Class Field Towers, Solvable Galois Representations and Noether's

Home , Inverse Galois problem

Class ﬁeld towers, solvable Galois representations and Noether’s problem in Galois theory

Jonah Leshin

B.A., Northwestern University; Evanston, IL, 2008

M.A., University of Cambridge; Cambridge, UK, 2009

A dissertation submitted in partial fulﬁllment of the

requirements for the degree of Doctor of Philosophy

in Mathematics at Brown University

PROVIDENCE, RHODE ISLAND

by Mathematics as satisfying the

dissertation requirement for the degree of Doctor of Philosophy.

Date

Joseph Silverman, Ph.D., Advisor

Recommended to the Graduate Council

Date

Michael Rosen, Ph.D., Reader

Date

Stephen Lichtenbaum, Ph.D., Reader

Approved by the Graduate Council

Date

Peter Weber, Dean of the Graduate School

iii Vitae

The author grew up in Newton, Massachusetts. He received his B.A. in Mathematics in 2008 from Northwestern University. The following year, he attended the University of Cambridge on a Fulbright Scholarship, where he earned an M.S. in Mathematics. He enrolled in the Ph.D. program at Brown University in the fall of 2009 and completed his

Ph.D. in the spring of 2014.

iv Acknowledgements

First and foremost, I would like to thank my advisor, Joe Silverman. Whenever I was stuck, Joe was there with a multitude of suggestions. I am grateful for his guidance, encouragement, and ultimately his investment in my development as a mathematician. I am also thankful to Mike Rosen, who was always happy to speak with me and so often suggested exactly the right reference.

I am grateful to have spent the past ﬁve years at Brown. I am indebted to many fellow and former graduate students for their helpful conversations. In particular, I would like to acknowledge E. Mehmet Kiral, Florian Sprung, and Wade Hindes. In the course of all the reading, writing, and problem solving that goes with a Ph.D., I have also managed to make several friendships that I hope will be life long. I have truly enjoyed being a part of the welcoming, collegial, Kabob and Curry frequenting math graduate student community.

Finally, it is my pleasure to thank my family – my parents Rosalyn and Michael, and my sisters Miriam and Rachel – for their unconditional love and support. My largest debt of gratitude is owed to my wife, Dahlia, for being right beside me through the trials and tribulations of everyday life.

v Abstract of “ Class ﬁeld towers, solvable Galois representations and Noether’s problem in Galois theory ” by Jonah Leshin, Ph.D., Brown University, May 2014

We begin by investigating the class field tower problem for Kummer extensions of cyclotomic fields. Specifically, given primes l and p satisfying certain mild hypotheses, we show

pl a b the existence of infinitely many primes q for which the field Q(ζl, p q ) has infinite class field tower, where a and b are appropriately chosen. As an application, we show that the √ 3 field Q(ζ3, 79 · 97) has infinite class field tower, giving an example of a “small” field with infinite class field tower.

Motivated by the class field tower problem, we next study the behavior of the root discriminant in a solvable number field extension. In particular, we use class field theory to show that for any fixed number field K, there are only finitely many extensions of K of

a given solvable length and bounded root discriminant. This theorem should be viewed in

light of the fact that all ﬁelds in a class ﬁeld tower have the same root discriminant.

In the next part of the thesis, we analyze the possibilities for the ﬁelds cut out by a

solvable three-dimensional representation of GQ ramiﬁed at a single ﬁnite prime. In doing so, we give bounds for the number of such representations with given Artin conductor. This

work is motivated by similar results in the case of two-dimensional Galois representations,

where Langlands-type techniques are available.

We then continue our study of Galois representations in the context of torsion points

on abelian varieties over number ﬁelds. We prove several facts about the image of the

representation of Galois acting on the l-torsion points of an abelian variety subject to

certain constraints.

Lastly, we study a variant of Noether’s problem in Galois theory. Building on work of

Lenstra [35], we give an upper bound for the degree of irrationality of ﬁelds of the form

A K(x1, . . . , x|A|) , where A is a ﬁnite abelian group, K is a ﬁeld, and A acts on the variables

xi via the regular representation. Contents

Vitae iv

Acknowledgments v

1 Class Field Towers 3 1.1 Introduction ...... 4 1.2 The embedding problem for Kummer extensions of cyclotomic fields . . . . 7 1.2.1 Proof of Theorem 3 ...... 8 1.2.2 The case l =3 ...... 13 1.2.3 Some other fields with infinite 3-class field tower ...... 13 1.3 p-Principal fields ...... 14 1.4 Solvable number field extensions of bounded root discriminant ...... 16 1.4.1 Discriminants and ramification groups ...... 17 1.4.2 Proof of Theorem 7 ...... 18 1.4.3 Further questions ...... 30

2 Artin representations 32 2.1 Introduction ...... 33 2.2 An application of Serre’s conjecture ...... 33 2.3 Counting Artin representations ...... 35 2.4 Deﬁnitions and preliminaries ...... 37 2.5 Imprimitive representations ...... 39 2.6 Primitive representations ...... 43 2.6.1 Representations with projective image isomorphic to P1 ...... 44 2.6.2 Representations with projective image isomorphic to P3 ...... 56 2.7 Comparison to the two-dimensional case ...... 61

3 Torsion points on elliptic curves and abelian surfaces 63 3.1 Introduction ...... 64 3.2 3-Torsion ...... 65 3.3 Roots of unity in torsion ﬁelds ...... 66 3.4 A local-global property ...... 67

vi 4 Noether’s Problem in Galois Theory 69 4.1 Introduction ...... 70 4.2 Notation and Lenstra’s Setup ...... 74 4.3 Bounding the Degree of Irrationality from Above ...... 76 4.3.1 The General Case ...... 79 4.3.2 An Example of Theorem 23 ...... 80

vii 1 Commonly used notation

• ζl Primitive lth root of unity.

• µl Group of lth roots of unity.

• OK Ring of integers of a number ﬁeld K.

ab • K Maximal subﬁeld of K that is abelian over Q.

• ClK Ideal class group of a number ﬁeld K.

• hK Class number |ClK | of K.

• Km Ray class ﬁeld of K modulo m.

m • ClK Ray class group of K modulo m.

• dK/F Relative discriminant from K to F .

• dK | dK/Q |.

• DK/F Relative diﬀerent from K to F .

• G(L/K) Galois group of L over K.

• GK G(K/K).

• Frobp Frobenius element at p in GK , where p is a prime of K.

n • vp(I) Maximal integer n such that p |I.

• e(P/p) vP(p).

• f(P/p) Residue degree of P over p.

• A[l] Kernel of the multiplication by l map on an abelian variety A.

• ρA,l Representation of GK acting on A[l], where A is deﬁned over K.

• Sn Symmetric group on n letters. 2

• Cn Cyclic group of order n.

• Z(G) Center of a group G. Chapter One

Class Field Towers 4

1.1 Introduction

Let K be a number ﬁeld. In general, the ring of integers OK of K is not a unique factorization domain (UFD). One way to rectify this problem is to look instead at factorization of ideals in OK into prime ideals. Since OK is a Dedekind domain, any ideal factors uniquely into a product of prime ideals. We can measure the extent to which OK fails to have unique factorization with the ideal class group ClK := IK /PK of K. Here IK denotes the group of fractional ideals of K and PK those fractional ideals that are principal. (From hereon, we may abuse notation and use language such as “ideals of K” as shorthand for ideals of

OK ). Since OK is a Dedekind domain, it is a UFD if and only if it is a principal ideal domain, i.e., ClK = 1. Suppose that ClK 6= 1, but we insist on unique factorization on the level of prime elements for all elements of OK . Although this is impossible in K, we might hope that K can be embedded in some larger number ﬁeld L with unique factorization, so that every x ∈ K can at least be factored uniquely as an element of L (up to units). This question–the existence of such an L– is known as the “embedding problem,” and has an equivalent reformulation in the language of class ﬁeld theory, which we now describe.

Let H denote the maximal abelian extension of K that is unramiﬁed at all primes of

K. By class field theory, H is a finite extension of K, called the Hilbert class field of K.

The Hilbert class ﬁeld enjoys several nice properties, including the following two.

Theorem 1. [44] Let K be a number ﬁeld and H the Hilbert class ﬁeld of K.

1. There is an isomorphism ClK → G(H/K).

2. Every ideal of OK becomes principal in OH .

From the ﬁrst statement of Theorem 1, we see that OK has unique factorization if and only if H = K. Now set K = K(0) and let K(1) denote the Hilbert class ﬁeld of K.

For i ≥ 1, we recursively deﬁne K(i) to be the Hilbert class ﬁeld of K(i−1). The tower

K(0) ⊆ K(1) ⊆ K(2) ⊆ ... is called the Hilbert class ﬁeld tower of K. We say the tower is 5

finite if the union ∞ [ K(∞) := K(i) i=0 is a finite extension of K, meaning the tower stabilizes for large i. If K(∞) is an infinite

extension of K, we say K has inﬁnite class ﬁeld tower. The following lemma illustrates

the crucial role played by the Hilbert class ﬁeld tower in the context of the embedding

problem.

Lemma 1. The embedding problem for a number ﬁeld K has an aﬃrmative solution if

and only if K has ﬁnite class ﬁeld tower (therefore, the embedding problem is also known

as the “class ﬁeld tower problem”).

Proof. If K has ﬁnite class ﬁeld tower, then K(i) = K(i+1) for some i, which means K(i) has

unique factorization. Conversely, suppose that K may be embedded in some ﬁeld H with

ClH = 1. Suppose for contradiction that K has inﬁnite class ﬁeld tower. Then there exists

(i−1) (i) (i) (i−1) (i) i with K ⊆ H, but K * H. Since K /K is unramiﬁed abelian, HK /H is as

well, and by construction the extension is non-trivial, contradicting ClH = 1.

The general embedding problem – the question of whether every number field may be embedded in a number field of class number one – was proposed by Furtwangler and publicized further by Hasse [25] in an exposition on class field theory. In 1964, Golod and Shafarevich solved the problem by demonstrating the existence of number fields with infinite class field tower [18]. A critical step in their proof is the following.

Theorem 2. Let G be a non-trivial ﬁnite p-group and let G act trivially on Fp. Let 1 2 d2 d = dimFp H (G, Fp) and r = dimFp H (G, Fp). Then r > 4 .

For a prime p, we define the p-Hilbert class field of K to be the maximal abelian unramified extension of K of p-power degree over K. We may then analogously define (∞) the p-Hilbert class field tower Kp of K. One shows that the number d in Theorem 2 is the minimal number of elements needed to generate G, and r is the number of relations 6 needed among such a generating set in a presentation for G. By working with the group (∞) G = G(Kp /K) and using Theorem 2, Golod and Shafarevich are able to provide examples of fields with infinite p-class field tower. For a nice exposition of a proof of Golod and

Shafarevich’s work, see [55].

Every known example of a number field K with infinite class field tower comes from a

field with infinite p-class field tower in the sense that there exists a number field L ⊆ K(∞) such that L has infinite p- class field tower for some prime p. Although p-class field towers tend to be more tractable than general class field towers, we remark that there exist fields

K with infinite class field tower, but finite p-class field tower for any fixed prime p. An √ √ example of such a field is Q( −239, 4049) [40].

Recall that the root discriminant of a number ﬁeld K is deﬁned to be

1 [K:Q] rd(K) := dK .

The root discriminant is a useful quantity when studying class ﬁeld towers. Given a tower of number ﬁelds L/K/F , we have the following equality of ideals of F :

[L:K] dL/F = NK/F (dL/K )(dK/F ) , (1.1)

where dL/F denotes the relative discriminant. It follows from (1.1) that if L is an extension of K, then rd(K) ≤ rd(L), with equality if and only if dL/K = 1, i.e., L/K is unramiﬁed at all ﬁnite primes. Thus we have obtained the following proposition.

Proposition 1. If K has infinite class field tower, then all fields K(i) have the same root discriminant. 7

1.2 The embedding problem for Kummer extensions of cy-

clotomic ﬁelds

The theorem of Golod and Shafarevich has motivated the construction of number fields with infinite class field tower that are subject to various constraints. Golod and Shafarevich show that any number field satisfying certain ramification conditions must have infinite class field tower. For example, any imaginary quadratic extension of the rationals ramified √ at seven or more primes, such as Q( −3 · 5 · 7 · 11 · 13 · 17 · 19), has infinite 2-class field tower. If an extension of a number field K can be embedded in a field with class number

1, then clearly K can as well. Therefore, if K has infinite class field tower, then any finite

extension of K does as well. Furthermore, if K is ramiﬁed at suﬃciently many primes,

where the number of primes depends only on [K : Q], then K has infinite class field tower [3]. Thus a task of interest becomes finding number fields of small size with infinite class

ﬁeld tower. The size of a number ﬁeld K might be measured by the number of rational primes ramifying in K, the size of the rational primes ramifying in K, the root discriminant of K, or any combination of these three.

With regard to number of primes ramifying, Schmithals [57] gave an example of a quadratic number field with infinite class field tower in which a single rational prime ramifies. Odlyzko’s discriminant bounds [47] imply that any number field with infinite class

ﬁeld tower must have root discriminant at least Ω := 4πeγ ≈ 22, where γ is Euler’s constant, and this number can be improved to 2Ω if we assume the Generalized Riemann Hypothesis;

−1 √ Martinet showed that the number field Q(ζ11 + ζ11 , 46), with root discriminant ≈ 92.4, has infinite class field tower [39]. Note that the only primes ramifying in this field are 2, 11, and 23.

Let l be an odd prime and let ζl denote a primitive lth root of unity. By Kummer √ l theory, a cyclic extension of degree l of Q(ζl) is of the form Q(ζl, a), for some a ∈ Q(ζl). In this section we investigate the embedding problem for such ﬁelds. We use a theorem of 8

Schoof to produce a class of Kummer extensions of Q(ζl) with infinite class field tower and ramification at three rational primes. Our main theorem is the following. √ Theorem 3. Let l and p be distinct primes and suppose that the class number h of Q(ζl, l p) is at least 3 if l ≥ 5, and that h ≥ 6 if l = 3. For infinitely many primes q, there exists √ a b l δ ∈ {p q }1≤a,b≤l−1 such that Q(ζl, δ) has infinite class field tower.

√ 3 As a direct consequence of the proof of Theorem 3, we find that Q ζ3, 79 · 97 has infinite 3-class field tower.

1.2.1 Proof of Theorem 3

Our construction is analogous to that of Schoof [58], Theorem 3.4. From hereon, for a prime l, deﬁne

l 2 Al = {a : a ∈ Z/l Z}.

We begin with a lemma.

Lemma 2. Let l be a prime and n an integer prime to l. The prime (ζl − 1) of Q(ζl) is √ l unramiﬁed (and splits completely) in Q( n, ζl) if and only if n ∈ Al.

Proof. This can also be deduced from [22, Theorem 119]. We provide our own independent proof for completeness.

√ l Let F = Q(ζl),M = F ( n). Let l = (ζl −1) be the unique prime of F above l. Suppose that l were inert in M. Then there would only be a single prime of M, and therefore a √ √ single prime of Q( l n), lying over l. The extension Q( l n)/Q cannot be unramified at l since its compositum with its conjugates contains ζl. But the extension cannot be totally ramified either since that would imply that M/Q has ramification degree l(l − 1) above l.

Therefore, either M/Q is totally ramified above l, or the ramification degree is l − 1, in which case the rational prime l splits into l primes in M. Suppose that we are in the 9 case of the latter, so each corresponding local extension of M/Q above l is totally ramified √ 0 l of degree l − 1. It follows that any prime l of Q( n) above l either splits completely in √ √ l ˜ l M (the case Q( n)l0 = M˜l, where l|l) or is totally ramified in M (the case Q( n)l0 = Ql). √ Thus, there must be two primes above l in Q( l n), one of which splits completely in M and has ramification degree l − 1 over l, and one of which ramifies completely in M and is unramified over l with residue degree 1. We have established:

l totally ramified in M ⇔ l totally ramified in M √ ⇔ l totally ramified in Q( l n)

⇔ no lth root of n is contained in Ql.

Deﬁne f(x) = xl − n, and let f¯ denote its reduction modulo l3. A root α of f¯ satisﬁes

0 2 |f(α)|l < |f (α)|l , so by Hensel’s lemma, f(x) has a solution in Ql if and only if n is an 3 2 lth power in Z/l Z, which is equivalent to n being an lth power in Z/l Z.

√ Let p be any prime different from l, and let h be the class number of Q(ζl, l p) with H its Hilbert class field. Let q be a rational prime that splits completely in H, so by class field theory, q is a prime that splits completely into principal prime ideals in √ Q(ζl, l p). In particular, q ≡ 1 (mod l), and thus by Lemma 2, (1 − ζl) is totally ramified √ 2 √ √ in Q(ζl, l q) unless q ≡ 1 (mod l ). Set F = Q(ζl),E = F ( l p, l q). In what follows, we √ a b l find δ = δp,q ∈ {p q }1≤a,b≤l−1 so that E is unramified over K = Kδ := F ( δ) (see Figure 1.1).

Case I: Suppose that p∈ / Al.

√ 2 ∗ In this case, (ζl − 1) ramiﬁes totally in F ( l p) by Lemma 2. By viewing (Z/l Z) as a b Z/lZ × Z/(l − 1)Z, we see there exists a and b with 1 ≤ a, b ≤ l − 1 such that p q ∈/ Al. Set

δ = paqb. 10

Figure 1.1: Field Diagram for Theorem 5.

√ L = H( l q)

H √ √ E = F ( l q, l p)

√ √l √ F ( l p) K = F ( δ) F ( l q)

F = Q(ζl)

We claim that the ramiﬁcation degree e(E, l) of l in E is l(l − 1). Suppose for contradiction that this is not so, in which case we must have e(E, l) = l2(l − 1). It follows from

Lemma 2 that this is impossible if q ∈ Al, so assume q∈ / Al. This means that the field E ˜ has a single prime l lying above l, and that E˜l/Ql is totally ramified. Since q ≡ 1 (mod l) 2 c c 0 √ but q 6≡ 1 (mod l ), there exists c such that pq ∈ Al. Set γ = pq , and let E = Q(ζl, l γ). 0 The extension E /Q(ζl) is unramified above above (ζl − 1) by Lemma 2, a contradiction.

We claim that E/K is unramified. Since E is the splitting field over K of either xl − p or xl − q, the relative discriminant of E/K must be a power of l. Therefore, the only possible primes of K that can ramify in E are those lying above l. It is necessary and sufficient to show that e(K, l) = l(l − 1). By the definition of δ and Lemma 2, we know

(ζl − 1) is totally ramiﬁed in Kδ, from which it follows that e(K, l) = l(l − 1).

Case II: Suppose that p ∈ Al.

If q∈ / Al, Case I with the roles of p and q now reversed allows us to pick δ so that

E/Kδ is unramiﬁed. If q ∈ Al, then E/F is unramiﬁed above l, so for any choice of

a b δ ∈ {p q }1≤a,b,≤l−1, E/Kδ is unramiﬁed. 11

We are now ready to invoke a theorem of Schoof [58]. First we set notation. Given any number field H, let OH denote the ring of integers of H. Let UH be the subgroup of the idèlegroup of H consisting of elements with valuation zero at all finite places. Given a

finite extension L of H, we have the norm map N = NUL/UH : UL → UH , which is just the ∗ restriction of the norm map from the idèlesof L to the idèlesof H. We may view OH as a subgroup of UH by embedding it along the diagonal. Given a finitely generated abelian group A, let dl(A) denote the dimension of the Fl-vector space A/lA.

Theorem 4. [Schoof][58] Let H be a number ﬁeld. Let L/H be a cyclic extension of prime

degree l, and let ρ denote the number of primes (both ﬁnite and inﬁnite) of H that ramify

in L. If

q ∗ ∗ ∗ ρ ≥ 3 + dl OH /(OH ∩ NUL/UH UL) + 2 dl(OL) + 1 , then L has inﬁnite l-class ﬁeld tower.

√ We apply Schoof’s theorem to the extension L := H( l q) over H, where H, as above, is √ the Hilbert class ﬁeld of F ( l p). All hl(l −1) primes in H above q ramify completely in the √ ﬁeld H( l q). Thus ρ ≥ hl(l − 1), with strict inequality if and only if the primes above l in

∗ 1 2 ∗ 1 H ramify in L. By Dirichlet’s unit theorem, dl(OL) = 2 hl (l −1) and dl(OH ) = 2 hl(l −1). Thus, after some rearranging, we see that if h and l satisfy

1 3 r1 1 h(l − 1) ≥ + 2 h(l − 1) + , 2 l 2 l2

then L will have inﬁnite l-class ﬁeld tower. If l = 3, the minimal such h is given by h = 6.

If l ≥ 5, the minimal such h is given by h = 3. Since L/K is an unramified (as both L/E and E/K are unramified) solvable extension, it follows that K has infinite class field tower as well.

This proves the following version of our main theorem.

√ Theorem 5. Let p and l be distinct primes and suppose the class number h of Q(ζl, l p) 12 satisfies h ≥ 3 if l ≥ 5, and satisfies h ≥ 6 if l = 3. Let q be a prime that splits completely √ √ a b l into principal ideals in Q(ζl, l p). Then there exists δ ∈ {p q }1≤a,b≤l−1 such that Q(ζl, δ) has infinite class field tower.

1 Remark 1. Given p and l in Theorem 5, the density of such q is l(l−1)h by the Chebotarev density theorem.

c Remark 2. If δ ∈ Al then δ ∈ Al for all powers c. Thus, the proof of Theorem 5 goes

c through with δ replaced by δ , and we always generate l − 1 extensions of Q with Galois

group Z/lZ o Z/(l − 1)Z unramified outside {l, p, q} with infinite class field tower.

In the proof of Theorem 5, we were assuming that

∗ ∗ dl(OH ) = dl(OH ∩ NUL/UH UL).

∗ Let x be an arbitrary element of OH . We attempt to construct y = (yw) ∈ UL such that Ny = x. Consider ﬁrst the primes of H that are unramiﬁed in L. Let v be such a

prime and suppose {w1, . . . , wa} (a = 1 or l) are the primes above v in L. Because v is

∗ ∗ unramiﬁed, the local norm map N : O → O is surjective, so we can pick yv ∈ Lw1 Lwi Hv

such that Nyv = x. Put 1 in the wi components of y for i ≥ 2 if a = l.

Now let v be a prime of H that ramifies (totally) in L. If v splits completely in √ pl ∗ pl ∗ l H( OH ), then OH ∈ Hv. Letting w be the prime above v in L, we set yw = x. Putting the ramified and unramified components of y together gives the desired element.

The inequality needed for an inﬁnite class ﬁeld tower is then

3 r1 1 h(l − 1) ≥ + 2 h(l − 1) + , l 2 l2 which is satisﬁed by h ≥ 2 if l = 3, and is satisﬁed with no restriction on h if l ≥ 5.

pl ∗ Suppose now that the primes of H that ramify in L split completely in H( OH ). If p ∈ Al and q∈ / Al, then ramiﬁcation considerations show that the primes above l in H 13 ramify in L; otherwise, the only primes in H ramifying in L are those above q. This gives us the following result.

Theorem 6. Let p be a prime with p∈ / Al. If l ≥ 5, then for infinitely many primes q, √ a b l there exists δ ∈ {p q }1≤a,b≤l−1 such that Q(ζl, δ) has infinite class field tower. If l = 3, √ the conclusion holds if we also assume that the class number of Q(ζl, l p) is at least 2.

Proof. For such p, the set of desired primes q consists of all rational primes splitting

p3 ∗ completely in H( OH ).

1.2.2 The case l = 3

We apply Theorem 5 in the case l = 3 to explicitly produce an inﬁnite class ﬁeld tower.

√ 3 The ﬁeld Q(ζ3, 79) has class number 12, and 97 splits completely into a product of principal ideals in this ﬁeld [67], so we obtain

√ 3 Corollary 1. The field K = Q ζ3, 79 · 97 has a solvable unramified extension with infinite 3-class field tower, and thus K has infinite class field tower.

1.2.3 Some other fields with infinite 3-class field tower

It is a theorem of Koch and Venkov [70] that a quadratic imaginary field whose class group has p-rank three or larger has infinite p-class field tower, where p is an odd prime. The table [28] of class groups of imaginary quadratic fields, although not constructed with the intent of producing number fields with infinite class field tower and small root discriminant, enables us to find a multitude of imaginary quadratic fields whose class group has 3-rank at least three, and thus have infinite 3-class field tower. From [28], we may conclude that the imaginary quadratic field with infinite 3-class field tower having smallest root discriminant √ is Q( −3321607), with root discriminant ≈ 1822.5. The field mentioned in Corollary 1 14 with infinite 3-class field tower has root discriminant equal to the root discriminant of √ 3 Q ζ3, 79 · 97 , which is ≈ 1400.4.

One may creatively use Schoof’s theorem (Theorem 4) to construct various examples of number fields with infinite l-class field tower and small root discriminant. Below we outline an example for the case l = 3 that was communicated to the author by the referee of the paper in which Section 1.2 of this thesis appears.

Let H be the subfield of the cyclotomic field Q(ζ600) fixed by the order four automor- 7 phism ζ600 7→ ζ600. By construction, the rational prime 7 splits completely in H into 40

primes pi. Now, let K be the unique cubic subfield of Q(ζ7). All the pi ramify in HK, so the inequality in Theorem 4 implies that the 3-class field tower of HK is infinite. One

checks that the root discriminant of HK is ≈ 391.1.

1.3 p-Principal ﬁelds

In this section, we pose a question about factorization that is related to class ﬁeld towers.

If L/K is an extension of number ﬁelds and hL = 1, then every element of OK can be

factored uniquely into prime elements of OL (up to units). Moreover, the class ﬁeld tower of K is ﬁnite, meaning that every element of K can be factored uniquely in some K(i)

(i) (recall that we set K = K0 and set K = HK(i−1) for i ≥ 1). What if, however, for a

number field K with infinite class field tower and a given element α ∈ OK , we merely ask

(i) whether there exists i = iα such that α factors uniquely into prime elements in K . That

is, we are asking whether there exists i such that the principal ideal αOK factors as a

product of principal prime ideals of OK(i) . This is equivalent to asking whether all primes P of K(i) lying above primes p of K that divide α are principal.

Given a prime p of a number ﬁeld K, we say that an extension L of K is p-principal

if all prime ideals in L above p are principal. It is natural to ask whether there exists a 15 number ﬁeld K and a prime p of K such that K(∞) contains a strictly increasing sequence

(i) of fields K = K0 ⊂ K1 ⊂ K2 ⊂ ... (for example, we could take Ki = K if K had infinite class field tower), none of which is p-principal.

Consider the following construction. Fix an odd prime l and a number field K = K0 satisfying Hypotheses I in [20]. For this section only, let HK denote the l-Hilbert class field of K. By Theorem B of [20], there exists a strictly increasing sequence of number fields

(∞) K = K0 ⊂ K1 ⊂ K2 ⊂ ... with [Ki : Ki−1] = l and Ki ⊆ K for all i. Furthermore, by taking the value t in Theorem B to be at least l, we may assume that the l-rank of ClKi is i+1 at least l for all i. We claim that we may assume Ki/K is Galois for all i: by induction,

suppose that Ki−1/K is Galois. Then HKi−1 /K is Galois since the Galois closure of

HKi−1 /K is the compositum of abelian unramiﬁed extensions of l-power degree over Ki−1,

and HKi−1 is the maximal such extension. Thus, G(HKi−1 /Ki−1) ¡ G(HKi−1 /K). In general, every normal subgroup H of a p-group G contains subgroups of all orders dividing

|H| that are normal in G. Therefore, there is a field Ki ⊆ HKi−1 with [Ki : Ki−1] = p and (i) Ki/K Galois. Note, the fields Ki are in general not the successive Hilbert class fields K discussed earlier.

We conjecture that with the setup above, there should exist primes p of K for which no Ki is p-principal. The intuition behind the conjecture is as follows. Let hi = hKi . For −1 a “random” prime p of K, the chance that p is not principal is 1 − h0 . If we consider

a prime P of K1 above p to be a random prime of K1, then the chance that P is not −1 principal is 1 − h1 . So the chance that a random prime p never factors into a product of

principal prime ideals of Ki for any i is

∞ Y −1 (1 − hi ). i=0

li+1 Since our Ki’s satisfy hKi > l , the product is positive, and thus there should exist primes p for which Ki is not p-principal for any i. 16

1.4 Solvable number ﬁeld extensions of bounded root dis-

criminant

As noted in Section 1.2, due to bounds by Odlyzko [47], a number ﬁeld with inﬁnite class

field tower must have root discriminant larger than Ω ≈ 22. What Odlyzko actually shows is that there exist only finitely many number fields with root discriminant less than Ω.

From here, it follows by Lemma 1 that if K has inﬁnite class ﬁeld tower, then rd(K) > Ω.

In particular, the set

ZN,K := {L : L/Q ﬁnite,L ⊇ K, and rd(L) ≤ N}

is infinite for N ≥ rd(K). The best known bound for N such that ZN,K is infinite for some K is N = 82.2, given by Hajir and Maire using tamely ramified class field towers [21].

We will not be concerned with specific values of root discriminants but rather with the following question: fix a number field K and an arbitrary (large) real number N > 0. How can infinite subsets of ZN,K arise? For example, fix a positive integer n. A group G is solvable of length n if Gn = 1, where Gn is the nth derived subgroup of G. If N were large enough, could the following set be infinite:

{L : L/Q ﬁnite, L/K solvable length n, and rd(L) ≤ N}?

The answer to the question is no, and this is the main theorem of this section.

Theorem 7. Fix a number ﬁeld K, a positive integer n, and a positive real number N.

The set

Yn,N,K := {L : L/Q ﬁnite, L/K solvable length n, and rd(L) ≤ N}

is ﬁnite.

Remark 3. • Taking n = 1 gives ﬁniteness for abelian extensions. The general solv- 17

able case follows by induction from the n = 1 case, and the proof of the n = 1 case

occupies the bulk of the paper. We set YN,K := Y1,N,K .

• Rather than considering the root discriminant of extensions L of K, we could equiva-

1/[L:K] lently consider the quantity (NK/QdL/K ) . This is evident by (1.1) from Section 1.1

• Odlyzko mentions in [48] that Theorem 7 is known to be true in the case K = Q, but he does not give a proof.

From here on, all ﬁeld extensions are assumed to be ﬁnite unless otherwise stated.

1.4.1 Discriminants and ramiﬁcation groups

Let L/K be a Galois extension of local ﬁelds, with K a ﬁnite extension of Qp. In [62],

Serre gives the following formula for the relative diﬀerent DL/K in terms of the ramiﬁcation groups Gi of L/K:

∞ X vL(DL/K ) = (|Gi| − 1), i=0 where vL denotes the normalized P-adic valuation of a fractional ideal of OL, P the unique maximal ideal of OL. If L/K is now a Galois extension of global ﬁelds with P a prime of L lying above a prime p of K, then

∞ X vp(dL/K ) = gf (|Gi| − 1), (1.2) i=0 where Gi are the ramiﬁcation groups of P/p, g is the number of primes of L above p, and f is the residue degree of P/p. So for a Galois extension K/Q, we obtain

∞ 1 X v (rd(K)) = (|G | − 1). p |G | i 0 i=0 18

Note that if we deﬁne the root discriminant of a ﬁnite extension Kp of Qp to be

v (d ) Qp Kp/Qp [K : ] rd(Kp) = p p Qp ,

v (rd(K)) then p p = rd(Kp).

1.4.2 Proof of Theorem 7

Fix a number ﬁeld K and a real number N > 0. Our ﬁrst goal is to show that the set

XN,K := {L : L/K abelian, L/Q Galois, and rd(L) ≤ N} is ﬁnite in the case when K/Q is Galois.

If E/F is a Galois extension of number ﬁelds ramiﬁed at a prime p of F with e = eE(p),

(e−1)fg then by (1.2), p | dL/K . It follows that if L is a number ﬁeld Galois over Q with √ rd(L) ≤ N, then L/Q can not ramify at any rational prime p with p > N.

Let S be the union of the real places of K and the set of primes of K lying above √ the rational primes p with p ≤ N. Suppose that XN,K is inﬁnite. Then there exists an increasing sequence of natural numbers nl such that [Lnl : K] = nl and Lnl ∈ XN,K . For a ﬁxed positive integer m, the maximal abelian extension of K of exponent m that is

unramiﬁed outside S is ﬁnite [65]. Thus we may assume that Lnl /K is cyclic for each l. We deal with the following two cases separately:

Case I: For every p ∈ S, lim sup e (p) < ∞, (the lim sup being indexed by l). Lnl

To ease notation, write Ll for Lnl . Write S = {pj}. Let fl = f(Ll/K) be the conductor

of Ll/K. As l gets arbitrarily large, there exists j such that the power aj of pj dividing 19

fl gets sufficiently large. This is because Ll/K can only be ramified at the primes in S, so its conductor is divisible by only these primes, which means Ll is contained in the ray class field of K modulo the product of the infinite real places of K and a power ml of the product of the finite primes ramifying in Ll/K. As [Ll : K] increases, the minimal such ml increases, which, by definition of the ray class field, implies aj increases for some j.

Set p = pj and a = aj. Let Pl be a prime of Ll lying above p. Write Lˆl for Ll , and Pl n to keep notation consistent write Kˆ for Kp. Deﬁne the group of n-units U of Kˆ to be the

n group of units of OKˆ that are congruent to 1 (mod p ). The p-contribution to fl is the ˆ conductor of the abelian extension of local ﬁelds Lˆl/Kˆ , which we denote by fl. Let θ be the

∗ ˆ bl local reciprocity map Kˆ → G(Lˆl/Kˆ ). Then fl = p , where bl is by deﬁnition the smallest

n n th n integer n satisfying θ(U ) = 1. As θ maps U onto the n upper ramiﬁcation group lG

b −1 b −1 of lG := G(Lˆl/Kˆ ), we see that lG l is non-trivial (in fact, lG l is the last non-trivial

b −1+ ramiﬁcation group in the sense that lG l = 1 for any > 0). For a discussion of the upper ramiﬁcation groups, see [62].

By the previous two paragraphs, l → ∞ implies a → ∞, which implies bl − 1 → ∞. As bl − 1 → ∞, the largest integral index cl for which the clth lower ramification group lGcl of lG is non-trivial tends to ∞. Now, let p be the rational prime over which p lies. We ˆ have lG ≤ G(Ll/Qp) :=lΓ and lGn ≤ lΓn for any n. (It follows from the definition of the ramification groups that lΓe(n+1)−1 =lGn, where e = e(p/p)). Therefore, as l increases, the largest n for which lΓn is non-trivial increases as well. From Section 2 we know that the exponent of p in rd(Lˆl) is

∞ 1 X (| Γ | − 1). (1.3) | Γ | l i l 0 i=0

The condition of Case I means that lΓ0 can be bounded independently of l. So as l → ∞,

(1.3) does as well, which in turn implies that rd(Lˆl), and therefore rd(Ll), tends to ∞, a contradiction. 20

Case II: : There exists p ∈ S such that lim sup e (p) = ∞. Lnl

Let m = m0m∞ be a modulus of K, where m0 is a product of ﬁnite primes and m∞ is a product of r real places of K. We have the following exact sequence from class ﬁeld theory

∗ ∗ m O → (O/m) → ClK → ClK → 1, (1.4)

∗ where OK are the units of the ring of integers O = OK , ClK is the ideal class group of K, m ∗ ∗ r ClK is the ray class group of K modulo m, and (O/m) is deﬁned to be (O/m0) × {±1} .

Let p be the rational prime lying below p. Because Ll/Q is assumed to be Galois, the assumption of Case II implies that lim sup e (p0) = ∞ for every prime p0 of K with Lnl 0 p ∩ Z = (p). For any rational prime q lying below a prime of S, deﬁne

Y q˜ = q. q∈S,q|q

For any modulus m, let Rm denote the ray class ﬁeld of K modulo m. Let n be the modulus Q q∈S q (note that whether a prime of K is contained in S depends only on the rational prime over which it lies). Ll is contained in Rns(l) for some positive integer s(l). Our intermediate goal is to show

lim sup[Ll ∩ Rp˜s(l) : K] = ∞. l→∞

Notation: Suppose {El/Fl}l is a set of number ﬁeld extensions indexed by l with

K ⊆ Fl for all l. We say El/Fl is (∗) if the ramiﬁcation above p of El/Fl is bounded independently of l.

Because lim sup eLl (p) = ∞, to prove our intermediate goal, it suﬃces to show that √ Ll/Ll ∩ Rp˜s(l) is (∗). Let T be the set consisting of the rational primes ≤ N and the inﬁnite real place of Q, i.e., the set of rational primes below the primes of S. Consider 21

Rns(l)

Ll Q q∈T Rq˜s(l)

Q Ll ∩ q∈T Rq˜s(l) Rp˜s(l)

Ll ∩ Rp˜s(l)

Figure 1.2: Diagram for Case II.

Q Figure 1.2. It follows from the exact sequence (1.4) that the p-part of [ q∈T Rq˜s(l) : Rp˜s(l) ], Q and therefore of [Ll ∩ q∈T Rq˜s(l) : Ll ∩ Rp˜s(l) ], is bounded independently of l. Therefore Q Q Ll ∩ q∈T Rq˜s(l) /Ll ∩ Rp˜s(l) is (∗). Thus it suﬃces to show that Ll/Ll ∩ q∈T Rq˜s(l) is (∗). We accomplish this with the following lemma.

Q Lemma 3. With notation as above, [Rns(l) : q∈T Rq˜s(l) ] is bounded independently of l.

Proof. By induction on the number of primes in T , it suﬃces to show that for any positive

integers a, b:

Y Y [Rp˜aq˜b : Rp˜a Rq˜b ] ≤ (Np − 1) (Nq − 1), p|p q|q

where q is a finite prime of T distinct from p, and the products are over primes of K (if q were the infinite place of Q, one finds that the right-hand side of the inequality would be r 2 1 , r1 being the number of real places of K). Define maps

∗ ∗ ∗ ∗ ∗ ∗ πp˜q˜ : O → (O/p˜q˜) , πp˜ : O → (O/p˜) , πq˜ : O → (O/q˜) .

Let HK be the Hilbert class ﬁeld of K. Looking at (1.4) and using that Rp˜a ∩ Rq˜b = HK 22 and that (O/p˜q˜)∗ =∼ (O/p˜)∗ × (O/q˜)∗, we see that

| Im(πp˜)|| Im(πq˜)| [Rp˜aq˜b : Rp˜a Rq˜b ] = . | Im(πp˜q˜)|

We have a decomposition

a ∗ ∼ Y ∗ a ∼ Y (O/p˜ ) = (O/p) × (1 + p)/(1 + p ) = Z/(Np − 1) × P, p|p p|p

Q a where P = p|p (1 + p)/(1 + p ) is a p-group. We have the analogous decomposition of b ∗ (O/q˜ ) . Let H = Im(πp˜q˜). Deﬁne the natural projections

φ :(O/p˜q˜)∗ → (O/p˜)∗, ψ :(O/p˜q˜)∗ → (O/q˜)∗,

so that Im(πp˜) = φ(H) and Im(πq˜) = ψ(H).

Since |φ(H)| and |ψ(H)| both divide |H| we obtain

|φ(H)||ψ(H)| ≤ gcd |φ(H)|, |ψ(H)|. |H|

We obtain the result of the lemma by noting that

∗ ∗ Y Y gcd |φ(H)|, |ψ(H)| gcd |(O/p˜) |, |(O/q˜) | (Np − 1) (Nq − 1). p|p q|q

This completes the proof of our intermediate goal. With this in hand, we now complete the proof that XN,K is ﬁnite under the assumption of Case II.

Set El = Ll ∩Rp˜s(l) , and let Pl be a prime of of El lying above p. Since K/Q is assumed

to be Galois, the Galois closure of El/Q is an abelian extension of K ramiﬁed only above p,

and it is contained in Ll since Ll/Q is Galois. But El is the maximal such ﬁeld; therefore 23

El/Q is Galois. In proving our intermediate goal, we showed that lim sup eEl (p) = ∞. Let

Eˆl = El and Kˆ = Kp, and consider the extension of local fields Eˆl/Kˆ with Galois group Pl s(l) ∗ s(l) lG. From the structure of (O/p˜ ) and the exact sequence (1.4) with m =p ˜ , we deduce that there exists C > 0 such that the prime-to-p part of |lG| is less than C for all l. Let Fl be the fixed field of the inertia subgroup of p of G(El/K), so that El/Fl is totally ramified above p. Let Ql be the prime of Fl below Pl and set Fˆl = Fl so that Eˆl/Fˆl is a totally Ql ramified extension of local fields. Set lH = G(Eˆl/Fˆl). If lJ denotes the p-Sylow subgroup of lH, then lim supl→∞ |lJ| = ∞. Recall that we are assuming that Ll/K is cyclic, so all groups in sight are cyclic. Let Mˆ l be the fixed field of lJ, with maximal ideal rl.

Eˆl Pl

Mˆ l rl

lG lH

Fˆl Ql

Kˆ p

Fix l for the moment and put r = rl. In this paragraph and the next, we summarize a few results from [59] and [62] and apply them to the situation at hand. For non-negative

n integers n, let U be the n-units of Mˆl, as deﬁned in the proof of Case I. For any real v ≥ 0, deﬁne U v = U n, where n − 1 < v ≤ n. For all v ≥ 0, the local reciprocity map

∗ v v θ : Mˆl → lJ maps U onto the upper ramiﬁcation group lJ . It follows that any jumps in the ﬁltration J v must occur at integral values of v (i.e., if J v+ 6= J v for any > 0,

v v+1 then v is an integer), and that for v ≥ 1, the quotient lJ /lJ of upper ramiﬁcation groups is isomorphic to a quotient of U v/U v+1. For integral n ≥ 1, we have isomorphisms n n+1 ∼ n n+1 ∼ + U /U = r /r = F ˆ , the last group being the additive group of the residue ﬁeld of Ml

Mˆ l, which is an elementary abelian p-group. Since lJ is cyclic, we ﬁnd that for v ≥ 1, the

v v+1 quotients lJ /lJ are either trivial or are cyclic groups of order p. 24

φ(u) For integral values of u, the function φ satisfying lJu =lJ is given by

u ! 1 X φ(u) = | J | − 1. (1.5) | J | l i l 0 i=1

By (1.5), φ(n + 1) − φ(n) ≤ 1, from which it follows that the lJn/lJn+1 are also either trivial or cyclic of order p. Thus each subgroup of lJ occurs as lJn for some n. Because jumps in the upper numbering occur at integral values, if n is a non-negative integer with lJn 6=lJn+1, then φ(n) ∈ Z. Using this fact along with (1.5) and the fact that every subgroup of lJ occurs as a ramiﬁcation group, we derive what is essentially the example on p.76 of [62]:

Lemma 4. For each i, 0 ≤ i ≤ t, there exists ni such that

i lJ(i) = lJni+1 = lJni+2 = ··· = lJni+p ,

t t−i where |lJ| = p and lJ(i) denotes the unique subgroup of lJ of order p . Less formally,

t−i i the subgroup of order p of lJ occurs as at least p lower ramiﬁcation groups of lJ.

v Remark 4. • The fact that jumps in the upper ramiﬁcation groups of lJ occur only at integral values of v also follows from the Hasse-Arf Theorem, which says that if

L/K is an abelian extension of local ﬁelds with perfect residue ﬁelds, then jumps in

the upper ramiﬁcation groups Gv of L/K occur only at integral values of v. The proof

does not require class ﬁeld theory [62].

n n+1 • One can more directly obtain that lJn/lJn+1 is isomorphic to a subgroup of U /U

n n+1 by considering the map lJn/lJn+1 ,→ U /U induced by σ 7→ σ(β)/β where β is a

uniformizer of Eˆl [62].

We use the information just obtained about the ramiﬁcation groups of lJ to obtain bounds for rd(Ll), which will complete the proof of Case II. Since El ⊆ Ll, it suf-

fices to show that lim supl→∞ rd(El) = ∞. By Section 2, it then suffices to show that ˆ ˆ ˆ lim supl→∞ rd(El) = ∞. The nth ramification group of El/Ml is a subgroup of the nth 25

ˆ ramiﬁcation group of El/Qp. By Section 2, it suﬃces to show that

∞ 1 X lim sup (| Γ | − 1) = ∞, (1.6) | Γ | l i l→∞ l 0 i=0

ˆ where lΓn are the ramiﬁcation groups of El/Qp. Let lγn denote the ramiﬁcation groups of

Eˆl/Mˆ l. It follows from Lemma 4 that

∞ 1 X (| γ | − 1) ≥ t − 1, (1.7) | γ | l i l l 0 i=0

tl where |lJ| = p . Note that lim supl→∞ tl = ∞. We also have

ˆ ˆ ˆ ˆ ˆ |lΓ0| = e(El/Fl)e(Fl/K)e(K/Qp).

ˆ ˆ ˆ The quantity e(K/Qp) is ﬁxed and e(Fl/K) = 1. By the deﬁnition of lJ,

1 1 | γ | = | J| ≥ | G(Eˆ /Fˆ )| = e(Eˆ /Fˆ ), l 0 l C l l C l l and thus we obtain the second inequality in

∞ ∞ ∞ 1 X 1 X 1 X (| Γ | − 1) ≥ (| γ | − 1) ≥ (| γ | − 1), | Γ | l i | Γ | l i ˆ l i l 0 i=0 l 0 i=0 C|lγ0|e(K/Qp) i=0 which, using (1.7), gives the desired result (1.6), completing Case II .

This completes the proof that XN,K is ﬁnite when K/Q is Galois. We use this to show that YN,K is ﬁnite.

We initially maintain the assumption that K/Q is Galois. Suppose L ∈ YN,K . We will show there is a constant A := AN;K , independent of L, such that rd(L˜) < AN;K , where L˜ denotes the Galois closure of L/Q. Suppose we have shown this to be true. The field L˜ is a compositum of abelian extensions of K, thus abelian over K. The finiteness of XAN;K ,K implies that there is some large number field FA, which can be chosen independently of L, such that L˜, and thus L, is contained in FA. So, upon showing the existence of AN;K , we 26

will have shown that YN,K is ﬁnite under the assumption that K/Q is Galois.

Existence of AN;K :

0 For any σ : L,→ C, σ maps a prime P of L ramifying in L/K to a prime P of σ(L) ramifying in σ(L)/K, where P and P0 lie over the same rational prime p. It follows that the rational primes ramifying in L/Q are the same as those ramifying in L/˜ Q, and using the fact that the discriminant is the norm of the diﬀerent, one checks that rd(L) = rd(σ(L)).

The degree [L : Q] is the number of embeddings σ : L,→ C. For each such σ, τσ(L) = σ(L) for all embeddings τ : σ(L) → C that ﬁx K. There are [σ(L): K] = [L : K] such τ, so

the number of distinct ﬁelds σ(L) as σ ranges over all embeddings L,→ C is at most

[L : Q]/[L : K] = [K : Q].

Lemma 5. Let E and F be number ﬁelds, both Galois over E ∩ F . Then

rd(EF )rd(E ∩ F ) ≤ rd(E)rd(F ).

Proof. For notational ease, put K = E ∩ F . Using (1.1) multiple times, we have

1 [F :Q] rd(F ) = NK/Q(dF/K ) rd(K), and 1 [EF :Q] rd(EF ) = (NK/Q dEF/K ) rd(K) 1 [EF :E] [EF :Q] = NK/Q NE/K dEF/E dE/K rd(K)

1 1 [EF :Q] [E:Q] = NE/Q(dEF/E) NK/Q(dE/K ) rd(K) 1 [EF :Q] = NE/Q(dEF/E) rd(E).

1 1 [EF :Q] [F :Q] Thus we must show NE/Q(dEF/E) ≤ NK/Q(dF/K ) .

[EF :F ] Let p be a prime of K above p. It suﬃces to show that the exponent a of p in dF/K is greater than or equal to the exponent b of p in NE/K (dEF/E). Let q be a prime of F above p and Q a prime of EF above q. Let P be the prime of E below Q. 27

EF Q

EF P q

E ∩ F = K p

Q p

It follows from the deﬁnition of ramiﬁcation groups that the natural restriction map

G(EFQ/EP) → G(Fq/Kp) takes G(Q/P),i (injectively) into G i+1 , where dne de- (q/p),d e −1e notes the least integer greater than or equal to n and e = e(Q/q). From this we ﬁnd that

X X α := (| G(Q/P),i | − 1) ≤ e (| G(q/p),i | − 1) := eγ. i≥0 i≥0

One checks that

a = γgF/K (p)fF/K (p)[EF : F ] and b = fE/K (p)gE/K (p)gEF/E(P)fEF/E(P)α.

We can rewrite b as

αfEF/F (q)gEF/F (q)fF/K (p)gF/K (p).

Using [EF : F ] = efEF/F (q)gEF/F (q) and α ≤ eγ, we ﬁnd that a ≥ b, as desired. This complete the proof of the lemma.

For our purposes, we only need the weaker result rd(EF ) ≤ rd(E)rd(F ). By the lemma and induction on n we obtain

n Y n rd(σ1(L) ··· σn(L)) ≤ rd(σ(Li)) = rd(L) . i=1

[K: ] [K: ] Thus rd(L˜) ≤ rd(L) Q , and we may take AN;K = N Q . 28

Thus, as long as K/Q is Galois, we have shown that YN,K is ﬁnite. We now show that the assumption K/Q Galois is unnecessary. Recall the following theorem from algebraic number theory (see, for example, [45] or [62]).

Theorem 8. Let E/F be a ﬁnite Galois extension of degree n and let p be a prime of F lying below a prime P of E with ramiﬁcation index e = e(P/p). Then

vp(dE/F ) ≤ n(1 + vp(e) − 1/e).

Let K˜ denote the Galois closure of K/Q and suppose L/K is abelian with rd(L) ≤ N.

We show there is a constant CN,K such that

rd(LK˜ ) ≤ CN,K rd(L).

Suppose we know this to be true. Since K/˜ Q is Galois, there exists a large number ﬁeld F such that E ⊆ F for any E ∈ Y . In particular, since LK/˜ K˜ is CN,K N,K˜ CN,K N,K˜ CN,K N,K˜ abelian, LK˜ ∈ Y . Thus CN,K N,K˜

rd(L) ≤ N =⇒ rd(LK˜ ) ≤ C N =⇒ L ⊆ LK˜ ⊆ F =⇒ Y is ﬁnite . N,K CN,K N,K˜ N,K

We show such a CN,K exists. It follows from (1.1) that

1 ˜ ˜ [LK:Q] rd(LK) = NL/Q dLK/L˜ rd(L). (1.8)

The primes ramifying in K/K˜ are ﬁxed with K, so there exists a ﬁnite set of rational

primes R, independent of L, such that every prime of L ramifying in LK/L˜ lies above a

prime in R. Let q be a prime of L ramifying in LK/L˜ lying above a rational prime q. 29

LK˜

q L K˜

q Q

˜ Let e = eLK˜ (q) (note LK/L is Galois). By Theorem 8, the exponent of q in dLK/L˜ is at most

[LK˜ : L] 1 + vq(e) − 1/e ≤ 2[LK˜ : L]vq(e).

We have the same bound for every prime q above q in L. Let eq = eL/Q(q), fq = fL/Q(q),

respectively (do not confuse e with eq). We obtain:

X ˜ vq(NL/Q dLK/L˜ ) ≤ 2[LK : L]fqvq(e) q|q X = 2[LK˜ : L] fqeqvq(e) q|q

≤ 2[LK˜ : L][L : Q][LK˜ : L],

P where we have used the fact that fqeq = [L : Q] and the obvious inequality

vq(e) ≤ [LK˜ : L] to obtain the ﬁnal inequality above. In light of (1.8), taking

Y 2[K˜ :K] CN,K = q q∈R

gives the result.

When n = 1, this completes the proof of Theorem 7 in its entirety. We obtain the

Theorem for general n by induction:

i Let L ∈ Yn,N,K . Let G denote the ith derived subgroup of G, so n is the smallest integer 30

n n−1 with G = 1. The quotient G/G corresponds to an intermediate ﬁeld L0 of L/K with n−1 ∼ G/G = G(L0/K). The fact that L0 ⊆ L implies rd(L0) ≤ rd(L), so L0 ∈ Yn−1,N,K .

By induction there are only a ﬁnite number of possibilities for L0. For each such L0,

n−1 G(L/L0) = G is abelian, so by Theorem 7 for n = 1, L can only be one of ﬁnitely many ﬁelds.

Remark 5. The proof of the Theorem 7 does not lend itself to an eﬀective bound of the size of Yn,N,K in terms of n, N and K.

Corollary 2. Fix N > 0, a positive integer n, and a number ﬁeld K. Then

#{L : L/K is solvable, G(L/K) ⊆ GLn(F) for some finite field F, and rd(L) ≤ N} is finite.

Proof. It is a theorem of Zassenhaus [68] that the length of any solvable subgroup of

GLn(F) has a ﬁnite bound depending on n, independent of F (in fact F need not even be ﬁnite). The result now follows from Theorem 7.

1.4.3 Further questions

As mentioned in the beginning of Section 1.4, a natural extension of Theorem 7 is to consider the size of various subsets of ZN,K . Kedlaya [31], generalizing work by Yamamoto [73], has shown that for any n, there exist infinitely many real quadratic number fields K admitting an unramified degree n extension L with G(L/K˜ ) isomorphic to the alternating group An, where L˜ denotes the Galois closure of L/K. It is unknown, however, whether or not a fixed (say real quadratic) number field may admit an unramified degree n extension with Galois closure having Galois group An for infinitely many n.

Fix a number field K. Suppose that ZN,K is infinite. One can ask whether ZN,K must contain an infinite class field tower. Maire [38] has proven the existence of infinite 31 unramified extensions of number fields with class number one. Let K be such a number

field with infinite degree unramified extension L. Maire constructs L by constructing an infinite class field tower of a finite unramified extension of K. This leaves open the question of whether a general infinite unramified extension L/K must contain an infinite class field tower–i.e., whether G(L/K) must have a pro-solvable subquotient. One can also ask the same question but replace the condition L/K unramified with the condition rd(L) < N, by which we mean rd(M) < N for every field M between L and K with M/K finite. The answers to such questions are likely beyond the scope of class field theory.

Since there are only finitely many number fields of bounded root discriminant of any degree, finding infinite subsets of ZN,K amounts to finding L ∈ ZN,K with arbitrarily large degree. One may alternatively consider how many number fields L of degree n over K with rd(L) < N exist, with n fixed and N varying. Significant work in this direction has been done by Ellenberg and Venkatesh [13] and others.

Finding large number ﬁelds with small root discriminant remains a task of interest.

In order to find number fields that do not arise as solvable extensions of well understood number fields, we can look to Galois representations attached to analytic objects. Using

Galois representations attached to a particular modular form, Dembélé[10] demonstrated the existence of a non-abelian Galois field K/Q with root discriminant less than 56 and 6 [K : Q] ≈ 2 × 10 . In the next chapter, we discuss a correspondence between Galois representations and modular forms before studying three-dimensional Galois representations with solvable image. Chapter Two

Artin representations 33

2.1 Introduction

One of the major quests in number theory is to understand the absolute Galois group

GQ of the rationals. Thanks to class field theory, finite, solvable extensions of Q are well understood, at least in theory. For example, in Section 1.4, we used class field theory to show there are only finitely many extensions of a number field K with bounded root discriminant and bounded solvable length. One goal of the Langlands program is to better understand non-solvable extensions of Q by relating Galois representations to analytic objects. The result of Dembéléat the end of Section 1.4 is an example of such a result.

The standard permutation representation of Sn on n letters gives an embedding of Sn into GLn(F ) for any field F . Thus, any finite group embeds into GLn for some n, and to understand all finite extensions of the rationals is to understand all finite-dimensional continuous representations of GQ, where we take the discrete topology on GLn(F ), which guarantees us that the image of a representation is finite.

One might be interested, for example, in characterizing all representations of GQ un- ramiﬁed outside ∞ and a single odd prime p since the restriction on ramiﬁcation allows for the use of local techniques.

2.2 An application of Serre’s conjecture

For the moment, let us take our ﬁeld F to be Fp. A modest goal is to characterize all

1-dimensional representations GQ → GL1(Fp), that is, the case n = 1. This is straightfor- ward, as we are merely asking for finite extensions of Q unramified outside p and ∞ with ∗ Galois group isomorphic to a subgroup of Fp. By class field theory, these are exactly the subfields of Q(ζp). Next we consider the case n = 2. Our representations are no longer necessarily solvable, so class field theory will not suffice. What the Langlands program does, however, is associate to each such representation an analytic object. More precisely, 34 we have Serre’s conjecture [63], now a theorem of Wintenberger and Khare [33, 34]. We give a special case of Serre’s conjecture in Theorem 9, below. Recall that a representation of GQ is said to be odd if the determinant of complex conjugation is −1. Let Fl ∈ GQ be Frobenius at l.

n Theorem 9. Let f = Σanq be a normalized (a1 = 1) cusp form of level 1, weight k, and nebentypus character . Then for every prime p, there exists a representation

ρf,p :GQ → GL2(Fp) unramiﬁed outside p such that for all primes l 6= p, the trace of Fl is k−1 al and the determinant of Fl is (l)l . Moreover, if we’re given an odd, irreducible representation ρ :GQ → GL2(Fp) unramiﬁed outside p, there exists a corresponding eigenform f such that ρf,p = ρ.

Suppose that we have an odd, irreducible, Galois representation ρ :GQ → GL2(Fp) unramified outside p, as in the statement of Serre’s conjecture. The fixed field K of the

kernel of ρ is a number field unramified outside p, and G(K/Q) is isomorphic to the image of ρ. Therefore, Serre’s conjecture provides information about the existence of such fields

K. For example, we have the following corollary.

Corollary 3. Let p ∈ {2, 3, 5, 7}. There do not exist any Galois extensions K of Q that simultaneously satisfy the following three conditions:

1. K/Q is unramiﬁed outside of p

2. G(K/Q) is isomorphic to a ﬁnite irreducible subgroup of GL2(Fp)

3. If p 6= 2, then complex conjugation is not contained in the center of G(K/Q).

Proof. We may view K as the kernel of the ﬁxed ﬁeld of a two-dimensional representation

ρ of GQ. Let c be the image of complex conjugation under ρ. Since the representation is assumed to be irreducible, the third condition in the statement of the corollary implies that c is not a scalar (by Schur’s Lemma [61, Section 2.2]). It follows that det c = −1, and our representation ρ satisﬁes the hypotheses of Serre’s conjecture. Thus, ρ comes from a 35 modular form. Moreover, after twisting said modular form by an appropriate power of the mod p cyclotomic character, we can obtain a modular form of weight p+1 [42]; however, it is well-known that there exist no cusp forms of level 1 and weight less than 12. Therefore, since we have assumed p < 11, the ﬁeld K in question cannot exist.

2.3 Counting Artin representations

We now turn our attention to the situation F = C. In this case, a representation of GQ is called an Artin representation. Using a field of characteristic zero allows us to consider representations that do not avoid ramification at any particular prime. One of the first

Langlands-type results in this situation is the Langlands-Tunnell Theorem.

Theorem 10. [6, Chapter VI] Let ρ :GQ → GL2(C) be an odd, continuous, irreducible, two-dimensional representation with solvable image. Then there exists a normalized eigen-

P∞ n form n=1 anq of weight 1 (and some level and nebentypus) such that the trace of Frobe- nius at l is al for every prime l.

In addition to being a powerful theorem in its own right, the Langlands-Tunnell Theo- rem was also instrumental in the proof of Fermat’s last Theorem. It was used to show the modularity of the representation of Galois acting on the 3-torsion of an elliptic curve. The key fact is the solvability GL2(F3), and thus the solvability of a lift of the image of Galois in GL2(F3) to GL2(C).

In the general case of two-dimensional Artin representations, the corresponding analytic objects are modular forms of weight 1 and Maass forms of weight 0, which correspond to odd and even two-dimensional Artin representations, respectively. Speciﬁcally, Deligne and

Serre [9] showed that given a cuspidal modular form of weight 1, level N, and Nebentypus

P n character with normalized Fourier expansion anq , there exists an Artin representation

ρ :GQ := Gal(Q/Q) → GL2(C) 36 of conductor N such that (via ρ) the trace of the conjugacy class of the Frobenius at p, Frobp, in GQ is ap, and the determinant of Frobp is (p). Conversely, given a two- dimensional Artin representation ρ, if we insist that either ρ has solvable image [6] or that

ρ is odd [32], then there is known to exist an automorphic form f such that f gives rise to ρ in the manner described above. Using this correspondence, Wong [72], building on work of

Duke [12], bounded the number of two-dimensional Artin representations of Q with Artin conductor N in terms of N.

Less is known in the case of three-dimensional Artin representations. We will bound the number of three-dimensional irreducible Artin representations with solvable image and prime power conductor. It is known that certain such representations come from automorphic representations of GL3(AQ), where AQ denotes the ad`elesof Q. We use purely algebraic techniques for our bounds, which therefore, in principle, could give bounds for the number of automorphic representations of GL3(AQ) with particular properties.

The main results of this section, which are made more precise in Theorems 11, 12, 14, and 15, are the following.

• The number of irreducible, imprimitive three-dimensional Artin representations of

m m+1 GQ with conductor dividing p is at most Cmp , where C is an explicitly given constant.

• The number of solvable, irreducible, primitive three-dimensional Artin representa-

m m 3 +50 tions of GQ with conductor dividing p is at most Dp , where D is an explicitly given constant.

The current chapter is divided into two main parts. The ﬁrst part looks at imprimitive representations. We use the fact that any imprimitive three-dimensional representation is monomial to bound the number of all imprimitive representations in one fell swoop. The 37 second part deals with primitive representations, where we do a case-by-case analysis of the three groups that can occur as the projective image of a primitive three-dimensional

Artin representation.

2.4 Deﬁnitions and preliminaries

Let V be a ﬁnite-dimensional vector space over an arbitrary ﬁeld, and let G be a group.

A representation ρ : G → GL(V ) is called imprimitive if there exist proper subspaces

m V1,...,Vm of V such that V = ⊕i=1Vi and ρ(G) permutes the Vi. If such a decomposition of V does not exist, then ρ is called primitive. If ρ is imprimitive, and if the corresponding

Vi are one-dimensional and permuted transitively by G, then ρ is said to be monomial (this is equivalent to the standard deﬁnition of monomial, which is that there exists a subgroup H ≤ G and a linear character χ of H such that ρ is induced from χ). Thus monomial representations of dimension greater than 1 are imprimitive, and the converse is true if the dimension of the representation is prime, and if G acts transitively on the Vi. A group is said to be monomial if each of its irreducible representations is monomial. A comprehensive treatment of these deﬁnitions and the general theory of representations of

ﬁnite groups can be found in [8].

An Artin representation of a number ﬁeld F is a continuous homomorphism

ρ : Gal(F /F ) → GLn(C).

We take the analytic topology on GLn(C), so Artin representations necessarily have finite image. The conductor N(ρ) of ρ is defined to be the product of its local conductors, whose definition we now recall. Let p be a prime of F . For the p-part Np(ρ) of N(ρ),

consider the restriction ρp of ρ to G(F p/Fp). The representation ρp factors through a ﬁnite

quotient G(K/Fp) of G(F p/Fp). Let gi denote the order of the ith ramiﬁcation group Gi of G(K/Fp). We then deﬁne Np(ρ) to be p raised to the power 38

X gi dimV/V Gi , g0 i≥0 where V is the representation space for ρ and V Gi is the subspace of elements ﬁxed by

Gi. If a is an ideal in the ring of integers O of a number ﬁeld, such as the conductor of an

Artin representation, and if p is a prime of O, then we take vp(a) to be the exponent of the highest power of p dividing a. For the remainder of Section 2.3 we make the following hypotheses on all Artin representations:

F = Q and n = 3.

We will always count representations up to isomorphism. The natural projection

π : GL3(C) → PGL3(C) will be denoted by π, and a lift of a projective representation

ψ : G → PGL3(C) of a group G is any representation ρ : G → GL3(C) such that ψ = π ◦ ρ.

A cyclic group of order n will be denoted by Cn, and Z(G) denotes the center of a group G. If L/K is a Galois extension of ﬁelds, then we say that E is a central extension of L/K

if E/K is Galois and G(E/L) ≤ Z G(E/K).

The symbol Eρ will denote the ﬁxed ﬁeld of the kernel of one of our irreducible three-

dimensional Artin representations ρ with solvable image, and Fρ will denote the ﬁxed ﬁeld of the kernel of π ◦ ρ (when the context is clear, the subscript ρ may be dropped). Thus ρ

descends to an injection G(Eρ/Q) ,→ GL3(C), which by abuse of notation we also denote by ρ, and Z G(E/Q) = G(E/F ), which consists of scalars (by Schur’s Lemma, since ρ is irreducible [61, Section 2.2]) and is thus cyclic.

We say an Artin representation ρ is ramified at p if the extension Eρ/Q is ramified at p. In sections 2.5 and 2.6, all representations are assumed to be unramified outside p∞,

where p is a ﬁnite prime and ∞ is the real place of Q. To avoid special cases, most of which are trivial, we will assume p ≥ 5.

In our computations, we will wish to bound the number of Ca or Ca × Ca extensions of 39 a number field K with given ramification conditions, where a is prime. We do this by first

ﬁnding the a-rank of the maximal extension L of K with the given ramiﬁcation conditions.

b Let ra,b(x) be the number of (Ca) extensions of K contained in a Galois extension L, where G(L/K) is abelian and has a-rank x. We will consider the cases a = 2 or 3, although the following formulas hold for any prime a:

ax − 1 (ax − 1)(ax − a) r (x) = , r (x) = . a,1 a − 1 a,2 (a2 − 1)(a2 − a)

When we have an extension L/K of number fields with L/Q and K/Q Galois, and L/Q ramified only at p, we will write e(L/K) to denote the ramification from K to L of the

primes above p in K. We shall say an extension of fields is unramified if it is unramified

at all ﬁnite primes. The ring of integers of a number ﬁeld K will be denoted by OK and

its absolute discriminant by dK . We let hK denote the class number of K, ClK the ideal

m class group of K, ClK the ray class group of K of modulus m, HK the Hilbert class field of K, and Km the ray class field of K of modulus m. If m is a product of finite primes of

K, we will write m∞ to denote the modulus that is the product of m multiplied by all the

real places of K.

2.5 Imprimitive representations

In this section we assume that ρ :GQ → GL3(C) is irreducible, imprimitive, and unramiﬁed outside p∞. Since ρ is monomial, there exists a ﬁeld L of degree 3 over Q and a character ∗ Q Q χ :GL → C such that ρ = χL, where χL denotes the induced representation of χ to GQ.

Suppose ﬁrst that L ⊂ Q(ζp) and let p be the prime of L above p. It follows from the corollary to Proposition 4, Chapter VI of [62] that

vp(N(ρ)) = vp(N(χ)) + 2. (2.1) 40

Let    m   Irreducible ρ :GQ → GL3(C) such that N(ρ) | p  cm = # , ∗ Q  and there exists χ :GL → C such that ρ = χL 

∗ m dm = #{χ :GL → C such that N(χ) | p }.

It follows from Mackey’s irreducibility criterion [61, Section 7.4] that the induction ρ to GQ ∗ of a non-trivial χ ∈ Hom(GL, C ) will be irreducible exactly when ρ restricted to GL is the sum of three distinct characters, each of which induces to give ρ by Frobenius reciprocity

[61, Section 7.2]. Therefore, using (2.1), we have

d c ≤ m−2 . m 3

m We first bound dm. By class field theory, any χ with N(χ) | p factors through G(Lpm /L), where Lpm is the ray class field of L of modulus pm. Class field theory gives

the exact sequence

∗ m ∗ pm∞ OL → (OL/p ∞) → ClL → ClL → 1, (2.2)

m ∗ m ∗ 3 where (OL/p ∞) is deﬁned to be (OL/p ) × h±1i , the exponent 3 coming from the

∗ m ∗ three real places of L. The image of OL in (OL/p ∞) has order at least 2, so we obtain

m 1 d = | G(Lp ∞/L)| ≤ 4h pm−1(p − 1) < 4 p pm−1(p − 1) = 2pm(p − 1), m L 2 where the ﬁrst inequality follows from (1.4), and the second inequality from [36, Corollary

4], for which we need the fact that L is totally real. Thus, we obtain the following theorem.

Theorem 11. With notation as above,

2pm−1 c < . m 3

For the remainder of this section, we suppose that L *Q(ζp), or equivalently, that 2 L/Q is not Galois. By the following lemma, p must split in L as p1p2. 41

Lemma 6. Let L/Q be a degree 3 extension ramiﬁed at a single ﬁnite prime p ≥ 5. Then

L/Q is Galois if and only if p is totally ramiﬁed in L.

Proof. The only if part is clear. For the converse, suppose that L/Q is totally ramiﬁed at p and let p be the prime of L above p. Let Lp be the completion of L at p and O = OLp be the ring of integers in Lp. Let f(x) be the minimal polynomial for α ∈ O, where α generates O as a Zp-algebra. We may assume α ∈ OL. The p-part of the discriminant 2 disc(f) of f equals the local discriminant dLp/Qp , which has p-part p (p 6= 3), the same 2 as the p-part of the global discriminant dL/Q. So we can write disc(f) = p D, with ∈ {±1} and p - D. The Galois closure K of L/Q is ramiﬁed only at p and is given by p √ √ K = L disc(f) = L D . For p 6= 2, this is only possible if D ∈ L, i.e., L/Q is Galois.

By the corollary to Proposition 4 of [62], we have

Q N(χL) = NormL/Q(N(χ)) dL .

a1 a2 a1+a2 It follows that if N(χ) = p1 p2 , then N(ρ) = p p. Let cm = cm,L be the number of m imprimitive ρ with conductor dividing p that are induced from GL. To estimate cm, we need to estimate

a1 a2 #{χ : N(χ) | p1 p2 , and a1 + a2 ≤ m − 1}.

pa1 pa2 ∞ Thus we need to bound P | G(L 1 2 /L)|. We have a1+a2=m−1

a1 a2 X p p ∞ X a1 a2 m−1 cm ≤ | G(L 1 2 /L)| ≤ 4hL (p − 1)(p − 1) < 4hLmp , (2.3)

a1+a2=m−1 a1+a2=m−1 where the ﬁrst inequality follows from the exact sequence (1.4) applied to the situation at hand.

Since L/Q is not Galois, GL is not normal in GQ, so it is possible that there may only ∗ be one χ ∈ Hom(GL, C ) that gives ρ when induced up to GQ. Since we are assuming that 42

H := G(Eρ/L) is not normal in G := G(Eρ/Q), the group H has three conjugate subgroups ∗ s inside G. Given a coset sH and χ ∈ Hom(H, C ), we can deﬁne the representation χ ∈ −1 ∗ s −1 Hom(sHs , C ) by χ (x) = χ(s xs). The formula for an induced character [61, Section 7.2] shows that χs induced up to G is isomorphic to χ induced up to G. The three conjugate

s subgroups H correspond to conjugates of L, each of degree three over Q. Thus, it remains

np to bound hL and 3 , where np is the number of non-Galois fields L of degree 3 over Q, ramified only at p∞. We do the latter first. Let L be such a number field. The Galois

closure K of L/Q is an S3 extension of Q ramiﬁed only at p∞. By Lemma 6, p splits as 2 p1p2 in L. Let q be a prime of K above p2. We have

e(q/p) = e(q/p2)e(p2/p) = e(q/p2) ≤ 2. (2.4)

On the other hand, e(q/p) is at least 2 since p is ramified in K, and thus e(q/p) = 2. √ √ Therefore, K/Q( p∗) is unramified, where Q( p∗) is the subfield of K of degree 2 over Q. Therefore, L and its two Galois conjugates contained in K combine to give one unramified √ √ ∗ ∗ C3 extension of Q( p ). Ramification at the infinite primes of Q( p ) can only contribute a power of 2 to the degree of the maximal abelian unramified (at all finite primes) extension √ of Q( p∗). Since we are interested in the 3-rank of this extension, we may ignore the √ 22.2p potential ramification at the infinite primes. By [43, page 95], we have hQ( p∗) < π2 , and thus the 3-rank of Cl √ is less than Q( p∗)

22.2p log , 3 π2 from which it follows that n 22.2p p < r log , 3 3,1 3 π2 which is less than 11.1p . π2

22.2p Putting this last bound together with (2.3), and using the bound hL < π3 (which 43 follows from [43, page 95]), we obtain the following.

Q m Theorem 12. The number of imprimitive ρ with ρ = χL, N(ρ) dividing p , and L/Q non-Galois is less than 985.7mpm+1 . π5

Remark 6. It is possible to obtain a slightly better bound by using a slightly better bound

√ for the 3-rank of ClQ( p∗). See, for example, [50].

2.6 Primitive representations

For this section, we suppose that ρ :GQ → GL3(C) is primitive. We continue with our assumption that ρ is unramiﬁed outside p∞ and has solvable image. Since ρ is primitive

with solvable image, the image of π ◦ ρ in PGL3(C) is isomorphic to one of the following groups [11, Theorem 4.8]:

P1 := C3 × C3 o C4,P2 := C3 × C3 o Q8,P3 := C3 × C3 o SL2(F3),

where Q8 is the quaternion group of order 8. We can view P1 as a subgroup of P2, and P2

as a subgroup of P3.

Remark 7. Counting four dimensional primitive representations is considerably more dif-

ﬁcult, as there are 28 ﬁnite primitive subgroups of PGL4(C) [1].

We claim that Im(π ◦ ρ) cannot be P2. If G(Fρ/Q) = P2, then Fρ (and thus Eρ) has ∼ a subfield L with L/Q Galois and G(L/Q) = C2 × C2. But there is no such field ramified

only at one prime (p 6= 2) of Q.

In the next two subsections, we consider the cases G(Fρ/Q) = P1 and G(Fρ/Q) = P3.

For each case, we first fix a number field F with G(F/Q) = Pi and bound the number of

representations ρ with Fρ = F . We then bound the number of such possible F – i.e., F 44

such that G(F/Q) = Pi and F/Q is unramiﬁed outside p∞.

We will use the fact that the Schur multiplier of G, H2(G, Z), is isomorphic to C3 for

G = P1 and G = P3. An alternative deﬁnition of the Schur multiplier of a group G is the largest group X such that there exists a group Γ with Γ/X =∼ G and X ⊆ Γ0∩Z(Γ) [27, page 151], where Γ0 is the commutator subgroup of Γ. Using this deﬁnition, one shows that the p-

Sylow subgroup of H2(G, Z) is isomorphic to a subgroup of the Schur multiplier of a p-Sylow subgroup of G. Using this fact, it is an exercise to show that H2(P1, Z) = C3. To show that

H2(P3, Z) = C3, we need to show that the 2-Sylow subgroup of P3, the quaternion group

Q8 of order 8, has trivial Schur multiplier and that the 3-Sylow subgroup (C3 × C3) o C3 of P3 has Schur multiplier C3. The former holds because every abelian subgroup of Q8 is cyclic, and one can show that all such group have trivial Schur multiplier. To show the latter, one can, for example, use the facts about the Schur multiplier of a semi-direct product that are proven in [14].

2.6.1 Representations with projective image isomorphic to P1

We count the number of Artin representations ρ unramified outside p∞ with projective ∼ image isomorphic to P1, i.e., G(Fρ/Q) = P1. Let L denote the fixed field of C3 × C3 ≤ P1.

Since L is unramiﬁed outside p∞, it is the unique degree four subﬁeld of Q(ζp). Writing

[Eρ : Fρ] = n, we have the tower of ﬁelds in Figure 2.1.

Eρ

Fρ

C3 × C3 L

Q Figure 2.1: Field Diagram for Section 2.6.1. 45

Lemma 7. Let K be a subﬁeld of Q(ζp), and let p be the unique prime of K above the p∞ rational prime p. Then K = HK (ζp).

p∞ p∞ Proof. Certainly HK (ζp) ⊆ K . Suppose the containment is strict. Then K /HK (ζp)

p∞ is totally ramiﬁed at the primes of HK (ζp) above p. It follows that K /Q(ζp) is tamely

ramiﬁed at the prime (1 − ζp) of Q(ζp) above p. But such an extension cannot exist since

(1−ζp) Q(ζp) = HQ(ζp) [23, Proposition 3.4].

p We have F := Fρ ⊆ L , where p is the unique prime of L above p. Suppose that the ramification degree e = e(F/L) of p in F is 3 or 9. Then by Lemma 7, the ramification in F/L must be obtained by adjoining the unique degree 3 (e = 3) or 9 (e = 9) subfield

0 ab of Q(ζp) to L. It follows that |P1| = [F : F ∩ Q ] = 3 (e = 3) or 1 (e = 9). This is a 0 contradiction since P1 = C3 × C3. Therefore, F/L is unramiﬁed and F ⊆ HL.

We have the following lemma of Tate:

Lemma 8. (Tate [60, Theorem 5]) For each prime p, let Ip be the inertia group of p in

GQp , which we view as a ﬁxed subgroup of GQ. Let ρ˜ :GQ → PGLn(C) be a projective representation. Suppose that for each p there exists a lift ρ :G → GL ( ) of ρ˜| such p Qp n C GQp that ρp|Ip trivial for all but ﬁnitely many p. Then there exists a unique lift ρ :GQ → GLn(C) of ρ˜ such that ρ|Ip = ρp|Ip for all p.

Returning to our notation before Lemma 8, since Fρ/Q is unramified outside p, for all primes l 6= p,ρ ˜|G factors through a cyclic extension of l. The Schur multiplier H2(G, ) Ql Q Z of a cyclic group G is trivial, so any projective representation of G lifts to an ordinary representation of G (see [24, Chapter 20], in particular Theorem 20.8, part c). We may thus liftρ ˜|G to a map ρl :G → GL3( ) such that the fixed field of the kernel ofρ ˜|G is Ql Ql C Ql the same as the fixed field of the kernel of ρl; in particular, both fixed fields are unramified at l.

Since Fρ/L is unramified, FρQ(ζp)/Q(ζp) is unramified. By class field theory, the 46

principal ideal (1 − ζp) splits completely in FρQ(ζp). It follows that the ideal p of L splits completely in F . Thus the kernel of the fixed field ofρ ˜| is a totally tamely ramified ρ GQp extension of , so cyclic. We may therefore liftρ ˜| to a map ρ such that the kernel of Qp GQp p the fixed field of ρ is the same as the kernel of the fixed field ofρ ˜| . p GQp

Putting this local data together, Lemma 8 now tells us that there exists a lift ρ ofρ ˜ such that the ramification in Fρ/Q is the same as the ramification in Eρ/Q (i.e., Eρ is unramified at all primes l 6= p and the ramification degree at p in Eρ/Q is the same as the ramification degree at p in Fρ/Q).

We will need the following theorem.

Theorem 13. (Fr¨ohlich [16, page 31; page 43, Corollary 2]) Let L/K be a Galois extension of number ﬁelds, and let N be a central extension of L/K. Then there is a surjective map

ab H2(G(L/K), Z) G N/L(N ∩ K ) , and there exists a central extension N 0 of L/K such that

0 0 ab H2(G(L/K), Z) → G N /L(N ∩ K )

0 is an isomorphism. Moreover, if K = Q, one can take N to be ramiﬁed at the same rational primes at which L is ramiﬁed.

Remark 8. The Schur multiplier H2(G, Z) of a group G is intimately related to the study of central extensions and projective representations of G. See, for example, [24, Chapter

20].

Lemma 9. For a prime p ≡ 1 (mod 4), let L ⊆ Q(ζp) be the unique ﬁeld with [L : Q] = 4.

Suppose that L admits an unramiﬁed C3 × C3 extension F . Then F/Q has a unique degree three unramiﬁed central extension.

Proof. We ﬁrst show the existence of such an extension. Letρ ˜ be a projective representation 47

of G(F/Q) with image isomorphic to P1. Let ρ be a lift ofρ ˜ such that the image of ρ|Ip has order 4, the existence of which is guaranteed by Lemma 8. Let N be the fixed field of ker ρ. Since the ramification degree of p in L is 4, N/L is unramified (recall we have

ab assumed that F/L is unramiﬁed) and N ∩ F Q = F . By Theorem 13, we must have ab [N : N ∩ F Q ] = [N : F ] ≤ 3. We cannot have N = F because the group P1 does not have any irreducible three-dimensional representations, so [N : F ] = 3. Thus N gives the desired extension.

Now for uniqueness. Suppose that N1 and N2 are distinct non-trivial unramiﬁed central

ab ab extensions of F/Q. We have H2(G(F/Q), Z) = C3 and F (Ni ∩ Q ) ⊆ Ni ∩ F Q , so it ab follows from Theorem 13 that C3 G(Ni/Ni ∩ F Q ), i = 1, 2. Since Ni/F and F/L are ab unramiﬁed, we have Ni ∩ F Q = F . Suppose N1 6= N2. Then N1N2/F is an unramiﬁed ab central extension of F/Q with N1N2 ∩ F Q = F and [N1N2 : F ] = 9, contradicting Theorem 13.

∼ Lemma 10. Let F/Q be a ﬁnite Galois extension with group G, and suppose H2(G, Z) = C3.

For a fixed number field M ⊂ Q(ζp∞ ), FM/Q admits at most three distinct non-trivial cen- ab tral extensions Ei with Ei ∩ Q = M and Ei/Q unramified outside p∞.

Proof. Let Ei be ﬁelds as in the statement of the lemma. By Theorem 13, for each i, we Q have [Ei : FM] = 3. The compositum Ei is a central extension of FM/Q. By Theorem 0 Q 0 13, there exists M ⊂ Q(ζp∞ ) containing M such that | G( Ei/F M )| ≤ 3. This means Q Ei/F M has rank at most 2, i.e., its Galois group is C3 or C3 ×C3. Such a ﬁeld extension

has at most (in the C3 × C3 case) four strictly intermediate extensions; three ﬁelds Ei with

ab 0 0 Ei ∩ Q = M and FM , (and we must have [M : M] = 3).

We now come to the main result of this section.

Proposition 2. Let ρ :GQ → GL3(C) be an Artin representation unramiﬁed outside p∞ 48

with projective image isomorphic to P1. Let F be the ﬁxed ﬁeld of ker π ◦ ρ. Let Q(ζp,3)

denote the maximal subﬁeld of Q(ζp) of 3-power degree over L, and let Q(ζp,30 ) be the

composite of all subfields of Q(ζp∞ ) with degree prime-to-3 over L. Then the fixed field of ker ρ either

i) lies between N and NQ(ζp,30 ), where N is the unique non-trivial unramiﬁed central

extension of F/Q, and [N : F ] = 3, or

ii) lies between E3 and E3Q(ζp,30 ), where E3 * F Q(ζp,3) is one of two possible degree

3 ramiﬁed extensions of FM3 that is central over F/Q, where M3 is a subﬁeld of

Q(ζp,3).

Proof. The existence and uniqueness of N are given by Lemma 9. For the entirety of the

proof, we consider representations ρ for which Fρ = F for some ﬁxed choice of F . Let E

be a candidate for the ﬁxed ﬁeld of ker ρ. By Theorem 13, there exists M ⊂ Q(ζp∞ ) such ab that [E : FM] ≤ 3, and E ∩ Q = M. We claim that E 6= FM. If E were contained in k ∼ FM with [E : F ] = 2 m, m odd, then we would have G(E/Q) = (C3 × C3) o (C2k × Cm),

with C2k ×Cm acting on C3 ×C3 through its C4 quotient. Thus C3 ×C3 ×Cm is an abelian k normal subgroup of G(E/Q) of index 2 . By [8, Corollary 53.18], the degree of every k irreducible representation of G(E/Q) divides 2 . In particular, G(E/Q) has no irreducible three-dimensional representations, which is a contradiction.

The proof will be complete upon establishing the following two claims.

Claim I: If E/F M is unramiﬁed, then we are in Case i) of the proposition.

Claim II: If E/F M is ramiﬁed, then we are in Case ii) of the proposition. 49

Proof of Claim I. We are assuming that E/F M is unramiﬁed. Being a composite of two central extensions of F , NM is a central extension of F/Q as well (see Figure 2.3). NM

N FM EN

MF NM E

L NFM

F Q

Figure 2.2: Field diagrams for proof of Proposition 2.

If E 6= NM, then EN is an unramiﬁed central extension of FM. Since F/L is unram-

ab iﬁed, if EN/F M is unramiﬁed, then EN ∩ FMQ = FM with [EN : FM] = 9, which contradicts Theorem 13. So E = NM, and we have only to show that M ⊆ Q(ζp,30 ).

Suppose that 3 | [M : Q]. Then NM/F , having two intermediate subﬁelds of degree 3 over

F , is not cyclic. Thus we must have M ⊆ Q(ζp,30 ).

Proof of Claim II. Let M3 = M ∩Q(ζp,3) and M30 = M ∩Q(ζp,30 ), so M = M3M30 . Since 2 E/F M3 is cyclic (recall that E/F is cyclic) and [E : FM3] is divisible by 3 but not by 3 ,

there exists a degree 3 extension E3 of FM3 such that E = E3M. Note that E3M = E3M30 .

An elementary ramiﬁcation argument shows that E/F M3, and thus E3/F M3, is totally ramiﬁed.

Deﬁne

   Central extensions J of F/ containing M such that   Q 3    ΣM3 = J/F M3 is ramiﬁed at p and only at p,    ab   [J : FM3] = 3, and J ∩ Q = M3.  50

We are assuming E3/F M3 is ramiﬁed, so E3 ∈ ΣM3 . From Lemma 10, we have #ΣM3 ≤ 3.

To complete the proof of Claim II, we need to show that #ΣM3 ≤ 2. From the proof of 0 Lemma 10, the compositum J of all elements of ΣM3 is contained in a C3 × C3 extension 0 of FM3. We may assume #ΣM3 = 3, and thus that J is equal to a C3 × C3 extension 0 0 0 of FM3. The proof of Lemma 10 also tells us that FM ⊂ J , where M ⊂ Q(ζp∞ ) and 0 [M : M3] = 3. Since F/L is unramiﬁed, FM3/M3 is unramiﬁed as well, from which it

0 0 follows that FM /F M3 is ramified, as M /Q is totally ramified. If #ΣM3 = 3, then all 0 0 four intermediate subfields of J /F M3 are ramified over FM3, and thus J /F M3 is totally ramified. This is a contradiction, however, since the corresponding extensions of local fields

at primes above p would be totally tamely ramiﬁed but not cyclic.

The next goal is to count the number of representations GQ → GL3(C) with projective

image isomorphic to P1 and given prime power conductor. We must have p ≡ 1 (mod 4), and we will also assume (as is necessary for such a representation to exist), that the 3-rank

k of the class group ClL of L is at least 2 (recall the deﬁnition of r3,2 from Section 2.4).

Then L admits r3,2(k) unramiﬁed extensions F with G(F/L) = C3 × C3.

Let J = G(N/Q), where N is as in Case i) of Proposition 2. Since the 3-Sylow subgroup

G(N/L) := A of J is normal in J, the group J splits as A o C4. If A is abelian, then J is metabelian (i.e., the second commutator group of G is trivial), which gives a contradiction

since metabelian groups are monomial [8, Theorem 52.2], and ρ is primitive. Suppose that

A is isomorphic to the non-trivial semi-direct product C9 o C3. A generator of C4 acts on J/Z(J) as an automorphism of order 4 and thus must act on A as an automorphism of

order a multiple of 4. But an exercise in group theory shows that | Aut(C9 o C3)| = 54, so no such automorphism exists. Thus A must be the other non-abelian group of order 27,

(C3 × C3) o C3. A computer calculation [17] shows that the group J has eight irreducible three-dimensional representations. Let S denote the set of these representations. As an

example, we determine the conductors of these representations; that is, the conductors of

the representations of GQ for which the fixed field of the kernel is N. Recall the definition

of the Artin conductor N(ρ) of a representation ρ from Section 4.1. We have J0 = C4, 51

Ji = 0, i ≥ 1. Using [17], one ﬁnds that for two of the eight aforementioned representations,

J0 J0 ∼ 3 dim V = 0, and for the other six, dimV = 1, where V = C is the representation space. So we get two representations with conductor p3 and six with conductor p2.

Suppose now that we are in Case i) of Proposition 2, so the ﬁxed ﬁeld of the kernel of a representation ρ is of the form NM, with M ⊂ Q(ζp,30 ). The irreducible three-dimensional representations of G(NM/Q) are given by ψ ⊗ χ, where ψ is an element of S and χ is a character of G(M/L), of which there are [M : L]. By abuse of notation, we are allowing

ρ and χ to also denote their respective lifted representations to the group G(NM/Q);

furthermore, we are identifying characters of G(M/L) with characters of G(M/Q) modulo

characters of G(L/Q). To see this, note that the sum of the squares of the degrees of the

irreducible representations of G(NM/Q) is equal to the order of the group. Consider the

irreducible representations of G(NM/Q) the form ψ⊗χ, where ψ ranges over all irreducible

representations of G(N/Q) and χ ranges over all irreducible representations of G(M/L).

The sum of the squares of the degrees of these representations is the order of G(NM/Q). If we can show that these representations are distinct as ψ and χ vary, then it will follow

that these are all the irreducible representations of G(NM/Q). This amounts to showing 0 ∼ 0 that for distinct representations ψ and ψ of G(N/Q) we cannot have ψ ⊗ χ = ψ . One can see this by considering the restrictions of both representations to G(NM/N).

Thus we ﬁnd that the number of irreducible representations of ρ of dimension three

with ker ρ having ﬁxed ﬁeld NM is 8[M : L].

Continuing with Case i) of Proposition 2, we now ﬁx a C3 × C3 extension F of L and

m n count the number of ρ for which N(ρ) = p and Fρ = F . Write [M : Q] = 4cp , with p−1 n c | 4 , so [M : L] = p c. Let G = G(NM/Q), and let Gi, with order gi, denote the lower ramification groups of NM/Q. Technically, we mean the lower ramification groups of the extension NMp/Qp of local fields where p is a prime of NM lying above p. Since NM/Q is Galois and ramified only at p, there is no ambiguity created by this language, and we use it henceforth when we have a Galois extension of Q ramified at a single rational prime. As 52

soon as M * N, we get non-trivial central elements in Gi whenever Gi 6= 1. Such elements act by scalars and thus do not ﬁx any non-zero subspace of V . So for such ρ, we have

X gi vp(N(ρ)) = · 3. g0 i≥0,gi6=1

n n We must compute the gi. We know g0 = 4cp and g1 = p .

Lemma 11. Let E/Q be a Galois extension unramified outside p∞ and tamely ramified at p. Let M be subfield of Q(ζpn ) that is not contained in Q(ζpn−1 ). Then for i ≥ 1, the ith lower ramification groups of G(EM/Q) have the same order as the respective lower ramification groups of Q(ζpn )/Q.

0 i Proof. Let G = G(EM/Q) and G = G(Q(ζpn )/Q), with upper ramiﬁcation groups G and 0i 0 i G , respectively, and let H = G(EM/M) and H = G(Q(ζpn )/M). Since G is a p-group, for all i > 0,

(G/H)i = GiH/H = Gi/Gi ∩ H = Gi (2.5) and

(G0/H0)i = G0iH0/H0 = G0i/G0i ∩ H0 = G0i, (2.6) where the ﬁrst equality in both lines is a property of upper ramiﬁcation groups (see [62,

Chapter IV] for details about the upper and lower ramification groups). The order of the lower ramification groups is determined by the order of the upper ramification groups.

i 0 0 i 0 Since (G/H) = (G /H ) , (2.5) and (2.6) imply that |Gi| = |Gi| for all i ≥ 1 .

0 Let G := G Q(ζpn+1 )/Q . As shown in [62, Chapter IV.4], for any 0 ≤ k ≤ n,

0 k n−k n−k+1 |Gi| = p for p ≤ i ≤ p − 1, i ∈ Z.

0 k n−k n−k+1 By Lemma 11, |Gi| = |Gi| for all i ≥ 1. So |Gi| = p for p ≤ i ≤ p − 1 and 53

n 0 ≤ k ≤ n, and |Gi| = 1 for i ≥ p . Thus if ker ρ has ﬁxed ﬁeld NM,

p − 1 p(p − 1) pn−1(p − 1) p − 1 v N(ρ) = 3 1 + + + ··· + = 3 + 3n , p 4c 4cp 4cpn−1 4c and there are 8pnc representations with kernel ﬁxing NM.

We are now in a position to estimate the number am,i of ρ for which we are in Case

m i) of Proposition 2 and for which N(ρ) = p and Fρ = F . If m ≡ 1 or 2 (mod 3), then p−1 am,i = 0. Suppose m ≡ 0 (mod 3). Write m = 3k + 3, and let d = gcd( 4 , k). Then

k p − 1 k p − 1 X k p − 1 a = 8p + ··· + 8p d = 8 p j . m,i 4 4d 4j j|d

p−1 The ﬁrst term in the sum, for example, corresponds to the case c = 4 , n = k, in which k p−1 p−1 case there exist 8p 4 representations ρ with vp N(ρ) = 3 + 3n 4c = 3 + 3k.

If f and g are real valued functions deﬁned on an inﬁnite subset of N, then we write f ∼ g to mean limm→∞ f(m)/g(m) = 1. For the total number Am,F,i of representations ρ

m in Case i) of Proposition 2 with Fρ = F and N(ρ) | p , we have

X k+1 m Am,F,i ∼ 2p ∼ 2p 3 , m−3 0≤k≤ 3 where we are viewing Am,F,i as a function of p.

Suppose now that we are in Case ii) of Proposition 2. We can write the fixed field of ker ρ as E3M30 , with M30 ⊂ Q(ζp,30 ) and E3 a degree 3 ramified extension of FM3 not a contained in F Q(ζp,3), with M3 ⊆ Q(ζp,3). Let [M3 : L] = 3 . The group G(E3/Q) splits

as B o C4, where B = G(E3/L) is a 3-group.

∼ Claim: B = (C3a+1 ×C3)oC3, where a generator of the last C3 factor acts on C3a+1 ×C3 a by ﬁxing the C3a+1 factor and mapping (0, 1) 7→ (3 , 1). Here Z(B) is the ﬁrst C3a+1 factor, 54

which is also the center of G(E3/Q).

∼ Proof of Claim. The same reasoning that showed A = (C3 × C3) o C3 in the discussion ∼ immediately following the proof of Proposition 2 shows that G(E3/M3) = (C3 × C3) o C3. 0 0 Since E3/F is cyclic, it follows that if F is a ﬁeld strictly between L and F , then G(E3/F )

0 has rank two (see Figure 2.3). The group G(E3/F ) is abelian since its center has index at 0 ∼ 0 most 3. Thus, G(E3/F ) = C3a+1 × C3, and it is normal in B since F /L is Galois. Since ∼ 0 G(E3/M3) = (C3 × C3) o C3, it has a subgroup C of order 3 not contained in G(E3/F ), 0 and thus B splits as G(E3/F ) o C, as claimed.

E3M30

FM3 M3M30

F M3 F 0

Figure 2.3: Field diagram for Claim in Case ii) of Proposition 2.

∼ We would now like to count the number of irreducible representations of G(E3/Q) =

B o C4. While counting representations from Case i) of Proposition 2, we saw that

∼ a 0 ((C3 × C3) o C3) o C4 = ((3 C3a+1 × C3) o C3) o C4 := B o C4

has eight irreducible three-dimensional representations. If ψ is an irreducible representation

0 a a a+1 a of B o C4, and ψ maps the generator 3 ∈ 3 Z/3 Z of the 3 C3a+1 factor to the scalar a+1 α ∈ GL3(C), then mapping the generator 1 ∈ Z/3 Z of the C3a+1 factor to any of a 3 th roots of α gives a well-defined irreducible representation of B o C4, and distinct 55 choices of 3ath roots of α give distinct representations. Using the fact that the sum of the squares of the degrees of the irreducible representations is the order of the group, we see that we obtain all irreducible representations of B o C4 in this way. It follows that a the number of irreducible three-dimensional representations of G(E3/Q) is 8 · 3 . Recall that in Case ii), the fixed field of the kernel of our given representation ρ is of the form

E3M30 , where M30 ⊆ Q(ζp,30 ). The number of irreducible three-dimensional representations a of G(E3M30 /Q) is thus 8 · 3 · [M3M30 : M3].

n a Write [E3M30 : Q] = 108p c, with (c, p) = 1. Then our number 8 · 3 · [M3M30 : M3] n from above is equal to 8p c. Let Gi, with order gi, be the lower ramiﬁcation groups

G of G(E3M30 /Q). For all i ≥ 0, Gi has non-trivial central elements, so dimV/V i = 3

whenever gi 6= 1. So again we obtain

X gi vp (N(ρ)) = · 3, (2.7) g0 i≥0,gi6=1

n and we have to compute the gi. We have g0 = 12p c. It follows from Lemma 11 that the gi for Case ii) are the same as those for Case i). We are thus able to compute, analogously to Case i), p − 1 v N(ρ) = 3 + 3n . p 12c

In the same way that we bounded am,i, we can now bound the number am,ii of ρ for which

m we are in Case ii) of Proposition 2 and for which N(ρ) = p and Fρ = F . By (2.7), am,ii = 0 unless m ≡ 0 (mod 3), so write am,ii = 3k + 3. A similar series of computations to those in Case i) gives

k p − 1 k p − 1 X k p − 1 p − 1 a = 8p + ··· + 8p d = 8 p j , d = gcd , k . m,ii 4 12d 12j 4 j|d

When we are in Case ii), for a fixed M3 and F , there are two possible degree 3 extensions of FM3 that are candidates for the kernel of the fixed field of the representation ρ, so we 56 obtain

X 4 k+1 4 m A ∼ p ∼ p 3 . m,F,ii 3 3 m−3 0≤k≤ 3

m To bound the total number Bm of primitive representations ρ with N(ρ) | p and

Im(π ◦ ρ) = P1, we need to multiply Am,F,i + Am,F,ii by the number of unramiﬁed C3 × C3 extensions of L, which, as previously discussed, is r3,2(k), where k is the 3-rank of ClL. 22.2p3 22.2p3 From [43, page 95], the 3-part of hL is less than π4 , so k ≤ log3( π4 ) := Rp.

We introduce the following (non-standard) notation. Suppose that A and B are both positive, real-valued functions of n.

A(n) We write A / B to mean lim sup ≤ 1. n→∞ B(n)

Using this notation we obtain the following theorem.

Theorem 14. With notation as above and viewing the expressions below as functions of p, we have

4928.4 m +9 B (A + A )r (R ) ∼ p 3 . m / m,F,i m,F,ii 3,2 p 35π8

2.6.2 Representations with projective image isomorphic to P3

Recall that P3 := C3 × C3 o SL2(F3) (Figure 2.4). Using the notation from the previous ∼ section, we look to count the number of ρ with G(Fρ/Q) = P3 and given conductor. Our main tool will be the following proposition.

Proposition 3. Let ρ :GQ → GL3(C) be an Artin representation unramified outside p∞ with projective image isomorphic to P3. Let F be the fixed field of ker π◦ρ. If M ⊆ Q(ζp∞ ),

then there are at most three degree 3 central extensions of FρM/Q ramified over FρM and at most one that is unramified over FρM. There exists M ⊆ Q(ζp∞ ) such that of the four possible degree 3 central extensions of FρM/Q, the fixed field of ker ρ must be the unramified extension. 57

C3 × C3

C2 Q8 K SL3(F3) C2 × C2 L

C3 Q

Figure 2.4: Field diagram for Section 2.6.2.

∼ Proof. We have H2(P3, Z) = C3. Let E be a candidate for the ﬁxed ﬁeld of ker ρ and let ab M = E ∩ Q . By Lemma 10 and its proof, FM/Q admits a maximum of three distinct ab non-trivial central extensions Ei with Ei ∩ Q = M, each of which has degree 3 over FM, 0 0 in addition to the degree 3 central extension FM with M ⊆ Q(ζp∞ ).

0 The group G(FM /Q) has an abelian normal subgroup that is not cyclic, and it follows from [26, Corollary 6.13] that such a group has no faithful primitive representations, so

FM 0 cannot be the ﬁxed ﬁeld of ker ρ (the same argument shows that FM 00 cannot be

00 the fixed field of ker ρ for any M ⊂ Q(ζp∞ )). Since we cannot have a totally ramified

C3 × C3 extension of FM that is Galois over Q, we must have either at least one of 0 the aforementioned fields Ei be unramified over FM or have FM /F M unramified. If

0 FM /F M were unramiﬁed, then ramiﬁcation considerations force e := e(F/F0) to be 3 or

9, where F0 is the ﬁxed ﬁeld of the C3 × C3 normal subgroup of P3. This is impossible,

however, since we would have e(F/Q) = 3e, so the extension of Qp given by completing

F at a prime above p would have a C3e subgroup, and P3 does not have such a subgroup.

Thus one of the Ei’s is unramified over FM. If two of the fields Ei were unramified, then

the composite of these fields would give a C3 ×C3 unramified central extension of FM that contains FM 0, forcing FM 0/F M to be unramified, which we just saw gives a contradiction.

Thus, FM/Q may have up to four central extensions of degree 3, and at this point, three of them (all but FM 0) are possibilities for E. By Theorem 13, E is a degree 3 58

central extension of FM/Q for some M ⊂ Q(ζp∞ ) since C3 surjects onto G(E/F M). It remains to show that E/F M must be unramiﬁed. Suppose that E/F M is rami-

a fied. Write [FM : F ] = 3 b with 3 - b, and let M3 ⊂ Q(ζp∞ ) be the field satisfying a [FM3 : F ] = 3 . The extension FM3/F is totally ramified since F/F0 is unramified. We ∼ have G(FM3/Q) = (C3 × C3) o (Q8 o C3a+1 ) with the C3a+1 factor acting through its C3 quotient. The 3-part of the inertia group at a prime above p in E is cyclic of order 3a+2. ∼ It follows that G(E/Q) = (C3 × C3) o (Q8 o C3a+2 ) with the C3a+2 factor acting through

its C3 quotient. The left (C3 × C3) factor is then an abelian normal subgroup of G(E/Q), which contradicts the primitivity of the representation.

Remark 9. Since E/F is abelian (it is cyclic) and E/F M is unramiﬁed, there exists a ﬁeld

E0 between E and F such that E0/F is unramiﬁed of degree 3 and E/E0 is totally ramiﬁed.

Since FM3/F is totally ramiﬁed, E0 cannot be contained in FM3 (so E0 * FM), and thus we have E = E0M. Since E/F is cyclic, we also have that [FM : F ] is not divisible by 3.

Remark 10. Unlike in the P1 case, we may not use Lemma 8 here to guarantee the

existence of an unramiﬁed degree 3 central extension of F , because the extension of Qp given by completing at a prime above p in F is not necessarily cyclic, so we are not assured

of a lift ρ :G → GL ( ) of ρ˜| such that the fixed field of the kernel of ρ is unramified p Qp 3 C GQp p over the fixed field of the kernel of ρ˜| . GQp

Maintaining the notation from Figure 2.3, let L = F ab be the degree 3 extension of

Q contained in Q(ζp). The field L then admits a C2 × C2 extension, which is unramified by Lemma 7. In the proof of Proposition 3 we saw that F/F0 must be unramified, so e(F/Q) = 3 or 6.

Continuing in the manner of the P1 case, we now bound, for a ﬁxed F/Q with ∼ m G(F/Q) = P3, the number of ρ for which N(ρ) = p and Fρ = F . Let E = Eρ, so

E is of the form E0M for some M ⊆ Q(ζp∞ ), where E0 is as in Remark 9. Given M, by n p−1 Proposition 3 there is one possible choice for E. Write [M : L] = p c, with c | 3 . We n have e(E/Q) = 3p cx, where x = 1 or 2. Using Lemma 11 to compute analogously to the 59

P1 case, we ﬁnd p − 1 v (N ) = 3 + 3n . p ρ 3cx

An exhaustive search using [17] shows that there are three non-split central extensions of P3 of order 3|P3| = 648, and using [17], we can check that each of them has seven irreducible three-dimensional representations.

Remark 11. We are considering central extensions of P3 by C3 up to group isomorphism,

2 not up to isomorphism of central extensions, the latter of which are classiﬁed by H (P3,C3).

The three-dimensional irreducible representations of G(E/Q) are all the ψ ⊗ χ, as ψ

runs over the three-dimensional irreducible representations of G(E0/Q) and χ runs over all ∼ n characters of G(E/E0) = G(M/L). Thus G(E/Q) has 7p c irreducible three-dimensional m representations. We estimate the number am,F of representations ρ with N(ρ) = p and

Fρ = F . We must have 3 | m, and so we write m = 3k + 3. Proceeding as in the P1 case,

we see that the leading term in the expression for am,F is

7pk(p − 1) . (2.8) 3x

∼ As we did with P1, we must now bound the number of fields F with G(F/Q) = P3 and ab F/Q ramified only at p∞. As above, let L = F , the degree 3 subfield of Q(ζp). The field 0 L then admits a Q8 extension F0, which has three intermediate fields Li, each of which is a C2 extension of L, and unramified over L by Lemma 7. Let K be the composite of the

Li. The ﬁeld L is totally real, and so by the following lemma, K is as well.

Lemma 12. Let l be a totally real ﬁeld and let k/l be a Galois extension with group Q8. Then the unique intermediate ﬁeld that is degree 4 over l is totally real.

Proof. The ﬁeld in question is unique because Q8 has a unique subgroup of order 2. Let

0 0 0 0 k = k ∩ R. If k = k, then we are done. Otherwise, [k : k ] = 2, and k is the unique 60 intermediate ﬁeld of degree 4 over l; furthermore, k0 is totally real since it is Galois over the totally real ﬁeld k.

Returning to the notation before Lemma 12, K/L is unramified and K is totally real, so the number of choices for K is at most r2,2(k), where k is the 2-rank of ClL. Using [36, 1 Corollary 4], we have |ClL| < 2 p, so k < log2 p − 1. The field F0 is a degree 2 extension of p∞ K, and the extension F0/K is at most tamely ramified, so F0 ⊆ K , where p is the prime of L above p, viewed as an ideal of K. By class field theory, p splits into a product of four

p∞ primes in K since it is principal in L. Using the exact sequence (1.4) applied to ClK , we p∞ ﬁnd that G(K /K) has 2-rank at most 4 + 12 + the 2-rank of ClK . The order of ClK

22.2p8 12 22.2p8 is at most π12 [43, page 95]. Set β = log2(2 · π12 ). If F0/K is unramiﬁed above p, then there are at most

r2,1(β)r2,2(log2 p − 1) (2.9)

choices for F0, and if F0/K is ramiﬁed above p, then there are at most

(r2,1(4 + β) − r2,1(β)) r2,2(log2 p − 1). (2.10)

The ﬁnal step to bound the number of possible Fρ is to look at unramiﬁed C3 × C3

extensions of F0 (which give our ﬁeld Fρ). This amounts to ﬁnding r3,2(l), where l is the

3-rank of ClF0 . Using [43, page 95] and the fact that ramiﬁcation in the inﬁnite places does not contribute to the 3-rank, we obtain

22.2p16 l ≤ log (2.11) 3 π24

if F0/K is unramiﬁed, and 22.2p20 l ≤ log (2.12) 3 π24

if F0/K is ramiﬁed.

We have x = 1 or 2, depending on whether F0/K is ramiﬁed. To obtain our bound 61

∼ for the number am (where m = 3k + 3) of ρ ramiﬁed only at p∞ with G(Fρ/Q) = P3 and N(ρ) = pm, we need to combine (2.8) with (2.9) and (2.11), and combine (2.8) with (2.10) and (2.12). Doing so, we ﬁnd that the leading term of am in p, which comes only from the x = 2 case, is bounded by 605134.5pk+51 . π50

Finally, to conclude the section on P3, we ﬁnd that for the number Bm of three-

m dimensional Artin representations with projective image P3 and conductor dividing p , we have the following theorem.

Theorem 15. Using the notation above,

m +50 605134.5p 3 B . m / π50

2.7 Comparison to the two-dimensional case

For comparison to the two-dimensional case, we look at [72]. Looking at Theorem 3 in

[72] and multiplying by φ(pm) = pm−1(p − 1) to account for all possible χ, we see that the bounds for the number of two-dimensional representations with conductor pm and projective image isomorphic to A4 or S4 (and conjecturally for A5) is on the order of

2m− 4 m 1 1 1 Cp log p for = 6 , 8 , or 12 and C a constant, both and C depending on the projective image of the representation. Letting p → ∞ for ﬁxed m, we see that the bounds in [72] are better than our bounds for the imprimitive case. By going through the proofs of Theorems 14 and 15 in this paper, we may bound the number of ρ with Im π ◦ ρ = P1,

m m N(ρ) | p , and with Im π ◦ ρ = P3, N(ρ)|p , respectively, as m → ∞ for ﬁxed p. These bounds can be given by easily computable constants multiplied by the bounds given in

Theorems 14 and 15. Therefore, our bounds as m → ∞ are stronger than those in [72], and our bounds are stronger as either p or m tends to ∞ for the primitive case. Of course,

Wong’s bounds hold for representations of any conductor N, not only for representations 62 of prime power conductor. Chapter Three

Torsion points on elliptic curves and abelian surfaces 64

3.1 Introduction

In the previous chapter, we studied Galois representations abstractly by analyzing the possibilities for the field cut out by a representations with given properties. One source of concrete Galois representations is the l-torsion points of an abelian variety. The group of l-torsion points of an abelian variety A of dimension g defined over a number field

2g K is isomorphic to (Z/lZ) . The Galois group GK acts on the l-torsion A[l], giving a

representation ρ = ρA,l :GK → GL2g(Fl). The existence of the Weil pairing on A[l] forces

the image of ρ to land in the general symplectic group GSP2g(Fl). Recall that this is the

subgroup of matrices M ∈ GL2g(Fl) that preserve an alternating bilinear form up to scalar

multiplication. In general, one expects the image of ρ to be the full group GSP2g(Fl). For example, we have the following celebrated theorem of Serre.

Theorem 16. [64] Let E be an elliptic curve deﬁned over a number ﬁeld K without complex

multiplication. Then for all but ﬁnitely many l, the image of ρE,l is the full GL2(Fl) (note:

GSP2(Fl) = GL2(Fl)).

It is natural to investigate the scenarios under which the image of ρA,l is not the full

GSP2g(Fl). We do not exclude the situation when A has complex multiplication.

Definition 1. Let K be a field, g ≥ 1 an integer, and l a prime. Define

   K− isomorphism classes [A] of abelian varieties A of dimension g      ∞ S(K, g, l) = deﬁned over K such that K(A[l ])/K(µl) is pro-l      and is unramiﬁed outside l. 

Note that the existence of the Weil pairing implies that µl ⊆ K(A[l]). To say that

∞ K(A[l ])/K(µl) is unramiﬁed outside l is equivalent to saying that the l-adic Tate module

Tl(A) is unramiﬁed outside l. By the N´eron-Ogg-Shafarevich criterion [65], such an abelian variety A has good reduction away from l. The Shafarevich conjecture (proved by Faltings

[15]) says that there are only finitely many abelian varieties defined over K and with good 65 reduction outside l. Therefore, the set S(K, g, l) is finite. Note that if A ∈ S(K, g, l), then

K(A[l])/K(µl) has order a power of l.

In [53], Rasmussen and Tamagawa conjecture that for ﬁxed K and g, the set

S(K, g) := {([A], l):[A] ∈ S(K, g, l)} is ﬁnite. Furthermore, they conjecture that the bound for S(K, g) should depend only on

g and [K : Q]. See [52] and [53] for circumstances under which the conjecture is true and for the original motivation of the authors.

3.2 3-Torsion

In [51], Rasmussen provides an example of an abelian surface A with A ∈ S(Q, 2, 3). Motivated by this ﬁnding, we prove the following result.

Theorem 17. Let A/Q be an abelian surface with Q(A[3])/Q(ζ3) an extension unramiﬁed √ 3 outside 3 and of 3-power order. Suppose that µ9 *Q(A[3]). Then Q(A[3]) ⊆ Q(ζ3, −3).

Proof. Let K = Q(A[3]). We know that the Galois representation

∼ ρ3 :GQ → Aut(A[3]) = GL4(Z/3Z)

factors through G := G(K/Q) and that the image lands in GSP4(Z/3Z). By assump-

tion, H := G(K/Q(ζ3)) is a 3-group. One works out that the 3-Sylow subgroup P3 of

GSP4(Z/3Z) is isomorphic to

3 (Z/3Z) o Z/3 := (< x > × < y > × < z >)o < w > with

w :(x, y, z) 7→ (x, y, z + wx + wy).

0 0 Let K denote the maximal subfield of K that is abelian over Q(ζ3), so K is the fixed field 66

0 of the commutator H of H. Let p := (1 − ζ3) be the prime of Q(ζ3) above 3. The ﬁeld 0 pn n K is contained in the ray class ﬁeld K of Q(ζ3) of modulus p for some n. Using the pn results of [23], we see that for all n, the group G(K /Q(ζ3)) has order a power of 3 and 0 has rank at most two. Since G(K /Q(ζ3)) is a subquotient of P3, it has exponent 3 and

thus must be isomorphic to either Z/3Z or Z/3Z × Z/3Z.

0 0 0 Since H is characteristic in H and H ¡ G, we have that H ¡ G, which means K /Q

is Galois. Using [54], one checks that the only Galois number ﬁelds of degree 6 over Q √ √ 6 3 unramiﬁed outside 3 and ∞ are Q(ζ9) and Q( −3) = Q( −3, ζ3). Since µ9 * K by √ 0 0 6 assumption, the only possibility for K is K = Q( −3).

0 We have that K is the only strictly intermediate ﬁeld between Q(ζ3) and K that is 0 abelian over Q(ζ3) since for any other such ﬁeld L must be contained in K . Therefore, H 3 is a subgroup of (Z/3Z) o Z/3 whose maximal abelian quotient is Z/3Z. The only way 0 this is possible, however, is if H is Z/3Z itself, which implies K = K .

3.3 Roots of unity in torsion ﬁelds

In Theorem 17, we had to assume that µ9 is not contained in the 3-torsion ﬁeld of A. In general, one might ask about the largest integer n = n(l, g) such that there exists

an abelian variety A/Q of dimension g with µln ⊆ Q(A[l]). It is easy to see that there

exist elliptic curves E deﬁned over Q such that the Q(E[2]) contains µ4. For example, 2 2 consider the curves E : y = (x − a)(x + 1), a ∈ Q. Thus n(2, 1) ≥ 2. Furthermore, since

G(Q(µ8)/Q) = Z/2Z × Z/2Z and G(Q(E[2])/Q) ≤ S3 for any elliptic curve E, we see that n(2, 1) = 2. For g = 1 and arbitrary l we have the following proposition.

Proposition 4. With the notation above, n(l, 1) = 1 for l ≥ 3. In other words, if E is an

elliptic curve deﬁned over Q, then µl2 *Q(E[l]).

Proof. Suppose for contradiction that µl2 ⊆ Q(E[l]). Then the intertia group at l has 67 order l(l − 1)– and note that this is the largest possible inertia since the highest power of l dividing the order of GL2(Fl) is l. Therefore, the group G(Q(µl2 )/Q) lifts isomorphically to G(Q(E[l])/Q). Pick a prime p 6= l such that Frobp is a generator of this inertia group.

Let Mp denote the image of Frobp under ρE,l inside GL2(Fl). Since Mp has order l(l−1), in matrix form it is conjugate to an upper triangular matrix that is constant on the diagonal.

Such a matrix evidently has square determinant modulo l; however, the determinant of Mp

is the mod l cyclotomic character applied to Frobp, so det Mp ≡ p (mod l). On the other

hand, we have assumed that Frobp generates G(Q(µl2 )/Q), and therefore that p generates 2 ∗ (Z/l Z) , so it cannot be a square modulo l. This contradiction completes the proof.

3.4 A local-global property

One way to gain information about torsion points on abelian varieties over number global

fields is to look locally. For example, if A is an abelian variety defined over a number field

n K, then we know that the l -torsion of A(K) injects into A(Fp) for any prime p of K with residue characteristic diﬀerent from l and at which A has good reduction. Fix a number

ﬁeld K and an abelian variety A. Given a prime l, let

   Primes p of K such that A has good reduction at  Sl0 =  p and p is prime to l. 

n We can ask whether the converse holds– that is, suppose l |#A(Fp) for all primes p ∈ Sl0 . n Does this imply that l |#A(K)tor? The answer is no. For example, consider the elliptic

2 3 curve E : y = x + x. Using modular arithmetic, one checks that 4|#E(Fp) for all odd ∼ ∼ primes p, and that E(F3) = Z/4Z but E(F5) = Z/2Z × Z/2Z. Since E(Q)tor injects into ∼ both E(F3) and E(F5), we conclude that E(Q)tor = Z/2Z. We do however, have an isogeny from E to the curve E0 : y2 = x3 − 4x, given by

y2 y(1 − x2) (x, y) 7→ , . x2 x2 68

0 0 Moreover, 4|E (Q)tor since E has full 2-torsion.

Recall that if A0 is isogenous to A over K, then they have the same primes of good reduction and the same number of points modulo p at all primes in Sl0 [65]. Therefore, it

n is natural to ask whether the hypothesis l |#A(Fp) might imply the weaker condition that 0 n 0 there exists a K-isogenous A for which l |#A (K)tor. This question was ﬁrst answered by Katz in [30] by rephrasing the problem in purely group-theoretic terms. For the case of elliptic curves and for the case of abelian surfaces when n = 1, he answered the question

in the aﬃrmative. He gave counterexamples in dimensions 3 and higher. Since Katz’s

original paper, Cullinan [7] has given group theoretic descriptions of all three-dimensional

counterexamples to Katz’s question in the n = 1 case.

Here, we use a lemma of [53] to prove the following theorem, which gives an affirmative answer for n = 1 when A satisfies the additional condition defining the set S(K, g, l).

Theorem 18. With notation as in Section 3.1, suppose that A ∈ S(K, g, l) and that

0 0 l|#A(Fp) for all p ∈ Sl0 . Then there exists a K-isogenous A such that l|#A (K)tor.

Proof. The assumption l|#A(Fp) for all p ∈ Sl0 means that det(1 − Frobp) ≡ 0 (mod l)

for all p ∈ Sl0 , where we have identiﬁed 1 − Frobp with ρA,l(1 − Frobp). By Lemma 3 in

[53], there exists a basis for A[l] such that the image of ρA,l is upper triangular and the ith

j diagonal entry is of the form χ i , 0 ≤ ji ≤ l − 2, where χ is the mod l cyclotomic character. Therefore, we have

Y ji det(1 − Frobp) = (1 − p ), i

for all p ∈ Sl0 , where p is the residue characteristic of p. Under the assumption that

j l|#A(Fp) for all p ∈ Sl0 , for each p, p i ≡ 1 (mod l) for some i. Use Dirichlet’s theorem on primes in arithmetic progressions to pick p such that p has multiplicative order l − 1 in

∗ j (Z/lZ) . For such p, we must have ji = 0, which means that χ i is the trivial character. Thus, the semi-simpliﬁcation of A[l] contains the trivial representation, which, by [30],

implies that A is K-isogenous to an abelian variety with a rational l-torsion point. Chapter Four

Noether’s Problem in Galois Theory 70

4.1 Introduction

The inverse Galois problem for a ﬁeld K and a ﬁnite group G asks whether there exists a

Galois extension L/K with group G. In Chapter 2, we considered the number of solvable

Galois extensions of the rationals subject to certain ramiﬁcation conditions. In 1956,

Shafarevich proved that any finite solvable group can be realized as the Galois group of some number field over the rationals (see [46] for a nice exposition). For non-solvable groups, the inverse Galois problem is, in general, open. One way in which we can obtain non-solvable Galois extensions of the rationals is via the representation of the torsion points on abelian variety defined over Q, as we saw in the previous chapter. For example, if we consider an arbitrary elliptic curve E/Q without complex multiplication, Serre’s open image theorem tells us for all sufficiently large l (conjecturally l > 37), the image of the representation in GL2(Fl) of GQ acting on E[l] is all of GL2(Fl). Thus, the kernel of such a representation at such a prime l is a GL2(Fl) extension of Q, and for l ≥ 5, GL2(F5) is not solvable.

Another tool we have to attack the inverse Galois problem works as follows. We can

n embed G in GLn(K) for some n, so G acts faithfully on V = K , and there is a faithful

G action of G on the field K(V ) = K(x1, . . . , xn). If the fixed field K(V ) is a purely transcendental extension of K (of transcendence degree n), then as a consequence of Hilbert irreducibility, we obtain the existence of a field (infinitely many fields, in fact) L/K with

G(L/K) = G.

A purely transcendental extension F of K is said to be rational over K. Given a finite group G and field K, consider the regular representation VG := hxgig∈G of G over K. Noether’s problem, first posed by Emmy Noether, asks whether the field

G K(G) := K(VG) is rational over K. By the discussion above, an aﬃrmative answer to Noether’s problem for 71

G and K implies an affirmative answer to the inverse Galois problem for G and K. Swan was the first to give an example of a group G and field K for which Noether’s problem had a negative answer. He proved [69] that Q(Z/47Z) is not rational over Q by showing that Noether’s problem for Z/pZ was equivalent to asking whether a prime ideal above p in Z[ζp−1] is principal, where ζp−1 is a primitive (p − 1)st root of unity. Building on Swan’s work, Lenstra gave necessary and sufficient conditions under which Noether’s problem has an affirmative answer for all finite abelian groups over any field [35]. Saltman pioneered the study of Noether’s problem over the complex numbers, giving an example of a group

9 G of order p for any prime odd prime p for which C(G) is not rational [56]. A good deal of work using techniques ranging from explicit computation to Galois cohomology and spectral sequences has been done on various cases of Noether’s problem over algebraically closed ﬁelds, including p-groups [49, 4], direct products and wreath products [29], and the alternating groups An [37]; however, there remain many open cases, such as the rationality

of C(A6).

Our aim in this chapter is to measure the extent to which K(A) may fail to be rational

for an arbitrary ﬁnite abelian group A and ﬁeld K. To this end, we introduce the following quantity.

Deﬁnition 2. Let V be a variety of dimension n over a ﬁeld K. The degree of irrational-

ity of V , which we denote by Irr(V ), is the minimal degree of a dominant rational map

n V 99K P .

In the case that V is a curve, Irr(V ) is the gonality of V , a quantity that has been studied

extensively and about which many questions remain.The quantity was ﬁrst introduced by

Heinzer and Moh in [41] in the context of function ﬁelds of one variable. Yoshihara and

others have studied Irr(V ) when V is an algebraic surface, beginning with [74]. The degree

of irrationality of hypersurfaces has also been studied [2].

In our case, we fix a finite abelian group A and a field K, and let V = VK,A be the |A| variety (up to birational equivalence) with function field K(A). That is, V = AK /A. 72

Definition 3. For a field K and a finite group G, define Irr(K,G) = Irr(VK,G).

An equivalent deﬁnition of Irr(K,G), with which we work primarily, is

Irr(K,G) = min {[K(G): L]: L/K is rational}. K⊆L⊆K(G)

That is, Irr(K,G) is the minimum degree of K(G) over any ﬁeld that is rational over

K. Since our original field K(xh : h ∈ G) is finitely generated over K, K(G) is finitely generated as well, so a transcendence basis S for K(G) is finite. Since K(G) is finitely generated and algebraic over K(S), it is finite over K(S). Therefore, the quantity Irr(K,G) is well defined. For example, Swan’s result that Q(Z/47Z) is not rational may be written as Irr(Q, Z/47Z) ≥ 2. For any finitely generated field E over F , we can similarly define the degree of irrationality Irr(E) of E to be the minimum over all transcendence bases S for E/F of [E : F (S)].

In the case G = Z/pZ, for example, we may embed G in the symmetric group Sp, which acts naturally on W := Kp by permuting coordinates. The algebraic independence of the elementary symmetric polynomials gives the rationality of K(W )Sp . The existence of this rational subﬁeld of K(G) implies Irr(K,G) ≤ (p − 1)!

Our main theorem of this chapter is the following.

Theorem 19. Let A be a ﬁnite abelian group and let K be a ﬁeld such that [K(ζs): K] is cyclic for every prime power s that divides the order of A and is prime to the characteristic of K. Then Irr(K,A) is less than an explicit quantity involving only the arithmetic of certain cyclotomic number rings, which is given in the statement of Theorem 23.

Deﬁnition 4. If E is a transcendental ﬁeld extension of F and S is a transcendence basis for E/F such that [E : F (S)] = Irr(E), then we will call S a maximal transcendence basis for E (over F ).

For an abelian group A and an algebraically closed ﬁeld of characteristic prime to |A|, 73

Noether’s problem is known to have an aﬃrmative answer, due to Fischer.

Theorem 20. [66] Let A be an abelian group of exponent e and K a ﬁeld of characteristic prime to e that contains the eth roots of unity µe. Then K(A) is rational.

a Proof. Let |A| = a. The group A acts on K(x1, . . . , xa). Let V = ⊕i=1Kxi be the

regular representation of A. Since A is abelian and K contains µe, V can be diagonalized–

i.e., V has a basis {y1, . . . , ya} such that for any g ∈ A, g · yi = χi(g)yi, for a character

∗ χi ∈ Aˆ = Hom(A, K ). Let M be the multiplicative free abelian group on the yi, and deﬁne

a group homomorphism ψ : M → Aˆ by sending yi 7→ χi. The kernel of ψ is a free abelian

group of rank a, generated, say, by {z1, . . . , za}. By construction, each zi ∈ K(A). If f

A is any element of K(y1, . . . , ya) = K(A), then since g acts by scalars on each monomial

term of f, we must have f ∈ K(z1, . . . , za). Therefore, K(A) = K(z1, . . . , za), and the zi are algebraically independent since there are a of them generating a ﬁeld of transcendence

degree a.

We note that the question of irrationality in Noether’s problem leads to several other natural questions. As alluded to earlier, one can ask about the degree of irrationality of an arbitrary variety. Additionally, rather than just considering the rationality of K(G), we can ask whether K(G) satisﬁes the weaker condition of stable rationality–that is, whether

K(G) becomes rational upon adding ﬁnitely many indeterminates– or the even weaker condition of retract rationality (see [56] or [5] for a deﬁnition and discussion of retract rationality). These conditions have been studied for the general case of a quotient variety

V/G when G is any linear algebraic group acting on a vector space V , assuming such a quotient makes sense. See [5] for a nice survey of this.

This chapter is organized as follows. In the next section, we introduce notation and review the results of Lenstra [35] that we will need for Theorem 19 (throughout, our presentation of Lenstra’s material is suitably adapted for our purposes). In Section 4.3,

n we modify Lenstra’s method to obtain Theorem 19 in the case G = Z/p Z, from which we 74 deduce Theorem 19 for a general abelian group A.

4.2 Notation and Lenstra’s Setup

Let K be a ﬁeld, π a group of automorphisms of K, and M a π-module that is a ﬁnitely generated free Z-module with Z-basis x1, . . . , xm. We use multiplication for the group op- eration of M, so elements of M are monomials in the xi. The group ring K[M] is then iso-

±1 ±1 morphic to the ring of Laurent polynomials in m variables over K– that is, K[x1 , . . . , xm ], and its quotient ﬁeld is the rational ﬁeld K(M) = K(x1, . . . , xm). The group π acts on K[M] by

σ σ σ (Σαimi) = Σαi mi , αi ∈ K, mi ∈ M, which extends to an automorphism of the ﬁeld K(M). The units of K[M] are monomials– that is, K∗M.

We introduce a few other pieces of notation. By ζm we denote a primitive mth root of

∗ unity. For a ﬁeld K, we have the natural injection G(K(ζm)/K) ,→ (Z/mZ) , which allows

G(K(ζm)/K) to act on (Z/mZ). We will take the set of divisors of a positive integer n to be all positive divisors of n. Lastly, the function φ refers to Euler’s φ function.

n We begin by studying our irrationality question in the case A = Z/p Z, where p is an odd prime. We also assume that gcd(char(K), p) = 1. Let L = K(ζpn ) and for i, 0 ≤ i ≤ n, let

mi = [K(ζpi ): K].

Let pi be a prime ideal of Z[ζmi ] lying above the rational prime p. The additive group of

the ideal pi is a free Z-module of rank r := φ(mi). Let π = G(L/K). If x1, . . . , xr is a

Z-basis for pn, written multiplicatively, then π acts on the monomials in the xj and thus

on the ﬁeld L(pn), acting on L by Galois automorphisms. 75

Let d be a divisor of mn and let πd be the quotient group of π of order d. The group

πd can be identiﬁed with G(Kd/K) for a subﬁeld Kd ⊆ L. We have a ring homomorphism

ψd : Z[π] → Z[πd] → Z[ζd], (4.1)

where the ﬁrst map is induced by the natural quotient map π πd, and the second map

is deﬁned by sending a generator of πd to ζd. This allows us to view any Z[ζd]-module as a

Z[π]-module. For any divisor d of mn, following Lenstra [35], we deﬁne a functor Fd from

the category of π-modules to the category of torsion-free Z[ζd]-modules by

Fd(M) = M ⊗π Z[ζd] /{additive torsion},

where we view Z[ζd] as a π-module via the map ψd.

n n We can make Z/p Z into a π-module by identifying π with a subgroup of Aut(Z/p Z). n There exists a unique map Z[π] → Z/p Z of π-modules taking 1 7→ 1. Let J = J(p) denote

be the kernel of this map, a free Z-module of rank mn. From [35, Proposition 3.6] and [35, Introduction], we deduce the following proposition.

∼ ∼ Proposition 5. Fd(J) = pi if d = mi, and Fd(J) = Z[ζd] if d 6= mi for any i.

The utility of the functor Fm is demonstrated in the following theorem of Lenstra, suitably adapted here for our purposes.

Theorem 21. [35, Proposition 2.4] Let M be a ﬁnitely generated, projective π-module.

π π The ﬁelds L(M) and L(⊕d|mn Fd(M)) are isomorphic, where π acts separately on each

direct summand of ⊕d|mn Fd(M) via the maps ψd in (4.1).

Remark 12. One checks that Fd respects direct sums, a fact we will use hereon without further reference.

We also have the following proposition. 76

n Proposition 6. [35, Proposition 5.3] The ﬁeld K(Z/p Z) is isomorphic to a purely transcendental extension of L(J)π.

We may apply Theorem 21 since by [35, Proposition 3.3] J is a projective π-module.

n Putting together Propositions 5 and 6 with Theorem 21, we ﬁnd that K(Z/p Z) is π isomorphic to a rational extension of the ﬁeld Lpn , where

  M M Lpn := L  pi ⊕ Z[ζd] . 0≤i≤n d|mn, and d6=mi for any i

From here, we can proceed as Lenstra does in [35, Theorem 2.6] to show that if pi is

π n principal for all i, then Lpn , and thus K(Z/p Z), is rational over K:

Suppose that each pi is principal, so a free Z[ζmi ]-module. One checks [35, Proposi-

tion 2.3] that Fd(Z[π]) = Z[ζd], for every d dividing mn. Therefore, the summands of

⊕d|mn Fd(Z[π]) and ⊕d|mn Fd(J) agree as π-modules. Thus

∼ ⊕mFm(Jp) = ⊕mFm(Z[π]),

π ∼ π and by applying Theorem 21 twice, it follows that L(J) = L(Z[π]) . But Z[π] is a

Z[π]-permutation module– that is, a free Z-module with a Z-basis that is permuted by π– π π and for any ﬁnitely generated Z[π]-permutation module N, L(N) is rational over L [35, π n π Theorem 1.4]. Therefore, L(J) , and thus L(Z/p Z), is rational over L = K, as desired.

4.3 Bounding the Degree of Irrationality from Above

Suppose now that at least one of the ideals pi is not principal. Lenstra [35] shows this

n n implies that K(Z/p Z) is not rational. In this case, we wish to bound Irr(K, Z/p Z) from 77 above. Recall from the introduction that

n n Irr(K, Z/p Z) = min{[K(Z/p Z): E]: E/K is rational}. E

Let Ii be a principal ideal of the ring Z[ζmi ] contained in the ideal pi. Consider the ﬁeld   M M LI := L  Ii ⊕ Z[ζd] , 0≤i≤n d|mn,d6=mi a subﬁeld of Lpn . Since Ii is a free Z[ζmi ]-module of rank 1,

M M ∼ M Ii ⊕ Z[ζd] = Fd (Z[π]) 0≤i≤n d|mn,d6=mi d|mn

π π as Z[π]-modules. Therefore, LI is isomorphic to L(⊕d|mn Fd(Z[π])) , which, as shown in π the case when the pi were assumed to be principal, is a rational extension of L = K. In the ﬁgure below, the Tj denote a ﬁnite list of indeterminates.

Figure 4.1: Field diagram for Theorem 19.

n ∼ π K(Z/p Z) = Lpn (Tj) d rational π ∼ π π L(J) = Lpn LI (Tj) d rational

π LI

rational K

π π Our next task is to give an upper bound for [Lpn : LI ], which will serve as our upper n π ∼ π bound for Irr(K, Z/p Z) (see Figure 4.1). Since L(J) = Lpn , by Proposition 6, we can n π identify K(Z/p Z) with Lpn (Tj) for indeterminates Tj. Since π acts faithfully on LI, from π π elementary ﬁeld theory, we have [Lpn : LI ] = [Lpn : LI]. Using elementary ﬁeld theory and using the fact that if M and N are two free Z-modules, then L(M ⊕ N) = L(M)(N), we

ﬁnd that [Lp : LI] = [L(p): L(I)]. 78

Lemma 13. Let M and N be free Z-modules with N ⊆ M and |M : N| < ∞. Let K be a ﬁeld. Then [K(M): K(N)] = |M : N|.

Proof. By induction on |M : N|, we may assume that M/N is cyclic of prime order p. The result will follow from the following two claims.

• Claim I If φ is an automorphism of M, then [K(M): K(N)] = [K(M): K(φ(N))].

Proof We may extend φ by linearity to an automorphism of K(M). We have

φ(K(N)) = K(φ(N)), and since φ sends elements of K(M) linearly independent over

K(N) to elements of K(M) linearly independent over φ(K(N)), we have

[K(M): K(N)] ≤ [K(M): φ(K(N))]. We obtain the reverse inequality using

φ−1.

• Claim II Let N,N 0 be submodules of M of index p and let φ : N → N 0 be an

isomorphism. Then φ may be extended to an automorphism φ˜ of M.

Proof We can do this by picking an element x ∈ M that lies in neither N nor N 0

and setting φ˜(x) = x.

These two claims allow us to assume that if M is generated by elements x1, x2, . . . , xr, then p N is generated by x1, x2, . . . , xr, and in this case it is clear that [K(M): K(N)] = p.

From Lemma 13 and the discussion preceding it, we obtain the following Theorem.

Theorem 22. Let K be a ﬁeld and p and odd prime. Then

n Y Irr(K, Z/p Z) ≤ min |pi : Ii|. I principal, I ⊆p 0≤i≤n i i i 79

4.3.1 The General Case

n We now extend our results about Irr(K, Z/p Z) to Irr(K,A), where A is an arbitrary abelian group.

Lemma 14. Let A = G1 × G2, where G1 and G2 are abelian groups. Then

Irr(K,A) ≤ Irr(K,G1) · Irr(K,G2).

Proof. Suppose that Irr(K,Gi) ≤ ci, for i = 1, 2. If Si is a maximal transcendence basis for

K(Gi), i = 1, 2, then by considering the transcendence basis S1 ∪ S2 for the free composite

0 K(G1)K(G2) := K , we see that Irr(K(G1)K(G2)) ≤ c1c2. Furthermore, by the proof of Theorem 1.3 in [29], K(A) is rational over K0. If T is a maximal transcendence basis for

0 0 K(A) over K (so K(A) = K (T )), then by considering the transcendence basis S1 ∪S2 ∪T

for K(A) over K, we see that Irr(K,A) ≤ c1c2.

Following Lenstra, write

A = P ⊕ B, where the order of P is a power of char(K) and the order of B is prime to char(K), and

P = 0 if char(K) = 0. Write ∼ f B = Z/2 Z ⊕s∈Ω Z/sZ, (4.2) where Ω is the set of prime powers giving the elementary divisor decomposition of the odd part of B. From hereon we make the following assumption:

For every prime power s dividing |B|, the group πs := G(K(ζs)/K) is cyclic.

The only situation this may preclude is s = 2n, n ≥ 3 and char(K) = 0. Given our

f assumption, by the Main Theorem of [35], K(Z/2 Z) and K(P ) are both rational over

K. Let cs = Irr(K, Z/sZ). By Lemma 14 and induction on the number of factors in the 80 elementary divisor decomposition of A, we obtain

Y Irr(K,A) ≤ cs. s∈Ω

Finally, using the upper bound for cs obtained in Theorem 22 we obtain our main theorem.

Theorem 23. Let A be a ﬁnite abelian group and K a ﬁeld. Write

f A = P ⊕ B = P ⊕ Z/2 Z ⊕ (⊕s∈ΩZ/sZ),

ns as above. Let s = p and let m = [K(ζ i ): K]. Let p be a prime ideal of [ζ ] lying s s,i ps s,i Z ms,i above p. Then Y Y Irr(K,A) ≤ min |ps,i : Ii|. Ii principal, Ii⊆ps,i s∈Ω 0≤i≤ns

4.3.2 An Example of Theorem 23

In this section we provide a numerical example of Theorem 23. Consider the case when

K is a ﬁeld with [K(ζp): K] = p − 1, and A = Z/pZ. By Theorem 23, we see that

Irr(K, Z/pZ) can be bounded above by

min |p : I|, I principal, I⊆p where p is a prime of Z[ζp−1] lying above p. Let t be an integer whose reduction mod- ∗ ulo p generates (Z/pZ) . We may take I to be the principal ideal (ζp−1 − t) inside p := (I, p) ⊆ Z[ζp−1]. Since Fp has all (p − 1)st roots of unity, p has norm p. By [19], for p satisfying log2(p − 1) ≥ 24, there exists a primitive root t modulo p such that

1 |t| ≤ p1/2, (4.3) 2 81 and if g(p) denotes the least primitive root modulo p in absolute value, then

g(p) = O(p1/4). (4.4)

We have |p : I| = N(I)/N(p) = N(I)/p, where N = NormQ(ζp−1)/Q. Let γ = G(Q(ζp−1)/Q)). We have p−1 Y 1 2 N(I) = (t − ζσ ) ≤ (t + 1)φ(p−1) ≤ p1/2 + 1 , p−1 2 σ∈γ

p−1 where we are using that φ(p − 1) ≤ 2 . We thus obtain the following numerical version 24 of Theorem 23 in the special case A = Z/pZ,[K(ζp): K] = p − 1, and p > 2 :

p−1 1 1 2 Irr(K, /p ) ≤ p1/2 + 1 . Z Z p 2

If we use (4.4), we obtain

p−9 Irr(K, Z/pZ) = O(p 8 ). (4.5)

We close with a series of remarks.

Remark 13. The upper bound for Irr(Q, Z/pZ) given by ﬁnding the principal ideal of minimal index inside p tends to inﬁnity with p since for any ideal I ⊆ p, we can write

I = pI0 for some ideal I0, so that |p : I| = N(I)/N(p) = N(I0) ≥ p since the norm of a prime ideal q lying over a rational prime q is qf , where qf ≡ 1 (mod p − 1).

Remark 14. As mentioned in the introduction, we also have the weaker bound

Sp Irr(K, Z/pZ) ≤ (p − 1)!, given by the ﬁeld K(x1, . . . , xp) . The extension

Sp π K(Z/pZ)/K(x1, . . . , xp) is not Galois for p ≥ 4. Let E = LI (T ), where T is an in- determinate, as in Figure 4.1. This is the ﬁeld that gives the bound in Theorem 23. It is

unknown to the author for which p the extension K(Z/pZ)/E might be Galois.

h Remark 15. Let hn denote the class number of Q(ζn). Since p p−1 is principal, letting

h = hp−1, we obtain the bound

min |p : I| ≤ |p : ph| = N(p)h−1 = ph−1. (4.6) I principal, I⊆p 82

+ Recall that the class number hn of the maximal real subﬁeld of Q(ζn) divides hn. Much is + − + known about the quotient hn/hn := hn , while comparatively little is known about hn (these quantities are discussed in detail in [71], for example). From [71, Theorem 4.20], we have

− 1 φ(n) hn ∼ n 4 . (4.7)

Taking n = p − 1 and using the bound (4.7) in place of h in (4.6) (which is cheating,

+ of course, since (4.7) does not account for hn ) already gives a much larger bound for

minIs principal, Is⊆as |as : Is| than we obtain via (4.5).

To the author’s knowledge, there is no known asymptotic formula (or even non-trivial

+ lower bound) for hn . Bibliography

[1] Proceedings of the 2001 International Symposium on Symbolic and Algebraic Com- putation, New York, 2001. Association for Computing Machinery (ACM). Held in London, ON, 2001. [2] Francesco Bastianelli, Renza Cortini, and Pietro De Poi. The gonality theorem of noether for hypersurfaces. Available on arXiv math.AG/1102.4550v2. [3] A. Brumer. Ramification and class towers of number fields. Michigan Math. J., 12:129–131, 1965. [4] Huah Chu and Ming-chang Kang. Rationality of p-group actions. J. Algebra, 237(2):673–690, 2001. [5] Jean-Louis Colliot-Thélèneand Jean-Jacques Sansuc. The rationality problem for fields of invariants under linear algebraic groups (with special regards to the Brauer group). In Algebraic groups and homogeneous spaces, Tata Inst. Fund. Res. Stud. Math., pages 113–186. Tata Inst. Fund. Res., Mumbai, 2007. [6] Gary Cornell, Joseph H. Silverman, and Glenn Stevens, editors. Modular forms and Fermat’s last theorem. Springer-Verlag, New York, 1997. Papers from the Instructional Conference on Number Theory and Arithmetic Geometry held at Boston University, Boston, MA, August 9–18, 1995. [7] John Cullinan. Local-global properties of torsion points on three-dimensional abelian varieties. J. Algebra, 311(2):736–774, 2007. [8] Charles W. Curtis and Irving Reiner. Representation theory of finite groups and associative algebras. AMS Chelsea Publishing, Providence, RI, 2006. Reprint of the 1962 original.

[9] Pierre Deligne and Jean-Pierre Serre. Formes modulaires de poids 1. Ann. Sci. Ecole´ Norm. Sup. (4), 7:507–530 (1975), 1974.

[10] Lassina Dembélé. A non-solvable Galois extension of Q ramified at 2 only. C. R. Math. Acad. Sci. Paris, 347(3-4):111–116, 2009. [11] Igor V. Dolgachev. Finite subgroups of the plane Cremona group. In Algebraic geometry in East Asia—Seoul 2008, volume 60 of Adv. Stud. Pure Math., pages 1–49. Math. Soc. Japan, Tokyo, 2010.

83 84

[12] W. Duke. The dimension of the space of cusp forms of weight one. Internat. Math. Res. Notices, (2):99–109, 1995. [13] Jordan S. Ellenberg and Akshay Venkatesh. The number of extensions of a number field with fixed degree and bounded discriminant. Ann. of Math. (2), 163(2):723–741, 2006. [14] Leonard Evens. The Schur multiplier of a semi-direct product. Illinois J. Math., 16:166–181, 1972. [15] G. Faltings. Endlichkeitssätzefürabelsche Varietäten über Zahlkörpern. Invent. Math., 73(3):349–366, 1983. [16] Albrecht Fröhlich. Central extensions, Galois groups, and ideal class groups of number fields, volume 24 of Contemporary Mathematics. American Mathematical Society, Providence, RI, 1983. [17] The GAP Group. GAP – Groups, Algorithms, and Programming, Version 4.7.2, 2013.

[18] E. S. Golod and I. R. Safareviˇc.Onˇ the class field tower. Izv. Akad. Nauk SSSR Ser. Mat., 28:261–272, 1964. [19] E. Grosswald. On Burgess’ bound for primitive roots modulo primes and an application to Γ(p). Amer. J. Math., 103(6):1171–1183, 1981. [20] Farshid Hajir. On the growth of p-class groups in p-class field towers. J. Algebra, 188(1):256–271, 1997. [21] Farshid Hajir and Christian Maire. Tamely ramified towers and discriminant bounds for number fields. II. J. Symbolic Comput., 33(4):415–423, 2002. [22] Erich Hecke. Lectures on the theory of algebraic numbers, volume 77 of Graduate Texts in Mathematics. Springer-Verlag, New York, 1981. Translated from the German by George U. Brauer, Jay R. Goldman and R. Kotzen. [23] Jing Long Hoelscher. Ray class groups of quadratic and cyclotomic fields. Int. J. Number Theory, 6(5):1169–1182, 2010. [24] Bertram Huppert. Character theory of finite groups, volume 25 of de Gruyter Exposi- tions in Mathematics. Walter de Gruyter & Co., Berlin, 1998. [25] E. Inaba. Review of “On the class field tower”. [26] I. Martin Isaacs. Character theory of finite groups. AMS Chelsea Publishing, Provi- dence, RI, 2006. Corrected reprint of the 1976 original [Academic Press, New York; MR0460423]. [27] I. Martin Isaacs. Finite group theory, volume 92 of Graduate Studies in Mathematics. American Mathematical Society, Providence, RI, 2008. [28] Michael J. Jacobson Jr., Shantha Ramachandran, and Hugh C. Williams. Sup- plementary tables for “Numerical results on class groups of imaginary quadratic fields”, 2006. http://page.math.tu-berlin.de/ kant/ants/Proceedings/ramachandran- 74/ramachandran-74-tables.pdf. [29] Ming-Chang Kang and Bernat Plans. Reduction theorems for Noether’s problem. Proc. Amer. Math. Soc., 137(6):1867–1874, 2009. 85

[30] Nicholas M. Katz. Galois properties of torsion points on abelian varieties. Invent. Math., 62(3):481–502, 1981. [31] Kiran Kedlaya. A construction of polynomials with squarefree discriminant. Available on arXiv math.NT/1103.5728v2. [32] Chandrashekhar Khare. Remarks on mod p forms of weight one. Internat. Math. Res. Notices, (3):127–133, 1997. [33] Chandrashekhar Khare and Jean-Pierre Wintenberger. Serre’s modularity conjecture. I. Invent. Math., 178(3):485–504, 2009. [34] Chandrashekhar Khare and Jean-Pierre Wintenberger. Serre’s modularity conjecture. II. Invent. Math., 178(3):505–586, 2009. [35] H. W. Lenstra, Jr. Rational functions invariant under a finite abelian group. Invent. Math., 25:299–325, 1974. [36] StéphaneR. Louboutin. Upper bounds for residues of Dedekind zeta functions and class numbers of cubic and quartic number fields. Math. Comp., 80(275):1813–1822, 2011.

[37] Takashi Maeda. Noether’s problem for A5. J. Algebra, 125(2):418–430, 1989. [38] Christian Maire. On infinite unramified extensions. Pacific J. Math., 192(1):135–142, 2000. [39] Jacques Martinet. Petits discriminants. Ann. Inst. Fourier (Grenoble), 29(1):xv, 159–170, 1979. [40] Cameron W. McLeman. A Golod-Shafarevich equality and p-tower groups. ProQuest LLC, Ann Arbor, MI, 2008. Thesis (Ph.D.)–The University of Arizona. [41] T. T. Moh and W. J. Heinzer. A generalized Lüroth theorem for curves. J. Math. Soc. Japan, 31(1):85–86, 1979. [42] Hyunsuk Moon. The non-existence of certain mod p Galois representations. Bull. Korean Math. Soc., 40(3):537–544, 2003. [43] Hyunsuk Moon. The number of monomial mod p Galois representations with bounded conductor. Tohoku Math. J. (2), 55(1):89–98, 2003. [44] JürgenNeukirch. Algebraic number theory, volume 322 of Grundlehren der Mathema- tischen Wissenschaften [Fundamental Principles of Mathematical Sciences]. Springer- Verlag, Berlin, 1999. Translated from the 1992 German original and with a note by Norbert Schappacher, With a foreword by G. Harder. [45] JürgenNeukirch. Algebraic number theory, volume 322 of Grundlehren der Mathema- tischen Wissenschaften [Fundamental Principles of Mathematical Sciences]. Springer- Verlag, Berlin, 1999. Translated from the 1992 German original and with a note by Norbert Schappacher, With a foreword by G. Harder. [46] JürgenNeukirch, Alexander Schmidt, and Kay Wingberg. Cohomology of number fields, volume 323 of Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Sciences]. Springer-Verlag, Berlin, 2000. [47] A. M. Odlyzko. Bounds for discriminants and related estimates for class numbers, regulators and zeros of zeta functions: a survey of recent results. Sém.Théor.Nombres Bordeaux (2), 2(1):119–141, 1990. 86

[48] A. M. Odlyzko. Bounds for discriminants and related estimates for class numbers, regulators and zeros of zeta functions: a survey of recent results. Sém.Théor.Nombres Bordeaux (2), 2(1):119–141, 1990. [49] Emmanuel Peyre. Unramified cohomology and rationality problems. Math. Ann., 296(2):247–268, 1993. [50] L. B. Pierce. The 3-part of class numbers of quadratic fields. J. London Math. Soc. (2), 71(3):579–598, 2005. [51] Christopher Rasmussen. An abelian surface with constrained 3-power torsion. In Galois-Teichmüller theory and arithmetic geometry, volume 63 of Adv. Stud. Pure Math., pages 449–456. Math. Soc. Japan, Tokyo, 2012. [52] Christopher Rasmussen and Akio Tamagawa. Arithmetic of abelian varieties with constrained torsion. Available on arXiv math.NT/1302.1477. [53] Christopher Rasmussen and Akio Tamagawa. A finiteness conjecture on abelian varieties with constrained prime power torsion. Math. Res. Lett., 15(6):1223–1231, 2008. [54] D. Roberts and J.Jones. http://hobbes.la.asu.edu/NFDB/. [55] Peter Roquette. On class field towers. In Algebraic Number Theory (Proc. Instructional Conf., Brighton, 1965), pages 231–249. Thompson, Washington, D.C., 1967. [56] David J. Saltman. Noether’s problem over an algebraically closed field. Invent. Math., 77(1):71–84, 1984. [57] Bodo Schmithals. Konstruktion imaginärquadratischer Körper mit unendlichem Klassenkörperturm. Arch. Math. (Basel), 34(4):307–312, 1980. [58] RenéSchoof. Infinite class field towers of quadratic fields. J. Reine Angew. Math., 372:209–220, 1986. [59] J.-P. Serre. Local class field theory. In Algebraic Number Theory (Proc. Instructional Conf., Brighton, 1965), pages 128–161. Thompson, Washington, D.C., 1967. [60] J.-P. Serre. Modular forms of weight one and Galois representations. In Algebraic number fields: L-functions and Galois properties (Proc. Sympos., Univ. Durham, Durham, 1975), pages 193–268. Academic Press, London, 1977. [61] Jean-Pierre Serre. Linear representations of finite groups. Springer-Verlag, New York, 1977. Translated from the second French edition by Leonard L. Scott, Graduate Texts in Mathematics, Vol. 42. [62] Jean-Pierre Serre. Local fields, volume 67 of Graduate Texts in Mathematics. Springer- Verlag, New York, 1979. Translated from the French by Marvin Jay Greenberg. [63] Jean-Pierre Serre. Sur les représentations modulaires de degré2 de Gal(Q/Q). Duke Math. J., 54(1):179–230, 1987. [64] Jean-Pierre Serre. Abelian l-adic representations and elliptic curves, volume 7 of Research Notes in Mathematics. A K Peters Ltd., Wellesley, MA, 1998. With the collaboration of Willem Kuyk and John Labute, Revised reprint of the 1968 original. [65] Joseph H. Silverman. The arithmetic of elliptic curves, volume 106 of Graduate Texts in Mathematics. Springer, Dordrecht, second edition, 2009. 87

[66] Bhama Srinivasan and Judith D. Sally, editors. Emmy Noether in Bryn Mawr, New York, 1983. Springer-Verlag. [67] W. A. Stein et al. Sage Mathematics Software. The Sage Development Team. http://www.sagemath.org. [68] D. A. Suprunenko. Matrix groups. American Mathematical Society, Providence, R.I., 1976. Translated from the Russian, Translation edited by K. A. Hirsch, Translations of Mathematical Monographs, Vol. 45. [69] Richard G. Swan. Invariant rational functions and a problem of Steenrod. Invent. Math., 7:148–158, 1969. [70] B. B. Venkov and H. Koh. The p-tower of class fields for an imaginary quadratic field. Zap. Nauˇcn.Sem. Leningrad. Otdel. Mat. Inst. Steklov. (LOMI), 46:5–13, 140, 1974. Modules and representations. [71] Lawrence C. Washington. Introduction to cyclotomic fields, volume 83 of Graduate Texts in Mathematics. Springer-Verlag, New York, second edition, 1997. [72] Siman Wong. Automorphic forms on GL(2) and the rank of class groups. J. Reine Angew. Math., 515:125–153, 1999. [73] Yoshihiko Yamamoto. On unramified Galois extensions of quadratic number fields. Osaka J. Math., 7:57–76, 1970. [74] Hisao Yoshihara. Degree of irrationality of an algebraic surface. J. Algebra, 167(3):634– 640, 1994.