On Some Norm Equations Over Cyclotomic Fields · ABSTRACT For an odd prime p, the equation xp + yp...

On Some Norm

Equations over

Cyclotomic FieldsPeter LombaersTese de Doutoramento apresentada à

Faculdade de Ciências da Universidade do Porto

Matemática

2019

Peter Lombaers: On Some Norm Equations Over Cyclotomic Fields , July2019, Porto

supervisor:António Machiavelo

A B S T R A C T

For an odd prime p, the equation xp + yp = zp can be rewritten as(x+ y)Np(x+ yζp) = zp, where ζp is a primitive p-th root of unity andNp is the norm map of the field Q(ζp). That this norm equation doesnot have non-trivial integer solutions is known since Andrew Wiles’proof of Fermat’s last theorem. However, very little is known abouttwo natural generalisations of this equation given by the equationsNp(x+ yζp) = zp, and (a0 + a1 + · · ·+ ak)Np(a0 + a1ζp + · · ·+ akζk

p) =

zp, for some integer 1 ≤ k ≤ p− 3.In this thesis, we look at the cyclotomic methods that were used to

attack the equation xp + yp = zp, and we apply them to those two gen-eralisations. Naturally, this leads to a number of new observations andproblems. First of all, we fix the coefficients a0, . . . , ak and investigatetheir behaviour as p goes to infinity. This leads to an upper boundon p in certain cases. Next, we show that Np(a0 + · · ·+ akζk

p) can beexpressed in terms of linear recurrence sequences depending only onthe coefficients ai. This allows us to show that, for fixed a, b, thereare only finitely many primes p such that Np(aζ2

p + bζp − a) = zp hasnon-trivial integer solutions.

We review Kummer’s method of attack to the equation Np(x +

yζp) = zp by the use of a logarithmic derivative, and we show that theIwasawa p-adic logarithm can be used to obtain the same results. Wereview the theory of unramified extensions of degree p of Q(ζp), andwe see how this leads to a proof of Herbrand’s theorem. Combiningknowledge of the units of Z[ζp] with computational results, we showthat if Np(x + yζp) = zp, then max(|x|, |y|) > 5 × 108, for regularprimes p ≥ 5. Finally, we use the p-adic logarithm to give a unifiedway of proving congruence identities related to Wolstenholme’s the-orem. With the same method, we can also prove a number of othercongruences, which at first sight seem quite unrelated to each other.

iii

R E S U M O

Para um primo p ímpar, a equação xp + yp = zp pode ser reescritacomo (x + y)Np(x + yζp) = zp, em que ζp é uma raiz primitiva p-ésima da unidade e Np é a norma do corpo Q(ζp). Que esta equaçãonão tem soluções inteiras não-triviais é sabido desde a prova deAndrew Wiles do último teorema de Fermat. No entanto, pouco sesabe acerca das duas generalizações seguintes dessa equação: Np(x +

yζp) = zp e (a0 + a1 + · · ·+ ak)Np(a0 + a1ζp + · · ·+ akζkp) = zp, para

algum inteiro 1 ≤ k ≤ p− 3.Nesta tese, olhamos para os métodos ciclotómicos que foram usados

para atacar a equação xp + yp = zp, e aplicamo-los a essas duas gener-alizações. Naturalmente, isso conduz a uma série de novas observaçõese problemas. Em primeiro lugar, fixamos os coeficientes a0, . . . , ak einvestigamos o seu comportamento quando p tende para infinito. Istoconduz a um limite superior para p, em certos casos. Em seguida,mostramos que a norma Np(a0 + · · ·+ akζk

p) pode ser expressa emtermos de sequências de recorrência linear que dependem apenas doscoeficientes ai. Isso permite-nos mostrar que, para a, b fixos, há apenasum número finito de primos p tais que Np(aζ2

p + bζp − a) = zp temsoluções inteiras não-triviais.

Revemos o método de Kummer para atacar a equação Np(x +

yζp) = zp usando derivadas logarítmicas e mostramos que o log-aritmo p-ádico de Iwasawa pode ser usado para alcançar os mesmosresultados. Revemos a teoria das extensões não-ramificadas de graup de Q(ζp), e vemos como isso conduz a uma demonstração do teo-rema de Herbrand. Combinando conhecimento das unidades de Z[ζp]

com resultados computacionais, mostramos que se Np(x + yζp) = zp,então max(|x|, |y|) > 5 × 108, para primos regulares p ≥ 5. Final-mente, usamos o logaritmo p-ádico para dar uma maneira unificadade demonstrar identidades envolvendo congruências relacionadascom o Teorema de Wolstenholme. Com o mesmo método podemostambém demonstrar várias outras congruências que, à primeira vista,parecem não estar relacionadas entre si.

iv

A C K N O W L E D G M E N T S

This thesis would not have been possible without the help and supportof a number of people. First of all I would like to thank my supervisor.António, it’s been a great pleasure to work with you. You’ve alwaysbeen friendly, enthousiastic and helpful, and I feel lucky to have hadyou as my supervisor. Next, I would like to thank my friends here inPorto and the friends from the Netherlands that came to visit. Youmade my time here a great experience, whether it was to enjoy thegood things, or complain about the bad things. Special mention shouldgo to Atefeh, without you I would have been taking the stairs for thepast four years, and to Nikos, you were always there, listening to myexplanations of what I was working on, so that I could understandit better myself. I also want to thank my family for their love andsupport while I was far away from home. Finally, Saul, without you Iwould not survived the final year, but you turned it into a great year.

I would like to acknowledge the financial support by the FCT(Fundação para a Ciência e a Tecnologia) through the PhD grantPD/BD/128063/2016, and by CMUP (Centro Matemática - Universi-dade do Porto) (UID/MAT/00144/2013), which is funded by FCT (Por-tugal) with national (MEC) and European structural funds (FEDER),under the partnership agreement PT2020.

v

C O N T E N T S

1 introduction 1

1.1 The main problem 1

1.2 Kummer’s approach 2

1.3 Description of the chapters 4

1.4 Notation 6

2 limits of N-th roots 9

2.1 The Mahler measure 9

2.2 An upper bound on p 11

3 recurrence sequences 15

3.1 Norms in terms of a recurrence sequence 15

3.2 A general recurrence theorem 21

4 from norms to singular integers 27

4.1 Singular integers 27

4.2 Self-prime integers 28

5 singular integers 35

5.1 Idempotents and the Herbrand-Ribet theorem 35

5.2 Kummer’s logarithmic derivative 37

5.3 The p-adic logarithm 40

6 kummer extensions 47

6.1 Kummer extensions 47

6.2 Ramification 49

6.3 Units in cyclotomic fields 52

6.4 Unramified extensions 55

7 regular primes 61

7.1 Theoretical results 61

7.2 Computational results 63

8 wolstenholme’s theorem 67

8.1 Logarithms and binomial coefficients 68

8.2 Logarithms and cyclotomic integers 73

bibliography 81

vii

1I N T R O D U C T I O N

1.1 the main problem

In this thesis we look at two generalisations of Fermat’s last theorem.They arise naturally when we approach Fermat’s last theorem viathe classical, cyclotomic method. Let us first recall the statement ofFermat’s last theorem:

Theorem 1.1 (FLT). For each integer n ≥ 3 there are no non-zero integersx, y, z such that

xn + yn = zn. (1.1)

The first succesful proof was given by Andrew Wiles, using a con-nection with elliptic curves. Before that, the traditional method toattack FLT was via cyclotomic fields. We first note that we may assumein Theorem 1.1, without loss of generality, that n = 4 or n = p is anodd prime. Fermat himself already wrote a proof for the case n = 4(see chapter 1 of [29] for a good account of the early history of FLT).Therefore we can restrict our attention to the case n = p is an oddprime.

Since p is odd, we know that xp + yp is divisible by x + y. Let ζp bea primitive p-th root of unity. Then we have the equality

xp + yp =p−1

∏i=0

(x + yζ ip).

Let us denote by Np = NQ(ζp)/Q the norm function of the p-th cy-

clotomic field Q(ζp), hence we have ∏p−1i=1 (x + yζ i

p) = Np(x + yζp).Proving FLT is thus equivalent to showing that there are no non-zerointegers x, y, z such that

(x + y)Np(x + yζp) = zp. (1.2)

The first generalisation we shall consider is a slightly less generalversion of a conjecture by George Gras in [12]. He referred to it as‘Strong Fermat’s last theorem’.

Conjecture 1.2 (SFLT). For each prime p ≥ 5, the only solutions in pairwisecoprime integers x, y, z to the equation

Np(x + yζp) = zp (1.3)

are given by z = 1 and x + yζp ∈ {±1,±ζp,±(1 + ζp)}.

1

2 introduction

Note that in [12], Conjecture 1.5 states that also Np(x + yζp) =

pzp only has the trivial solutions. The second generalisation we willconsider is:

Problem 1.3. Let p ≥ 5 be a prime and let 1 ≤ k ≤ p− 3 be an integer.What are the integer solutions to the following two equations?

Np(a0 + a1ζp + · · ·+ akζkp) = zp (1.4)

(a0 + a1 + · · ·+ ak)Np(a0 + a1ζp + · · ·+ akζkp) = zp (1.5)

For k = 1, equation (1.5) gives us back FLT and equation (1.4),with coprime a0, a1, is precisely Conjecture 1.2. If k = p − 2 thena0 + a1ζp + · · ·+ ap−2ζ

p−2p is a generic element of Z[ζp], and thus (1.5)

has many solutions. For example, Np(uγp) will be a p-th power of aninteger if u is a unit of Z[ζp] and γ is any element of Z[ζp]. This isalso the reason why we restrict to p ≥ 5.

Though Conjecture 1.2 got a little attention on its own (see [2, 12,13]), almost all results come from theorems connected to FLT which canbe applied to Conjecture 1.2 and Problem 1.3 as well. It is interestingto compare this situation with Catalan’s conjecture, which states thatthe only positive integer solution to the equation

xn − 1 = yq

with n ≥ 2 is given by 32− 23 = 1. This has been proved completely bycyclotomic methods by Preda Mihailescu. The corresponding ‘strong’version is given by the Nagell-Ljunggren equation

xn − 1x− 1

= yq.

Contrary to SFLT, the Nagell-Ljunggren equation has a rich history ofitself (see [3] for a survey).

In this thesis we shall look at the approach Ernest Kummer took toFLT and improvements of that approach, and we shall see how thistranslates to Problem 1.3. Naturally this leads to a number of freshproblems which we shall look at. To make this more concrete, we givean overview of the different steps of Kummer’s approach to FLT, andindicate how these steps relate to the chapters of this thesis.

1.2 kummer’s approach

Suppose x, y, z are non-zero integers such that xn + yn = zn. We mayassume without loss of generality that n = p is an odd prime. Theanalog of Conjecture 1.2 for n not a prime would be to consider theequation

n−1

∏i=1

(x + yζ in) = zn

1.2 kummer’s approach 3

and we can define something similar for Problem 1.3. In chapter 2 welook at a way to give a lower bound for max(|x|, |y|). This methodalso works for n not a prime. In chapter 3 we look at a way to write

n−1

∏i=0

(a0 + a1ζ in + · · ·+ akζ ik

n )

in terms of recurrence sequences. This method also does not requirethat n is a prime number.

We may assume without loss of generality that x, y, z are pairwisecoprime. For FLT, we can assume this without loss of generality, butfor the equation (1.3) of SFLT this is not the case. In fact if a, b are anytwo integers, then we can write Np(a + bζp) = cdp for some integersc, d (with possibly d = 1). Then we see that Np(ac + bcζp) = cpdp isa solution to equation (1.3). So there are infinitely many solutions toConjecture 1.2 if we do not assume that x, y, z are coprime. We willnot consider the case where x, y, z have a common divisor any further.

We separate the proof into two cases. In the first case p - xyz, andin the second case p | xyz. These cases are traditionally referred to asCase 1 and Case 2, and we will write FLT1 and FLT2 accordingly. Wecan make a similar case distinction for SFLT, and we will refer to theseas SFLT1 and SFLT2.

Assume we are in Case 1, so p - xyz. Because x and y are coprime,we find that x + yζ i

p is coprime to x + yζjp if i 6≡ j mod p. Hence we

have (x + yζp)Z[ζp] = Ip for some ideal I of Z[ζp]. The fields Q(ζp)

do not have unique factorisation into irreducible elements for p ≥ 23.To get around this problem, Kummer used ideals and the class group.The fact that for p - xyz and x, y coprime we have

Np(x + yζp) = zp =⇒ (x + yζp)Z[ζp] = Ip (1.6)

does not carry over to the situation with more variables of Problem1.3. In chapter 4 we try to find conditions under which (1.6) holds forelements of the form a0 + a1ζp + · · ·+ akζk

p.Apply an annihilator η ∈ Z[Gal(Q(ζp)/Q)] of the class group of

Q(ζp) to get (x + yζp)η = uγp for some unit u ∈ Z[ζp] and someγ ∈ Z[ζp]. Kummer originally proved FLT for regular primes, soprime p for which p does not divide the class number of Q(ζp). Inthis case, the fact that Ip is a principal ideal implies that I must beprincipal itself. If the prime p is irregular, then I might not be principal,but certain products of Galois conjugates of I will be principal. Thetwo main theorems in this direction are Stickelberger’s theorem andthe Herbrand-Ribet theorem. Stickelberger’s theorem was proved forcyclotomic fields already by Kummer. Both of these theorems giveannihilators of the class group of Q(ζp). In chapter 6 we give a proofof the easy direction of the Herbrand-Ribet theorem using class fieldtheory and an explicit construction of an extension of Q(ζp).

4 introduction

Take the logarithmic derivative and look modulo p. The idea inthis step is that the logarithmic derivative of a p-th power should becongruent to 0 modulo p. So if x + yζp = uγp, then after taking thelogarithmic derivative, we only care about the unit u. We can then usethe knowledge we have of the units of Z[ζp] to derive a congruencecondition modulo p that x, y and z must satisfy. For example, Kummercould conclude that if (x, y, z) is a solution to FLT1, then they mustsatisfy:

Bp−3xy(x− y)(x + y)3 ≡ 0 mod p (1.7)

Here Bp−3 is the (p− 3)-rd Bernoulli number. In chapter 5 we willshow that instead of the logarithmic derivative, we can also use thep-adic logarithm. Although not obvious at the start, this will basicallygive the same information as we can obtain by using the logarithmicderivative. This method is not restricted to FLT and SFLT, but it appliesto general cyclotomic elements, like in Problem 1.3. In chapter 8 welook closer at the condition Bp−3 ≡ 0 mod p. Primes that satisfythis condition are called Wolstenholme primes. We use the p-adiclogarithm to get characterisation of these primes.

Use the symmetry of xp + yp = zp to conclude that p | xyz. Ifwe have xp + yp = zp, then we also get xp + (−z)p = (−y)p andyp + (−z)p = (−x)p. From equation (1.7) we concluded that p | x− y,so by using the symmetry of the equation we find that p | x + z andp | y + z. So then 0 ≡ xp + yp − zp ≡ 3x mod p. So p | xyz after andwe can go to Case 2. This kind of symmetry is only something we canuse in the case of FLT. For SFLT we have nothing similar. Also for themore general Problem 1.3 we do not have this symmetry.

In Case 2, from a solution to FLT2, construct a smaller solutionand use a descent argument to get a contradiction. Case 2 of FLT isconsiderably more difficult than Case 1. The approaches to Case 2 usea descent, and all rely on the symmetry of xp + yp = zp. Thereforethey do not carry over to SFLT or to Problem 1.3. Though we will notbe able to proof SFLT for a specific prime, we will be able to give alower bound on max(|x|, |y|).

1.3 description of the chapters

In chapter 2, we fix α ∈ Q(ζp), and look at the behaviour of

limp→∞

p√

Np(α).

We see that if α = f (ζp) for some polynomial f ∈ Z[X], then this limitconverges to the Mahler measure of f . We then use knowledge of thislimit, in Theorem 2.5, to obtain the lower bound max(|x|, |y|) ≥ 500,for x, y that satisfy Np(x + yζp) = zp.

1.3 description of the chapters 5

In chapter 3, we show that we can write norms in terms of sums oflinear recurrence sequences. More specifically, in Theorem 3.4, we seethat if f ∈ Z[X] is of degree k, then we can write

n−1

∏i=0

f (ζ in) =

k

∑j=0

A(j)n ,

where for each j, the sequence (A(j)n )n∈N is a linear recurrence se-

quence, depending only on the coefficients of f . There are manyresults on perfect powers in linear recurrence sequences. We use theseto show that for fixed a, b, there are only finitely many primes p, suchthat Np(aζ2

p + bζp − a) = zp has a solution.In chapter 4, we try to find out when the following implication

holds:

Np(α) = zp =⇒ αZ[ζp] = Ip for some ideal I of Z[ζp].

This question does not arise in the case of FLT, but only arises forProblem 1.3, so there is not much previously known. We give someconditions on α that ensure the implication is true. We also try to findout when α is coprime to its Galois conjugates. For x + yζp this occursprecisely when x and y are coprime, and p - x + y, but for other α thesituation is more difficult.

In chapter 5, we take a look at Kummer’s idea to study the equationx + yζp = uγp, using the logarithmic derivative. We see, in detail, howthis leads to a proof of FLT1 for primes which satisfy p - Bp−3. Nextwe see that, instead of the logarithmic derivative, we can also usethe p-adic logarithm to study these kind of equations. However, inTheorem 5.9, we show that the information we obtain with the p-adiclogarithm is, curiously, the same as the information we obtain withthe logarithmic derivative.

In chapter 6, we give a review of the theory of Kummer extensions.We look at the ramification of primes in these extensions. Then we lookat the unit group of Z[ζp], and use these units to construct unramifiedextensions of Q(ζp) of degree p. This leads to a proof Herbrand’sdirection of the Herbrand-Ribet theorem. All these results are well-known, but we hope that presentation of this chapter is concrete, andties in well with the logarithmic derivative of chapter 5.

In chapter 7, we look at SFLT for regular primes. If p is regular, wehave extra knowledge of the units, and this allows us to show thatxy(x− y) must be divisible by a number of small primes. In Theorem7.8 we use this to show that if p ≥ 5 is regular and Np(x + yζp) = zp,then max(|x|, |y|) > 5× 108. This method does not work for Problem1.3.

In chapter 8, we look at Wolstenholme’s theorem and generalisationsof it. Wolstenholme’s theorem states that, for prime p ≥ 5, we have(2p−1

p−1 ) ≡ 1 mod p3. A prime p that satisfies this equation modulo

6 introduction

p4 is called a Wolstenholme prime. This is equivalent to the factthat p divides Bp−3, and hence the link with the previous chapters.There are a number of results that generalise Wolstenholme’s theoremand characterise Wolstenholme primes modulo higher powers of p.We use the the p-adic logarithm to give easier proofs of some ofthese generalisations, and to extend previously known results. Forexample, we give a characterisation of Wolstenholme primes in termsof harmonic sums modulo p12. In the second section, we apply thep-adic logarithm to norms of certain cyclotomic integers. This givesus p-adic equations for the trace of these integers. Looking modulospecific powers of p, we obtain a number of previously known results,and a number of new results. For example, in Proposition 8.17, wegive another proof of the identity

p−1

∑k=1

1k

(2kk

)≡ 0 mod p2.

Using the relation with recurrence sequences of chapter 3, we provethe identity

p−1

∑k=1

(−1)k−1 1k

(2kk

)≡

(1− L2p)(L2

p − 3)2p

mod p2,

where (Ln)n∈N is the sequence of Lucas numbers, given by L0 =

2, L1 = 1 and Ln+2 = Ln+1 + Ln. The most interesting part aboutthis is that these results can be found using the same method, wherepreviously each identity required different methods to proof them.

1.4 notation

We use the following notation in this thesis. For us, p will alwaysdenote an odd prime, except when mentioned otherwise. For n ≥ 3an integer, we let ζn be a primitive n-th root of unity, so it is a root ofthe n-th cyclotomic polynomial Φn(X). In most chapters, n = p is anodd prime. If the prime p is fixed throughout a chapter, we will dropthe subscript p and just write ζ = ζp and Φ(X) = Φp(X). We willwork with the cyclotomic fields Q(ζn). In chapters 4, 5, 7 and section3 and 4 of chapter 6, we will fix n = p throughout the chapter, thenwe will denote K = Q(ζp). We assume that the reader has had a firstcourse in algebraic number, similar to the first chapter of [16]. The ringof integers of Q(ζn) is Z[ζn] and if we use the notation K = Q(ζp),then we will write OK for the ring of integers of K. We denote theelements of the Galois group Gal(Q(ζn)/Q) by σa, for a ∈ (Z/nZ)∗,where σa is determined by σa(ζn) = ζa

n. If p is fixed and K = Q(ζp),then we will also write Gal(K/Q) = G. The norm of K is given byNp(α) = ∏σ∈G σ(α) and the trace is given by Trp(α) = ∑σ inG σ(α). Ifit is obvious which prime p we are talking about, we will drop the

1.4 notation 7

subscript p and just write N and Tr. For us, π will always denotethe element 1− ζ of K, so that we have πp−1OK = pOK as ideals.We will denote (α) for the principal ideal generated by α if it isclear in what ring we are working. If p is a prime ideal of OK, thenordp(α) is the maximal n ∈ Z such that pn divides α. So, for exampleord(π)(p) = p− 1. The field Qp is the field of p-adic numbers. Thep-adic logarithm will always be denoted by logp and the ordinarylogarithm will be denote by log.

2L I M I T S O F N - T H R O O T S

Let α ∈ Q(ζp) and write α = f (ζp) for some polynomial f ∈ Z[X].Then we have

Np(α) =p−1

∏i=1

f (ζ ip),

and thus Np(α) = zp for some integer z ∈ Z if and only if

p

√√√√p−1

∏i=1

f (ζ ip) ∈ Z.

In this chapter we shall fix f and investigate the behaviour of theabove expression as p grows. In the first part we see how this relatesto the Mahler measure of a polynomial. We give a concrete formulafor the limit when the polynomial f has degree 2. In the second partof the chapter we use this to give an upper bound on p in terms of aand b (or equivalently a lower bound on a and b in terms of p) whenthere is a solution to Np(a + bζp) = cp. Concretely, Theorem 2.4 saysthat if Np(a + bζp) = cp, then max(|a|, |b|) > 500.

2.1 the mahler measure

Let f be a polynomial with integer coefficients. Note that ∏n−1i=1 f (ζ i

n)

makes sense for any positive integer n, but will only be equal to thenorm of an element of a cyclotomic field if n is an odd prime. Alsonote that, since f (ζ i

n) is the complex conjugate of f (ζn−in ), the number

∏n−1i=1 f (ζ i

n) is always positive, so we have ∏n−1i=1 f (ζ i

n) = ∏n−1i=1 | f (ζ i

n)|.Here |x| is the absolute value of x as an element of C.

Theorem 2.1. Suppose f (X) ∈ Z[X] factors as f (X) = a(X − α1)(X −α2) . . . (X− αk) over the complex numbers, and such that a, α1, . . . , αk 6= 0.Suppose |αi| 6= 1 for all i = 1, . . . , k. Then we have

limn→∞

n

√√√√n−1

∏i=1

f (ζ in) = |a| ∏

|αi |>1|αi| = |a|

k

∏i=1

max(1, |αi|).

9

10 limits of N -th roots

Proof. First of all, we see

n−1

∏i=1

f (ζ in) =

n−1

∏i=1| f (ζ i

n)|

= |a|n−1k

∏j=1

n−1

∏i=1|ζ i

n − αj|

= |a|n−1k

∏j=1

∣∣∣∣∣1− αnj

1− αj

∣∣∣∣∣=|a|n| f (1)|

k

∏j=1|1− αn

j |.

Now take the n-th root and let n go to infinity. We get

limn→∞

n

√|a|n| f (1)| = |a| lim

n→∞

1n√| f (1)|

= |a|.

Also, we have

limn→∞

n√|1− αn

j | =

1 if |αj| < 1

|αj| if |αj| > 1.

Combining this we end up with

limn→∞

n

√√√√n−1

∏i=1

f (ζ in) = |a| ∏

|αj|>1|αj|.

The condition |αi| 6= 1 is there to ensure that this limit actually con-verges. If |α| = 1 and α 6= 1, then limn→∞

n√

1− αn does not converge.Note that, even for f (X) ∈ Z[X], the condition ‘|α| 6= 1 for any rootof f ’ is not equivalent to the condition ‘ f (ζn) 6= 0 for all n ∈N’. Forexample, the polynomial x4− 2x3− 2x + 1 has two roots with absolutevalue 1 and two roots with absolute value different from 1.

Definition 2.2. For f (X) = a(x− α1)(x− α2) . . . (x− αk) ∈ C[X] definethe Mahler measure M( f ) of f as the number

M( f ) := |a|k

∏j=1

max(1, |αj|).

In the article [21], D. H. Lehmer considered a method to find largeprimes. Suppose f (X) is a monic polynomial with integer coefficientsand with roots α1, . . . , αk. Then, for each positive integer n, the num-bers ∆n( f ) = ∏k

j=1(αnj − 1) will be integers. The prime factors of

these numbers should satisfy certain congruence conditions, which

2.2 an upper bound on p 11

might make it easier to find large primes. For example, if we takef (X) = X− 2 then we are considering the prime factors of the number2n − 1, so this is related to Mersenne primes.

This lead Lehmer to consider the behaviour of these integers ∆n asn goes to infinity. If f has no roots which are a root of unity, then

limn→∞

∆n+1

∆n= M( f ).

For his purposes it was important that M( f ) is as small as possible.A theorem of Kronecker says that M( f ) = 1 if and only if f is theproduct of cyclotomic polynomials. Excluding these, the polynomialwith smallest Mahler measure Lehmer could find was the polynomial

f (X) = x10 + x9 − x7 − x6 − x4 − x3 + x + 1

which satisfies M( f ) ≈ 1.176. Lehmer’s conjectures states that thereis a constant µ > 1 such that M( f ) ≥ µ for all f which are notproducts of cyclotomic polynomials. This is still an open problem. Anon-cyclotomic polynomial with smaller Mahler measure than thenone above has not been found. Later, Kurt Mahler considered theproblem of multivariable polynomials, and the name Mahler measuregot associated to M( f ).

2.2 an upper bound on p

Let f (X) ∈ Z[X] be a polynomial which does not have a root withabsolute value 1. Then by Theorem 2.1 we know that

limn→∞

n

√√√√n−1

∏i=1

f (ζ in) = M( f ).

If M( f ) is not an integer, then we know that there is some n0 ∈N such

that for all n ≥ n0 the number n√

∏n−1i=1 f (ζ i

n) is not an integer. If M( f )

is an integer, but ∏n−1i=1 f (ζ i

n) 6= M( f ) from some n onwards, thenagain we can find such an integer n0. If n = p ≥ n0 is a prime then

we see that p√

∏p−1i=1 f (ζ i

p) =p√

Np( f (ζp)). So we find that Np( f (ζp))

cannot be a p-th power for all primes p ≥ n0.Let us find an explicit expression for n0 in a simple example. The

simplest example is the case where f (X) = aX + b is a linear polyno-mial with |a| 6= |b. In this case we have Np( f (ζp)) =

ap+bp

a+b . Note that ifNp(aζp + b) is a p-th power, then also Np(−aζp − b) and Np(a + bζp)

are p-th powers, so we can assume without loss of generality thata > |b| > 0. Although the following proposition immediately followsthe proof Fermat’s last theorem, we show this example anyway as anillustration of the method.

12 limits of N -th roots

Proposition 2.3. Let a > |b| > 0 and let p be a prime such that

p >log(2)

log(a + 1)− log(a).

Then there is no integer c such that ap + bp = cp.

Proof. By assumption we have

p >log(2)

log(a + 1)− log(a)>

log(2)log(a)− log(a− 1)

and thus we see that log(2) < p log( aa−1 ) and log(2) < p log( a+1

a ). So2(a− 1)p < ap and 2ap < (a + 1)p. Hence we find:

(a− 1)p < ap − (a− 1)p ≤ ap + bp < 2ap < (a + 1)p.

Therefore a− 1 < p√

ap + bp < a + 1. If b 6= 0 then p√

ap + bp 6= a, sowe conclude that p

√ap + bp is not an integer.

A slight variation on the previous proposition is the following.

Theorem 2.4. Let a > |b| > 0 and let p be a prime such that

p >log(2) + log(a)

log(a)− log(a− 1).

Then there is no integer c such that ap+bp

a+b = cp.

Proof. Note that 1− a− b ≤ 0, and thus ap(1− a− b) + bp < 0. Thismeans that for all p we have

ap + bp

a + b< ap.

Since we assume that p > log(2)+log(a)log(a)−log(a−1) , we find that (p− 1) log(a)−

p log(a− 1) > log(2), and thus ap−1

(a−1)p > 2. Consequently

ap > 2a(a− 1)p ≥ (a− 1)p(a + b + 1)

and therefore

ap + bp > ap − (a− 1)p > (a− 1)p(a + b).

We conclude that a− 1 < p√

ap+bp

a+b < a, so p√

ap+bp

a+b is not an integer.

This allows us to computationally give a lower bound on max(|a|, |b|).We fix an a > 0, and we check for all primes p < log(2)+log(a)

log(a)−log(a−1) and

1− a ≤ b ≤ a− 1 coprime to a that ap+bp

a+b is not a p-th power. Note, forp = 3, that aζ3 + b is a general element of Z[ζ3], so we will certainlyfind many third powers of the form a3+b3

a+b . The interesting case is p > 3.We did these computations using a simple Pari-GP [25] script. Thisgave us the following result.

2.2 an upper bound on p 13

Theorem 2.5. If p ≥ 5 is a prime and a, b are coprime integers such thatap+bp

a+b = cp, then max(|a|, |b|) > 500.

This computational process does not scale very efficiently. We willsee in chapter 7 that if we assume that p is a regular prime, thenwe can greatly improve this lower bound. In Lemma ?? we showedwhat the Mahler measure of a quadratic polynomial is. We wouldlike to make a similar analysis like that of Theorem 2.4 for Np(aζ2

p +

bζp + c). If f (X) = aX2 + bX + c, can we have p√

Np( f (ζp)) = M( f )

or p√(a + b + c)Np( f (ζp)) = M( f )? If not, there is an integer n0 such

that for all primes p ≥ n0 we know that p√

Np( f (ζp)) and (a + b +

c) p√

Np( f (ζp)) are not integers. Can we give this integer n0 explicitlyin terms of a, b and c? The situation is more complicated than in thetwo variable case, and we do not yet have an answer to these questions.

3R E C U R R E N C E S E Q U E N C E S

In this chapter we investigate a connection between norms of cyclo-tomic integers and linear recurrence sequences. We start with thesimplest case, when the cyclotomic integer is of the form aζ2

p + bζp + c.In Theorem 3.1 we will see that

n−1

∏i=0

(aζ2in + bζn + c) = an + cn − Hn,

where the numbers (Hn)n∈N are given by a linear recurrence relation.When n = p is an odd prime, this means that we have expressed(a + b + c)Np(aζ2

p + bζp + c) in terms of a recurrence sequence.The method of linear forms in logarithms can be used to investigate

the perfect powers that occur in a linear recurrence sequence. As acorollary, we will see that for fixed non-zero integers a and b, there areonly finitely many primes p such that bNp(aζ2

p + bζp − a) = zp hasa solution. In the second part of this chapter we generalise Theorem3.1 to hold for arbitrary cyclotomic elements. This does not make theproblem of finding solutions to Np(α) = zp easier, but it is interestingin its own right. The main result is given by Theorem 3.4, and we willsee some explicit examples for low variable cases.

3.1 norms in terms of a recurrence sequence

We start with the simplest example of a norm that can be expressed interms of recurrence sequences.

Theorem 3.1. For each integer n ≥ 3, and all a, b, c ∈ C with a 6= 0, wehave:

n−1

∏i=0

(aζ2in + bζ i

n + c) = an + cn − Hn,

where (Hn)n∈N is defined by the recurrence relation

Hn+2 = −bHn+1 − acHn

and the initial values H0 = 2 and H1 = −b.

Proof. Let f (X) = aX2 + bX + c = a(X − ω1)(X − ω2), for someω1, ω2 ∈ C. So we have ω1ω2 = c

a and ω1 + ω2 = − ba . Since Xn − 1 =

∏n−1i=0 (X− ζ i

n), we find thatn−1

∏i=0

(aζ2in + bζ i

n + c) = ann−1

∏i=0

(ζ in −ω1)(ζ

in −ω2)

= an(ωn1 − 1)(ωn

2 − 1)

= an + cn − (aω1)n − (aω2)

n.

15

16 recurrence sequences

Note that aω1 and aω2 are roots of the polynomial (X− aω1)(X−aω2) = X2 + bX + ac. Hence, if we define Hn := (aω1)

n + (aω2)n,then from the theory of linear recurrence relations, we find that Hn

satisfies Hn+2 = −bHn+1 − acHn, and has initial values H0 = 2 andH1 = −b.

We give another proof of Theorem 3.1 using determinants of ma-trices. It is a longer proof, but the recurrence relation appears verynaturally when computing the determinants.

Proof. By general properties of circulant matrices (see for example [7],page 72) we have:

n−1

∏i=0

(a0 + a1ζ in + a2ζ2i

n + · · ·+ an−1ζ(n−1)in ) =

∣∣∣∣∣∣∣∣∣∣∣∣∣

a0 a1 · · · an−2 an−1

an−1 a0 · · · an−3 an−2...

.... . .

......

a2 a3 · · · a0 a1

a1 a2 · · · an−1 a0

∣∣∣∣∣∣∣∣∣∣∣∣∣n

Here |M| is the determinant of the matrix M and we use the subscriptn to remember that it is an n× n matrix. In particular, we find

(−1)n−1n−1

∏i=0

(aζ2in + bζ i

n + c) =n−1

∏i=0

(aζ in + b + cζ−i

n )

=

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

b a 0 · · · 0 c

c b a · · · 0 0

0 c b · · · 0 0...

......

. . ....

...

0 0 0 · · · b a

a 0 0 · · · c b

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣n

(3.1)

We will define:

G(a,b,c)n = Gn :=

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

b a 0 · · · 0 0

c b a · · · 0 0

0 c b · · · 0 0...

......

. . ....

...

0 0 0 · · · b a

0 0 0 · · · c b

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣n

3.1 norms in terms of a recurrence sequence 17

This is the same matrix as in (3.1), but with 0 in the bottom left andupper right corners. Then we see that, for n ≥ 2, the determinant in(3.1) is equal to:

bGn−1 − a

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

c a 0 · · · 0 0

0 b a · · · 0 0

0 c b · · · 0 0...

......

. . ....

...

0 0 0 · · · b a

a 0 0 · · · c b

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣n−1

+ (−1)n+1c

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

c b a · · · 0 0

0 c b · · · 0 0

0 0 c · · · 0 0...

......

. . ....

...

0 0 0 · · · c b

a 0 0 · · · 0 c

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣n−1

=bGn−1 − acGn−2 + (−1)n+1a2

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

a 0 0 · · · 0 0

b a 0 · · · 0 0

c b a · · · 0 0...

......

. . ....

...

0 0 0 · · · a 0

0 0 0 · · · b a

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣n−2

+(−1)n+1c2

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

c b a · · · 0 0

0 c b · · · 0 0

0 0 c · · · 0 0...

......

. . ....

...

0 0 0 · · · c b

0 0 0 · · · 0 c

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣n−2

− acGn−2

=bGn−1 − 2acGn−2 + (−1)n+1(an + cn)

The numbers Gn satisfy a recurrence relation. Indeed, for n ≥ 2, wehave


Gn =

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

b a 0 · · · 0 0

c b a · · · 0 0

0 c b · · · 0 0...

......

. . ....

...

0 0 0 · · · b a

0 0 0 · · · c b

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣n

= bGn−1 − a

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

c a 0 · · · 0 0

0 b a · · · 0 0

0 c b · · · 0 0...

......

. . ....

...

0 0 0 · · · b a

0 0 0 · · · c b

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣n

= bGn−1 − acGn−2

Together with the initial values G0 = 1 and G1 = b, these determinethe numbers Gn. Hence, we find

n−1

∏i=0

(aζ2in + bζ i

n + c) = an + cn + (−1)n+1(bGn−1 − 2acGn−2).

Now define Hn := (−1)n(bGn−1 − 2acGn−2). It is easy to see that thenumbers Hn satisfy the recurrence relation Hn+2 = −bHn+1 − acHn.Using the initial values of the sequence (Gn)n∈N, we see:

H3 = −bG2 + 2acG1 = −b3 + 3abc

H2 = bG1 − 2acG0 = b2 − 2ac

Hence we find:

H1 = −H3 + bH2

ac= −b

H0 = −H2 + bH1

ac= 2

Finally we conclude that

n−1

∏i=0

(aζ2in + bζ i

n + c) = an + cn − Hn.

The determinants Gn from the previous proof are a specific instanceof the following, more general proposition, which also follows fromthe results in [16].

3.1 norms in terms of a recurrence sequence 19

Proposition 3.2. Let a, bi ∈ C for i ≥ 0. Set A0 := 1, and for n ≥ 1,define:

An :=

∣∣∣∣∣∣∣∣∣∣∣∣∣∣

b0 b1 b2 · · · bn−1

a b0 b1...

0 a b0...

.... . . b1

0 · · · 0 a b0

∣∣∣∣∣∣∣∣∣∣∣∣∣∣For any ci ∈ C, i ≥ 0, we have the equality:∣∣∣∣∣∣∣∣∣∣∣∣∣∣

c0 c1 c2 · · · cn−1

a b0 b1 · · · bn−2

0 a b0...

.... . . b1

0 · · · 0 a b0

∣∣∣∣∣∣∣∣∣∣∣∣∣∣= c0An−1− ac1An−2 + · · ·+(−1)nancn−1A0.

In particular, if bi = 0 for i ≥ k, then the matrices An satisfy the recurrencerelation

An = b0An−1 − ab1An−2 + · · ·+ (−1)k−1ak−1bk−1An−k.

Proof. We use induction on n. It is obvious for n = 0. Now assumethat it holds for matrices of size (n− 1)× (n− 1). Then we see∣∣∣∣∣∣∣∣∣∣∣∣∣∣

c0 c1 c2 · · · cn−1

a b0 b1 · · · bn−2

0 a b0...

.... . . b1

0 · · · 0 a b0

∣∣∣∣∣∣∣∣∣∣∣∣∣∣= c0An−1 − a

∣∣∣∣∣∣∣∣∣∣∣∣∣∣

c1 c2 c3 · · · cn−1

a b0 b1 · · · bn−3

0 a b0...

.... . . b1

0 · · · 0 a b0

∣∣∣∣∣∣∣∣∣∣∣∣∣∣= c0An−1−a(c1An−2 − · · ·+ (−1)n−1an−1cn−1A0)

= c0An−1−ac1An−2 + · · ·+ (−1)nancn−1A0

The last statement of the proposition is obtained by putting ci = bi forall i ≥ 0.

Note that the recurrence relation of the determinants Gn followsfrom this proposition by putting c0 = b0, c1 = b1 and ci = bi = 0 fori ≥ 2.

Let Gn be a non-degenerate (the coefficients of the recurrence rela-tion are non-zero) recurrence sequence of order 2. Shorey and Stewart[31], and independently Pethö [26], proved that there are effectivelycomputable constants C1, C2, depending only on the coefficients andstarting values of the sequence Gn, such that if n, y, q are integers withq > 1 satisfying

Gn = yq,


then max(|y|, n, q) < C1 if |y| > 1 and n < C2 if |y| ≤ 1. This result isobtained by using the method of linear forms in logarithms. Generally,the constants are very large. For example, if Gn is the Fibonacci se-quence, then one finds that q < 5.1× 1017. See [27, 28] for a survey byPethö on diophantine equations involving linear recurrence sequences.As an immediate consequence of this, we find:

Theorem 3.3. Let a, b be non-zero integers. There is an effectively com-putable constant C, depending only on a and b, such that if p is a prime andz an integer such that

bNp(aζ2p + bζp − a) = zp,

then p < C.

Proof. If p is an odd prime, then

p−1

∏i=0

(aζ2ip + bζ i

p + c) = (a + b + c)Np(aζ2p + bζp + c).

So if a = −c, then Theorem 3.1 shows that we have

bNp(aζ2p + bζp − a) = −Hn.

Here Hn is the second order linear recurrence sequence given by theinitial values H0 = 2, H1 = −b and the recurrence relation

Hn+2 = −bHn+1 + a2Hn.

So, taking C = max(C1, C2), we conclude that if bNp(aζ2p + bζp − a) =

zp, then we must have p < C.

For fixed a and b, if we want to solve the equation bNp(aζ2p + bζp −

a) = zp, this theorem tells us we only need to check a finite numberof primes p. However, in practice the constant C is too large to checkall primes below C and more clever methods are necessary. In thebreakthrough article [4], Bugeaud, Mignotte and Siksek show thatthe perfect powers in the sequence Fn of Fibonacci numbers are givenby F0 = 0, F1 = F2 = 1, F6 = 8 and F12 = 144. Furthermore, theyshowed that the only perfect powers in the sequence (Ln)n∈N ofLucas numbers are given by L1 = 1 and L3 = 4. They use a sharperversion of the linear forms in logarithms method together with themodular method, which originates in Wiles’ work on Fermat’s lasttheorem. These two methods combined reduce the bounds on q andn far enough for the problem to be resolved computationally. Notethat, because Np(ζ2

p − ζp − 1) = Lp by Theorem 3.1, we find as animmediate corollary that there are no primes p such that

Np(ζ2p − ζp − 1) = zp.

It is interesting that the modular method that is used in the resolutionof the Fermat equation (x + y)Np(xζp + y) = zp, and which is notcyclotomic in nature, can also be used to solve the equation Np(ζ2

p −ζp − 1) = zp.

3.2 a general recurrence theorem 21

3.2 a general recurrence theorem

Now we will show that Theorem 3.1 does not just hold for cyclotomicintegers of the form aζ2 + bζ + c, but can in fact be generalized to anycyclotomic integer. For j = 1, . . . , k, let ej(X1, . . . , Xk) be the elementarysymmetric polynomial of degree j in k variables. So e1(X1, . . . , Xk) =

X1 + X2 + . . . + Xk, e2(X1, . . . , Xk) = ∑1≤r<s≤k XrXs, and in general,

for 1 ≤ j ≤ k, we have ej(X1, . . . , Xk) = ∑1≤i1<i2<···<ij≤k

(∏

jr=1 Xir

).

We will set e0(X1, . . . , Xk) = 1.

Theorem 3.4. Let f (X) = akXk + ak−1Xk−1 + · · · + a1X + a0 ∈ C[X]

with ak 6= 0 and let ω1, . . . , ωk be its roots. For j = 0, . . . , k and n ∈N wedefine

A(j)n := (−1)(n+1)k−jan

k ej(ωn1 , . . . , ωn

k ).

For each n ≥ 3 we have

n−1

∏i=0

f (ζ in) =

k

∑j=0

A(j)n .

For each j, the sequence (A(j)n )n∈N is given by a recurrence relation

A(j)n+(k

j)=

(kj)−1

∑s=0

g(j)s A(j)

n+s

with g(j)0 , . . . , g(j)

(kj)−1∈ Z[a0, . . . , ak] and initial values A(j)

n ∈ Z[a0, . . . , ak].

Proof. Let f (X) = ak(X−ω1)(X−ω2) · · · (X−ωk) for certain ω1, . . . , ωk ∈C. Then we have:

ak−1

ak= −e1(ω1, . . . , ωk)

ak−2

ak= e2(ω1, . . . , ωk)

...a0

ak= (−1)kek(ω1, . . . , ωk)

Moreover we see:

n−1

∏i=0

f (ζ in) = an

k

n−1

∏i=0

(ζ in −ω1) · · · (ζ i

n −ωk)

= (−1)nkank (ω

n1 − 1) · · · (ωn

k − 1)

= (−1)nkank

(k

∑j=0

(−1)k−jej(ωn1 , . . . , ωn

k )

)

=k

∑j=0

(−1)(n+1)k−jank ej(ω

n1 , . . . , ωn

k )


For each j, we will show that(ej(ω

n1 , . . . , ωn

k ))

n∈Nand thus also(

A(j)n

)n∈N

satisfies a recurrence sequence as in the statement of thetheorem. In general, if

t

∑i=0

ciXi = ct

t

∏j=1

(X− γj)

then from the theory of linear recurrence sequences we know thatthe numbers Cn := ∑t

j=1 γnj satisfy the recurrence relation ctCn+t =

−∑t−1i=0 ciCn+i. In our case, we would like the numbers Cn to be equal

to ej(ωn1 , . . . , ωn

k ). To achieve this, we need to set the γj equal tomr(ω1, . . . , ωk), where mr(X1, . . . , Xk), r = 1, . . . , (k

j) are the mono-mials of the polynomials ej(X1, . . . , Xk). Therefore we consider thepolynomial

E(X) =

(kj)

∏r=1

(X−mr(ω1, . . . , ωk)).

Let hs ∈ C be such that

E(X) =

(kj)

∑s=0

hsXs.

Then each hs is given by a symmetric polynomial in ω1, . . . , ωk. By thefundamental theorem of symmetric polynomials, this means that

hs = Ps (e1(ω1, . . . , ωk), . . . , ek(ω1, . . . , ωk))

for some polynomial Ps ∈ Z[Y1, . . . , Yk].Setting Cn := ej(ω

n1 , . . . , ωn

k ), we see that

Cn =

(kj)

∑r=1

mr(ωn1 , . . . , ωn

k ),

and therefore we conclude that the numbers (Cn)n∈N satisfy the recur-rence relation

Cn+(kj)= −

(kj)−1

∑s=0

hsCn+s

Hence, if An := (−ak)nej(ω

n1 , . . . , ωn

k ) and gs = (−a)(kj)−shs, then we

get

An+(kj)= −

(kj)−1

∑s=0

gs An+s.

To finish the proof, we need to show that the coefficients gs and thevalues An are in fact elements of Z[a0, . . . , ak], for all s, n = 0, . . . , (k

j)−1. Note that they are all elements of Z[ a0

ak, . . . , ak−1

ak, ak], and that the


power of ak appearing in the denominator of hs is equal to the degreeof the polynomial Ps. Hence if deg P

(kj)−s ≤ s, then gs ∈ Z[a0, . . . , ak].

Similarly, if

ej(ωn1 , . . . , ωn

k ) = Q(e1(ω1, . . . , ωk), . . . , ek(ω1, . . . , ωk))

then we see that the power of ak appearing in the denominator ofej(ω

n1 , . . . , ωn

k ) is equal to the degree of Q. So if deg Q ≤ n, thenAn ∈ Z[a0, . . . , ak]. Now we use the following lemma.

Lemma 3.5. Let h(X1, . . . , Xk) be a symmetric polynomial in k variables,which has degree n as a polynomial in X1. Then there is some polynomialq ∈ Z[Y1, . . . , Yk] of degree n (as a polynomial in k variables) such that

h(X1, . . . , Xk) = q(e1(X1, . . . , Xk), . . . , ek(X1, . . . , Xk)).

Proof of the Lemma. The proof is just the usual proof of the funda-mental theorem of symmetric polynomials, but keeping track of thedegree of the polynomial q we end up with. For the usual proof, seefor example [6], page 178. We use induction on the leading term ofh in lexicographic ordering. Suppose the leading term is given byh0Xl1

1 Xl22 · · ·X

lkk for some h0 ∈ C. By assumption we have l1 = n. Be-

cause we use lexicographic ordering, and because h is symmetric, wehave l1 ≥ l2 ≥ . . . ≥ lk. We define

h(Y1, . . . , Yk) = h0Yl1−l21 Yl2−l3

2 · · ·Ylk−1−lkk−1 Ylk

k .

Then we see deg h = l1 = n. Moreover, we see that

h(X1, . . . , Xk)− h(e1(X1, . . . , Xk), . . . , ek(X1, . . . , Xk))

is a symmetric polynomial whose leading term is smaller in lexico-graphic ordering. By induction we see that h = q(e1, . . . , ek), for somepolynomial q ∈ Z[Y1, . . . , Yk] of degree n.

Because e1(ωn1 , . . . , ωn

k ) has degree n as a polynomial in ω1, thelemma tells us that Q has degree n as a polynomial in k variables. Fur-thermore, because in the definition of E(X) the monomials mr(ω1, . . . , ωk)

have degree at most 1 as a polynomial in ω1, we find that h(k

j)−s(ω1, . . . , ωk)

has degree s as a polynomial in ω1. Consequently, the lemma tells usthat Ps has degree s as a polynomial in k variables. So, indeed, the coef-ficients gs and the starting values An are elements of Z[a0, . . . , ak].

We can give some more information on the sequences (A(i)n )n∈N.

They satisfy some symmetry:

Corollary 3.6. Suppose that a0 6= 0 in Theorem 3.4, let f (X) = a0Xk +

a1Xk−1 + · · ·+ ak−1X + ak be the reciprocal polynomial and let (A(j)n )n∈N

be the corresponding recurrence sequences. Then we have the followingsymmetrical property:

A(j)n = (−1)(n+1)k A(k−j)

n


Proof. Note that f (X) = a0(X− 1ω1)(X− 1

ω2) · · · (X− 1

ωk) and thus

A(k−j)n = (−1)nk+jan

0 ej−k

(1

ωn1

, . . . ,1

ωnk

).

Because we have ω1ω2 · · ·ωk = (−1)k a0ak

we find:

ej(ωn1 , . . . , ωn

k ) = ∑1≤i1<...<ij≤k

ωni1 · · ·ω

nij

= ∑1≤i1<...<ik−j≤k

ωn1 · · ·ωn

kωn

i1· · ·ωn

ik−j

= (−1)nk an0

ank

ej−k

(1

ωn1

, . . . ,1

ωnk

)So we conclude that

A(j)n = (−1)(n+1)k−jan

k ej(ωn1 , . . . , ωn

k )

= (−1)k−jan0 ek−j

(1

ωn1

, . . . ,1

ωnk

)= (−1)(n+1)k A(k−j)

n .

Furthermore, we can give (A(i)n )n∈N explicitly for small i.

Corollary 3.7. In Theorem 3.4 we have A(0)n = (−1)(n+1)kan

k , A(k)n = an

0and for j = 1 and j = k− 1 we have the recurrences:

A(1)n+k = −

k−1

∑s=0

(−1)k(k−s)ak−1−sk as A(1)

n+s

A(k−1)n+k = −

k−1

∑s=0

ak−1−s0 ak−s A(k−1)

n+s

Proof. From the proof of Theorem 3.4 we see that, by definition, A(0)n =

(−1)(n+1)kank . For the case j = 1 we see that

E1(X) = E(X) =k

∏r=1

(X−ωr) = Xk +k−1

∑s=0

as

akXs.

So A(1)n = (−1)(n+1)k−1an

k e1(ωn1 , . . . , ωn

k ) satisfies the recurrence

A(1)n+k = −

k−1

∑s=0

(−1)k(k−s)ak−1−sk as A(1)

n+s.

The cases j = k− 1 and j = k now follow by the symmetry propertyof the previous corollary.


We will finish this chapter with a number of examples in the casewhere we have few variables. As a first example, we again look atf (X) = aX2 + bX + c. Theorem 3.4 tells us that

n−1

∏i=0

f (ζ in) = an + A(1)

n + cn,

where A(1)0 = −2, A(1)

1 = −ae1(ω1, ω2) = b, and where A(1)n satisfies

the recurrenceA(1)

n+2 = −bA(1)n+1 − acA(1)

n .

So, as expected, we obtain Theorem 3.1 again.In the case of f (X) = aX3 + bX2 + cX + d, we find that

n−1

∏i=0

f (ζ in) = (−1)n+1an + A(1)

n + A(2)n + dn.

Here (A(1)n )n∈N is given by A(1)

0 = 3, A(1)1 = b, A(1)

2 = b2 − ac and therecurrence

A(1)n+3 = bA(1)

n+2 − acA(1)n+1 + a2dA(1)

n .

The sequence (A(2)n )n∈N is given by A(2)

0 = −3, A(2)1 = c, A(2)

2 =

bd− c2 and the recurrence

A(2)n+3 = −cA(2)

n+2 − bdA(2)n+1 − ad2A(2)

n .

Finally, we shall look at the example f (X) = aX4 + bX3 + cX2 +

dX + e. Theorem 3.4 shows that we have

n−1

∏i=0

f (ζ in) = an + A(1)

n + A(2)n + A(3)

n + en.

For A(1)n and A(3)

n we can use Corollary 3.7 to get the recurrencerelation. We see that the sequence (A(1)

n )n∈N is given by the recurrencerelation

A(1)n+4 = −bA(1)

n+3 − acA(1)n+2 − a2dA(1)

n+1 − a3eA(1)n .

Let ω1, ω2, ω3, ω4 be the roots of f (X), and let us set ej(ωi) := ej(ω

i1, ωi

2, ωi3, ωi

4)

and ej := ej(ω). With this notation, we have A(j)n = (−1)janej(ω

n), and

thus the the initial values of A(1)n are given by:

A(1)0 = −e1(ω

0) = −4

A(1)1 = −ae1(ω) = b

A(1)2 = −a2e1(ω

2) = −a2(e21 − 2e2) = −b2 + 2ac

A(1)3 = −a3e1(ω

3) = −a3(e31 − 3e1e2 + 3e3) = b3 − 3abc + 3a2d


By symmetry, we find that the sequence (A(3)n )n∈N is given by the

recurrence

A(3)n+4 = −dA(3)

n+3 − ceA(3)n+2 − be2A(3)

n+1 − ae3A(3)n .

The initial values are given by:

A(3)0 = −4

A(3)1 = d

A(3)2 = −d2 + 2ce

A(3)3 = d3 − 3cde + 3be2

To find the recurrence relation that the sequence (A(2)n )n∈N satisfies,

we need to calculate the polynomial E(x) as in the proof of Theorem3.4:

E(x) = (x−ω1ω2)(x−ω1ω3)(x−ω1ω4)(x−ω2ω3)(x−ω2ω4)(x−ω3ω4)

= x6 − e2x5 + (e1e3 − e4)x4 + (2e2e4 − e21e4 − e2

3)x3

+ (e1e3e4 − e24)x2 − e2e2

4x + e34

= x6 − ca

x4 +bd− ae

a2 x4 +2ace− ad2 − b2e

a3 x3

+bde− ae2

a3 x2 − ce2

a3 x +e3

a3

As a result, we find that the recurrence relation is given by:

A(2)n+6 = cA(2)

n+5 + (ae− bd)A(2)n+4 + (ad2 + b2e− 2ace)A(2)

n+3

+ (a2e2 − abde)A(2)n+2 + a2ce2A(2)

n+1 − a3e3A(2)n

Finally, for the initial values A(2)n = ane2(ωn) for n = 0, . . . , 5, we

find:

A(2)0 = 6

A(2)1 = c

A(2)2 = c2 + 2ae− 2bd

A(2)3 = c3 − 3ace + 3ad2 + 3b2e− 3bcd

A(2)4 = c4 + 6a2e2 − 8abde− 4ac2e−+4ad2c + 4b2ce + 2b2d2 − 4bc2d

A(2)5 = c5 + 5a2ce2 + 5a2d2e + 5ab2e2 − 5abcde− 5ac3e− 5abd3

− 5b3de + 5ac2d2 + 5b2c2e + 5b2cd2 − 5bc3d

4F R O M N O R M S T O S I N G U L A R I N T E G E R S

4.1 singular integers

In the classical approach to Fermat’s last theorem the following lemmais used.

Lemma 4.1. Let x, y be coprime integers such that p does not divide x + y.Then x + yζ is coprime to its Galois conjugates.

Proof. Suppose q is a prime ideal of OK that divides both x + yζ andx + yζ i for some i 6≡ 0, 1 mod p. Then q also divides

x + yζ − (x + yζ i) = y(ζ − ζ i).

Since p does not divide x + y, we know that q 6= (1− ζ). Hence q

divides y. Similarly we see that q has to divide

ζ−1x + y− (ζ−ix + y) = x(ζ−1 − ζ−i)

and therefore q also divides x. But this contradicts the fact that x and yare coprime, so we conclude that x + yζ and x + yζ i are coprime.

This lemma allows us to conclude that if p does not divide x + y andN(x + yζ) = zp, then (x + yζ) = Ip for some ideal I of OK. Followingthe terminology of [29], chapter 9, we define:

Definition 4.2. An element α ∈ OK is called a singular integer if (α) = Ip

for some ideal I of OK.

Now suppose we have an element α ∈ OK satisfying N(α) = zp.The main question of this chapter is: Are there conditions that allowus to conclude that α is a singular integer? For a rational prime q, wedenote the primes of OK dividing q by qi for i = 1, . . . , g(q), whereg(q) f (q) = p− 1 and f (q) is the relative degree of q. With this notation,the factorisation of α into prime ideals looks like

(α) = ∏q prime

g(q)

∏i=1

qordqi (α)

i .

We see that N(α) = zp if and only if

g(q)

∑i=1

ordqi(α) ≡ 0 mod p, for all rational primes q. (4.1)

On the other hand we see that α is a singular integer if and only if

ordqi(α) ≡ 0 mod p, for all primes qi of OK. (4.2)

Given α = ap−2ζ p−2 + · · · + a1ζ + a0 we set α(X) := ap−2Xp−2 +

· · ·+ a1X + a0.

27

28 from norms to singular integers

Lemma 4.3. If N(α) = zp, and for each rational prime q the polynomialsα(X) and Φ(X) share at most one irreducible factor modulo q, then α issingular.

Proof. We want to show that ordq(α) ≡ 0 mod p for all prime idealsq of OK. Since (1− ζ) is the only prime above p, condition (4.2) isimmediate if q = (1− ζ). Now let q 6= p be a rational prime and letthe factorisation into irreducibles of Φ(X) modulo q be given by

Φ(X) ≡g(q)

∏i=1

hi(X) mod q.

Then the prime ideals above q are given by qi := (q, hi(ζ)). If qi dividesα, then there are polynomials r, s ∈ Z[X] such that

α = r(ζ) · q + s(ζ) · hi(ζ).

So there is some polynomial t(X) ∈ Z[X] such that

α(X) = r(X) · q + s(X) · hi(X) + t(X) ·Φ(X).

Because hi(X) is a factor of Φ(X) modulo q, we see that hi(X) isalso a factor of α(X) modulo q. So each qi dividing α gives a distinctirreducible common factor of α(X) and Φ(X). By assumption thereis at most one such common factor for each prime q, so there is atmost one prime ideal qi dividing α. Since N(α) = zp, we find that α issingular.

For example, if α = aζ + b, then α(X) has degree 1 and thus α(X)

and Φ(X) can have at most one common irreducible factor per prime q.If α = aζ2 + bζ + c then α(X) has degree 2. So if q 6≡ 0, 1 mod p, thenq does not split completely in Q(ζp) and thus f (q) = deg hi(X) ≥ 2.Therefore there can be at most one common factor of α(X) and Φ(X)

modulo q. So only at the primes which are congruent to 1 modulo pis it possible that the condition of Lemma 4.3 is not satisfied.

Of course, the above does not give a clear-cut way to say wether agiven element α with N(α) = zp is singular or not. For a specific primep and element α it is possible to check it, but we do not have a generalcondition in terms of the coefficients of α like we did in Lemma 4.1.

4.2 self-prime integers

In the case of α = x + yζ, we showed that α is singular by showing thatα is coprime to its Galois conjugates. For a general element α ∈ OK

this is a sufficient condition in order to get (4.1) =⇒ (4.2). That’swhy we shall look a bit closer that this property. The goal is to findreasonably simple conditions that tell us when an element is coprimeto its conjugates. Let us start with the basic definition.

4.2 self-prime integers 29

Definition 4.4. An element α ∈ OK is called self-prime if α is coprime toσi(α) for all i = 2, . . . , p− 1.

Being self-prime is a property that we can check locally, at all theprime ideals of OK. We give some general properties of self-primecyclotomic integers. The following lemma shows that the self-primecondition puts pretty big restrictions on α.

Lemma 4.5. If α ∈ OK is selfprime then all prime divisors of N(α) arecongruent to 1 modulo p.

Proof. Suppose q divides N(α) and q is not congruent to 1 modulop. Then q does not split completely in K/Q. Hence for any prime q

above q there is some σ ∈ Gal(K/Q) such that σ(q) = q. But then if qdivides α, it also divides σ(α).

If some α is self-prime, then we automatically know some otherself-prime cyclotomic integers.

Lemma 4.6. Let u be a unit in OK and let σ be an element of Gal(K/Q). Ifα is self-prime, then also uα and σ(α) are self-prime.

Proof. This is immediate from the definition.

Lemma 4.7. Suppose Q ⊆ L ( K. If α is an element of L then α is notself-prime.

Proof. Since L is a proper subfield of K, there is some non-trivial σ ∈Gal(K/Q) that fixes K. But then σ(α) = α so α is not self-prime.

We will proceed by giving a characterisation of self-prime integersusing the characteristic polynomial. Denote by χα(X) the charateristicpolynomial of α. So we have

χα(X) = ∏σ∈Gal(K/Q)

(X− σ(α))

= Xp−1 + gp−2(α)Xp−2 + · · ·+ g1(α)X + g0(α).

The gi(α) are the elementary symmetric polynomials in σi(α), i =

1, . . . , p− 1. They are invariant under the action of the Galois groupand thus they are integers. We have

gp−2(α) = −p−1

∑i=1

σi(α) = −Tr(α)

g0(α) = (−1)p−1p−1

∏i=1

σi(α) = N(α)

g1(α) = −p−1

∑i=1

∏p−1j=1 σj(α)

σi(α)= −Tr(α−1)N(α)

The following proposition gives a characterisation of self-primeintegers.


Proposition 4.8. An element α ∈ OK is self-prime if and only if g0(α) =

N(α) and g1(α) are coprime.

Proof. Suppose q is a prime lying above some prime integer q, whichdivides both α and σi(α) for some i 6= 1. Then obviously q dividesN(α). Since g1(α) is the sum of the products of p− 2 different σi(α),each term in this sum is divisible by q. Hence q also divides g1(α).

Conversely suppose that q divides both g0(α) and g1(α). Since itdivides g0(α), there is some q above q which divides α. Since q dividesg1(α) and α, it also divides ∏

p−1j=2 σj(α). Consequently q divides σj(α)

for some j 6= 1.

We only need only p− 1 powers of ζ to get a basis of K over Q. Forexample, we can take 1, . . . , ζ p−2 as a basis and express α ∈ OK interms of this basis as α = a0 + a1ζ + · · ·+ αp−2ζ p−2 for some integersai. The linear map x 7→ αx then corresponds to the matrix

Mα =

a0 a1 a2 . . . ap−2

−ap−2 a0 − ap−2 a1 − ap−2 . . . ap−3 − ap−2

ap−2 − ap−3 −ap−3 a0 − ap−3 . . . ap−4 − ap−3...

......

...

a2 − a1 a3 − a1 a4 − a1 . . . a0 − a1

.

Then we have the equality

det(XI −Mα) = χα(X).

However, by choosing p − 1 elements out of 1, ζ, . . . , ζ p−1 to forma basis, we lost some of the symmetry which these elements have.Instead we could consider the matrix

M′α =

a0 a1 a2 . . . ap−2 0

0 a0 a1 . . . ap−3 ap−2

ap−2 0 a0 . . . ap−4 ap−3...

......

......

a2 a3 a4 . . . a0 a1

a1 a2 a3 . . . 0 a0

.

In a way, this matrix corresponds to the presentation of α as a0 +

a1ζ + · · ·+ ap−2ζ p−2 + 0 · ζ p−1. Denote α(1) = a0 + a1 + . . . + ap−2. Byadding the top p− 1 rows to the bottom row, we get only α(1)’s onthe bottom row. If we then substract the last column from the firstp− 1 columns, we find that det M′α = α(1)det Mα. Similarly we seethat we have

det(XI −M′α) = (X− α(1))det(XIp−1 −Mα) = (X− α(1))χα(X).


In particular, if we write det(XI −M′α) = Xp + hp−1(α)Xp−1 + · · ·+h1(α)X + h0(α), then we have:

hp−1(α) = gp−2(α)− α(1) = −(Tr(α) + α(1))

h0(α) = −α(1)g0(α) = −α(1)N(α)

h1(α) = −α(1)g1(α) + g0(α) = (1− α(1)Tr(α−1))N(α)

It makes sense to denote α(1) = a0 + a1 + · · ·+ ap−2 when we thinkof it as α = f (ζ) for some polynomial f ∈ Z[X]. However, when weconsider the Galois group, perhaps it makes more sense to think ofit as a map σ0 : OK → Z given by σ0(α) = a0 + a1 + · · ·+ ap−2. Withthis notation we have

hp−1(α) = −p−1

∑i=0

σi(α)

h0(α) = −p−1

∏i=0

σi(α)

h1(α) =p−1

∑i=0

∏p−1j=0 σj(α)

σi(α)

We can use the hi(α) instead of the gi(α) in Proposition 4.8.

Proposition 4.9. Let α ∈ OK. Then α is self-prime if and only if h0(α) andh1(α) are coprime.

Proof. Suppose α is not self-prime. By Proposition 4.8 there is someprime q dividing both g0(α) and g1(α). Then, using the expressionsabove, we see that q divides h0(α) and h1(α).

Conversely, suppose that q divides h0(α) and h1(α). From the firstwe see that q divides α(1) or N(α). But if q divides α(1) then thesecond gives us that q divides N(α). So we know that q divides N(α)

and α(1)g1(α).If q divides N(α) and g1(α) we are done, by Proposition 4.8. Hence

suppose that q divides N(α) and α(1). Then there is some prime idealq lying above q that divides both α and α(1). Write α = a0 + a1ζ +

· · ·+ ap−2ζ p−2. Then we get

q | α(1)− α = (1− ζ)(a1 + a2 + · · ·+ ap−2).

So either q = p or q divides a1 + a2 + · · · + ap−2. If q = p then weknow that α is not self-prime. If q divides α(1) and a1 + · · · ap−2 thenq divides a0. Similarly, using α(1)− ζ−iα instead of α(1)− α, we canshow that q divides ai for any i. But if q divides all ai then it willdivide σ(α) for any σ, so α is not self-prime.

In general the hi(α) are a bit nicer to work with, because they aremore symmetrical. We can use this to give another proof of Lemma4.1.


Lemma 4.10. Let α = a + bζ. Then α is self-prime if and only if a and bare coprime and p does not divide a + b.

Proof. If a and b have a non-trivial common divisor, then obviouslyα is not self-prime. Moreover, if p divides a + b, then we know byLemma 4.5 that α is not self-prime. Hence assume that a and b arecoprime and p does not divide a + b. We have

det(XI−M′α) =

∣∣∣∣∣∣∣∣∣∣∣∣∣

X− a −b 0 . . . 0

0 X− a −b . . . 0

0 0 X− a . . . 0...

......

. . ....

−b 0 0 . . . X− a

∣∣∣∣∣∣∣∣∣∣∣∣∣= (X− a)p− bp.

This means that h0(α) = −(ap + bp) and h1(α) = pap−1. So if qdivides h1(α) then q = p or q divides a. If q divides a and h0(α) thenalso q divides b, which contradicts the coprimality of a and b. If q = pthen 0 ≡ −h0(α) ≡ ap + bp ≡ a + b mod p, which does not hold byassumption.

In the case of three variables we get the following.

Theorem 4.11. Let α = a + bζ + cζ2. Then α is self-prime if and only if(a + b + c)N(α) and ap − cp are coprime and p does not divide a + b + c.

Proof. By Lemma 4.9 we know that α is self-prime if and only if h0(α)

and h1(α) are coprime. Note that we have h0(α) = −∏p−1i=0 (aζ−i + b +

cζ i) and:

h1(α) =p−1

∑j=0

p−1

∏i=0i 6=j

(aζ−i + b + cζ i)

= − ∂

∂b

p−1

∏i=0

(aζ−i + b + cζ i)

= − ∂

∂bh0(α)

So if a rational prime q divides h0(α) and h1(α), then it must alsodivide the resultant of h0(α) and d

db h0(α) when considered as a polyno-mial in b. But this resultant is precisely the discriminant discb(h0(α))


of h0(α) with respect the variable b. Since h0(α) = ∏p−1i=0 (b− (−aζ−i +

cζ i)) we find

discb(h0(α)) = (−1)p(p−1)

2

p−1

∏i,j=0i 6=j

(aζ−i + cζ i − aζ−j − cζ j)

= (−1)p−1

2

p−1

∏i,j=0i 6=j

(a(ζ−i − ζ−j) + c(ζ i − ζ j))

= (−1)p−1

2

p−1

∏i,j=0i 6=j

ζ−i(1− ζ i−j)(a− cζ i+j)

Note that we have:

p−1

∏i,j=0i 6=j

ζ−i =p−1

∏i=0

p−1

∏j=0j 6=i

ζ−i =p−1

∏i=0

ζ i = 1

p−1

∏i,j=0i 6=j

(1− ζ i−j) =p−1

∏i=0

p−1

∏k=1

(1− ζk) =p−1

∏i=0

p = pp

p−1

∏i,j=0i 6=j

(a− cζ i+j) =p−1

∏i=1

p−1

∏k=0

(a− cζk) =p−1

∏i=1

(ap − cp) = (ap − cp)p−1

We conclude that

discb(h0(α)) = (−1)p−1

2 pp(ap − cp)p−1.

So h0(α) and h1(α) are coprime if and only h0(α) and discb(h0(α))

are coprime, which happens if and only if h0(α) and p(ap − cp) arecoprime. We know that p divides h0(α) = −(a + b + c)N(α) if andonly p divides a + b + c, so we end up with the statement of thetheorem.

As an example, let p = 23 and consider the elements ζ223 + ζ23 − 4

and 2ζ223 − ζ23 + 3. We have N(ζ2

23 + ζ23 − 4) = 472 · 15927174689, but1+ 423 is not divisible by 47, so this element is self-prime. It is divisibleby (47, ζ + 20)2 and not by any other prime above 47. On the otherhand N(2ζ2

23 − ζ23 + 3) = 472 · 139 · 75211 and 223 − 323 is divisible by47. So this element is not self-prime. Indeed, it is divisible by both theprime ideals (47, ζ23 + 30) and (47, ζ23 + 40).

For dealing with the Diophantine equation N(a + bζ + cζ2) = zp,this condition is not so useful. We can not say much about how theprimes that divide z depend on the coefficients a, b and c. Ideally wewould like to have a more simple condition to determine whether a +bζ + cζ2 is self-prime, a condition which does not involve calculatingthe norm of the element.

5S I N G U L A R I N T E G E R S

If α ∈ OK is a singular integer and η ∈ Z[G] annihilates the classgroup, then we get an equation

αη = uγp,

for some unit u ∈ O∗K and some γ ∈ OK. In [20], Kummer’s hadthe great idea to study these kind of equations using the logarith-mic derivative. Intuitively this makes sense, because the logarithmicderivative should turn the p-th power γp into something which iscongruent to 0 modulo p, so we can hope to obtain information on α

modulo p. Kummer used this to prove that if the first case of Fermat’slast theorem fails, then p must divide the Bernouilli numbers Bp−3

and Bp−5. His methods have later been generalised, to find many moreconditions modulo p that should be satisfied if the first case fails. Seefor example [29].

It is also possible to obtain information modulo p from the equationαη = uγp by taking the p-adic logarithm. It seems that this has notbeen done before, but it turns out that the information we get is thesame as the information we can obtain with Kummer’s logarithmicderivative. In this chapter we first review Kummer’s method. In thefirst part will see how it leads to congruence conditions that α mustsatisfy modulo p. In the second part of the chapter we use the p-adic logarithm instead. In a very similar way as with the logarithmicderivative, we obtain information on α modulo p. In fact, in Theorem5.9 we give a formula that exactly gives the relation between the valueof logp(α) and the logarithmic derivative of α.

5.1 idempotents and the herbrand-ribet theorem

We start with some algebraic prerequisites we use in this chapter, andthe next. This can be found in chapter 6.3 of [36], but we include ithere to make the presentation more self-contained.

Let G = Gal(K/Q) ∼= (Z/pZ)∗, and let Zp be the ring of p-adicintegers. By Hensel’s lemma, for each a ∈ (Z/pZ)∗, there is a uniquep-th root of unity ρa ∈ Zp satisfying ρa ≡ a mod p. Define ω : G →Zp by ω(σa) = ρa.

Let Zp[G] be the group ring of G over Zp. We define the orthogonalidempotents of Zp[G] to be the elements

ε i :=1

p− 1

p−1

∑a=1

ωi(σa)σ−1a

35

36 singular integers

for 0 ≤ i ≤ p− 2. A quick check shows that they satisfy:

• ε2i = ε i

• ε iε j = 0 if i 6= j

• 1 = ∑p−2i=0 ε i

• σaε i = ωi(σa)ε i

If M is any Zp[G]-module, then from these properties it follows thatwe have

M =p−2⊕i=0

ε i M.

Moreover, using the last property, we see that

ε i M ={

x ∈ M | σa(x) = ωi(σa)x}

.

As an example, let us look at C = CK, the class group of the field K.The Galois group G acts on C in the obvious manner. We will oftendenote this action multiplicatively, so cσ instead of σ(c), for c ∈ C andσ ∈ G. This makes sense, because C is abelian. Let Cp be the p-sylowsubgroup of C. If c ∈ Cp and ∑∞

i=0 ai pi ∈ Zp, then we define

c∑∞i=0 ai pi

=∞

∏i=0

cai pi

This product is well-defined because Cp is a finite p-group (all itselements have order a power of p). In this way, Cp is a Zp[G]-module.Hence we have

Cp =p−2⊕i=0

Cεip .

If we are only interested in the group C[p] := {c ∈ C | cp = 1}, thenthe action of Zp[G] is completely determined by the first coefficientof the p-adic number. So, we only care about the elements of Zp[G]

modulo p, and we can see it as an action of Fp[G]. In this case, thecorresponding idempotents are ε i = −∑

p−1a=1 aiσ−1

a ∈ Fp[G].Kummer showed that Cp is non-trivial if and only if one of the

Bernoulli numbers B2, B4, . . . , Bp−3 is divisible by p. The Herbrand-Ribet theorem makes this relation, between the class group andBernoulli numbers, more precise.

Theorem 5.1 (Herbrand-Ribet). Let 3 ≤ i ≤ p− 2 be odd. The group Cεip

is non-trivial if and only if p divides Bp−i.

Showing that p divides Bp−i if Cεip is non-trivial is the easier direction,

and was proven by Herbrand [15]. In the next chapter we shall give aproof of this direction. The other direction is more difficult and is dueto Ribet [30] (see also chapter 15 of [36]).

5.2 kummer’s logarithmic derivative 37

5.2 kummer’s logarithmic derivative

Kummer’s method is well explained in [10] and [32], and we followthese expositions. For α = a0 + a1ζ + · · · + ap−2ζ p−2 we define theformal power series α(eX) = a0 + a1eX + · · ·+ ap−2e(p−2)X. For 1 ≤n ≤ p− 2 we define the logarithmic derivative `n : OK \ πOK → Fp

by

`n(α) :=[(

∂

∂X

)n

log α(eX)

]X=0

mod p.

Lemma 5.2. For all 1 ≤ n ≤ p− 2, the maps `n are well-defined.

Proof. Note that(

∂∂X

)nlog α(eX) will be of the form gα(eX)/α(eX)k for

some polynomial gα ∈ Z[X] and some integer k. If π does not divideα, then α(1) is not divisible by p, and therefore we get an element ofFp after evaluating at X = 0. Because the maps are defined on the setOK \ πOK, we indeed get a map to Fp.

We also need to check that the maps `n are independent of the waywe write α. So we need to check that `n(α) = `n(α + f (ζ)Φ(ζ)), forf ∈ Z[X]. We see that

(∂

∂X

)n

log(α(eX) + f (eX)Φ(eX)) ≡(

∂

∂X

)n

log(α(eX))

mod(

Φ(eX),∂

∂XΦ(eX), . . . ,

(∂

∂X

)n

Φ(eX)

).

Now notice that Φ(X) ≡ (X − 1)p−1 mod p, which implies that[(∂

∂X

)iΦ(eX)

]X=0≡ 0 mod p for all 0 ≤ i ≤ p− 2. So, indeed, we

have `n(α) = `n(α + f (ζ)Φ(ζ)), and `n(α) is well-defined as a mapOK \ πOK → Fp.

We continue by showing some basic properties that this logarithmicderivative satisfies.

Lemma 5.3. The maps `n satisfy the following properties:

1. `nσj = jn`n for all 1 ≤ n ≤ p− 2 and 1 ≤ j ≤ p− 1.

2. `n(α1α2) = `n(α1) + `n(α2) for all 1 ≤ n ≤ p − 2 and α1, α2 ∈OK \ πOK.

3. `1(ζj) = j and `n(ζ j) = 0 for j = 0, . . . , p− 1 and 2 ≤ n ≤ p− 2.

4. `n(γ) = 0 for γ ∈ OK ∩R and 1 ≤ n ≤ p− 2 odd.

5. `n(αεi) = `n(α) if n = i, and else `n(αεi) = 0, for 1 ≤ n ≤ p− 2and 1 ≤ i ≤ p− 1.


Proof. For the first one, we prove by induction on n that there is someAn(X) ∈ Q(X) such that(

∂

∂X

)n

log α(ejX) = jn An(ejX).

For n = 1 it is easy to see that we need to take A1(X) = Xα′(X)α(X)

. Nowsuppose the claim holds for n− 1. Then(

∂

∂X

)n

log α(ejX) =∂

∂Xjn−1An−1(ejX) = jnejX A′n−1(e

jX).

So we can take An(X) = XA′n−1(X) and the claim follows by induction.For the proof of the first property, we now note:

`nσj(α) ≡ jn[

An(ejX)]

X=0≡ jn

[An(eX)

]X=0≡ jn`n(α) mod p.

The second property is immediate from the general properties ofthe logarithmic derivative. For the third property, note that `n(ζ j) =[(

∂∂X

)njX]

X=0. The fourth follows from the first, because if γ is real

then`n(γ) = `n(σ−1(γ)) = (−1)n`n(γ).

The final property follows from the following computation:

`n(αεi) ≡ −

p−1

∑j=1

ji`n(σ−1j (α)) ≡ −

(p−1

∑j=1

jn−i

)`n(α) mod p.

Now use the fact that ∑p−1j=1 jn−i ≡ −1 mod p if p− 1 | n− i, and else

it is congruent to 0 modulo p.

Example 5.4. If α = x + yζ with x, y ∈ Z and p - (x + y) then the firstfew values of `n(α) are given by:

n `n(x + yζ)

1 yx+y

2 xy(x+y)2

3 xy(x−y)(x+y)3

4 xy(x2−4xy+y2)(x+y)4

5 xy(x−y)(x2−10xy+y2)(x+y)5

If α = x + yζ + zζ2 with x, y, z ∈ Z and p - x + y + z then the first fewvalues of `n(α) are given by:

n `n(x + yζ + zζ2)

1 y+2zx+y+z

2 xy+4xz+yz(x+y+z)2

3 (x−z)(xy+8xz−y2+yz)(x+y+z)3

4 x3y+16x3z−4x2y2−13x2yz−64x2z2+xy3+8xy2z−13xyz2+16xz3+y3z−4y2z2+yz3

(x+y+z)4

5.2 kummer’s logarithmic derivative 39

The following theorem was proven by Kummer in [20] in the casewhere α is of the from x + ζy. The case of a general singular integerwas dealt with by Granville [9].

Theorem 5.5. Let α ∈ OK \ πOK be a singular integer. For all odd integers3 ≤ n ≤ p− 2 we have

Bp−n`n(α) ≡ 0 mod p.

Proof. Suppose that α ∈ OK and αOK = Ip. Applying εn for odd3 ≤ n ≤ p − 2 gives us αεn = Iεn p. If Iεn is not principal, then bythe Herbrand-Ribet theorem we know that p divides Bp−n. If Iεn isprincipal, then αεn = uγp for some u ∈ O∗K and γ ∈ OK. We knowthat u = ζtv for some integer t and some v ∈ OK ∩R. By using theproperties in Lemma 5.3 of the maps `n we see that:

`n(α) ≡ `n(αεn) ≡ `n(ζ

t) + `n(v) + `n(γp) ≡ 0 mod p.

Kummer used Theorem 5.5 to prove that, if the first case of Fer-mat’s last theorem fails, then p must divide both Bp−3 and Bp−5.We will recall his argument here. Assume xp + yp = zp, with p -xyz, gcd(x, y, z) = 1. Then x + yζ is singular and therefore

Bp−3`3(x + yζ) ≡ Bp−3xy(x− y)(x + y)3 ≡ 0 mod p.

Note that p - z implies that x + y is not divisible by p, so this iswell-defined. We find that if p does not divide Bp−3, then p dividesx− y. Now we use the symmetry of our equation. We reorder it asxp + (−z)p = (−y)p and do the same procedure as before. We findthat p divides x + z. But then

0 = xp + yp − zp ≡ x + y− z ≡ 3x mod p

which gives a contradiction for p > 3.If p does not divide Bp−5 then we have

`5(x + yζ) =xy(x− y)(x2 − 10xy + y2)

(x + y)5 ≡ 0 mod p.

Again, we can use the symmetry of the initial equation to find that

(x− y)(x2 − 10xy + y2) ≡ (x + z)(x2 + 10xz + z2)

≡ (y + z)(y2 + 10yz + z2) ≡ 0 mod p.

If x − y ≡ x + z ≡ 0 mod p then 3x ≡ 0 mod p. If x − y ≡ x2 +

10xz + z2 ≡ 0 mod p, then from the first congruence we find 2x ≡ z,and from the second we find 25z2 ≡ 0 mod p. Finally, assume that

x2 − 10xy + y2 ≡ x2 + 10xz + z2 ≡ y2 + 10yz + z2 ≡ 0 mod p.


Adding the first part to the middle part, and using that y − z ≡x mod p, because xp + yp = zp, one finds that 12x2 + y2 + z2 ≡ 0mod p. Similarly, by adding the first part to the last part we findx2 + 12y2 + z2 ≡ 0 mod p and therefore 11(x + y)(x− y) ≡ 0 mod p.So for p 6= 3, 5, 11 this shows that p must divide Bp−5.

5.3 the p-adic logarithm

Given an equation of the form αη = uγp, we can also use the (Iwasawa)p-adic logarithm to obtain information on α modulo p. We recall thedefinition and some basic properties. See, for example, [19] for proofsof these facts. Let Cp be the completion of the algebraic closure of thefield Qp and let | · |p be the p-adic valuation. For x ∈ Cp with |x|p < 1,the following series converges:

logp(1− x) := −∑i≥1

xi

i

If y ∈ Cp is any non-zero element (not necessarily with |y|p < 1), thenit can be written in the form y = prξx, where r = ordp(X) ∈ Q, ξ

is a root of unity, and x ∈ Cp satisfies |x− 1|p < 1. Then we definelogp(y) := logp(x).

Just like we had the properties of Lemma 5.3 for the logarithmicderivative, for the p-adic logarithm we have the following.

Lemma 5.6. For all x, y ∈ OK − πOK we have:

1. logp(xy) = logp(x) + logp(y)

2. σ(logp(x)) = logp(σ(x)) for all σ ∈ G

3. logp(ξ) = 0 for any root of unity ξ

4. logp(u) = logp(σ−1(u)) for all u ∈ O∗K

5. logp(x) ≡ 0 mod p if x ∈ Z \ pZ.

Proof. These are basic properties of the p-adic logarithm, see for ex-ample [19]. For the fourth property, note that u = ζrv for some integerr and real unit v. For the last one, we use that if p - x, then there is a(p− 1)-st root of unity congruent to x modulo p.

Because of the first two properties, we will sometimes write η logp(α)

instead of logp(αη) for η ∈ Z[G].

Theorem 5.7. Let α ∈ OK \ πOK be a singular integer. For all odd 3 ≤n ≤ p− 2, we have

Bp−nεn logp(α) ≡ 0 mod p.

5.3 the p-adic logarithm 41

Proof. Just like in Theorem 5.5, we see that if p does not divide Bp−n,then we must have αεn = uγp for some u ∈ O∗K and γ ∈ OK. Usingthe properties in Lemma 5.6 we see that

εn logp(α) ≡ logp(u) ≡ σ−1 logp(u)

≡ σ−1εn logp(α) ≡ (−1)nεn logp(α) mod p.

Because n is odd, this implies that εn logp(α) ≡ 0 mod p.

We can use εn logp(α) ≡ 0 mod p to get information about α mod-ulo p. As an example we take α = x + yζ and n = 3. Note that wehave

logp(x+ yζ) = logp(x+ y)+ logp(1−y

x + yπ) ≡ logp(1−

yx + y

π) mod p,

because logp(x + y) ≡ 0 mod p. So, by definition of the p-adic loga-rithm, we have

logp(x + yζ) ≡ −∑i≥1

yi

(x + y)iπi

imod p.

The terms with i = p− 1 and i ≥ p + 1 are all congruent to 0 modulop. Moreover, we have

0 = logp(ζ) = logp(1− π) = −∑i≥1

πi

i,

and therefore we see that

πp

p≡ −

p−2

∑i=1

πi

imod p.

Hence, we conclude that

logp(x + yζ) ≡p−2

∑i=1

(yp

(x + y)p −yi

(x + y)i

)πi

i

≡p−2

∑i=1

(y

x + y− yi

(x + y)i

)πi

imod p,

and thus

logp(x + yζ) ≡ xy(x + y)2

π2

2+

xy(x + 2y)(x + y)3

π3

3mod π4.

Now we apply the idempotent ε3. We have

ε3πi ≡ −p−1

∑j=1

j−3σjπi ≡ −

p−1

∑j=1

j−3(1− ζ j)i mod p.


Because we have

1− ζ j = 1− (1− π)j =j

∑r=1

(−1)r+1(

jr

)πr,

we find that

(1− ζ j)2 ≡ j2π2 − 2j(

j2

)π3 ≡ j2π2 + (j2 − j3)π3 mod π4,

(1− ζ j)3 ≡ j3π3 mod π,

and therefore:

ε3π2 ≡ −p−1

∑j=1

j−3 (j2π2 + (j2 − j3)π3) ≡ −π3 mod π4

ε3π3 ≡ −p−1

∑j=1

j−3 j3π3 ≡ π3 mod π4.

So, finally, we see that:

ε3 logp(x + yζ) ≡ xy(x + y)2

ε3π2

2+

x2y + 2xy2

(x + y)3ε3π3

3

≡(− xy

2(x + y)2 +x2y + 2xy2

3(x + y)3

)π3

≡ − xy(x− y)6(x + y)3 π3 mod π4.

In other words, we found that

ε3 logp(x + yζ) ≡ −`3(x + yζ)π3

6mod π4.

In particular, ε3 logp(x + yζ) ≡ 0 mod π4 if and only if p dividesxy(x− y). So, by using Theorem 5.7, we again find that if the first caseof Fermat’s last theorem fails, then p divides Bp−3.

Remark 5.8. If θ ∈ Z[G] is in the Stickelberger ideal (see chapter 6 of [36]),and α is singular, then we can show, in pretty much the same manner as inTheorem 5.7, that

θ logp(α) ≡ 0 mod p.

In this way we can obtain congruence conditions for the first case of Fermat’slast theorem. The process is similar to what we did above, but the detailsare more messy. For a clear exposition on the relationship between thesecongruence conditions for FLT1, the Stickelberger ideal, and congruences forBernoulli numbers, see the PhD-thesis [18].

That `3(α) appeared in the p-adic expansion of ε3 logp(α) is nocoincidence, as the following theorem shows. Let [rs] be the unsignedStirling numbers of the first kind. They are defined by the equation

x(x + 1) . . . (x + r− 1) =r

∑s=1

[rs

]xs.


Note that we have(ab

)=

1b!

a(a− 1) . . . (a− b + 1) =1b!

b

∑s=1

(−1)b+s[

bs

]as.

Theorem 5.9. Let α ∈ OK \ πOK and 1 ≤ n ≤ p− 2. Then we have

εn logp(α) ≡ (−1)n`n(α)

(p−2

∑r=n

1r!

[rn

]πr

)mod p.

Proof. The idea of the proof is that εn logp(α) and `n(α) are both givenby the n-th coefficient of certain power series. In the first part of theproof we show what these power series are in both cases. In the secondpart, we calculate the n-th coefficient of the series. This will lead tothe formula in the statement of the theorem.

Suppose that we have a power series ∑i≥0 fiXi with coefficientsfi ∈ Fp. Then for any 1 ≤ n ≤ p− 2 we have[(

∂

∂X

)n

f (X)

]X=0

= n! fn. (5.1)

Similarly, for the polynomial g(X) = ∑p−2i=0 giXi we have

p−1

∑j=1

j−ng(j) =p−2

∑i=0

gi

p−1

∑j=1

ji−n.

We know that ∑p−1j=1 ji−n ≡ 0 mod p if i 6≡ n mod p− 1, and ∑

p−1j=1 ji−n ≡

−1 mod p if i ≡ n mod p− 1. Consequently, we find

p−1

∑j=1

j−ng(j) ≡ −gn mod p. (5.2)

We see that both[(

∂∂X

)nf (X)

]X=0

and ∑p−1j=1 j−ng(j) give us the n-th

coefficient.Now consider α ∈ OK \ πOK and suppose we have α = a0 + a1ζ +

· · ·+ ap−2ζ p−2 for some integers ai. Since α(1) is not divisible by p,we see:

logp(σj(α)) = logp

(p−2

∑k=0

akζ jk

)

= logp (α(1)) + logp

(1−

p−2

∑k=1

ak

α(1)(1− ζ jk)

)

≡ logp

(1−

p−2

∑k=1

ak

α(1)(1− ζ jk)

)mod p.

Similarly, since[(

∂∂X

)nlog(α(1))

]X=0

= 0 we get[(∂

∂X

)n

log(α(eX))

]X=0

=

[(∂

∂X

)n

log(1−p−2

∑k=1

ak

α(1)(1− ekX))

]X=0

.


So, from formula (5.1) we see that `n(α) is n! times the n-th coef-ficient of log(1− ∑

p−2k=1

akα(1) (1− ekX)) as a power series in X. On the

other hand

εn logp(α) ≡ −p−1

∑j=1

j−nσj(logp(α)) mod p

is, by formula (5.2), equal the n-th coefficient of logp(1−∑p−2k=1

akα(1) (1−

ζ jk)) as a polynomial in Fp[j]. The rest of the proof will consist ofcalculating the n-th coefficient in both cases.

Suppose that we have

log

(1−

p−2

∑k=1

ak

α(1)(1−Yk)

)= ∑

i≥0biYi (5.3)

for some coefficients bi ∈ Q, as a formal power series in Y. SubstitutingY = eX in (5.3), we get

log

(1−

p−2

∑k=1

ak

α(1)(1− ekX)

)= ∑

i≥0bieiX.

Then we see that

∑i≥0

bieiX = ∑i≥0

bi

(∑s≥0

(iX)s

s!

)= ∑

s≥0

(∑i≥0 biis

s!

)Xs.

So in the end we find

`n(α) ≡ n!(

∑i≥0 biin

n!

)≡ ∑

i≥0biin mod p.

On the other hand, substituting Y = ζ j in (5.3), we get

logp

(1−

p−2

∑k=1

ak

α(1)(1− ζ jk)

)= ∑

i≥0biζ

ij.

Because ∑p−2k=1

akα(1) (1− ζ jk)) ≡ 0 mod p, this series actually converges.

We want to write ∑i≥0 biζij as an element of Fp[j]. First note that

(ijr) ≡ (ij+p

r ) mod p, and therefore

ζ ij = ζ ij+p =ij+p

∑r=0

(−1)r(

ij + pr

)πr ≡

p−2

∑r=0

(−1)r(

ijr

)πr mod p.

This ensures that we do not have to differentiate between the casewhere ij > p and the case where ij < p.

Moreover, we have(ijr

)= (−1)r

r

∑c=1

(−i)c

r!

[rc

]jc.


So, combining these equations, we get

ζ ij ≡p−2

∑r=0

r

∑c=1

(−i)c

r!

[rc

]πr jc ≡

p−2

∑c=1

p−2

∑r=c

(−1)c ic

r!

[rc

]πr jc mod p.

Finally, we have

∑i≥0

biζij ≡ ∑

i≥0bi

p−2

∑c=1

p−2

∑r=c

(−1)c ic

r!

[rc

]πr jc

≡p−2

∑c=1

(−1)c

(∑i≥0

biic

)(p−2

∑r=c

1r!

[rc

]πr

)jc.

We know that εn logp(α) is equal to the coefficient of jn, and thus wefind

εn logp α ≡ (−1)n

(∑i≥0

biin

)(p−2

∑r=n

1r!

[rn

]πr

)

≡ (−1)n`n(α)

(p−2

∑r=n

1r!

[rn

]πr

)mod p.

The theorem again gives us ε3 logp(x+ yζ) ≡ `3(x+ yζ)π3

6 mod π4.We get some easy consequences for the p-adic expansion of logp(α).

Corollary 5.10. For α ∈ OK \ πOK and 1 ≤ n ≤ p− 2 we have:

1. εn logp(α) ≡ 0 mod πn

2. εn logp(α) ≡ 0 mod πn+1 if and only if εn logp(α) ≡ 0 mod p ifand only if `n(α) ≡ 0 mod p

3. logp(α) ≡ 0 mod πn if and only if εk logp(α) ≡ 0 mod p for all1 ≤ k ≤ n− 1.

Proof. For the first part, just note that(

∑p−2r=n

1r! [

rn]π

r)≡ 0 mod πn.

In fact,(

∑p−2r=n

1r! [

rn]π

r)≡ [nn]

πn

n! ≡πn

n! mod πn+1. So εn logp(α) ≡ 0

mod πn+1 if and only if `n(α) ≡ 0 mod p if and only if εn logp(α) ≡ 0mod p. For the last part, note that

logp(α) ≡p−2

∑k=0

εk logp(α) mod p.

Since εk logp(α) ≡ 0 mod π2 for k ≥ 2, we see that logp(α) ≡ 0mod π2 if and only if ε1 logp(α) ≡ 0 mod p. By looking modulo π3,we see that logp(α) ≡ 0 mod π3 if and only if ε1 log( α) and ε2 logp(α)

are congruent to 0 modulo p. Continuing in similar fashion, we findthat logp(α) ≡ 0 mod πn if and only if εk logp(α) ≡ 0 mod p for all1 ≤ k ≤ n− 1.


This shows that we cannot hope to get more information aboutsingular integers using the p-adic logarithm, than we could get byusing the logarithmic derivative. Looking at εn logp(α) mod p and at`n(α) basically amounts to the same thing.

6K U M M E R E X T E N S I O N S

Our discussion of the classical approach to Fermat’s last theoremnaturally led us to consider singular integers. That is α ∈ Q(ζp) suchthat αZ[ζp] = Ip for some ideal I of Z[ζp]. There is an intimateconnection between singular integers in a number field and cyclicextensions of this field. In this chapter we explore this connection.

In the first two sections, we discuss Kummer extensions of a fieldand the ramification of primes in these extensions. This can, for exam-ple, be found in [1] and [14]. In the last two sections we restrict ourattention to extensions of cyclotomic fields. These sections are for alarge part inspired by [36]. The last section concerns the Hilbert classfield and the Herbrand-Ribet Theorem. We give a proof of Herbrand’sdirection of the Herbrand-Ribet theorem. The structure is very similarto Herbrand’s original article [15].

Though the contents of this chapter are not new, it still seemsgood to have this chapter in the thesis. Singular integers in Q(ζp)

are very naturally connected to unramified extensions of Q(ζp). Notincluding a chapter on this relation would seems like a waste, afterwe have talked about singular integers so much. We tried to makethis chapter as concrete as possible. In general, in class field theorythe constructions are quite abstract, and concrete constructions of theHilbert class field of a number field are difficult. At least in the case ofQ(ζp), we can give explicit units generating the unramified extensions.Moreover, we can use the results of the previous chapter, to explainhow the Galois group of Q(ζp) acts on the Galois group of the Hilbertclass field.

6.1 kummer extensions

The theory of Kummer extensions can be found in many sources. Aclassic text is the third chapter of [1].

Let n be a natural number fixed throughout this chapter, and K bea number field, which contains a primitive n-th root of unity ζ = ζn.So in this section, and in the next section, K does not denote thecyclotomic field Q(ζp). From the third section onwards, K = Q(ζp)

like in the rest of this thesis. We will classify all cyclic extensions of Kof degree n.

Lemma 6.1. If a ∈ K∗ has order n in the group K∗/(K∗)n then K( n√

a)/Kis a cyclic extension of degree n. Conversely, if L/K is a cyclic extension ofdegree n, then there is some a ∈ K such that L = K( n

√a).

47

48 kummer extensions

Proof. Suppose a has order n in K∗/(K∗)n and let α = n√

a. Since Kcontains the n-th roots of unity, the extension K(α)/K is Galois. Wehave

a[K(α):K] = NK(α)/K(a) =(

NK(α)/K(α))n

,

and because a has order n in K∗/(K∗)n, we find that n divides [K(α) :K]. Hence we find [K(α) : K] = n and Xn − a is irreducible. Finally,we note that if σ ∈ Gal(K(α)/K), then σ(α) = ζ iα for some i ∈ Z/nZ

and σ is completely determined by this i. This means we get aninjective group homomorphism from Gal(K(α)/K) to the group of n-th roots of unity, given by σ 7→ σ(α)/α. Since |Gal(K(α)/K)| = n, thishomomorphism must be an isomorphism and consequently K(α)/Kis a cyclic extension. This completes the proof of the first part.

Now suppose that L/K is a cyclic extension of degree n and let σ

be a generator of the corresponding Galois group. We will explicitlyconstruct an a ∈ K such that L = K( n

√a). By the normal basis theorem

(see, for example, section 3.2.5 of [5]), there is some γ ∈ L suchthat L = K(γ) and γ, σ(γ), σ2(γ), . . . , σn−1(γ) forms a basis for L/K.Define α = ∑n−1

i=0 ζ iσi(γ) and note that α 6= 0 because the elementsσi(γ) are linearly independent over K. We find

σ(α) =n−1

∑i=0

ζ iσi+1(γ) =n−1

∑i=0

ζ i−1σi(γ) = ζ−1α.

This implies that σ(αn) = αn, so αn ∈ K. Similarly, σ(αr) 6= αr, soαr 6∈ K, for 1 ≤ r ≤ n− 1. Let a := αn. Then ar = cn for some c ∈ Kif and only if for some integer j, αr = ζ jc ∈ K. Since αr is not in Kfor 1 ≤ r ≤ n− 1 we find that a has order n in K∗/(K∗)n. Hence, thefirst part of the lemma tells us that K( n

√a) is a cyclic extension of K of

degree n contained in L, and therefore it must by equal to L.

The following lemma tells us when two elements a and b generatethe same cyclic extension.

Lemma 6.2. Let a, b ∈ K∗ both have order n in K∗/(K∗)n. Then K( n√

a) =K( n√

b) if and only if there is some c ∈ K, and r ∈ Z relatively prime to n,such that a = brcn.

Proof. Suppose that a = brcn. Then

K( n√

a) = K( n√

brcn) = K(( n√

b)r).

Obviously we have ( n√

b)r ∈ K( n√

b). Conversely, since r and n rela-tively prime, there exist integers k, l such that rk + nl = 1. Then wesee

n√

b = (n√

b)rk+nl = (n√

b)rkbl ∈ K( n√

br).

Hence K( n√

a) = K( n√

b).For the other direction, note that any element of Gal(K( n

√a)/K) is

completely determined by what it does to n√

a or to n√

b. Suppose that

6.2 ramification 49

σ( n√

a) = ζ n√

a. Then σ generates Gal(K( n√

a)/K), and thus σ( n√

b) =ζr n√

b for some r relatively prime to n. The powers of n√

a form a basis ofK( n√

a)/K, so there are elements ci ∈ K such that n√

b = ∑n−1i=0 ci(

n√

a)i.This means we also have

n−1

∑i=0

ciζr( n√

a)i = ζr n√

b = σ(n√

b) =n−1

∑i=0

ciζi( n√

a)i.

Since the powers of n√

a are linearly independent over K, we find thatonly cr can be non-zero, and thus n

√b = cr( n

√a)r. Raising this to the

n-th power gives b = cnr ar, as desired.

Combining these two lemmas gives us the following theorem.

Theorem 6.3. There is a bijection between the cyclic subgroups of K∗/(K∗)n

of order n and the cyclic extensions of K of degree n. It is given by

〈a〉 7→ K( n√

a).

Proof. This is a direct consequence of the previous two lemmas. Lemma6.2 tells us that the map is well-defined and injective, and Lemma 6.1tells us that the map is surjective.

In fact we can get a more general theorem. We say that a groupG has exponent n if for every g ∈ G we have gn = 1. We call a fieldextension L/K a Kummer n-extension if K contains the n-th rootsof unity and L/K is Galois with Gal(L/K) a finite abelian group ofexponent n.

Theorem 6.4. There is a bijection between the finite subgroups of K∗/(K∗)n

of and the Kummer n-extensions of K. It is given by

W 7→ K( n√

W)

where K( n√

W) denotes the field obtained by adjoining the n’th roots of allelements of W to K.

Proof. This can found, for example, as Theorem 8.1 of chapter 5 of [17].Since we will be mainly interested in cyclic extensions, we will notreproduce the proof here. The main idea is to write an abelian groupof exponent n as the direct sum of cyclic groups of order dividing n.Then the result basically follows from the previous theorem.

6.2 ramification

Let K be like in the previous section, and suppose we are given acyclic Kummer extension L/K of degree n. What can we say aboutthe splitting behaviour of the primes of K? We know that L is givenby K( n

√a) for some a ∈ K∗, and that a is a generator of a subgroup of

K∗/(K∗)n of order n.


Lemma 6.5. The discriminant of L divides nnan−1, so a prime p of K whichdoes not divide na is unramified. In this case the residue degree is the leastnumber f such that xn ≡ a f mod p is solvable in K.

Proof. Let OK be the ring of integers of K. We know that OK[

n√

a]

is a submodule of the ring of integers OL of L. Since OK[

n√

a]

hasdiscriminant nnan−1, we see that discriminant of L must divide thisnumber. Let p be a prime of K not dividing na. Then it is unramified,so its residue degree is equal to the degree of the field extensionKp( n√

a)/Kp. If xn − a f has a root modulo p, then Hensel’s lemma saysthat this roots lifts to a root in Kp. So we see that

[Kp( n√

a) : Kp

]is at

most f . On the other hand,

NKp( n√a)/Kp( n√

a)n = NKp( n√a)/Kp(a) = a[Kp( n√a):Kp].

Therefore f is at most[Kp( n√

a) : Kp

], and thus they must be equal.

In particular, the lemma above tells us that the prime p splits com-pletely if and only if the equation xn ≡ a mod p is solvable. So, justas splitting of primes in quadratic extensions is related to quadraticresidues, so is splitting of primes in cyclic extensions related to n-thpower residues. Now let us consider the primes that divide a.

Theorem 6.6. Let p be a prime of K that divides a and let r be the exact powerof p dividing a ∈ K. If r is relatively prime to n, then p is totally ramified inK( n√

a). Suppose that p - n and gcd(r, n) = s. Then p is unramified in theextension K( s

√a) and the prime factors of p in K( s

√a) are totally ramified in

K( n√

a).

Proof. Suppose that p is a prime of K such that pr | a, but pr+1 - a forsome integer r ≥ 1. First note that we may assume 0 ≤ r ≤ n − 1.Indeed, if r = s + tn for some integers s, t with 0 ≤ s ≤ n− 1 then wecan pick an element b ∈ p−t and use the fact that K( n

√a) = K( n

√abn).

Suppose that r = 1 and that p factors in L/K as p = ∏gi=1 P

ei . Then

for each i we see that Pi | p+ ( n√

a) and therefore

Pni | (p+ ( n

√a))n.

So, in fact, pn/e = ∏gi=1 P

ni divides (p+ ( n

√a))n. If e < n this implies

that p2 | a, which is in contradiction with the fact that r = 1. Thus wemust have e = n.

Hence, in the case where r = 1, we see that p is totally ramifiedand in fact p = (p+ ( n

√a))n. Alternatively, we could have used the

fact that xn − a is an Eisenstein polynomial if r = 1 to show that p istotally ramified.

If gcd(r, n) = 1, we can always reduce to the case r = 1 by using thesame trick as above. Pick some h, g such that gr + hn = 1 mod p andpick an element b ∈ ph which is not in ph+1. Then K( n

√a) = K( n

√agbn),

and therefore we are back in the case r = 1. The final part immediatelyfollows by applying the previous parts to the extension K( s

√a).

6.2 ramification 51

As a corollary, we find a relation between ramification of primes inextensions of K and singular integers.

Corollary 6.7. Let a ∈ K∗ be relatively prime to n. Then aOK = In forsome ideal I of OK if and only if K( n

√a)/K is unramified at the primes

not dividing n, and this happens if and only if K( n√

a) = K( n√

u) for someu ∈ O∗K.

Proof. If a is singular, then the previous theorem tells us that K( n√

a)/Kis unramified outside n. The converse holds as long as a and n arerelatively prime. This deals with the first ‘if and only if’. For thesecond one, we obviously have that K( n

√u)/K is unramified outside

n for u ∈ O∗K. In the other direction, if (a) = In, we can pick someb ∈ K with ordp(b) = −ordp(I) for all prime ideals p of K. ThenK( n√

a) = K( n√

abn) and abn is a unit.

So, the extensions of K of degree n which are unramified outside nare precisely the extensions of the form K( n

√u) for some unit u.

This leaves only the ramification at the primes of K which dividen. However, no general results are known, except for the case wheren = p is a prime integer. In this case, the splitting behaviour of theprimes p above p in K( p

√a)/K is intimately related to solutions of the

equation xp ≡ a mod pb for integers b. The following can be foundas exercises 9.1-9.3 of [36] or in section 38 of [14].

Let n = p be a prime. We know that (p) = (1− ζp)p−1 in the fieldextension Q(ζp)/Q. Hence if p is a prime of K above p, then p divides1− ζ. If the exact power dividing 1− ζ is pt, then the exact power of pdividing p is pt(p−1). Assume that a ∈ OK is such that ξ p = a has nosolution inOK, and hence K( p

√a) is a cyclic extension of K. By Fermat’s

little theorem we have aN(p) ≡ a mod p. So ξ p ≡ a mod p always hasa solution in OK. In fact, ξ p ≡ a mod pb may have a solution in OK

for b > 1. The following theorem shows that the maximum value of bis related to the ramification above p in K( p

√a)/K.

Theorem 6.8. Let p be a prime of K dividing p and let t = ordp(1− ζ).Let a ∈ OK such that it is not a p-th power in OK. Let L = K( p

√a) and

let b ∈ Z>0 ∪ {∞} be maximal such that ξ p ≡ a mod pb has a solutionin OK. (With b = ∞ we mean that ξ p ≡ a mod pb has a solution for anyb > 0.) Then p splits completely in L/K if and only if b > tp if and only ifb = ∞. The prime p is inert if and only if b = tp, and p is totally ramified ifand only if b < tp.

Proof. Suppose p splits completely, and let P be a prime factor of p inL. Because p splits completely, we know that for any b ∈N we have[OL/Pb : OK/pb] = 1. Hence there is some ξ ∈ OK such that ξ ≡ p

√a

mod Pb. But then we see that ξ p ≡ a mod pb.Conversely, suppose that ξ p ≡ a mod ptp+1 for some ξ ∈ OK. Let

ρ ∈ K be such that ordp(ρ) = −t and ordq(ρ) ≥ 0 for all other primes


q of K. We consider the element β = ρ( p√

a− ξ). Then β is a root ofthe polynomial

(x+ ρξ)p− ρpa = xp +

(p1

)ρξxp−1 + · · ·+

(p

p− 1

)ρp−1ξ p−1x+ ρp(ξ p− a).

Since pt(p−1) divides p and ordp(ρ) = −t, we find that (pi )ρ

i is inOK for i = 1, . . . , p − 1. Since ξ p − a ∈ ptp, we see that ρp(ξ p − a)is in OK. So the polynomial has coefficients in OK, and thus β is inOL. Its norm is ρp(ξ p − a), which is divisible by p precisely whenξ p ≡ a mod ptp+1. If σ ∈ Gal(L/K) is given by σ( p

√a) = ζ p

√a then

β − σ(β) = ρ p√

a(1 − ζ) 6∈ p. Thus, if P is the prime factor of p

containing β then σ(P) 6= P. Because L/K is Galois and of primedegree, this means that p splits completely.

So, we have shown that p splits completely if and only if b ≥ tp + 1.If b is exactly tp then β is still an element of OL. Moreover, note thatσi(β)− σj(β) is relatively prime to p. Hence also

∏1≤i 6=j≤p−1

(σi(β)− σj(β))

is relatively prime to p. We conclude that the discriminant ideal ∆L/Kcontains an element relatively prime to p, and thus p does not divideit. Therefore p is unramified in L/K. By the first part of the theoremwe know that p does not split, so p must be inert in L/K.

We can finish the proof by showing that if b < tp, then p is totallyramified. Assume that we have b < tp. By Fermat’s little theorem, b isat least 1. We first show that b is not divisible by p. Suppose there issome 0 < c ≤ t− 1 and ξ ∈ K such that

a ≡ ξ p mod pcp.

Let λ ∈ OK satisfy ordp(λ) = c. Then for any κ ∈ OK we find

(ξ + λκ)p ≡ ξ p + λpκp mod pcp+1,

since all other terms are divisible by λp. Now a− ξ p ∈ ppc and λp 6= 0as an element of ppc/ppc+1. Hence we can pick κ in such a way thatξ p + λpκp ≡ a mod ppc+1, and therefore b is at least pc + 1.

So we know that b = cp + v for some c ≤ t− 1 and 1 ≤ v ≤ p− 1.Let ρ ∈ K satisfy ordp(ρ) = −c and ordq(ρ) ≥ 0 for other primes q. Inthe same way as above, we see that β := ρ( p

√a− ξ) is an element of

OL which is not divisible by p. Now, NL/K(β) = ρp(ξ p− a) is divisibleexactly by pv, but p does not divide β. This means that p cannot be inertin L/K, and by the first part p cannot split in L/K. Since the extensionis of prime degree, we conclude that p must be totally ramified.

6.3 units in cyclotomic fields

In this section we apply the results of the previous two sections toextensions of degree p of cyclotomic fields. Our main reference for this

6.3 units in cyclotomic fields 53

part is chapter 8 of [36]. From now on, we set ζ = ζp and K = Q(ζ)

and we set G = Gal(K/Q). In the previous section we saw that anextension L/K is unramified outside of p if and only if L = K( p

√u),

for some unit u ∈ O∗K, so in this section we will investigate the unitsof K. To understand the unramified extensions, we only need to careabout the units up to p-th powers, thus we are interested in the groupO∗K/(O∗K)p. The subgroups of this group correspond to the extensionsof K unramified outside p.

To ease the notation, we will set E := O∗K, K+ = K∩R = Q(ζ + ζ−1)

and E+ := OK+ . By Dirichlet’s unit theorem, we know that E and E+

are both groups whose non-torsion part has rank p−32 . We also know

that for every u ∈ E there is some integer i such that ζ iu is an elementof E+.

In general, we do not have a nice description of the fundamentalunits generating E, but we do have a nice discription for a subgroupof E of finite index. Define the group of cyclotomic units to be thesubgroup C of E generated by−ζ and the units 1−ζ i

1−ζ for i = 1, . . . , p− 1.

Note that 1−ζ−i

1−ζ = −ζ i 1−ζ i

1−ζ which means that the unit

ξi := ζ1−i

21− ζ i

1− ζ

is a real unit. Let C+ := C ∩ E+, so that C+ is generated by the ξi fori = 2, . . . , p−1

2 .To get more information about the units we need to consider the

action of the Galois group G. Since multiplying a unit by the p-thpower of another unit will give us the same extension of K, we aremostly interested in the group Ep := E/Ep. Similarly, we will setE+

p = E+/(E+)p, Cp := C/Cp and C+p := C+/(C+)p. Note that, since

any unit u ∈ E can be written as u = ζrv, for some r ∈ Z and v ∈ E+,we have Ep = 〈ζ〉 ⊕ E+

p . All these above groups have an Fp[G]-modulestructure, with the obvious action of G, so we can use the idempotentsof Fp[G] to decompose these group (see the beginning of chapter 5 foran introduction). Then we see

Ep =p−2⊕i=0

Eεip .

Note that ε0 = ∑p−1a=1 σa, and thus Eε0

p = N(Ep) = 1. Since Eεip consists

of the elements u such that uσa = uai, for all 1 ≤ a ≤ p− 1, we see that

the roots of unity are elements of Eε1p . Furthermore, if v is real and i is

odd, thenvεi = vσ−1εi = v(−1)iεi = v−εi ,

and therefore vεi = 1. This means that we have:


Proposition 6.9.

Ep = 〈ζ〉 ⊕p−3⊕i=2

i even

Eεip and E+

p =p−3⊕i=2

i even

Eεip

The proof is immediate from the comments above.

Lemma 6.10. As an Fp[G]-module, C+p is generated by ξg, where g is a

generator of (Z/pZ)∗.

Proof. If a = gr, then:

ξa = ζ1−gr

21− ζgr

1− ζ=

r−1

∏j=0

ζgj−gj+1

21− ζgj+1

1− ζgj =r−1

∏j=0

ξσ

jg

g .

Define Ei := ξεig . By the previous lemma, we now know that

C+p

εi = 〈Ei〉 and C+p =

p−3⊕i=2

i even

〈Ei〉.

For each even 2 ≤ i ≤ p− 3, we see that 〈Ei〉 is a subgroup of Eεip .

Moreover, we see that 〈Ei〉 6= Eεip if and only if Ei is the p-th power of

some unit in E. Although we will not prove it, at this point it is niceto note the following theorem, which is Theorem 8.2 of [36].

Theorem 6.11. The index of the real cyclotomic units inside the group of allreal units is given by

[E+ : C+] = h+

where h+ is the class number of K+.

As a consequence, we see that if p does not divide h+, then E+p = C+

pand thus none of the units Ei is a p-th power of a unit in E. Vandiver’sconjecture states that p never divides h+, which has been verified forp < 4000000, though it is not clear whether this means the conjectureshould be true (see the discussion on page 158 of [36]).

We have defined the elements Ei as elements of E/Ep. In the theorembelow, we talk about the extension K( p

√Ei)/K. Here we pick any lift

of Ei to E, and then take the p-th root. The extension we get clearlydoes not depend on the particular lift we have chosen.

Theorem 6.12. For each even 2 ≤ i ≤ p− 3 and r ∈ Z/pZ, if the exten-sion K( p

√ζrEi)/K is an unramified extension, then p divides the Bernoulli

number Bi.

6.4 unramified extensions 55

Proof. If K( p√

ζrEi)/K is the trivial extension, then there is some η ∈OK such that ηp = ζrEi. If K( p

√ζrEi)/K is a unramified extension

of degree p then we have ηp ≡ ζrEi mod πp for some η ∈ OK, byTheorem 6.8. In both cases we see that

0 ≡ p logp(η) ≡ logp(ζrEi) ≡ logp(Ei) mod p.

Since Ei = ξεig , we can use Theorem 5.9 of chapter 5 to see that

logp(Ei) ≡ ε i logp(ξg) ≡ ì(ξg)

(p−2

∑r=i

1r!

[ri

]πr

)mod p.

So logp(Ei) ≡ 0 mod p if and only if ì(ξg) ≡ 0 mod p.

Because ξg = ζ1−g

21−ζg

1−ζ , and because ì is multiplicative, we find

ì(ξg) = ì

(1− ζg

1− ζ

).

Hence, ì(ξg) is equal to i! times the coefficient of Xi of the power

series log egX−1ex−1 . Note that we have:

ddX

logegX − 1eX − 1

=gegX

egX − 1− eX

eX − 1

= g(

1egX − 1

+ 1)−(

1eX − 1

+ 1)

=1X

(gX

egX − 1− X

eX − 1

)+ g− 1.

By definition of the Bernoulli numbers we have XeX−1 = ∑∞

m=0BmXm

m! . Socombinining all of this, we see that

ì(ξg) ≡ (gi − 1)Bi mod p.

Because g is a generator of (Z/pZ)∗ and i < p− 1, we know thatgi − 1 is never congruent to 0 modulo p. Hence ì(ξg) ≡ 0 mod p ifand only if Bi ≡ 0 mod p.

6.4 unramified extensions

We will use class field theory to see how unramified extensions of Krelate to the class group of K. This allows us to prove the Herbrand’sdirection of the Herbrand-Ribet theorem, by constructing a certainunramified extension of K.

Suppose that L/K is an extension of degree p, which is unramifiedoutside p. Then, by Corollary 6.7, L = K( p

√u) for some unit u ∈ E.

Because of the results of the previous section, if Vandiver’s conjectureholds, we must have

L ⊂ M := K(ζp2 , p√E2, . . . , p

√Ep−3).


Because the Galois group Gal(M/K) is an elementary p-group (all ele-ments other than the identity have order p), we call M a p-elementaryextension of K. In fact, under Vandiver’s conjecture, M is the maximalp-elementary extension of K that is unramified outside p.

Since M/K is an abelian extension, there is an action of G onGal(M/K) given by conjugation: If τ ∈ Gal(M/K) and σ is any lift ofσ ∈ G to M, then we define

τσ := στσ−1.

This is well-defined (it does not depend on which lift of σ we choose)because M/K is abelian. Because Gal(M/K) is an elementary p-group,this action extends to an action of Fp[G] on Gal(M/K) by putting

τ∑p−1a=1 naσa := (τn1)σ1 · · · (τnp−1)σp−1 .

In this way, we get a decomposition of Gal(M/K) using the orthogonalidempotents ε i ∈ Fp[G]. The following lemma describes the differentcomponents of Gal(M/K) of this decompostion.

Lemma 6.13. For i = 0, . . . , p− 2 and τ ∈ Gal(M/K) we have:

1. τε0(ζp2) = τ(ζp2) and τεi(ζp2) = ζp2

2. τεi( p√Ej) = p

√Ej for j 6= p− i

3. τεi( p√Ep−i) = τ( p

√Ep−i) for i ≥ 3 odd

Proof. Let σa be a lift of σa to M. Then

σa(ζp2)p = σa(ζp) = ζap,

and therefore we see that

σa(ζp2) = ζa+spp2 ,

for some s ∈ Z/pZ. If τ ∈ Gal(M/K), then τ(ζp) = ζp and thus

τ(ζp2) = ζ1+tpp2 ,

for some t ∈ Z/pZ. Combining this, we see that

σ−1a τai

σa(ζp2) = σ−1a τai

(ζa+spp2 )

= σ−1a (ζ

a(1+aitp)+spp2 )

= ζ1+aitpp2 .

So we conclude that

τεi(ζp2) = ζ1−(

∑p−1a=1 ai

)tp

p2 .


Hence τε0(ζp2) = τ(ζp2) and τεi(ζp2) = ζp2 for all i = 1, . . . , p − 2.That deals with the first part of the lemma.

For the second and third part, note that Ej = ξε jg , and thus σa(Ej) =

E aj

j vp1 for some v1 ∈ E. Hence, we find that

σa( p√Ej) = v2

p√Ej

aj

,

for some v2 ∈ E. We know that τ( p√Ej) = ζt p

√Ej, for some t ∈ Z/pZ,

and therefore

σ−1a τai

σa

(p√Ej

)= σ−1

a τai(

v2p√Ej

aj)= σ−1

a

(v2ζaj+it p

√Ej

aj)= ζaj+i−1t p

√E .

We conclude that

τεi( p√Ej) = ζ

−(

∑p−1a=1 aj+i−1

)t p√Ej.

So τεi fixes p√Ej if j 6= p − i, and if j = p − i, then τεi( p

√Ej) =

τ( p√Ej).

The lemma tells us that the restriction map induces an isomorphismsGal(M/K)εi → Gal(K( p

√Ep−i)/K) for i = 3, . . . , p − 2 odd, and it

induces an isomorphism Gal(M/K)ε0 → Gal(K(ζp2)/K).Now we will use some results from class field theory. A classic

reference for this is [1]. Another book is [17], which gives a ratherconcrete approach to class field theory, and fits quite well with thischapter. We give a short summary of the things we need.

Class field theory gives us the existence of the Hilbert class fieldH = H(K) of K. It is the maximal abelian unramified (at all theprimes, also the infinite ones) extension of K. It has the propertythat Gal(H/K) is isomorphic to the class group C = CK of K. Theisomorphism is given by the Artin map: If q is a prime of K, whichdoes not ramify in H/K and Q is a prime of H above q, then thereis a unique ϕQ ∈ Gal(H/K) such that ϕQ(x) ≡ xNK/Q(q) mod Q forall x ∈ OH. This map ϕQ is called the Frobenius automorphismcorresponding to Q. If Q′ is another prime of H above q, then ϕQ andϕQ′ will be conjugate elements. Since Gal(H/K) is abelian, this meansthat ϕQ = ϕQ′ , so we can denote it by ϕq. This means we have a mapq 7→ ϕq from the set of unramified primes of K to Gal(H/K). Thisinduces a map ϕ : CK → Gal(H/K), called the Artin map. Class fieldtheory tells us that this map is in fact an isomorphism.

Just like before for the field M, there is an action of G on Gal(H/K),given by conjugation. We have another action of G on Gal(H/K) viathe class group. If τ ∈ Gal(H/K) and σ ∈ G we can define

τσ := ϕσϕ−1(τ).


The following lemma tells us that these actions are in fact the same.

Lemma 6.14. Let σ be a lift of σ ∈ G to H and let τ ∈ Gal(H/K). Thenwe have

ϕσϕ−1(τ) = στσ−1.

Proof. Let [q] ∈ C be such that ϕ(q) = τ. Then by definition we have

τσ−1(x) ≡ σ−1(x)NK/Q(q) mod Q.

Hence we see

στσ−1(x) ≡ xNK/Q(q) mod σ(Q).

So ϕσϕ−1(τ) = στσ−1.

Because C is a finite abelian group, we can write it like the directsum of its Sylow subgroups. In particular, we can write C = Cp ⊕D,where Cp is the p-part of C and D is the complement of Cp in C.We define Hp = Hϕ(D), the subfield of H fixed by all elements ofϕ(D). Then the restriction map from H to Hp induces an isomorphismϕ(Cp)→ Gal(Hp/K). We call Hp the p-part of H.

We know there is an action of G on C and Gal(Hp/K). Becausethese are p-groups (all elements have order a power of p), this actionextends to an action of Zp[G]. Now we are ready to prove the followingtheorem.

Theorem 6.15 (Herbrand). For i = 3, . . . , p− 2 odd, if the group Cεip is

non-trivial, then p divides the Bernoulli number Bp−i.

Proof. Let 3 ≤ i ≤ p− 2 be odd. Then we see:

Cεip is nontrivial ⇐⇒ C1−εi

p 6= Cp

⇐⇒ Hϕ(Cp)1−εi

p /K is a nontrivial extension

⇐⇒ ∃ L/K with [L : K] = p and τε j(L) = L

∀ τ ∈ Gal(Hp/K), ∀ j 6= i

If Ep−i is the p-th power of some unit of E, then as we saw in theproof of Theorem 6.12, p must divide Bp−i. Otherwise E

εp−ip = 〈Ep−i〉.

From Lemma 6.13 we conclude that L ⊂ K(ζp2 , p√Ep−i). Since [L :

K] = p, we find

L = K( p√

ζrEp−i),

for some r ∈ Z/pZ. Since L is contained in the Hilbert class field, L/Kmust be an unramified extension. So, by Theorem 6.12, the Bernoullinumber Bp−i is divisible by p.

This proof is quite similar to Herbrand’s proof in [15]. The otherdirection, showing that if p | Bp−i, then Cεi

p is non-trivial, is dueKenneth Ribet [30]. He used techniques from algebraic geometry to


construct unramified extensions. Later, a more elementary proof wasfound (see chapter 15 of [36]).

This finishes the chapter on extensions of Q(ζp). We see that thesingular integers we investigated in previous chapters, give us theextensions of Q(ζp) that are unramified outside of p. The proof of The-orem 6.15 can be seen as a nice application of Kummer’s logarithmicderivative.

7R E G U L A R P R I M E S

In this chapter we look at the equation

N(x + yζ) = zp (7.1)

for pairwise coprime integers x, y, z and p ≥ 5 a regular prime. Thefact that p is regular means we have extra knowledge of the units of K,which we try to exploit. First we show that there is no solution to (7.1)in the ‘first case’, i.e. when p - xy(x − y). This follows immediatelyfrom what we looked at in chapter 5. However, because p is regular,we get extra information.

Then we will see that if there is a solution to (7.1) for a regular primep, then xy(x− y) must be ‘big’ in the sense that it must be divisibleby all the ‘small’ primes congruent to 1 modulo p. More concretely, inCorollary 7.4, we show that if q = np + 1 is prime, with 2 ≤ n ≤ p− 3,then q must divide xy(x − y). In the computational section we usethis, together with extra data, to show that if p ≥ 5 is a regular primeand x and y are coprime integers such that N(x + yζ) = zp, thenmax(|x|, |y|) > 5× 108 (Theorem 7.8).

7.1 theoretical results

We are interested in the equation

N(x + yζ) = zp

for pairwise coprime integers x, y, z. Note that, for any integers a andb, if we write N(a + bζ) = w, then we see that N(wa + wbζ) = wp isa solution to (7.1). So, if we would not restrict to the case of coprimeintegers, then we would immediately have infinitely many solutions.We will henceforth restrict to the case of coprime integers x and y.Note that if a prime divides any two of x, y and z, then this prime willalso divide the third one. So assuming that x and y are coprime is thesame as assuming that x, y and z are pairwise coprime.

Lemma 7.1. If p is an odd prime and x and y are coprime integers such thatN(x + yζ) = zp, then p does not divide z.

Proof. Since pOK = πp−1OK, we see that p divides z if and only ifπ divides x + yζ, and this happens if and only if p divides x + y. Ifp divides z then pp divides N(x + yζ) and thus ordπ(N(x + yζ)) ≥p(p− 1). Since ordπ(x + yζ) = ordπ(x + yζ i) for any i = 1, . . . , p− 1,we see that ordπ(x + yζ) ≥ p, and thus p divides x + yζ. Because

61

62 regular primes

1, ζ, . . . , ζ p−2 form a basis of OK as a Z-module, we find that p dividesboth x and y. This is in contradiction with our assumption that x andy are coprime.

If p = 3, then any element of Z[ζ3] is of the form x + yζ3, soN(x + yζ3) = z3 if and only if x + yζ3 = uα3 for some unit u ∈ Z[ζ3]

and α ∈ Z[ζ3]. In particular, the case p = 3 has many solutions.

Theorem 7.2. Let p be an odd, regular prime and let x and y be coprimeintegers such that N(x + yζ) = zp. Then p divides xy(x− y).

Proof. From chapter 4 we know that if x and y are coprime, the ele-ments x + yζ i and x + yζ j are coprime whenever i 6≡ j mod p. Hence(x + yζ)OK = Ip for some ideal I of OK. Because p is regular, I = γOK

for some γ ∈ OK. Thus we find that

x + yζ = uγp (7.2)

for some unit u of OK. Now use the logarithmic derivative, or thep-adic logarithm, as in chapter 5.

We can be more specific about the unit u appearing in equation(7.2).

Theorem 7.3. If p is an odd, regular prime and x, y are coprime integerssatisfying N(x + yζ) = zp, then one of the following holds for some γ ∈ OK:

1. p | x and x + yζ = ζγp

2. p | y and x + yζ = γp

3. p | x− y and x + yζ = (1 + ζ)γp

Proof. We have x + yζ = uγp0 for some unit u ∈ O∗K and γ0 ∈ OK. In

Lemma 7.1 we saw that the assumption of the theorem imply thatp does not divide x + y. Therefore we have γ

p0 ≡ g mod p for some

integer g not divisible by p. So g has an inverse modulo p, and we get

g−1x + g−1yζ ≡ u mod p.

By Theorem 7.2, we know that p divides xy(x − y). If p dividesx, then ζ−1u is congruent to an integer modulo p. Since p is regular,Kummer’s lemma tells us that ζ−1u = vp for some unit v. Puttingγ = vγ0, we find that x + yζ = ζγp.

If p divides y, then u ≡ g−1x mod p and thus u = vp for some unitv. Again putting γ = vγ0, we get that x + yζ = γp.

Finally, suppose that p divides x− y. Then we have x+ yζ ≡ x(1+ ζ)

mod p. Since 1 + ζ is a unit we find that there is a unit v such thatu

1+ζ = vp. Putting γ = vγ0, we obtain x + yζ = (1 + ζ)γp.

7.2 computational results 63

Bringing the units in Theorem 7.3 to the left hand side, we findthat one of xζ−1 + y, x + yζ or x+yζ

1+ζ = 11+ζ (x − y) + y must be the

p-th power of an element of OK. In particular, this should be the caselocally at primes q of OK. Theorem 7.3 gives us a way to show thatcertain ‘small’ primes have to divide xy(x− y).

Corollary 7.4. Let p be an odd, regular prime and x, y coprime integerssatisfying N(x + yζ) = zp. Suppose that 2 ≤ n ≤ p − 3 is such thatq := np + 1 is a prime. Then p divides x (resp. y, x − y) if and only if qdivides x (resp. y, x− y).

Proof. Since q is congruent to 1 modulo p, it splits in K. So we have

qOK = (q, ζ − a1)(q, ζ − a2) · · · (q, ζ − ap−1)

for some distinct a1, . . . , ap−1 ∈ (Z/qZ)∗.If p divides x, then by Theorem 7.3 we know that xζ−1 + y = γp for

some γ ∈ OK. Looking modulo (q, ζ − ai), we see that xa−1i + y is a

p-th power in Z/qZ. The number of p-th powers in Z/qZ is equal ton + 1 (the p-th powers in (Z/qZ)∗ together with 0), so this numberis less than p− 1. In particular, there must be some j 6= i such thatxa−1

i + y ≡ xa−1j + y mod q, and therefore q must divide x.

Conversely, if q divides x, then q does not divide y or x− y becausex and y are coprime. So by what we just showed, p does not divide yor x− y. Hence p must divide x.

The proof for the cases where p divides y or x− y goes completelyanalogous.

7.2 computational results

For a prime p we define

Mp := p ∏1≤n≤p−3

np+1 prime

(np + 1).

With this notation, Corollary 7.4 tells us that if p is regular, then oneof x, y and x− y is divisible by Mp. Heuristically one would expectthere to be approximately

1p− 1

p2

2 log p≈ p

2 log p

primes less than p2 which are congruent to 1 modulo p. This wouldmean that Mp grows quite rapidly as p gets bigger.

Giving a theoretical lower bound on Mp other than Mp ≥ p does notseem feasible, though. Denote by P(a, d) the smallest prime congruentto a modulo d. Linnik’s theorem (see [38]) states that for any coprime,positive integers a, d with 1 ≤ a ≤ d− 1 we have

P(a, d) < cdL

64 regular primes

for some effictively computable constants c and L. It is conjecturedthat L ≤ 2, but the current record is L = 5, given in [38]. A lowerbound on Mp better than p would imply that P(1, p) < p2.

For relatively small primes p we can take a computational approach.Though not very efficient, we can still use the lower bound Mp ≥ p.This allows us to calculate all primes p such that Mp < B for a givenupper bound B, in finite time. We did this for the bound B = 109,using PARI/GP [25]. This leads to the following lemma.

Lemma 7.5. The only primes p such that Mp < 109 are given by:

p Mp Prime Factors

2 2 2

3 3 3

5 55 5 · 11

7 203 7 · 29

11 1508639 11 · 23 · 67 · 89

13 7130461 13 · 53 · 79 · 131

17 57332993 17 · 103 · 137 · 239

19 831041 19 · 191 · 229

Note that these are all regular primes, seeing as the first irregularprimes are 37, 59 and 67. Let q = np + 1 be a prime (not necessarilywith n ≤ p− 3) and let s be a primitive root modulo q. Even whenwe can not apply Corollary 7.4, we can still get some computationalresults. The p-th powers modulo q are 0 and spi for i = 1, . . . , q−1

p . The

p-th roots of unity modulo q are given by sq−1

p i for i = 0, . . . , p − 1.If x + yζ = γp, then, by looking modulo a prime of K above q, we

see that x + ysq−1

p i should be a p-th power modulo q for any 1 ≤ i ≤p − 1. Similarly if xζ−1 + y = γp then xs−

q−1p i + y should be a p-th

power modulo q, for all 1 ≤ i ≤ p− 1. Finally, if x−y1+ζ + y = γp, then

(x− y)(1 + sq−1

p i)−1 + y has to be a p-th power modulo q.For example, if p = 5 and q = 31, then s = 3 is a primitive root

modulo 31. Using PARI/GP, we see that there are no a, b ∈ (Z/31Z)∗

such that a + 36ib is fifth power modulo 31, simultaneously for i =1, 2, 3 and 4. Hence if x + yζ5 = γ5 or xζ−1 + y = γ5, then xy isdivisible by 31. There are also no a, b ∈ (Z/31Z)∗ such that a + (1 +36i)−1b is a fifth power modulo 31. So if x−y

1+ζ + y = γ5 then 31 divides(x− y)y.

We did the same computations for primes the 5 ≤ p ≤ 19 and somesmall primes q ≡ 1 mod p. The results are the given in the followingtables.

7.2 computational results 65

Lemma 7.6. Let x, y be coprime integers satisfying N(x + yζ) = zp. Forprimes p in the table below, if p divides xy, then xy is also divisible by theprimes in the right column of the following table:

p q such that q | xy

5 11, 31, 41, 61, 101, 251, 271, 751

7 29, 43, 71, 113, 127, 197, 211, 239, 281, 337, 379, 421, 449, 463,

491, 547, 631, 659, 673, 701, 743, 757, 827, 883, 911, 953, 967

11 23, 67, 89, 199, 331, 353, 397, 419, 463,

617, 661, 683, 727, 859, 881, 947, 991

13 53, 79, 131, 157, 313, 443, 521, 547, 599, 677, 859, 911, 1093, 1171

17 103, 137, 239, 307, 409, 443, 613,

647, 919, 953, 1021, 1123, 1259, 1327

19 191, 229, 419, 457, 571, 761, 1103, 1217, 1483, 1559, 1597

If p divides x− y then (x− y)y is divisible by the following primes:

p q such that q | (x− y)y

5 11, 31, 41, 61, 71, 101, 151, 191, 241, 311

7 29, 43, 71, 113, 127, 211, 239, 281, 337, 379, 449, 463,

491, 547, 617, 631, 659, 673, 701, 743, 757, 827, 883, 953

11 23, 67, 89, 199, 331, 353, 397, 419, 463,

617, 661, 683, 727, 859, 881, 947, 991

13 53, 79, 131, 157, 313, 443, 521, 547, 599, 677, 859, 911, 1093, 1171

17 103, 137, 239, 307, 409, 443, 613,

647, 919, 953, 1021, 1123, 1259, 1327

19 191, 229, 419, 457, 571, 761, 1103, 1217, 1483, 1559, 1597

Remark 7.7. We checked this for q < 1000, and, where necessary, for moreprimes q in order to ensure that the product of the primes q corresponding top is at least 5× 108. For the primes p ≥ 7 it should be possible to find morecorresponding q, given extra computational time. For p = 5 the primes q inthe list are all q < 10000.

Using the data from the previous two lemmas we can give a lowerbound on the absolute value of x and y.

Theorem 7.8. Let p ≥ 5 be a regular prime and let x and y be coprimeintegers such that N(x + yζ) = zp. Then max(|x|, |y|) > 5× 108.

Proof. By Corollary 7.4 we know that one x, y and x − y should bedivisible by Mp. So we see

2 max(|x|, |y|) > max(|x|, |y|, |x− y|) ≥ Mp.

Lemma 7.5 tells us that for p ≥ 23 we have Mp ≥ 109 and thusmax(|x|, |y|) > 5× 108. For the remaining primes we use Lemma 7.6.

66 regular primes

For p = 5, if 5 divides xy, then xy is divisible by

5 · 11 · 31 · 41 · 61 · 101 · 251 · 271 · 751 = 22000998843422555.

Hence we find

max(|x|, |y|) >√|xy| ≥

√22000998843422555 > 1.4× 109.

If 5 divides x− y then (x− y)y is divisible by

5 · 11 · 31 · 41 · 61 · 71 · 101 · 151 · 191 · 241 · 251 · 311 = 61600621624429072505.

Therefore we see

max(|x|, |y|) > 12|(x− y)y| ≥ 1

2

√61600621624429072505 > 4× 1010.

In both cases we have max(|x|, |y|) > 5 · 108. The calculations for theprimes p = 7, . . . , 19 are done in the same manner.

We conclude with some remarks on the equation N(a + bζ + cζ2) =

zp. In this case, it is not easy to give conditions for when a + bζ + cζ2

is singular, as we saw in chapter 4. If it is singular, then from chapter 5,Example 5.4, we can conclude that p divides (a− c)(ab+ bc+ 8ac− b2).This is of course not as strong as Theorem 7.2. We know that a + bζ +

cζ2 = uγp for some unit u and γ in OK. Even if we assume that pdivides a− c, we cannot deduce anything like we did in Theorem 7.3.So it seems that these methods are not strong enough to attack theequation N(a + bζ + cζ2) = zp.

8W O L S T E N H O L M E ’ S T H E O R E M

In 1862, Joseph Wolstenholme [37] proved the following theorem:

Theorem 8.1. For any prime p ≥ 5 we have:(2p− 1p− 1

)≡ 1 mod p3, (8.1)

p−1

∑k=1

1k≡ 0 mod p2, (8.2)

p−1

∑k=1

1k2 ≡ 0 mod p. (8.3)

Let us set H(i)n = ∑n

k=11ki and Hn = H(1)

n . So Wolstenholme’s the-

orem says Hp−1 ≡ 0 mod p2 and H(2)p−1 ≡ 0 mod p. In 1900, James

Glaisher [8] showed:

Theorem 8.2. For any prime p ≥ 5 we have

Hp−1 ≡ −13

p2Bp−3 mod p3.

This means that a prime p divides the Bernoulli number Bp−3 if andonly if Hp−1 ≡ 0 mod p3. So the condition p | Bp−3 that we found inchapter 5 can also be charaterised in terms of harmonic sums.

Over the years many generalisations of Theorem 8.1 have beeninvestigated. For a nice survey, see the article [23]. One generalisationis to consider (2p−1

p−1 ) modulo higher powers of p and express it in

terms of harmonic sums H(i)p−1, see for example [24, 35]. This leads to

statements such as ([23], p. 6):(2p− 1p− 1

)≡ 1 + 2pHp−1 +

23

p3H(3)p−1 + 2p2 (Hp−1

)2 (8.4)

+43

p4Hp−1H(3)p−1 +

25

p5H(5)p−1 mod p9.

In the article [11], the author used the p-adic logarithm and exponen-tial functions to study congruence conditions for binomial coefficients.In this chapter we will show that many Wolstenholme type theoremssuch as (8.4) can be proven in a uniform manner by using the proper-ties of the p-adic logarithm and the p-adic exponential function. Usingthis method, we can extend identities such as (8.4) to arbitrary high

67

68 wolstenholme’s theorem

powers of p. This leads to Theorem 8.6 below. As an application, inCorollary 8.7 we prove the following congruence:(

2p− 1p− 1

)≡ 1 + 2pHp−1 +

23

p3H(3)p−1 +

25

p5H(5)p−1 +

27

p7H(7)p−1

+29

p9H(9)p−1 + 2p2 (Hp−1

)2+

29

p6(

H(3)p−1

)2

+43

p4Hp−1H(3)p−1 +

45

p6Hp−1H(5)p−1

+43

p3 (Hp−1)3

+43

p5 (Hp−1)2 H(3)

p−1 mod p12.

Following the article [22], we call a prime p a Wolstenholme primeif it satisfies (2p−1

p−1 ) ≡ 1 mod p4. The author of that article showsthat the primes 16843 and 2124679 are the only two Wolstenholmeprimes smaller than 109. Moreover he conjectured there are infinitelymany Wolstenholme primes. Using the p-adic logarithm, we obtaincharacterisations of Wolstenholme primes in terms of harmonic sums.This leads to a characterisation of Wolstenholme primes modulo p12 inTheorem 8.11. Thereby we confirm a conjecture from [23] (Remark 24),which says that a prime p is Wolstenholme if and only if it satisfies(

2p− 1p− 1

)≡ 1 + 2pHp−1 +

23

p3Hp−1 +25

p5H(5)p−1 mod p8.

In the second part of this chapter we apply the logarithmic methodof the first part to norms of integers of cyclotomic fields. The ana-log of Lemma 8.4 is given by Theorem 8.15. As an application, inProposition 8.17 we prove the identity

p−1

∑k=1

1k

(2kk

)≡ 0 mod p2.

This is the weaker, modulo p2, version of a result from [35]. Moreover,in Theorem 8.20, we get the congruence

p−1

∑k=1

(−1)k−1 1k

(2kk

)≡

(1− L2p)(L2

p − 3)2p

mod p2,

where Lp is the p-th Lucas number. This improves on a result in [34].Finally we find a congruence involving the integers of the sequenceA092765 of the OEIS.

8.1 logarithms and binomial coefficients

In this section we shall see how the relation between binomial co-efficients and harmonic sums becomes clear if you use the p-adiclogarithm. For basic properties of the p-adic logarithm, see section 3

of chapter 5 and the references there. In addition to this, we will needthe following lemma.

8.1 logarithms and binomial coefficients 69

Lemma 8.3. Let x ∈ Zp satisfy x ≡ 1 mod p, and let 0 < k < p− 2.Then

x ≡k

∑i=0

(logp(x)

)i

i!mod pr(k+1)

if and only if x ≡ 1 mod pr.

Proof. For any r ≥ 1, we have x ≡ 1 mod pr if and only if logp(x) ≡ 0mod pr. Moreover, we have

x = expp logp(x) = ∑i≥0

(logp(x)

)i

i!,

so it is enough to show that for any y ≡ 0 mod p we have

∑i≥k+1

yi

i!≡ 0 mod pr(k+1)

if and only if y ≡ 0 mod pr. This amounts to showing that νp

(yi

i!

)>

νp

(yk+1

(k+1)!

), because then we have νp

(∑i≥k+1

yi

i!

)= νp

(yk+1

(k+1)!

). But

νp

(yk+2

(k+2)!

)> νp

(yk+1

(k+1)!

)follows from the fact that k < p− 2, whilst

for i > k + 2 the claim is obvious.

Our main tool will be the next lemma.

Lemma 8.4. In Qp we have the following three identities:

∞

∑i=1

pi

iH(i)

p−1 = 0, (8.5)

∞

∑i=1

(−1)i−1 pi

iH(i)

p−1 = logp

(2p− 1p− 1

), (8.6)

2∞

∑i=1

p2i−1

(2i− 1)H(2i−1)

p−1 = logp

(2p− 1p− 1

). (8.7)

Proof. Note that, since p is odd, 1 = ∏p−1k=1

k−pk . Hence we get:

0 = − logp(1) = −p−1

∑k=1

logp(1−pk)

=p−1

∑k=1

∞

∑i=1

pi

kii=

∞

∑i=1

pi

iH(i)

p−1.

For the second identity, we note (2p−1p−1 ) = ∏

p−1k=1

p+kk ≡ 1 mod p,

which allows us to proceed in exactly the same manner as for the firstidentity. Finally, the third identity is obtained by adding the first twoto each other.


The following lemma is useful for simplifying some of the expres-sions we will obtain.

Lemma 8.5. For any i ∈N we have:

H(i)p−1 ≡

0 mod p, if p− 1 - i

−1 mod p, if p− 1 | i

If i is odd, then:

H(i)p−1 ≡

0 mod p2, if p− 1 - i + 1ip2 mod p2, if p− 1 | i + 1

Proof. The first claim follows easily from the existence of a primitiveroot modulo p. For the second claim, note that we have

p−1

∑k=1

1ki =

(p−1)/2

∑k=1

ki + (p− k)i

ki(p− k)i

≡(p−1)/2

∑k=1

ki + (−k)i + ip(−k)i−1

ki(p− k)i mod p2.

So, if i is odd, we see that

H(i)p−1 ≡ −ip

(p−1)/2

∑k=1

1ki+1 ≡ −

ip2

H(i+1)p−1 mod p2.

Lemma 8.5 covers parts (8.2) and (8.3) of Wolstenholme’s theo-rem 8.1. Note that the equivalence of (8.2) and (8.3) also follows fromour Lemma 8.4 by looking at equation (8.5) modulo p3, after know-ing that H(3)

p−1 ≡ 0 mod p. The equivalence between (8.1) and theother two parts of Wolstenholme’s theorem follows by looking at (8.7)modulo p3.

For p = 3 we have (53) = 1 + 32, and by Wolstenholme’s theorem

we have (2p−1p−1 ) ≡ 1 mod p3 for p ≥ 5. Therefore, by the general

properties of the p-adic logarithm and exponential, we know thatexpp logp (

2p−1p−1 ) = (2p−1

p−1 ). Combined with Lemma 8.4, this allows us

to express (2p−1p−1 ) in terms of harmonic sums.

Theorem 8.6. Let p be an odd prime. In Qp we have the identity:(2p− 1p− 1

)= ∑

n≥0

1n!

(2 ∑

i≥1

p2i−1

2i− 1H(2i−1)

p−1

)n

(8.8)

8.1 logarithms and binomial coefficients 71

Proof. The proof follows immediately from the combination of Lemma 8.4and the definition of the exponential function:(

2p− 1p− 1

)= expp logp

(2p− 1p− 1

)= ∑

n≥0

1n!

(logp

(2p− 1p− 1

))n

= ∑n≥0

1n!

(2 ∑

i≥1

p2i−1

2i− 1H(2i−1)

p−1

)n

By looking modulo a specific power pk, and using Lemma 8.5 toeliminate the terms which are congruent to 0 modulo pk, we can obtainan extension of previously known results (see Remark 8.9 below).

Corollary 8.7. For any prime p ≥ 11 we have the identity:(2p− 1p− 1

)≡ 1 + 2pHp−1 +

23

p3H(3)p−1 +

25

p5H(5)p−1 +

27

p7H(7)p−1

+29

p9H(9)p−1 + 2p2 (Hp−1

)2+

29

p6(

H(3)p−1

)2

+43

p4Hp−1H(3)p−1 +

45

p6Hp−1H(5)p−1

+43

p3 (Hp−1)3

+43

p5 (Hp−1)2 H(3)

p−1 mod p12.

Proof. The proof is immediate by looking at (8.8) and using Lemma 8.5.

Remark 8.8. The lower bound on the prime p is there to ensure that certainterms dissapear by using Lemma 8.5. Naturally, for the smaller primes westill obtain a corresponding identity using (8.8), but there will be some extraterms in the above formula.

Remark 8.9. Looking modulo p6, we obtain for all primes p ≥ 5:(2p− 1p− 1

)≡ 1 + 2pHp−1 +

23

p3H(3)p−1 mod p6.

This was originally proved in [35]. Looking modulo p9, we get the identitymentioned in the introduction, i.e. for all primes p ≥ 7:(

2p− 1p− 1

)≡ 1 + 2pHp−1 +

23

p3H(3)p−1 +

25

p5H(5)p−1

+ 2p2 (Hp−1)2

+43

p4Hp−1H(3)p−1 mod p9.

So far we have been looking at identities that hold for all primes.Now we will shift our attention to Wolstenholme primes. By usingLemma 8.4, one can quickly prove the following well-known fact.


Theorem 8.10. For any prime p the following are equivalent:(2p− 1p− 1

)≡ 1 mod p4, (8.9)

Hp−1 ≡ 0 mod p3, (8.10)

H(2)p−1 ≡ 0 mod p2. (8.11)

Proof. From Lemma 8.3, we know that, for p > 3, (2p−1p−1 ) ≡ 1 mod p4

if and only if logp (2p−1p−1 ) ≡ 0 mod p4. Looking at (8.7) modulo p4, we

see that this happens if and only if Hp−1 ≡ 0 mod p3. Finally, looking

at (8.6) modulo p4 we see that Hp−1 ≡ p2 H(2)

p−1 mod p3, which provesthe theorem.

As in the article [22], we call a prime satisfying any of these condi-tions a Wolstenholme prime. We can use the same logarithmic method togive characterisations of Wolstenholme primes in terms of harmonicsums.

Theorem 8.11. A prime p is Wolstenholme if and only if:(2p− 1p− 1

)≡ 1 + 2pHp−1 +

23

p3H(3)p−1 +

25

p5H(5)p−1 +

27

p7H(7)p−1

+29

p9H(9)p−1 + 2p2H2

p−1 +43

p4Hp−1H(3)p−1

+45

p6Hp−1H(5)p−1 +

29

p6(

H(3)p−1

)2mod p12.

Proof. By Lemma 8.3 we know that p is Wolstenholme if and only if(2p− 1p− 1

)≡ 1 + logp

(2p− 1p− 1

)+

12

(logp

(2p− 1p− 1

))2

mod p12.

We substitute the expression (8.7) we obtained for the logarithm intothis equality. We obtain the result of the theorem after eliminating theterms which are 0 modulo p12, by using Lemma 8.5 and Theorem 8.10.Since the first primes are not Wolstenholme, we do not need to concernourselves with a lower bound for p like in the previous section.

Remark 8.12. By looking modulo p8 we obtain that p is a Wolstenholmeprime if and only if(

2p− 1p− 1

)≡ 1 + 2pHp−1 +

23

p3Hp−1 +25

p5H(5)p−1 mod p8.

This is identity (49) of [23]. In Remark 24 of that article, the author askswhether this identity characterizes Wolstenholme primes. The theorem aboveproves that this is indeed the case.

In [22] the author conjectures that there are infinitely many Wol-stenholme primes, but no primes p such that (2p−1

p−1 ) ≡ 1 mod p5. Forsake of reference we also give the characterisation for these primes interms of harmonic sums.

8.2 logarithms and cyclotomic integers 73

Proposition 8.13. For any prime p, the following are equivalent:(2p− 1p− 1

)≡ 1 mod p5,

Hp−1 ≡ 0 mod p4,

H(2)p−1 ≡ 0 mod p3.

Proof. For p > 5, we know (2p−1p−1 ) ≡ 1 mod p5 if and only if

logp

(2p− 1p− 1

)≡ 0 mod p5.

We know

logp

(2p− 1p− 1

)≡ 2pHp−1 mod p5,

and also

logp

(2p− 1p− 1

)≡ pHp−1 −

p2

2H(2)

p−1 mod p5,

which proves the equivalence of the three statements.

Just like in the previous two cases, we will extend this to a higherpower of p.

Proposition 8.14. A prime p satisfies (2p−1p−1 ) ≡ 1 mod p5 if and only if(

2p− 1p− 1

)≡ 1 + 2Hp−1 +

23

p3H(3)p−1 +

25

p5H(5)p−1 +

27

p7H(7)p−1 mod p10.

Proof. Again, the proof is a simple application of Lemmas 8.3 and8.4.

8.2 logarithms and cyclotomic integers

In thiss section we will be working in the p-th cyclotomic field K =

Q(ζp) again, with the corresonding notations.In exactly the same way as we did for binomial coefficients modulo

p, we can use the p-adic logarithm to obtain congruence relations fornorms of elements of OK. We will denote by νπ the valuation withrespect to the prime ideal πOK. The analog of Lemma 8.4 is given bythe following theorem.

Theorem 8.15. Let α ∈ OK be such that νπ(α) ≥ 1. Then we have, in Qp:

∞

∑k=1

1k

Tr(αk) = − logp N(1− α), (8.12)


If moreover we know that 1− α is a unit, then

∞

∑k=1

1k

Tr(αk) = 0, (8.13)

−2∞

∑k=1

Tr(α2k−1)

2k− 1= logp N(1 + α), (8.14)

and in particular we get:

p−1

∑k=1

1k

Tr(αk) ≡ 0 mod pνπ(α). (8.15)

Proof. Applying the p-adic logarithm to N(1− α) we see that:

logp(N(1− α)) =p−1

∑j=1

logp(1− σj(α))

= −p−1

∑j=1

∞

∑k=1

σj(αk)

i

= −∞

∑k=1

1k

Tr(αk),

which gives the first identity. If 1− α is a unit, then we have N(1− α) =

1, so that logp N(1− α) = 0. For the third identity, note that we have

logp(N(1 + α)) = ∑i≥1

(−1)i+1

iTr(αi).

Adding this to the second equation, we get

logp(N(1 + α)) = ∑i≥1

(−1) + (−1)i+1

iTr(αi) = −2 ∑

i≥1

Tr(α2k−1)

2k− 1.

Now we investigate the p-adic valuation of 1k Tr(αk). We know that

νπ(Tr(x)) ≥ νπ(x), so we find:

νπ

(1k

Tr(αk)

)≥ kνπ(α)− νπ(k).

If p - k, then of course νπ

( 1k Tr(αk)

)≥ kνπ(α). When νp(k) = t ≥ 1,

then, using Bernoulli’s inequality (1 + x)r ≥ 1 + rx, one gets

kνπ(α)− νπ(k) ≥ ptνπ(α)− t(p− 1)

≥ (1 + t(p− 1)) νπ(α)− t(p− 1)

≥ νπ(α) + t(p− 1)(νπ(α)− 1).

In particular, we see that for any k ≥ p one has

νπ

(1k

Tr(αk)

)> (p− 1)(νπ(α)− 1),


and since 1k Tr(αk) is a rational number, this means that

νp

(1k

Tr(αk)

)≥ νπ(α).

The congruence (8.15) then follows from equality (8.13).

The discrete Fourier transform will be useful for calculating thetrace of an element:

Lemma 8.16. Let p be an odd prime and for j = 0, . . . , p− 1, let f j be acomplex number. If gi = ∑

p−1j=0 f jζ

ij then f j =1p ∑

p−1i=0 giζ

−ij.

Proof. This is a simple calculation:

p−1

∑i=0

giζ−ij =

p−1

∑i=0

(p−1

∑k=0

fkζ ik

)ζ−ij =

p−1

∑k=0

fk

(p−1

∑i=0

ζ i(k−j)

)= p f j.

We are now ready to apply this to specific units.

Proposition 8.17. For any prime p ≥ 5 we have

p−1

∑k=1

1k

(2kk

)≡ 0 mod p2.

Proof. Consider the element ζ − 1 + ζ−1 = 1 + ζ−1(1− ζ)2, so that inthis case we have α = −ζ−1(1− ζ)2. Note that for p > 3 we have

ζ(ζ − 1 + ζ−1)(ζ + 1) = ζ3 + 1 =1− ζ6

1− ζ3 ,

so that 1− α is indeed a unit. We will calculate the trace of αk. Weknow

(1− ζ i)2k =∞

∑r=−∞

(2kr

)(−1)rζ ir =

p−1

∑j=0

(∑

r≡j mod p(−1)r

(2kr

))ζ ij.

Now we apply Lemma 8.16 with gi = (1− ζ i)2k to get

∑r≡j mod p

(−1)r(

2kr

)=

1p

p−1

∑i=1

(1− ζ i)2kζ−ij =1p

Tr(ζ−j(1− ζ)2k).

In particular, we have

Tr(αk) = p ∑r≡k mod p

(−1)k+r(

2kr

),

and for 0 < k < p this implies that

Tr(αk) = p(

2kk

).


Because νπ(α) = 2, (8.15) immediately gives us that

pp−1

∑k=1

1k

(2kk

)=

p−1

∑k=1

1k

Tr(αk) ≡ 0 mod p2.

We can easily get the result modulo p2 by using the fact that, for k > p,one has νπ(

αk

k ) ≥ 2(p + 1), and thus 1k Tr(αk) ≡ 0 mod p3. This means

that we have

p−1

∑k=1

1k

Tr(αk) ≡ − 1p

Tr(αp) = − ∑r≡0 mod p

(−1)p+r(

2pr

)= −

((2pp

)− 2)≡ 0 mod p3,

by Wolstenholme’s theorem.

Remark 8.18. These trace calculations can be used to find some identities ontheir own. For example we can write α = −ζ(1− ζ)2 from Proposition 8.17also as α = (1− ζ)(1− ζ−1). Doing the trace calculation with this secondform we get

Tr(αk) = p ∑r≡s mod p

(−1)r+s(

kr

)(ks

).

If k < p the only solution for r ≡ s mod p is r = s. Combining this withthe expression we found for Tr(αk) in Proposition 8.17, we find(

2kk

)=

k

∑r=0

(kr

)2

(which is also an easy consequence of Vandermonde’s identity).

Remark 8.19. In [35] the author shows, via a different method, that in factwe have

p−1

∑k=1

1k

(2kk

)≡ −8

3Hp−1 ≡ −

23

((2pp

)− 2)

mod p4.

In chapter 3 we saw that for certain cyclotomic elements, the normcan be expressed completely by a recurrence sequence. We apply thisin the following theorem.

Theorem 8.20. Let p be an odd prime, and let Lp be the p-th Lucas number,i.e. L0 = 2, L1 = 1 and Lk+2 = Lk+1 + Lk. Then we have:

p−1

∑k=1

(−1)k−1 1k

(2kk

)≡

(1− L2p)(L2

p − 3)2p

mod p2.

Proof. Again, let α = −ζ−1(1 − ζ)2 as in the previous proposition.Then 1 + α = 3− ζ − ζ−1. Note that we have

(ζ2 − ζ − 1)(ζ−2 − ζ−1 − 1) = 3− ζ2 − ζ−2


and therefore N(1 + α) = N(ζ2− ζ − 1)2. By using Theorem 3, we seethat N(ζ2 − ζ − 1) = Lp. So N(1 + α) = L2

p, and we can use this tocalculate logp(N(1 + α)). We find:

logp(N(1 + α)) ≡ N(1 + α)− 1− 12(N(1 + α)− 1)2

≡ L2p − 1− 1

2

(L2

p − 1)2

≡(1− L2

p)(L2p − 3)

2mod p3.

On the other hand, in the previous proposition we saw that Tr(α)k ≡0 mod p3 for k ≥ p, and thus (8.12) tells us that

logp(N(1+ α)) ≡p−1

∑k=1

(−1)k−1 1k

Tr(αk) ≡ pp−1

∑k=1

(−1)k−1 1k

(2kk

)mod p3.

Together these give the desired result.

Remark 8.21. In the article [34], the authors show that

p−1

∑k=1

(−1)k−1 1k

(2kk

)≡ 5

Fp−( p5 )

pmod p.

Here Fk is the k-th Fibonacci number. Using the the fact Fp−( p5 )≡ L2

p − 1mod p it is easy to see that Theorem 8.20 reduces to this result modulo p.

A prime p for which Fp−( p5 )≡ 0 mod p2 is called a Wall-Sun-Sun

prime. It has been shown that if the first case of Fermat’s last theorem fails,then p must be a Wall-Sun-Sun prime. It is interesting to compare this withthe case of Wolstenholme primes. From chapter 5 we know that if the firstcase of Fermat’s last theorem fails, then p must divide the Bernoulli numberBp−3, which happens if and only if (2p−1

p−1 ) ≡ 1 mod p4.

Now we apply our results to a different element α. The result weobtain is concerned with the following sequence. For k ≥ 0, define

ak :=k

∑s=0

(ks

)(k

2k− 3s

).

This is sequence A092765 of the OEIS [33], starting with

1, 0, 4, 6, 36, 100, 430, 1470, 5796, . . .

Proposition 8.22. For a prime p ≥ 7 we have

(p−1)/2

∑k=1

(−1)k ak

k≡ 0 mod p.


Proof. Consider the element α = −ζ2 + ζ + ζ−1 − ζ−2 = −ζ−2(1−ζ)(1− ζ3). Then

1− α = ζ2 − ζ + 1− ζ−1 + ζ−2 = ζ−2 1 + ζ5

1 + ζ

is indeed a unit. To calculate the trace we use the same trick as in theprevious example.

(1− ζ i)k(1− ζ3i)k =

(∞

∑r=−∞

(kr

)(−1)rζ ir

)(∞

∑s=−∞

(ks

)(−1)sζ3is

)

=∞

∑r,s=−∞

(−1)r+s(

kr

)(ks

)ζ i(3s+r)

=p−1

∑j=0

(∑

3s+r≡j mod p(−1)r+s

(kr

)(ks

))ζ ij

By Lemma 8.16 we get

∑3s+r≡j mod p

(−1)r+s(

kr

)(ks

)=

1p

p−1

∑i=1

(1− ζ i)k(1− ζ3i)kζ−ij

=1p

Tr(ζ−j(1− ζ)k(1− ζ3)k).

Because νπ(α) = 2 we know that νπ

( 1k Tr(αk)

)> p− 1 for k > p−1

2 ,and thus νπ

( 1k Tr(αk)

)> 1. Now Theorem 8.15 tells us that

(p−1)/2

∑k=1

1k

Tr(αk) ≡ 0 mod p2.

For 0 ≤ k ≤ p−12 and 0 ≤ r, s ≤ k, the only solution to 3s + r ≡ 2k

mod p is r = 2k− 3s. Thus we find

0 ≡(p−1)/2

∑k=1

1k

Tr(αk)

≡ p(p−1)/2

∑k=1

1k

(∑

3s+r≡2k mod p(−1)k+r+s

(kr

)(ks

))

≡ p(p−1)/2

∑k=1

(−1)k

k

(k

∑s=0

(ks

)(k

2k− 3s

))mod p2

The previous results were all obtained by starting with an elementof the cyclotomic field, calculating its norm and trace, and then usingthe logarithm to obtain a congruence. Clearly one can get many moreresults in this way. However, a more interesting problem is the inverseproblem. Suppose there is a congruence, similar to those in this chapter,


one would like to prove. Is there a corresponding cyclotomic elementthat does the trick? Also, it would be interesting to see if there areother known identities that can be proved or improved using thelogarithmic method.

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On Some Norm Equations Over Cyclotomic Fields · ABSTRACT For an odd prime p, the equation xp + yp...

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