemma follows.

We state a classical result of Dickson

(Dickson) Let q be a power of the odd prime p and Fq = Fp(c) with c 6= 0. Then the

matrices

(1 10 1

)and

(1 0c 1

)generate SL2(q) unless q = 9.

Lemma 6.2.6. Let q be a power of an odd prime p with q 6= 9. Let Fq = Fp(τ) with τ 6= 2. Let

C(τ) be the class of SL2(q) containing σ3 :=

(τ − 1 1τ − 2 1

)−1.

1. If 2−τ is nonsquare in Fq then the class vector C = (C1, C2, C(τ)) is quasi-rigid in SL2(q).

2. If 2− τ is a square in Fq then the class vector C = (C1, C1, C(τ)) is quasi-rigid in SL2(q).

Proof. Let σ1 := U1. Consider σ2 :=

(1 0

τ − 2 1

). If 2 − τ is nonsquare in Fq then this is in

C2 and if 2− τ is square in Fq then it is in C1 by Lemma 6.2.5. Now(1 10 1

)(1 0

τ − 2 1

)=

(τ − 1 1τ − 2 1

)shows that ∃ a triple σ = (σ1, σ2, σ3) ∈ C such that σ1σ2σ3 = 1. By Dickson’s theorem σ1and σ2 generate SL2(q). It remains to show that for any other triple (σ′1, σ

′2, σ′3) with the same

properties ∃ γ ∈ Aut(SL2(q)) such that γ(σi) = σ′i and that γ is inner. If σ′1 = σ1 then theinner automorphism is conjugation by an element of Γ := CSL2(q)(σ1). So to show quasi-rigidneed to show that all σ′2 ∈ C2 when 2− τ is nonsquare, or all σ′2 ∈ C1 when 2− τ is a square,with σ1σ

′2 of trace τ such that 〈σ1, σ′2〉 = SL2(q) are conjugate under Γ.

Now σ1 is upper triangular, and 〈σ1, σ′2〉 = SL2(q), so σ′2 is not upper triangular. Hence, by

Lemma 6.2.4, σ′2 is conjugate to

(1 0c 1

)under Γ. Thus

trace(σ1σ′2) = τ ⇒ trace

(σ1

(1 0c 1

))= τ

But

σ1

(1 0c 1

)=

(1 10 1

)(1 0c 1

)=

(1 + c 1c 1

)has trace τ ⇒ c = τ − 2. That is σ′2 is conjugate to

(1 0

τ − 2 1

)under Γ, as required.

Corollary 6.2.7. PSL2(q) occurs regularly over Qab for each odd prime power q.

45

Proof. PSL2(q) = SL2(q)/{±1} . We have shown the existance of a quasi-rigid class vector

(C1, C2, C(τ)) or (C1, C1, C(τ)). Since each tuple of conjugacy classes in Qab-rational we canapply Theorem 6.2.3, since Z(SL2(q)) = {±1}. For the exceptional case, q = 9 we know thatPSL2(9) ∼= A6, and we have demonstrated a stronger result earlier for this, namely that A6

occurs regularly over Q.

Theorem 6.2.8. The group PSL2(p) occurs regularly over Q for each prime p 6≡ ±1 mod 24

Proof. Firstly, PSL2(2) ∼= S3 and PSL2(3) ∼= A4 occur regularly over Q as demonstrated inSection 6.1. Now consider p > 3. Let τ ∈ Fp such that τ 6= 2 and τ is not a square in Fp.Then by Lemma 6.2.6 the class vector C = (C1, C2, C(τ)) is quasi-rigid in SL2(p). Let m be aninteger prime to p. Then (Cm1 , C

m2 ) = (C1, C2) or (C2, C1) depending on whether m is square

mod p or not. Thus if C(τ) is rational then C is a quasi-rigid, rational class vector, whichimplies, by Theorem 6.2.3, that PSL2(p) occurs regularly over Q.

It remains to find τ ∈ Fp with 2 − τ nonsquare in Fp and C(τ) rational. Here C(τ) is theclass consisting of all elements of trace τ .If 2 is nonsquare mod p then set τ = 0. Then 2− τ = 2 is non-square by assumption. Now C(0)consists of elements of order 4, and these are conjugate to their inverse. Thus C(0) is rational.If 3 is nonsquare mod p then set τ = −1. Then 2 − τ = 3 is non-square by assumption. NowC(−1) consists of elements of order 3 and these too are conjugate to their inverse and thusC(−1) is rational.

But 2, 3 nonsquare mod p is equivalent to p 6≡ ±1 mod 24 by elementary number theory.

6.3 M12 as Galois group over Q

Another example of the power of the rigidity method is that it has been used to realise 24 of the26 so called sporadic groups over Q. Using Corollary 4.5.7 the quest to find rigid and rationalclass vectors in a group G, and hence to realise G as a Galois group over Q, can be reduced toexamination of the character table of G, most useful given Conway’s Atlas of finite groups [1].An example is presented in which the character table of the Mathieu group M12, a sporadicgroup, can be used to show that M12 occurs over Q. In most applications of the rigidity criteriato sporadic groups, and in general, first the character tables are used to verify the formula 4.5.7holds for a given class vector C, and then it remains to show that a given system in C generatesthe group. This is usually done by showing that H = 〈σ〉 cannot be contained in any maximalsubgroup of the group [7]

Recall Σ(C) = {σ ∈ C : σi ∈ Ci, σ1 · · ·σr = 1}, where C = (C1, . . . , Cr). C is rigid ifthere exists σ ∈ Σ(C) such that G = 〈σ〉 and if σ′ ∈ Σ(C) is another tuple with the sameproperties then there is g ∈ G such that gσig

−1 = σ′i for i = 1, . . . , r. That is to say that G acts

transitively on Σ(C) by conjugation if C forms a rigid class vector. Then Corollary 4.5.7 saysthis second condition is equivalent to σ satisfying the formula∑

χ∈Irr(G)

χ(σ1) · · ·χ(σr)

χ(1)r−2=|CG(σ1| · · · |CG(σr)|)|G|r−2|Z(G)|

(6.3)

46

From [4] the size of Σ(C) is calculated to be

|Σ(C)| = |G|r−1

|CG(σ1)| · · · |CG(σr)|∑Irr(G)

χ(σ1) · · ·χ(σr)

χ(1)r−2(6.4)

From which, if Z(G) = 1 and |Σ(C)| = |G| then rearranging gives (6.3). That is to say,if |Σ(C)| = |G| and Z(G) = 1 then C is a rigid tuple if there is σ ∈ C generating G. Thiscan be used to find so called candidates for rigid class vectors, given that the classes are rational.

In the vast majority of applications of the rigidity method, rigid tuples consist of three conju-gacy classes. I therefore restrict my attention to rigid triples in the arguments that follow.

Before consulting the Atlas [1] to see if one can find a σ ∈ Σ(C) generating G and satisfy-ing (6.3), the following lemma helps by eliminating some possible candidate tuples.

Lemma 6.3.1. If Out(G) 6= 1 and α ∈ Out(G) fixes C then C cannot be a rigid tuple.

Proof. Suppose C is a rigid class vector fixed by an α ∈ Out(G). Then C rigid ⇒ there existsσ = (σ1, σ2, σ3) ∈ Σ(C) such that G = 〈σ〉 and if σ′ ∈ Σ(C) has the same properties then thereexists g ∈ G such that gσig

−1 = σ′i, i.e. G acts transitively on Σ(C) by conjugation.

Now σ1σ2σ3 = 1 ⇒ σα1 σα2 σ

α3 = 1, and σαi ∈ Ci since α fixes C by hypothesis. Therefore

σα = (σα1 , σα2 , σ

α3 ) ∈ Σ(C). But G is transitive on Σ(C), therefore 〈σ〉 is centralised by an outer

automorphism, and so doesn’t generate G. This is a contradiction.

Hence if Out(G) 6= 1 then we need only consider class vectors not fixed by Out(G). NowOut(M12) = 2 [1]. The classes not fixed by Out(M12) are 4A and 8A, see Appendix B for charac-ter table [4]. There are 7 possible candidates for rigid class vectors containing one of these classesthey are (2A, 3A, 8A), (2A, 3B, 8A), (2A, 4A, 6A), (2B, 3A, 8A), (2B, 4A, 10A), (3A, 3B, 4A) and(3B, 4A, 4A) 1. It remains to find a candidate class vector that can generate G. However, ascan be seen from the character tables of the maximal subgroups of M12 in Appendix B, all ofthese triples generate a proper subgroup of G, and hence are not rigid 2.

Now consider the extension by the full outer automorphism group, M12.2. This contains acandidate triple (2A, 4C, 12A) which doesn’t generate a proper subgroup since no maximal sub-group of M12.2 contains representatives of both classes 4C and 12A. To see this notice thatψi(4C) = 0 for i = 1, . . . , 8, 11, 14 and ψi(12A) = 0 for i = 1, 2, 3, 4, 6, 7, 9, 10, 12, 13, so at leastone is alway 0 (Appendix B, Figure B.1).

Finally, use Hilbert’s result Lemma 6.1.1 to see that, since |M12.2 : M12| = 2, and we justdemonstrated that M12.2 contains a rigid triple of rational classes, then M12 occurs regularlyover Q.

6.4 The Ree Groups 2G2(q2) as Galois groups over Qab

The exceptional groups of Lie type consist of ten families of finite simple groups. Unlike theclassical Lie groups, they do not posses a convenient matrix representation. Currently there is

1All the information required to calculate |Σ(C)| is available in Appendix B. A computer can easily check allpossible triples including at least one of 4A, 8A to establish these candidates. However, I got these from [4]

2To see this notice that for every candidate triple (C1, C2, C3) the permutation character of the classes ofmaximal subgroups ψi(C1), ψi(C2), ψi(C3) 6= 0 for some i ∈ [1, . . . , 14]

47

no uniform way of realising all the exceptional groups of Lie type, infact it is not yet knownwhether they can all be realised as Galois groups over Q [7]. Rigidity has offered the mostfruitful results in this area, the problem being that it is ofen difficult to satisfy the rationalitycondition on rigid class vectors. As discussed earlier, when this is the case, one often uses theQab rigidity criterion. The following is an example gives a realisation of the Ree groups incharacteristic 3 over Qab.

A useful result when applying the rigidity criterion follows. It is helpful when showing thata system is not contained in a maximal subgroup of the group in question, and hence is used toshow that a given system can generate the group. It appears in many applications of rigidity(for instance [4], Lemma 1).

Lemma 6.4.1. Suppose G = 〈σ1, σ2, σ3 : σa1 = σb2 = σc3 = σ1σ2σ3 = 1〉 where a, b, c are pairwisecoprime. Then G = [G,G] = 〈x〉G

Proof. Consider the quotient G/

[G,G] . Then o(σ1) divides a = o(σ1) and bc = o(σ2)o(σ3).

Therefore σ1 is trivial, as a, b, c pairwise coprime. Similarly, σ2 and σ3 are trivial in G/

[G,G] .

Let H = 〈σ1〉G. Let σ′2, σ′3 be the images of σ2, σ3 in G /H . Then (σ′2)

b = 1, (σ′3)c = 1 and

σ′2σ′3 = 1 hence H = G.

The conjugacy classes of G =2 G2(q2) in characteristic 3 (let q2 = 32m+1) and the relevant

portion of the character table were determined in [13]. A complete list of maximal subgroupswas determined in [6]. The character table of 2G2(3

2m+1) is presented, with the same notationused in [13], where ξi are irreducible characters, C2 is the unique class of involutions, C3 isthe the class of 3-elements lying central in a Sylow 3-subgroup and Cs is one of the classescontaining semi-simple elements of order q2 −

√3q + 1 ∈ Z.

1 C2 C3 Csξ1 1 1 1 1

ξ5√36 q(q

2 − 1)(q2 +√

3q + 1) −12(q2 − 1) −1

6(3q2 +√

3q) −1

ξ7√36 q(q

2 − 1)(q2 +√

3q + 1) −12(q2 − 1) −1

6(3q2 +√

3q) −1

Lemma 6.4.2. The class vector C = (C2, C3, Cs) is rigid

Proof. This portion of the character table is the relevant portion as ξ1, ξ5, ξ7 are the onlycharacters that do not vanish anywhere on C2, C3, Cs, and hence are the only classes thatcontribute to the formula in Corollary 4.5.7. Let σ = (σ1, σ2, σ3) with σ1 ∈ C2, σ2 ∈ C3, σ3 ∈ Cs.Given that |2G2(q

2)| = q6(q2 − 1)(q6 + 1), can calculate,

|CG(σ1)| = q2(q4 − 1)

|CG(σ2)| = q6

|CG(σ3)| = q2 −√

3q + 1

with the class sizes from [13]. Now plugging the numbers in one verifies that the following holdstrue, ∑

χ∈Irr(G)

χ(σ1) · · ·χ(σs)

χ(1)s−2=|CG(σ1)| · · · |CG(σs)||G|s−2|Z(G)|

48

It remains to show that 〈σ〉 generates G. Let H := 〈σ〉. The maximal subgroups of G accordingto Kliedman in [6] are

L2(q2)× 2

(22 ×D q2+12

).3

[q6].(q2 − 1)

(q2 +√

3q + 1).6

(q2 −√

3q + 1).6

2G2(r2) with r2 = 32k+1 where

2m+ 1

2k + 1is prime

Now the orders of σi are pairwise coprime, so by Lemma 6.4.1, H = [H,H]. Now the onlynon-solvable maximal subgroups are L2(q

2) × 2 and 2G2(r2) [7]. But the order of L2(q

2) × 2is prime to the order of σ3, which would contradict Lagrange. So the only possible maximalsubgroup that H could be contained in is 2G2(r

2). But the largest order of an element in2G2(r

2) is r2 +√

3r + 1 ≤ 2r2 + 1 which is less than the order of σ3 = q2 −√

3q + 1 ≥ 12q2+1

[7]. Therefore H is not contained in a any maximal subgroup, and so H = G. That is, σ is agenerating system for G. Therefore, by Corollary 4.5.7, C is a rigid class vector.

Finally it remains to check when C forms a rational class vector, whence application of theRigidity Criterion for Q would give that G occurs as a Galois group over Q. Now C2 and C3

are rational classes. But Cs is not. Thus one cannot use the rigidity criterion over Q and so thebest result from this class vector is obtained using the rigidity criterion over Qab, and providesthe result that G occurs as a Galois group over Qab.

6.5 Realising the Monster over Q

The Monster groupM, or the Fischer-Griess Monster is the largest sporadic group, with order

246 · 320 · 59 · 76 · 112 · 133 · 17 · 19 · 23 · 29 · 31 · 41 · 47 · 59 · 71

and represents one of the most celebrated results in the classification of finite simple groups[1]. There is a famous paper by Thompson, [11] in which he introduces rigidity and uses toshow that the Monster group occurs as a Galois group over Q. This chapter concludes withThompson’s realisation of M , a magnificent example of the power of rigidity; imagine trying tofind a polynomial whose Galois group is M by hand! But one does exist.

M has no outer automorphisms, and contains two classes of elements of order two and threeclasses of elements of order 3 [1]. Using a computer and the ATLAS [1] one can check possiblecandidates for rigid tuples, as in the example of M12. These candidates are triples that satisfythe formula in Corollary 4.5.7(2). If there are any, it remains to show that M is generated byone of them. The aforementioned computation shows that there are candidates. Consider, asThompson does in [11], such a candidate C = (2A, 3B, 29A).

Let H = 〈σ1, σ2, σ3 : σ1 ∈ 2A, σ2 ∈ 3B, σ3 ∈ 29A, σ1σ2σ3 = 1〉. Now 2, 3, 29 are pairwisecoprime, hence by Lemma 6.4.1, H = [H,H]. Let H = H /L where L is a maximal normal

49

subgroup of H. By the classification of finite simple groups the possibilities for H are

L2(29)

L2(59)

Rv

Fi′24

since 29||H| [4]. The involutions in the class 2A are {3, 4, 5, 6}-transposition. That is forx, y ∈ 2A, o(xy) ≤ 6, but the classes of involutions in L2(29), L2(59) and Rv are not {3, 4, 5, 6}-transpositions [4]. Now suppose H = Fi′24. This is a contradiction as the character 196883a ofM (see [1], pg.220) does not restrict to a character in Fi′24 [4]. Therefore H is not Fi′24. Soif H = Fi′24 then L 6= 1. Now for x ∈ 29A, |CM(x)| = 3 · 29 from [1]. From [4], if E is anelementary abelian 2-group (cyclic and every element order 2) then |E| ≤ 224, o(2) = 28, o(3) =28, o(5) = 14, o(7) = 7( mod 29). Putting this together with the classification of finite simplegroups shows that H = 3 · F ′24 or M [4]. Suppose that H = 3 · F ′24. Then H = CM(u) foru ∈ 3A. Let σ1 ∈ 2A, σ2 ∈ 3B, σ3 ∈ 29A such that σ1σ2σ3 = 1. From this create anothergenerating set of a different class vector, σ1σ2(uu

−1)σ3 = 1 so consider

σ′1 = σ1 σ′2 = σ2u σ′3 = u−1σ3

Since u ∈ Z(H) o(u−1σ3) = 87. Also, o(σ2u) = 3. But, consideration of candidate triples byevaluation of [1] shows that

|Σ(2A, 3A, 87A)| = 0 |Σ(2A, 3B, 87A)| = 0

as in the example of M12, and as defined in Section 4.5. The only possible triple is thus(2A, 3C, 87A), i.e. σ′2 = σ2u ∈ 3C. Now by the classification of finite simple groups 〈σ2, u〉contains at least 2 Suzuki element, 2 Fischer elements and 2 Thompson elements [4]. CM(σ′2) =3× E. Now the character of 196883a of M restricts to 3× E as

(1 + ω + ω)⊗ 1a + (ω + ω)⊗ 248a + (1 + ω + ω)⊗ 4123a + 1⊗ 306281 + (ωω)⊗ 61256a

from [4]. For x ∈ 3A, 3B, 3C, 196883a(x) = 728, 53,−1 respectively. Thus the following classesare identified under this restriction, 3A in E with 3A in M, 3B and 3C in E with 3B in M.But 196883a(ux) = −1 for x ∈ 3A, 3B or 3C in E ⇒ ux ∈ 3C in M [4]. Therefore everyelementary alelian subgroup of order 9 in M contains 0 or 6 Thompson elements [4], and thusH is not 3 · F ′24 Therefore H =M and so condition 1 of Corollary 4.5.7 has also been satisfied.Therefore (2A, 3B, 29A) is a rationally rigid class vector in the Monster group, and thus thereexists a finite Galois extension of Q with Galois group equal to M.

50

Chapter 7

Conclusion

The aim of this project was to explore the concept of rigidity, and to understand how it is usedto address the Inverse Galois Problem. Sufficient background was developed to understandRiemann’s Existence Theorem and Hilbert’s Irreducibility Theorem. Rigidity was then definedand the role it plays set out. The summary of ideas is, given a group G with a rigid tuple,Riemann’s Existence Theorem realises G over C(x). If the tuple is rationally rigid one canthen descend to realise the group over Q(x), which, when combined with Hilbert’s IrreducibilityTheorem guarantees the existence of an extension of Q with Galois group G. That is, G has apositive solution to the Inverse Galois Problem.

By considering a variety of examples, the application of these ideas in practice is demonstrated.In the introduction a motivating example was considered, as D8 was realised as a Galois groupover Q by considering a well chosen polynomial. Looking back that example helps highlight thepower of these ideas; attempting to find a polynomial whose Galois group is the Monster, forinstance, is completely unfeasible, but given that the character table is known, it was possibleto show that it does have a positive solution to the Inverse Galois Problem.

To this day, rigidity has been used, much in the same way as demonstrated in the examples ofthis project, to realise a large number of groups over Q, including linear groups Ln(p), unitarygroups Un(p), symplectic groups S2n(p) and orthogonal groups O2n+1(p) with certain congru-ence conditions on p, many exceptional groups of Lie type and all but two of the sporadic groups.

That said, the story doesn’t end here as the Inverse Galois Problem is still open. One areaof focus is the extension problem. It seeks to use the composition series of a group to yield atower of extensions whose relative Galois groups are simple. The problem arises when trying tofit the extensions together in an appropriate way. The solvable groups are the easiest exampleof this, and Shafaravich was able to realise them all over Q using this idea [12]. However, thishas not yet been successfully generalised to non-solvable groups.

This project has looked at some of the most fruitful techniques used in considering the In-verse Galois Problem, and has demonstrated how they have been applied. Probably the mostfamous application of rigidity is Thompson’s realisation of the Monster over the rationals, andit is in showing this that the aim of the project is realised.

51

Bibliography

[1] Conway, J. H., Curtis, R. T., Norton, S. P., Parker, R. A. and Wilson, R. A. [1985], AnAtlas of Finite Groups, Oxford Univeristy Press, Oxford.

[2] Fraleigh, J. B. [1968], A First Course in Abstract Algebra, Addison-Wesley PublishingCompany, London.

[3] Hatcher, A. [2002], Algebraic Topology, Cambridge University Press, Cambridge.

[4] Hunt, D. [1986], ‘Rational rigidity and the sporadic groups’, Journal of Algebra 99, 577–592.

[5] Jacobson, N. [1985], Basic Algebra I, Freeman, New York.

[6] Kleidman, P. B. [1988], ‘The maximal subgroups of the chevally groups G2(q) with q odd,the Ree groups 2G2(q) and of their automorphism groups’, Journal of Algebra 117, 30–71.

[7] Malle, G. and Matzat, B. H. [1999], Inverse Galois Theorey, Springer, New York.

[8] Queen, C. S. [1996], ‘Factorial domains’, Proceedings of the American Mathematical Society124(1).

[9] Rotman, J. J. [1965], An Introdution to the Theory of Groups, Allyn and Bacon, Cambridge.

[10] Rotman, J. J. [1998], Galois Theory, Springer, New York.

[11] Thompson, J. G. [1984], ‘Some finite groups that appear as Gal(L/K), where K ⊂ Q(µn)’,Journal of Algebra 89, 437–499.

[12] Volklein, H. [1996], Groups as Galois Groups, Cambridge University Press, Cambridge.

[13] Ward, H. N. [1966], ‘On Ree’s series of simple groups’, Trans. Amer. Math. Soc. 121, 62–89.

[14] Zannier, U. [2009], ‘On the Hilbert Irreducibility Theorem’, Rend. Sem. Mat. Univ. Pol.Torino 67(1).

52

Appendix A

GAP Code

GAP is a system designed for use in computational group theory. As well as a programminglanguage it has a large database of algebraic objects. I have used GAP as one way of obtainingthe character table of specific groups. I present the code, in red, together with the GAP output,in green.

GAP attempts to present the character table in a similar way to the ATLAS [1]. The firstrows list, on the left, the primes dividing the order of the group, and then for each class theexponents of the prime factorisation of the centraliser order. The next row is the name of theconjugacy classes, and, where applicable, the further rows give all the primes p dividing thegroup order. Then the character table is presented, with irreducible character on the left ar-ranged according to degree. Zeros in the table are represented by a ‘·’. Algebraic irrationalitiesare denoted by a letter, and these are defined below (where applicable). The final computationis the size of the conjugacy classes.

53

Figure A.1: Character Table for SL2(8)

Figure A.2: Character Table for S4

54

Appendix B

Characters for M12

The following table from [4] displays the character table of M12, this is the 15 × 15 upper lefttable. The remaining 9 columns give the characters for M12.2. Following the convention from[1] the column headings are firstly the size of the conjugacy class, then the power maps, andthen the name of the class. The irreducible characters are χi with i = 1, . . . , 15. The lower14 × 21 table gives the permutation characters ψi with i = 1, . . . , 14, for the conjugacy classesof maximal subgroups.

55

Figure B.1: Character Table of M12 with additional information

56