the structure of the n-th roots of unity · 16 boaz cohen the p-adic number eld and by o p the ring...

International Mathematical Forum, Vol. 13, 2018, no. 1, 15 - 64HIKARI Ltd, www.m-hikari.com

https://doi.org/10.12988/imf.2018.712101

The Structure of the n-th Roots of Unity

in Residue Rings of Prime Ideals P over

p in Algebraic Number Fields

Part III: n-th Roots of Unity when n = pb

Boaz Cohen

Department of Pure MathematicsTel Aviv University

Ramat Aviv, Tel Aviv 69978, Israel

Copyright c© 2018 Boaz Cohen. This article is distributed under the Creative Commons

Attribution License, which permits unrestricted use, distribution, and reproduction in any

medium, provided the original work is properly cited.

Abstract

In this paper we continue the study initiated in parts I and II ofdetermining the solutions of the congruence xn ≡ 1 (mod M) over OK,where n is an arbitrary positive integer, K is an algebraic number fieldand M is an ideal in OK. In this part we shall determine the solutionsof the special case xp

b ≡ 1(modPa), where a and b are positive integers,P denotes a prime ideal in OK lying over the rational prime number pregardless of whether P‖(p) or P2 | (p).

Mathematics Subject Classification: 11R04, 11Sxx, 11A07

Keywords: Roots of unity, Residue rings, Prime ideals, Algebraic numberfields, p-adic fields

1 Introduction

Let K be an algebraic number field. Denote by OK the ring of integers ofK and by M an ideal in OK. In these settings, we denote by M(OK/M) acomplete residue system modulo M. Given a prime ideal P, we denote by KP

16 Boaz Cohen

the P-adic number field and by OP the ring of P-adic integers of KP. Readersare referred to [1] for more details regarding the exact definitions of the objectsmentioned above.

In this paper we continue the study initiated in [1] and [2], whose aim is todetermine the solutions over OK of the congruence xn ≡ 1 (mod M), where nis an arbitrary positive integer.

In Part I and Part II we determined the solutions of the congruence xn ≡1 (mod Pa) for a prime ideal P in two special cases: In Part I in the casewhere p - n and in Part II in the case where P‖(p), focusing in Part II on nof the form n = pb. In this paper we continue studying the case n = pb, forarbitrary P, regardless of whether P‖(p) or P2 | (p). Solving the congruencexp

b ≡ 1(modPa) for P with a general ramification index e will require differentmethods than those we used in Part II. Nevertheless, the results will be ade-quate for both cases. Our study will be carried out separately in three cases.These cases are described schematically as follows:

2 6 a 6 e

Theorem 3.2(see page 9)

e < a

p− 1 - e p− 1 | eTheorem 4.6(see page 19)

Theorem 4.9(see page 24)

An integral part of this work is the exploration of the solutions of the equationxn = 1 in the P-adic field KP. In Part I we studied the solutions of xn = 1when p - n and in Part II we studied the solutions of xn = 1 when P‖(p). Inthis paper the structure of the solutions of xp

b= 1 is fully determined. See

Theorem 5.2 (page 41) for more details.In this paper the notation and preliminaries summarized of Part I are

assumed. Furthermore, we shall use freely results from Part I and Part II. Forclarity, we shall refer to results from Part I and Part II by adding the letter “I”or “II”, respectively, to the number of the result. For example, Theorem I.4.3refers to Theorem 4.3 of Part I.

2 The lifting method

Given a prime ideal P in OK lying over the rational prime number p andpositive integers a and n, consider the congruence xn ≡ 1 (mod Pa). In orderto solve this congruence, it is worthwhile noticing that if xn ≡ 1(modPa), thenxn ≡ 1(modPa−1). This means that the solutions of xn ≡ 1(modPa) are to befound among those of xn ≡ 1 (mod Pa−1). Suppose that x ≡ α (mod Pa−1) is a

The structure of the n-th roots of unity 17

solution of xn ≡ 1 (mod Pa−1) and let γ ∈ P \ P2 be chosen arbitrarily. Now,since P‖(γ), we get that gcd((γa−1),Pa) = Pa−1. Thus, by Proposition II.2.1

x ≡ α (mod Pa−1)⇔ x ≡ α + θ1γa−1, . . . , α + θNPγ

a−1 (mod Pa),

where {θ1, θ2, . . . , θNP} is a complete residue system modulo P. It may happen,of course, that not every number α + θiγ

a−1 in the above list is a solution ofthe original congruence modulo Pa. Thus, it is desired to determine the θi’sso that α + θiγ

a−1 satisfies the congruence modulo Pa. Let us illustrate thisprocedure by two examples.

Example 2.1. We want to solve x3 ≡ 1 (mod P3) in Z for P = (7). Clearly,P = (7) is a prime ideal in Z with NP = 7, so we can choose the completeresidue system M = {0, 1, 2, . . . , 6} and the uniformizer γ = 7.

We first look for solutions modulo P, that is, solutions for x3 ≡ 1 (mod 7).Trying successively the numbers 0, 1, 2, . . . , 6, we find that

x ≡ 1, 2, 4 (mod P)

are the solution of x3 ≡ 1 (mod P). For each of 1, 2 and 4, the next step is todetermine the θ’s for which

1 + θγ ; 2 + θγ ; 4 + θγ

are solutions of x3 ≡ 1 (mod P2). Let us illustrate this process for 2 + θγ. Wewish to find the θ’s in {0, 1, 2, . . . , 6} for which (2 + θγ)3 ≡ 1 (mod P2). Since

(2 + θγ)3 = 8 + 12θγ + 6θ2γ2 + θ3γ3 ≡ 8 + 12θγ (mod P2),

the congruence we need to solve is 8 + 12θγ ≡ 1 (mod P2), that is 12θγ ≡−7 (mod P2). But γ = 7 so 12θ ≡ −1 (mod P) and hence θ ≡ 4 (mod P).Therefore θ = 4, which yields the following solution of x3 ≡ 1 (mod P2):

x ≡ 2 + 4γ (mod P2).

Similarly, applying this process for 1 + θγ and 4 + θγ, we obtain two moresolutions:

x ≡ 1 + 0γ (mod P2) ; x ≡ 4 + 2γ (mod P2).

We continue to determine the θ’s for which

1 + 0γ + θγ2 ; 2 + 4γ + θγ2 ; 4 + 2γ + θγ2

18 Boaz Cohen

satisfy x3 ≡ 1(modP3). Again, we shall illustrate this process for 2+4γ+θγ2:

(2 + 4γ + θγ2)3 ≡ 1 (mod P3)8 + 48γ + 96γ2 + 12θγ2 + γ3 · (someting) ≡ 1 (mod P3)

48γ + 12θγ2 + 96γ2 ≡ −7 (mod P3)m

48 + 12θγ + 96γ ≡ −1 (mod P2)12θγ + 96γ ≡ −49 (mod P2)

m12θ + 96 ≡ 0 (mod P)

mθ ≡ 6 (mod P).

Therefore θ = 6, which yields the following solution of x3 ≡ 1 (mod P3):

x ≡ 2 + 4γ + 6γ2 (mod P3).

Similarly, applying this process for 1 + 0γ + θγ2 and 4 + 2γ + θγ2, we obtaintwo more solutions:

x ≡ 1 + 0γ + 0γ2 (mod P3) ; x ≡ 4 + 2γ + 0γ2 (mod P3).

To conclude, the incongruent solutions of x3 ≡ 1 (mod P3) are

x ≡ 1 + 0γ + 0γ2 = 1 (mod P3)x ≡ 2 + 4γ + 6γ2 = 324 (mod P3)x ≡ 4 + 2γ + 0γ2 = 18 (mod P3).

Example 2.2. We want to solve x3 ≡ 1 (mod P3) in Z for P = (3). Clearly,P = (3) is a prime ideal in Z with NP = 3, so we can choose the completeresidue system M = {−1, 0, 1} and the uniformizer γ = 3.

We first look for solutions modulo P, that is, solutions for x3 ≡ 1 (mod 3).Trying successively the numbers −1, 0, 1, we find that x ≡ 1(modP) is the onlysolution of x3 ≡ 1 (mod P). The next step is to determine the θ’s for whichx ≡ 1 + θγ (mod P2) is a solution of x3 ≡ 1 (mod P2), that is, we wish to findthe θ’s in {−1, 0, 1} for which (1 + θγ)3 ≡ 1 (mod P2). Since

(1 + θγ)3 = 1 + 3θγ + 3θ2γ2 + θ3γ3 ≡ 1 + 3θγ ≡ 1 (mod P2),

the congruence we need to solve is 1 ≡ 1 (mod P2) which is trivially true.Hence, any θ from {−1, 0, 1} satisfies this congruence. Accordantly, we obtainthat the solutions modulo P2 are

x ≡ 1− 1γ (mod P2) ; x ≡ 1 + 0γ (mod P2) ; x ≡ 1 + 1γ (mod P2).


We continue to determine the θ’s for which 1 + 0γ + θγ2 is a solution ofx3 ≡ 1 (mod P3). Since

(1 + θγ2)3 = 1 + 3θγ2 + 3θ2γ4 + θ3γ6 ≡ 1 (mod P3),

it follows that any θ from {−1, 0, 1} satisfies is suitable. We do the same with1 + 1γ + θγ2. Since

(1 + γ + θγ2)3 = 1 + 3γ + 3(θ + 1)γ2 + γ3 · (something) ≡ 1 + 3γ (mod P3),

the congruence we need to solve is 1+3γ ≡ 1(modP3), that is 3γ ≡ 0(modP3),which is false. Thus, in this case there is no suitable θ. As one can verify,also the option 1 − γ + θγ2 does not yield any solution, so we may concludethat the incongruent solutions of x3 ≡ 1 (mod P3) are

x ≡ 1 + 0γ − 1γ2 = −8 (mod P3)x ≡ 1 + 0γ + 0γ2 = 1 (mod P3)x ≡ 1 + 0γ + 1γ2 = 10 (mod P3).

Generally, the process of solving this kind of congruences is called the“lifting method”. This process can be described schematically using a rootedtree. The root is unlabeled. The vertices of the tree in the a-th level, besidethe root, are the incongruent solutions of xn ≡ 1(modPa). The “children” of avertex in the a-th level are the solutions (if there are any) of xn ≡ 1(modPa+1)which were lifted by their parent vertex. For example, by continuing theprocess of solving for arbitrary a, the tree which corresponds to the congruencex3 ≡ 1 (mod 7a) over Z is shown in Figure 1.a and the tree corresponding tothe congruence x3 ≡ 1 (mod 3a) is shown in Figure 1.b

1

1

1

1

2

2+4γ

2+4γ+6γ2

2+4γ+6γ2+3γ3

4

4+2γ

4+2γ

4+2γ+3γ3

Figure 1a: The tree cor-responding to the solu-tions of x3 ≡ 1 (mod 7a).

1

1

1

1

1−γ

1−γ2

1−γ3

1+γ

1+γ2

1+γ3

Figure 1b: The tree cor-responding to the solu-tions of x3 ≡ 1 (mod 3a).

Comparing these two examples, we can see that in Figure 1a, in every level wehave 3 solutions and any solution is the only “descendant” of its predecessor.Moreover, each infinite “branch” converges to a solution of x3 = 1 in Q7. InFigure 1b, although in every level we have 3 solutions, we have only one infinite

20 Boaz Cohen

“branch” which converges to the solution x = 1 of x3 = 1 in Q3. It turns outthat this phenomenon takes place generally: infinite branches “converge” to asolution of the corresponding equation in the P-adic field KP.

Our goal is to apply the lifting method to the congruence xpb ≡ 1 (mod Pa)

for a general prime ideal P lying above p, which has a general ramificationindex e. The relation between p and e will play a crucial role in our results.We start with the following useful Proposition.

Proposition 2.3. Suppose that p is a rational prime number, P is a primeideal in the algebraic number field K lying above (p), M is a complete residuesystem modulo P, γ ∈ P \ P2 and a > 2, b > 1 are integers. In addi-tion, suppose that S = {α1, α2, . . . , αn} are the incongruent solutions of xp

b ≡1(modPa−1) and S′ ⊆ S is the set of elements of S satisfying xp

b ≡ 1(modPa).Then the incongruent solutions of xp

b ≡ 1 (mod Pa) are

x ≡ α′ + θγa−1 (mod Pa),

where α′ ∈ S′ and θ ∈M is arbitrary.

Proof. If ξ satisfies the congruence xpb ≡ 1 (mod Pa), then clearly ξp

b ≡1 (mod Pa−1), so there is α ∈ S such that ξ ≡ α (mod Pa−1). By assump-tion γ ∈ P \ P2, so P‖(γ) and hence gcd((γa−1),Pa−1P) = Pa−1. Thus, byProposition II.2.1 there is θ ∈M such that ξ ≡ α+ θγa−1 (mod Pa) (note thatProposition II.2.1 implies also that for each θ ∈M, the numbers α+ θγa−1 areincongruent). We shall show that α ∈ S′. Note that

(α + θγa−1)pb

= αpb

+ pbαpb−1θγa−1 +

pb∑j=2

(pb

j

)αp

b−jθjγj(a−1).

Since P‖(γ), we get Pj(a−1) | (γj(a−1)) for each j. Moreover, if j > 2, thena 6 2(a − 1) 6 j(a − 1) because a > 2, so γj(a−1) ≡ 0 (mod Pa) for everyj > 2. In addition, Pa | (pbγa−1) since b > 1. Therefore, we deduce that(α+θγa−1)p

b ≡ αpb(modPa), that is ξp

b ≡ αpb(modPa). But ξp

b ≡ 1(modPa),so αp

b ≡ 1 (mod Pa) and hence α ∈ S′. This shows that for α′ = α, we getξ ≡ α′ + θγa−1 (mod Pa).

Conversely, if α′ ∈ S′, then for each θ ∈M, α′+ θγa−1 satisfies the congru-ence xp

b ≡ 1(modPa). Indeed, as shown above (α′+θγa−1)pb ≡ (α′)p

b(modPa)

for each θ ∈ M, and since α′ satisfies xpb ≡ 1 (mod Pa), we deduce that

(α′ + θγa−1)pb ≡ 1 (mod Pa), as desired.

In view of Proposition 2.3, in order to solve the congruence xpb ≡ 1(modPa),

we first solve the congruence xpb ≡ 1 (mod Pa−1). This indicates the inductive

procedure, which we shall use. The following example illustrates how to usethis proposition.


Example 2.4. We want to solve x3 ≡ 1 (mod P4) in K = Q(√

3), whereP = (

√3). Here OK = Z + Z

√3, P2 = (3) and we can choose a uniformizer

γ =√

3.

Before solving this congruence, we need to characterize the elements of Pa.If a = 2k, then clearly P2k = (3k), so x + y

√3 ∈ P2k iff 3k | x and 3k | y. If

a = 2k + 1, then

P2k+1 = (3)kP = (3k√

3) = {3k√

3(x+ y√

3) : x, y ∈ Z}= {3k+1y + 3kx

√3 : x, y ∈ Z}.

Hence, x + y√

3 ∈ P2k+1 iff 3k+1 | x and 3k | y. In addition, since NP = 3,M = {0, 1,−1} can be taken as a complete residue system modulo P.

Now we can begin solving the congruence. By Proposition I.3.2 the congru-ence x3 ≡ 1 (mod P) has only one solution, namely x ≡ 1 (mod P). Turningnow to modulo P2, by Proposition 2.3 we need only to check whether x = 1satisfies x3 ≡ 1 (mod P2). This is clearly true and hence the solutions moduloP2 are x ≡ 1+θ1γ (modP2), where θ1 ∈M, that is x ≡ 1, 1+γ, 1−γ (modP2).To solve the congruence modulo P3, we need to check which of {1, 1 +γ, 1−γ}satisfies x3 ≡ 1(modP3). Clearly, x = 1 satisfies this congruence and thereforeproduces the solutions x ≡ 1 + θ2γ

2 (mod P3), θ2 ∈M.

For the options x = 1 + γ and x = 1− γ, we have

(1± γ)3 = (1±√

3)3 = 10± 6√

3 ≡ 1 (mod P3).

Hence, we obtain another six solutions x ≡ 1 ± γ + θ2γ2 (mod P3), θ2 ∈ M.

Combined together, we obtain that the solutions modulo P3 are x ≡ 1 + θ1γ +θ2γ

2 (mod P3), where θ1, θ2 ∈M.

For modulo P4, note that

13 = 1 (mod P4)

(1 + γ2)3 = 43 = 64 ≡ 1 (mod P4)

(1− γ2)3 = (−2)3 = −8 ≡ 1 (mod P4)

(1± γ)3 = (1±√

3)3 = 10± 6√

3 6≡ 1 (mod P4)

(1± γ + γ2)3 = (4±√

3)3 = 100± 51√

3 6≡ 1 (mod P4)

(1± γ − γ2)3 = (−2±√

3)3 = −26± 15√

3 6≡ 1 (mod P4).

This produces the following 9 solutions: x ≡ 1 + θ2γ2 + θ3γ

3 (mod P4), θ2, θ3 ∈M. The tree corresponding to this congruence is

22 Boaz Cohen

1

1−γ 1 1+γ

1−γ21−γ+γ21−γ1−γ−γ2 1 1+γ2 1+γ−γ2 1+γ 1+γ+γ2

1−γ31−γ2+γ31−γ21−γ2−γ3 1 1+γ3 1+γ2−γ3 1+γ2 1+γ2+γ3

1−γ41−γ3+γ41−γ31−γ3−γ4 1 1+γ4 1+γ3−γ4 1+γ3 1+γ3+γ4

Figure 2: the tree correspond to the solutions of x3 ≡ 1 (mod P4).

It turns out that continuing the above tree for any modulus Pa, produces onlyone infinite branch corresponding to the trivial solution x = 1 in KP. Lateron in Section 5 we shall prove criteria determining the number of such infinitebranches.

3 Solving xpb ≡ 1 (mod Pa) when 2 6 a 6 e

We approach now the solution of the congruence xpb ≡ 1 (mod Pa) in general.

Our first step covers the cases when a > 2 (by Proposition I.3.2, the case a = 1yields the trivial solution, so it will be ignored) and is bounded from above bythe ramification index e. Our main results in this section is Theorems 3.2. Weneed first the following proposition, which characterizes exactly the nilpotentelements in OK/P

a.

Proposition 3.1. Suppose that p is a rational prime number, P is a primeideal in the algebraic number field K lying above (p), M is a complete residuesystem modulo P, 0 ∈M, γ ∈ P \ P2 and a > 2, n > 2 are integers. Then theincongruent solutions of xn ≡ 0 (mod Pa) are

x ≡ θ∆γ∆ + θ∆+1γ

∆+1 + . . .+ θa−1γa−1 (mod Pa),

where ∆ = d ane and θ∆, θ∆+1, . . . , θa−1 ∈M are arbitrary.

Proof. First note that ∆ 6 a− 1. Indeed, since a > 2 and n > 2, we get that

∆ 6⌈a

2

⌉6a+ 1

2< a,

so ∆ 6 a− 1.To prove the proposition, we begin by showing that every number of the

form θ∆γ∆ + . . . + θa−1γ

a−1, where θi ∈ M, satisfies the congruence xn ≡0 (mod Pa). Indeed, ∆ = d a

ne > a

n, so n∆ > a. Hence

(θ∆γ∆ + . . .+ θa−1γ

a−1)n ≡ γ∆n(θ∆ + . . .+ θa−1γa−1−∆)n ≡ 0 (mod Pa).


Conversely, we shall show that if xn ≡ 0(modPa), then there exist elementsθ∆, . . . , θa−1 ∈M such that x ≡ θ∆γ

∆ + . . .+θa−1γa−1 (modPa). We prove this

assertion using induction on a. If a = 2, then assume that xn ≡ 0 (mod P2).Hence xn ≡ 0 (mod P), so x ≡ 0 (mod P) because OK/P is a field. By Propo-sition II.2.1 there is θ1 ∈M such that x ≡ θ1γ (mod P2), as desired. Take nowa > 2 and suppose that the assertion holds for 2, 3, . . . , a−1. By the inductionhypothesis, if xn ≡ 0 (mod Pa−1), then there are θ∆′ , . . . , θa−1 ∈M such that

x ≡ θ∆′γ∆′ + . . .+ θa−2γ

a−2 (mod Pa−1),

where ∆′ = da−1ne. By Proposition II.2.1, there is θa−1 ∈M such that

x ≡ θ∆′γ∆′ + θ∆′+1γ

∆′+1 + . . .+ θa−1γa−1

= θ∆′γ∆′ + γ∆′+1 (θ∆′+1 + . . .+ θa−1γ

a−2−∆′)︸︷︷︸ξ

(mod Pa).

Now, by assumption xn ≡ 0 (mod Pa), so

0 ≡ (θ∆′γ∆′ + γ∆′+1ξ)n = θn∆′γ

∆′n +n∑j=1

(n

j

)θn−j∆′ γ

∆′n+jξj (mod Pa).

Since ∆′ = da−1ne > a−1

n, we have ∆′n+j > ∆′n+1 > a for every 1 6 j 6 n, so

γ∆′n+j ≡ 0(modPa) and the above congruence reduces to θn∆′γ∆′n ≡ 0(modPa).

Now, if ∆′n > a, then da−1ne = ∆′ > a

n> a−1

nand therefore ∆′ = da−1

ne =

d ane, as desired. If ∆′n < a, then θn∆′γ

∆′n ≡ 0 (mod Pa) implies that θn∆′ ≡0 (mod Pa−∆′n). Note that

1 6 a− n∆′ = a− n⌈a− 1

n

⌉6 a− n

(a− 1

n

)= 1.

Thus a−n∆′ = 1 and the above congruence reduces to θn∆′ ≡ 0 (mod P) whichimplies that θ∆′ ≡ 0 (mod P). Moreover, since 0 ∈M, we deduce that actuallyθ∆′ = 0. Therefore, we get

x ≡ θ∆′+1γ∆′+1 + . . .+ θa−1γ

a−1 (mod Pa).

To complete the proof, note that since n | a− 1, we obtain⌈an

⌉=

⌈a− 1

n+

1

n

⌉=a− 1

n+

⌈1

n

⌉=a− 1

n+ 1 = ∆′ + 1.

Hence ∆′ + 1 = d ane, as desired.

We are ready now for the main result of this section.

24 Boaz Cohen

Theorem 3.2. Suppose that p is a rational prime number, P is a prime ideal inthe algebraic number field K lying above (p) with ramification index e > 2, Mis a complete residue system modulo P, 0 ∈M, γ ∈ P \ P2 and a ,b are positiveintegers. If 2 6 a 6 e, then the incongruent solutions of xp


x ≡ 1 + θ∆γ∆ + θ∆+1γ

∆+1 + . . .+ θa−1γa−1 (mod Pa),

where ∆ = d apbe and θ∆, θ∆+1, . . . , θa−1 ∈M.

Proof. Set y = x− 1. We need to solve the congruence (1 + y)pb ≡ 1 (mod Pa),

that is

1 + ypb

+

pb−1∑j=1

(pb

j

)yj ≡ 1 (mod Pa),

namely

ypb ≡ −

pb−1∑j=1

(pb

j

)yj (mod Pa).

But by Proposition II.2.2 p |(pb

j

)for each 1 6 j < pb, and since Pa | Pe | (p),

we deduce that ypb ≡ 0 (mod Pa). The theorem then follows by applying

Proposition 3.1.

Example 3.3. Let K = Q( 4√

2). As one can verify, P = ( 4√

2) is a prime idealover p = 2 with ramification index e = 4. In addition, its norm is NP = 2,so M = {0, 1} can be taken to be a complete residue system modulo P. Nowconsider the congruence

x2 ≡ 1 (mod Pa).

According to Theorem 3.2, ∆ = da2e and the solutions for 2 6 a 6 4 are

a = 2 : x ≡ 1 + θ1γ (mod P2)a = 3 : x ≡ 1 + θ2γ

2 (mod P3)a = 4 : x ≡ 1 + θ2γ

2 + θ3γ3 (mod P4)

where here we can choose γ = 4√

2 and θi ∈ {0, 1}. Therefore, the initial levelsof the corresponding tree are

1

1 1+γ

1+γ21

1 1+γ3 1+γ2 1+γ2+γ3

Figure 3: The tree corresponding to the solutions of x2 ≡ 1 (mod P4).


4 Solving xpb ≡ 1 (mod Pa) when e < a

Theorem 3.2 gives us the pb-th roots of unity modulo Pa for 2 6 a 6 e. Theformulas which we are going to develop later on in this section, deal with thecomplementary cases, namely for e < a. Note that the result of Theorem 3.2is valid for every value of p. The formulas for the cases e < a will dependupon the relationship between e and p. Our main results in this section areTheorems 4.6 and Theorem 4.9.

We begin this discussion by introducing a family of functions which willbe important for the understanding of the structure of the pb-th roots of unitymodulo Pa. Let p be a prime number and let e be a fixed positive integer. Forevery non-negative integer k, let us define the function Tk : R→ R to be

Tk(u) = kpu − eu.

We shall be interested in the integral minimum value of Tk, that is, in theminimum value of Tk(u) over u ∈ N0 = {0, 1, 2, 3, . . .}. By the real minimumvalue of Tk we mean the minimum value over u ∈ R. The integral minimumvalue of Tk will be denoted by mk. Note that for k = 0 we obtain the functionT0(u) = −eu which has (−∞) as the minimal value. This suggest using thenotation m0 = −∞ (with the convention that c+ (−∞) = −∞) which will bequite convenient in the sequel.

To illustrate the behavior of Tk we compute the values of Tk for (p, e) =(3, 4). The figure below describes its values graphically.

u

Tk(u)

k=1k=2

k=3

u T1(u) T2(u) T3(u)0 1 2 31 −1 2 52 1 10 193 15 42 694 65 146 2275 223 466 709

Figure 4: The functions Tk(u) = k3u − 4u for k = 1, 2, 3

26 Boaz Cohen

In Figure 4, the full dots represent the values of Tk(u) for integral u’s andthe blank dots represent the minimum value of Tk(u) for real u’s. The actualvalues appear in the table next to the figure. In this table, the integral minimalvalues are in boldface.

As Figure 4 indicates, the integral minimum value of Tk can be obtainedfor more then one u. For example, T2 obtains its integral minimal value foru = 0 and u = 1. Once we have found the real value u whose image is thereal minimum value of Tk, it is quite easy to find its integral minimal value:as Figure 4 indicates, if Tk gets its real minimal value at u, then its integralminimal value is obtained in due or in buc, possibly both.

In order to proceed, we shall need some additional notation. Let c be thenon-negative integer such that pc‖e. For any integer 0 6 t 6 c we define

Γt :=e

pt(p− 1)=

e

ϕ(pt+1).

Observe that 0 < Γc < Γc−1 < · · · < Γ1 < Γ0. In addition, note that Γt is aninteger iff p− 1 | e. For convenience we now establish the “Generic Notation”of this section.

Notation 4.1.

(1) p is a rational prime number

(2) P is a prime ideal in the algebraic number field K lying above (p) withramification index e > 1

(3) M is a complete residue system of OK modulo P with 0 ∈M

(4) γ ∈ P \ P2

(5) a and b are positive integers

(6) c is a non-negative integer satisfying pc‖e and r = min{c, b− 1}

(7) Γt =e

ϕ(pt+1)for integers 0 6 t 6 c.

(8) For each non-negative integer k, Tk denotes the real function Tk(u) =kpu − eu for u ∈ R

(9) mk denotes the integral minimal value of Tk over N0, including m0 =−∞

(10) E = max{2, e}

(11) ∆ =

⌈a

pb

⌉


(12) For each positive integer k, Ik is the set of non-negative integers ufor which Tk(u) = kpu − eu achieves its minimal integral value mk.

The following proposition summarizes some important features of the func-tion Tk and the set Ik, which will be useful later on.

Proposition 4.2. Assume Notation 4.1 and suppose that k > 0 is an integer.Then

(a) For k = 0, T0(u) achieves its minimal value m0 = −∞ for u = ∞.For k > 1, Tk(u) achieves its unique real minimal value at u = uk =logp(

ek ln p

).

Moreover, if k > 1, then Ik has the following properties:

(b) Ik = {u′k, u′′k}, where u′k, u′′k are non-negative integers satisfying 0 6 u′k 6

u′′k and Tk(u′k) = Tk(u

′′k) = mk. In particular, 1 6 |Ik| 6 2.

Moreover, if |Ik| = 2, then uk > 0, 0 6 u′k < uk < u′′k, u′k = bukc,u′′k = duke, u′′k = u′k + 1 and Tk(u

′k) = Tk(u

′k + 1). Furthermore,

p− 1 | e ande

p− 1= pu

′kk > k.

Consequently, if either uk 6 0 or p− 1 - e, then |Ik| = 1.

(c) If k 6= ep−1

and p(p− 1) - e, then Ik = {u′k}.

(d) If k = ep−1

, then Ik = {0, 1} and mk = k.

(e) If k > ep−1

, then Ik = {0} and mk = k.

(f) −∞ = m0 < m1 < m2 < · · · < mk for every k > 1.

Proof. (a) Since T0(u) = −eu, its minimal value is clearly m0 = −∞, whichis “achieved” for u =∞.

Suppose, now, that k > 1 and Tk(u) = kpu−eu. Then Tk(u) is a continuousreal function of u, T′k(u) = (k ln p)pu − e and T′′k(u) = k(ln p)2pu. The uniquereal solution of T′k(u) = 0 is uk = logp(

ek ln p

) and since T′′k(uk) > 0, Tk(u)achieves its unique real minimal value at u = uk.

From now on, assume that k is an integer satisfying k > 1.

(b) Tk(u) is a continuous function of u and by part (a), Tk has a uniquereal extreme point at u = uk, which is a minimum. Hence Ik = {u′k, u′′k}, whereu′k, u

′′k are non-negative integers satisfying 0 6 u′k 6 u′′k and Tk(u

′k) = Tk(u

′′k) =

mk. In particular, 1 6 |Ik| 6 2.

28 Boaz Cohen

Moreover, if |Ik| = 2, then uk > 0, 0 6 u′k < uk < u′′k, u′k = bukc,

u′′k = duke and u′′k = u′k + 1. Thus Tk(u′k) = Tk(u

′k + 1). Denote u′k = u. Then

kpu − eu = kpu+1 − e(u+ 1), which yields e = kpu(p− 1). Hence

p− 1 | e ande

p− 1= puk > k.

Consequently, if either uk 6 0 or p− 1 - e, then |Ik| = 1.

(c) If |Ik| = 2, then by (b) e = kpu′k(p − 1). If k 6= e

p−1, then pu

′k > 1

and p(p − 1) | e. Hence, if k 6= ep−1

and p(p − 1) - e, then we have reached a

contradiction. Therefore in this case |Ik| = 1, as required.

(d) If k = ep−1

, then

Tk(1) = kp− e =ep

p− 1− e =

e

p− 1= k = Tk(0),

and it suffices to show that Tk(u) > k for each integer u > 2. So let u > 2.Then u < pu−1 + pu−2 + . . .+ p+ 1 = pu−1

p−1and

eu

pu − 1<

e

p− 1= k.

Hence eu < kpu − k and k < Tk(u), as required.

(e) Suppose that k > ep−1

. Using the argument in part (d), it follows that

u 6 pu−1p−1

for integers u > 1. Hence

eu

pu − 16

e

p− 1< k,

implying that eu < kpu − k. Thus Tk(0) = k < Tk(u) for integers u > 1 andhence Ik = {0}. In particular, mk = k.

(f) By (a), m0 = −∞ and T1 achieves its real minimal value at u =logp(e/ ln p), so m0 = −∞ < m1. In order to complete the proof, it suffices toshow that mk −mk−1 > 1 for integers k > 2. So let k > 2. First observe thatTk(u) = Tk−1(u) + pu for each u. Let uk be a non-negative integer such thatmk = Tk(uk). Then

mk = Tk(uk) = Tk−1(uk) + puk > mk−1 + 1,

yielding mk −mk−1 > 1 for k > 2, as required. The proof of Proposition 4.2is complete.


We remark that by part (f) of Proposition 4.2, we have −∞ = m0 < m1 <m2 < · · · . Hence the quantity a − eb must satisfy mk < a − eb 6 mk+1 for aunique integer k > 0.

Before presenting our main results concerning the additional cases of a weare left with, we mention the following useful technical propositions.

Proposition 4.3. Assume Notation 4.1 and let α ∈ OK. In addition, letk > 0 be an integer and for each integer m > k choose an element θm ∈ OK.Using the θm, define the sequence ξk, ξk+1, ξk+2, . . . as follows: ξk = α and

ξn = α + θk+1γk+1 + θk+2γ

k+2 + . . .+ θnγn

for integers n > k + 1. Then the following statements hold:

(a) If mk < a− eb 6 mk+1, then

ξpb

n ≡ ξpb

k (mod Pa)

for every n > k.

(b) If a− eb = mk + 1, then

(α + θkγk)p

b ≡ αpb

+∑x∈I

(pb

px

)αp

b−px(θkγk)p

x

(mod Pa),

where I = [0, b] ∩ {u′k, u′′k}, 0 6 u′k 6 u′′k and {u′k, u′′k} is the set of non-negative integers for which Tk(u) = kpu−eu achieves its minimal integralvalue mk.

Proof. (a) It suffices to prove that ξpb

n ≡ ξpb

n−1 (mod Pa) for every n > k. Notethat

ξpb

n = (ξn−1 + θnγn)p

b

= ξpb

n−1 +

pb∑j=1

(pb

j

)ξp

b−jn−1 θ

jnγ

nj.

Fix j, 1 6 j 6 pb, and let u be the non-negative integer satisfying pu‖j. Since

P‖(γ) and pb−u |(pb

j

)by Proposition II.2.2, it follows that(pb

j

)ξp

b−jn−1 θ

jnγ

nj ≡ 0 (mod Pe(b−u)+nj).

Since j > pu and n > k implies n > k + 1, it follows that

nj − eu > (k + 1)pu − eu = Tk+1(u) > mk+1 > a− eb.

Thus a 6 e(b− u) + nj and therefore ξpb

n ≡ ξpb

n−1 (mod Pa) for every n > k, asrequired.

30 Boaz Cohen

(b) Note that

(α + θkγk)p

b

= αpb

+

pb∑j=1

(pb

j

)αp

b−jθjkγkj.

Fix j, 1 6 j 6 pb, and let u be the non-negative integer satisfying pu‖j. Then0 6 u 6 b and as above we get(

pb

j

)αp

b−jθjkγkj ≡ 0 (mod Pe(b−u)+kj).

Suppose, first, that u /∈ {u′k, u′′k}. Then kj − eu > kpu − eu = Tk(u) > mk,

yielding kj−eu > mk+1 = a−eb. Thus a 6 e(b−u)+kj and(pb

j

)αp

b−jθjkγkj ≡

0 (mod Pa).Suppose, next, that u ∈ {u′k, u′′k}, but j > pu. Then kj − eu > kpu − eu =

Tk(u) = mk, yielding kj−eu > mk+1 = a−eb. Hence again a 6 e(b−u)+kj

and(pb

j

)αp

b−jθjkγkj ≡ 0 (mod Pa).

Thus the only remaining summands modulo Pa are αpb

and those corre-sponding to j = pu, where u ∈ [0, b] ∩ {u′k, u′′k}, as required.

Proposition 4.4. Assume Notation 4.1. In addition, let k be the unique non-negative integer such that mk < a− eb 6 mk+1 and assume that Ik = {u′k, u′′k},where u′k 6 u′′k. Moreover, recall that E = max{2, e}. Then the followingstatements hold:

(1) If a = E, then k = ∆− 1, where ∆ = d apbe.

(2) If a > E, then k + 1 6 a− 1.

(3) If a > E and a− eb = mk + 1, then u′k 6 b.

(4) If a = E, then 0 6 k < Γr.

(5) If Γr 6 k, then u′k 6 b and consequently E 6 mk + 1 + eb.

Proof. (1) Suppose that a = E. Then either a = e or a = 2.

The case a = e: We wish to prove that mk < a − eb = e − eb 6 mk+1 fork = ∆− 1 = d e

pbe − 1. First we show that e− eb 6 m∆. In order to do so, we

need to prove that e − eb 6 ∆pu − eu for every integer u > 0. If u < b, then0 6 b− u− 1 and in particular 0 6 ∆pu + e(b− u− 1). Hence

e− eb 6 ∆pu − eu,


as desired.So suppose that u > b. Then 0 6 u− b and by the Bernoulli inequality we

get

1 + (u− b) 6 2u−b 6 pu−b.

Hence

e(1 + u− b) 6 epu−b =e

pbpu 6

⌈e

pb

⌉pu = ∆pu,

so e(1− b) 6 ∆pu − eu, as desired.Next we show that m∆−1 < e− eb. Note that 0 6 ∆− 1 = d e

pbe − 1 < e

pb.

Hence (∆− 1)pb < e and (∆− 1)pb− eb < e− eb. Thus T∆−1(b) < e− eb. Butclearly m∆−1 6 T∆−1(b), so m∆−1 < e− eb, as desired.

The case a = 2: Hence e 6 2, and ∆ = d 2pbe = 1. We wish to prove that

k = 0, that is m0 < a− eb = 2− eb 6 m1. Since m0 = −∞, it suffices to provethat 2− eb 6 T1(u) for every u > 0. Indeed, if u > 0, then

2(u− b) 6 2(u− 1) 6 2(2u−1 − 1) = 2u − 2.

Now e 6 2, so e(u− b) 6 2(u− b) 6 2u−2 6 pu−2, that is 2− eb 6 pu− eu =T1(u), as desired.

(2) Suppose that a > E = max{2, e}. We wish to prove that k+ 1 6 a− 1.Since a > E, one of the following cases must hold: either a = E or a > E.

If a = E, then k + 1 = ∆ by part (1) and hence

k + 1 =

⌈a

pb

⌉6⌈a

2

⌉6a+ 1

2< a

since a > 2. Thus k + 1 6 a− 1, as desired.Suppose, now, that a > E. If 0 6 k < e

p−1, then 0 6 k < e

p−16 e, yielding

k 6 e− 1. But e < a, so e 6 a− 1 and k 6 a− 2, as desired. If, on the otherhand, k > e

p−1, then parts (d) and (e) of Proposition 4.2 imply that mk = k

and mk+1 = k + 1. Since mk < a − eb 6 mk+1, it follows that a − eb = k + 1and hence k + 1 6 a− 1, as desired.

(3) Suppose that a > E = max{2, e} and a − eb = mk + 1. We wish toprove that u′k 6 b.

Suppose, to the contrary, that u′k > b. Then Tk(b) = kpb−eb > mk = a−1−eb, so kpb > a−1. Thus kpb > a > E = max{2, e} > e, so e 6 kpb 6 kpb(p−1).On the other hand, also Tk(u

′k−1) > Tk(u

′k), so kpu

′k−1−e(u′k−1) > kpu

′k−eu′k.

Thus e > kpu′k−1(p−1) > kpb(p−1), a contradiction. Hence u′k 6 b, as claimed.

32 Boaz Cohen

(4) Suppose that a = E. We wish to prove that 0 6 k < Γr. If a = 2, thenby part (2), k + 1 6 a− 1 = 1. Since k > 0, it follows that k = 0, as required.

If a = e, then part (1) and r 6 b− 1 imply that

k =

⌈e

pb

⌉− 1 <

e

pb6

e

pr+1<

e

pr(p− 1)= Γr,

as required.

(5) Suppose, to the contrary, that b < u′k. Then Tk(b) > Tk(u′k), that is

kpb − eb > kpu′k − eu′k, so

e(u′k − b) > k(pu′k − pb) > Γr(p

u′k − pb) = Γrpb(pu

′k−b − 1).

Set t = u′k − b. By assumption, t > 1 and since r 6 b− 1 it follows that

et >e

pr(p− 1)pb(pt − 1) = epb−r

pt − 1

p− 1> eppt−1 = ept.

So et > ept, that is t > pt, a contradiction.

It remains to prove that E 6 mk + 1 + eb. By the first part of the proofu′k 6 b. If u′k = b, then

E− 1 6 e < ep

p− 16 e

pb−r

p− 1= Γrp

b 6 kpb = kpu′k + e(b− u′k) = mk + eb,

as desired. If u′k < b, then E− 1 6 e 6 e(b−u′k) < e(b−u′k) + kpu′k = mk + eb,

again as desired. The proof of Proposition 4.4 is complete.

We continue with preliminary results:

Proposition 4.5. Assume Notation 4.1. In addition, let k be the unique non-negative integer such that mk < a − eb 6 mk+1 and suppose that one of thefollowing conditions holds:

(1) a = E.(2) a > E and k = 0.

Then the incongruent solutions of xpb ≡ 1 (mod Pa) are

x ≡ 1 + θk+1γk+1 + θk+2γ

k+2 + . . .+ θa−1γa−1 (mod Pa),

where θk+1, θk+2, . . . , θa−1 ∈M are arbitrary.


Proof. Notice that by Proposition 4.4(2), k + 1 6 a− 1 in both cases.(1) Suppose that a = E. We need to deal with two cases: a = 2 and a = e.

If a = 2, then by Proposition 4.4 k + 1 6 a − 1 = 1, so k + 1 = a − 1 = 1.We need to solve xp

b ≡ 1 (mod P2). By Proposition I.3.2, the only solution ofxp

b ≡ 1 (mod P) is x ≡ 1 (mod P). Clearly x = 1 satisfies xpb ≡ 1 (mod P2),

so by Proposition 2.3 the incongruent solutions of xpb ≡ 1 (mod P2) are x ≡

1 + γθ1 (mod P2) for θ1 ∈M, as required.So suppose that a = e. Then, by Theorem 3.2, the incongruent solutions

of xpb ≡ 1 (mod Pa) are

x ≡ 1 + θ∆γ∆ + θ∆+1γ

∆+1 + . . .+ θa−1γa−1 (mod Pa),

where ∆ = d epbe and θ∆, θ∆+1, . . . , θa−1 ∈M. By Proposition 4.4(a) ∆ = k+1,

as required. The proof of part (1) is complete.

(2) Suppose that a > E and k = 0. Since k = 0, we have a − eb 6 m1

and we must show that the incongruent solutions of xpb ≡ 1 (mod Pa) are

x ≡ 1 + θ1γ + θ2γ2 + . . .+ θa−1γ

a−1 (mod Pa), where θ1, θ2, . . . , θa−1 ∈M.By assumption E 6 a 6 m1 + eb. We shall apply induction on a. If a = E,

then the assertion holds by part (1). So suppose that a > E and the assertionholds for E,E + 1, . . . , a− 1. By our assumptions a− eb 6 m1 and a > E, soa − 1 − eb < m1 and a − 1 > E > 2. Hence the induction hypothesis impliesthat the incongruent solutions of xp

b ≡ 1 (mod Pa−1) are

x ≡ 1 + θ1γ + θ2γ2 + . . .+ θa−2γ

a−2 (mod Pa−1),

where θ1, θ2, . . . , θa−2 ∈M.Now we determine the θ’s for which

(1 + θ1γ + θ2γ2 + . . .+ θa−2γ

a−2)pb ≡ 1 (mod Pa).

Applying part (a) of Proposition 4.3 with k = 0, α = 1 and n = a − 2 > 1,we may conclude that the above congruence holds for every choice of the θ’s.Therefore, by Proposition 2.3, the incongruent solutions of xp

b ≡ 1 (mod Pa)are

x ≡ 1 + θ1γ + θ2γ2 + . . .+ θa−2γ

a−2 + θa−1γa−1 (mod Pa),

where θ1, θ2, . . . , θa−1 ∈M, as required. The proof of part (2) is complete.

Theorem 4.6. Assume Notation 4.1. In addition, let k be the unique non-negative integer such that mk < a− eb 6 mk+1 and suppose that the followingconditions hold:

(1) a > E = max{2, e}, and(2) either 0 6 k < Γr or p− 1 - e.

34 Boaz Cohen

Then the incongruent solutions of xpb ≡ 1 (mod Pa) are

x ≡ 1 + θk+1γk+1 + θk+2γ

k+2 + . . .+ θa−1γa−1 (mod Pa),

where θk+1, θk+2, . . . , θa−1 ∈M.

Proof. We shall prove the assertion by induction on k.By Proposition 4.4(1), m∆−1 < E− eb 6 m∆, where ∆ = dE/pbe. Since we

assume that a > E, it follows that

mk < a− eb 6 mk+1 for some k > ∆− 1.

Therefore k = ∆− 1 is the basis for our induction.So suppose that k = ∆ − 1. Since mk < E − eb 6 mk+1, it follows by

Proposition 4.4(4) that our assumption 0 6 k < Γr is satisfied. Our firstaim is to prove by induction on a that our assertion holds for all a satisfyinga > E and mk < a − eb 6 mk+1. As shown above, m∆−1 < E − eb 6 m∆

and by Proposition 4.5(1) our assertion holds for a = E. So suppose that a′

is an integer, a′ > E, mk < a′ − eb 6 mk+1 and our assertion holds for alla ∈ {E,E + 1, . . . , a′ − 1}.

If ∆ = 1, then k = 0 and our assertion holds by Proposition 4.5(2). Soassume that ∆ > 1. Since mk < E − eb 6 mk+1, a′ > E and mk < a′ − eb 6mk+1, it follows that mk + 1 < a′ − eb 6 mk+1. Hence

mk < (a′ − 1)− eb < mk+1

and by our inductive assumption concerning a′ − 1, the incongruent solutionsof xp

b ≡ 1 (mod Pa′−1) are

x ≡ 1 + θ∆γ∆ + θ∆+1γ

∆+1 + . . .+ θa′−2γa′−2 (mod Pa′−1),

where θ∆, θ∆+1, . . . , θa′−2 ∈ M. Notice that ∆ 6 (a′ − 1) − 1 = a′ − 2 byProposition 4.4(2). Now we determine the θ’s for which

(1 + θ∆γ∆ + θ∆+1γ

∆+1 + . . .+ θa′−2γa′−2)p

b ≡ 1 (mod Pa′).

Applying part (a) of Proposition 4.3 with α = 1 and n = a′ − 2, we mayconclude that

(1 + θ∆γ∆ + θ∆+1γ

∆+1 + . . .+ θa′−2γa′−2)p

b ≡ 1 (mod Pa′)

without any further restriction on the θ’s. Hence, by Proposition 2.3, theincongruent solutions of xp

b ≡ 1 (mod Pa′) are

x ≡ 1 + θ∆γ∆ + θ∆+1γ

∆+1 + · · ·+ θa′−2γa′−2 + θa′−1γ

a′−1 (mod Pa′),


as required. The proof for the basis k = ∆−1 of our induction is now complete.Suppose, now, that k satisfies k > ∆− 1 and either 0 6 k < Γr or p− 1 - e

holds. Moreover, assume that our assertion holds for all a satisfying a > Eand mk′ < a − eb 6 mk′+1, where k′ ∈ {∆ − 1,∆, . . . , k − 1}. Our aim is toprove that our assertion holds for all a satisfying a > E and

mk < a− eb 6 mk+1.

Our proof will be by induction on a. We wish to show, first, that a =mk + eb + 1 is the basis for our induction. Clearly a > mk + eb + 1. Ifa = mk + eb + 1, then a − eb = mk + 1, which implies that a − eb 6 mk+1.Moreover, a > mk + eb > m∆ + eb > E by Proposition 4.4(1), so a > E, asrequired. Thus, indeed, a = mk + eb+ 1 is the basis for our induction.

So suppose that a− eb = mk + 1. Since k > ∆ > 1, we have

mk = m(k−1)+1 with k − 1 > ∆− 1 > 0

and by our assumptions

mk−1 < m(k−1)+1 = (a− 1)− eb.

Since a− 1 > E, it follows by our inductive assumption concerning k− 1, thatthe incongruent solutions of xp


x ≡ 1 + θkγk + θk+1γ

k+1 + . . .+ θa−2γa−2 (mod Pa−1),

where θk, θk+1, . . . , θa−2 ∈ M. Notice that by Proposition 4.4(2) we have k 6a− 2. Now we determine the θ’s for which

(1 + θkγk + θk+1γ

k+1 + . . .+ θa−2γa−2)p

b ≡ 1 (mod Pa).

Applying part (a) of Proposition 4.3 with α = 1 + θkγk and n = a− 2, we may

conclude that

(1 + θkγk + θk+1γ

k+1 + . . .+ θa−2γa−2)p

b ≡ (1 + θkγk)p

b

(mod Pa).

Hence we need to consider the congruence

(1 + θkγk)p

b ≡ 1 (mod Pa).

By part (b) of Proposition 4.3 with α = 1, this congruence is equivalent to thecongruence ∑

x∈I

(pb

px

)(θkγ

k)px ≡ 0 (mod Pa),

where I = [0, b]∩ {u′k, u′′k}, 0 6 u′k 6 u′′k and {u′k, u′′k} is the set of non-negativeintegers for which Tk(u) = kpu − eu achieves its minimal integral value mk.

36 Boaz Cohen

We claim that if either 0 6 k < Γr or p − 1 - e, then I = {u′k}. ByProposition 4.4(3) we have u′k 6 b, so u′k ∈ I. If p − 1 - e, then I = {u′k} bypart (b) of Proposition 4.2. So suppose that 0 6 k < Γr. We must show thatu′′k 6 b implies u′k = u′′k.

Suppose, to the contrary, that u′k < u′′k 6 b. Then it follows by part (b) ofProposition 4.2 that kpu

′k(p − 1) = e. As pc‖e, it follows that u′k 6 c. Since

k < Γr, we have kpr(p− 1) < e and hence r < u′k 6 c. But r = min{c, b− 1},so r = b − 1 < u′k and u′k > b, contradicting u′k < u′′k 6 b. Hence u′k < u′′k 6 bis impossible, and it follows that if u′′k ∈ I, then u′k = u′′k. Thus I = {u′k}, asclaimed.

Hence, denoting u′k by u, it follows from the previous congruence that(pb

pu

)θp

u

k γkpu ≡ 0 (mod Pa).

Set α =(pb

pu

)γkp

u. Since mk = a− eb− 1 = kpu− eu, it follows that e(b− u) +

kpu = a−1, which implies, in view of Proposition II.2.2, that Pa−1‖(α). Henceθp

u

k ≡ 0 (mod P), and since OK/P is a field, we must have θk ≡ 0 (mod P). Butθk ∈M, so by our assumptions θk = 0.

Hence (1+θk+1γk+1+. . .+θa−2γ

a−2)pb ≡ 1(modPa) for all θk+1, . . . , θa−2 and

it follows by Proposition 2.3 that the incongruent solutions of xpb ≡ 1(modPa)

arex ≡ 1 + θk+1γ

k+1 + . . .+ θa−2γa−2 + θa−1γ

a−1 (mod Pa)

where θk+1, . . . , θa−2, θa−1 ∈ M, as required. The proof for the basis a =mk + eb+ 1 of our induction is now complete.

Suppose, now, that a > mk + eb + 1, where mk < a− eb 6 mk+1, and ourassertion holds for a′ satisfying mk < a′ − eb and a′ ∈ {mk + eb + 1,mk +eb + 2, . . . , a− 1}. Thus, since k > ∆− 1, we have a > mk + eb + 1 > E andmk + 1 < a− eb 6 mk+1. Since mk + 1 6 (a− 1)− eb < mk+1 and a− 1 > E,it follows by our inductive assumption for a− 1 that the incongruent solutionsof xp


x ≡ 1 + θk+1γk+1 + θk+2γ

k+2 + . . .+ θa−2γa−2 (mod Pa−1),

where θk+1, . . . , θa−2 ∈M. Notice that by Proposition 4.4(2) we have k + 1 6(a − 1) − 1 = a − 2. By applying part (a) of Proposition 4.3 with α = 1and n = a− 2 and then Proposition 2.3, it follows, like in the above proof fora = mk + eb + 1, that the incongruent solutions of xp

b ≡ 1 (mod Pa) are asrequired. The proof of the theorem is complete.

Using Theorems 3.2 and 4.6, we obtain a complete picture of the structureof pb-th roots of unity modulo Pa in the case when p − 1 - e. The followingexample illustrates these results.


Example 4.7. Suppose that K is a number field and let P be a prime ideal inK over p = 3 with ramification index e = 13. In addition, suppose that M is acomplete residue system modulo P with 0 ∈M and let γ ∈ P \P2. We want todescribe the solutions of a general congruence xp

b ≡ 1 (mod Pa) for all valuesof a > 2 and b. Since E = max{2, e} = 13, we need to distinguish between thecases 2 6 a 6 13 and 13 < a.

For 2 6 a 6 13, we invoke Theorem 3.2 to obtain the solutions

x ≡ 1 + θ∆γ∆ + . . .+ θa−1γ

a−1 (mod Pa),

where θi ∈M and

∆ =⌈ a

3b

⌉=

1 b = 1 2 6 a 6 32 b = 1 4 6 a 6 63 b = 1 7 6 a 6 94 b = 1 10 6 a 6 125 b = 1 a = 131 b = 2 2 6 a 6 92 b = 2 10 6 a 6 131 b > 3. 2 6 a 6 13

Now assume that a > 13. Here p − 1 - e and p - e, so in the notations ofTheorem 4.6, c = 0, r = 0 and hence Γ0 = 13

3−1= 6.5. Therefore, Theorem 4.6

can be invoked. In order to describe the solutions, we need first to find theintegral minimal values mk of Tk(u) = k3u − 13u.

If k > 6.5, that is, if k > 7, then by parts (d) and (e) of Proposition 4.2,mk = k. For 1 6 k 6 6, we shall compute the values mk directly fromthe definition. Let uk be the integral points where Tk(u) achieves its integralminimal value mk and let uk be the real points where Tk(u) achieves its realminimal value. By part (a) of Proposition 4.2 we get that uk = log3( 13

k ln 3).

The results are gathered in the table below:

k uk uk mk

1 2 2.24 −172 2 1.61 −83 1 1.24 −44 1 0.98 −15 1 0.78 26 1 0.61 5

By Theorem 4.6 the solutions of x3b ≡ 1 (mod Pa) for a > 13 are thereforex ≡ 1+θk+1γ

k+1 + . . .+θa−1γa−1 (modPa), where θi ∈M and k is the (unique)

non-negative integer such that mk + 1 6 a − eb 6 mk+1. Note that if k > 7,

38 Boaz Cohen

then as mentioned above mk = k, so in this case mk < a − eb 6 mk+1 iffa− eb = k + 1. Hence, if a− eb > 7, then the solutions are

x ≡ 1 + θa−ebγa−eb + . . .+ θa−1γ

a−1 (mod Pa),

where θi ∈M. To conclude, if a > 13, then the solutions of x3b ≡ 1 (mod Pa)are x ≡ 1 + θ∆γ

∆ + . . .+ θa−1γa−1 (mod Pa), where θi ∈M and

∆ =

0 + 1 a− eb 6 m1

1 + 1 m1 + 1 6 a− eb 6 m2

2 + 1 m2 + 1 6 a− eb 6 m3

3 + 1 m3 + 1 6 a− eb 6 m4

4 + 1 m4 + 1 6 a− eb 6 m5

5 + 1 m5 + 1 6 a− eb 6 m6

6 + 1 m6 + 1 6 a− eb 6 m7

a− eb m7 6 a− eb,

that is

∆ =

1 a− eb 6 −172 −16 6 a− eb 6 −83 −7 6 a− eb 6 −44 −3 6 a− eb 6 −15 0 6 a− eb 6 26 3 6 a− eb 6 57 6 6 a− eb 6 7

a− eb 7 6 a− eb.The solution of example 4.7 is now complete.

Now we turn to the cases where p−1 | e. This will complete the descriptionof the solution of xp

b ≡ 1 (mod Pa). In this analysis we shall use the followingvariation of Proposition I.2.2.

Proposition 4.8. Suppose that p is a rational prime number, P is a primeideal in the algebraic number field K lying above (p) with ramification indexe > 1, α, β ∈ OK and a, k are integers such that a > 1 and k > 0. If e

p−16 a,

thenα ≡ β (mod Pa) =⇒ αp

k ≡ βpk

(mod Pa+ek)

Proof. We shall prove that claim using induction on k. If k = 0, then theassertion is clear.

Suppose that k = 1. Note that if α ≡ β (mod Pa), then there is θ ∈ Pa

such that α = β + θ. Hence

αp = (β + θ)p = βp +

p−1∑j=1

(p

j

)βp−jθj + θp.


If 1 6 j 6 p − 1, then p |(pj

). In addition, since Pa | (θ) we get Paj | (θ)j.

Using the fact that Pe‖(p) we obtain therefore(pj

)βp−jθj ≡ 0(mod Pe+aj). Since

clearly e + a 6 e + aj we get αp ≡ βp + θp (mod Pa+e). By the assumptione/(p − 1) 6 a it follows that a + e 6 a + a(p − 1) = ap. Since Pap | (θ)p, weconclude that θp ≡ 0 (mod Pa+e). Therefore αp ≡ βp (mod Pa+e).

Take k > 1 and assume that the assertion is true for 1, 2, . . . , k− 1. If α ≡β (mod Pa), then αp

k−1 ≡ βpk−1

(mod Pa+e(k−1)) by the induction hypothesis.The assertion then follows using the case k = 1.

Theorem 4.9. Assume Notation 4.1. In addition, let k be the unique non-negative integer satisfying

mk < a− eb 6 mk+1

and suppose that the following conditions hold:

(1) a > E = max{2, e}, and(2) p− 1 | e

Then the following statements hold:(A) If 0 6 k < Γr, then the incongruent solutions of xp


x ≡ 1 + θk+1γk+1 + θk+2γ

k+2 + · · ·+ θa−2γa−2 + θa−1γ

a−1 (mod Pa),

where θk+1, . . . , θa−2, θa−1 ∈M are arbitrary.(B) If k > Γr, then the incongruent solutions of xp


x ≡ 1 + βΓrγΓr + . . .+ βkγ

k + θk+1γk+1 + . . .+ θa−1γ

a−1 (mod Pa),

where θk+1, . . . , θa−1 ∈ M are arbitrary and βΓr , . . . βk ∈ M satisfy one of thefollowing systems of congruences, depending upon the value of k:

(a) If Γr = k = Γ0, then

pbβΓ0γΓ0 +

(pb

p

)(βΓ0γ

Γ0)p ≡ 0 (mod Pa).

(b) If Γr 6 k < Γ0, then

(1 + βΓrγΓr + βΓr+1γ

Γr+1 + . . .+ βkγk)p

b ≡ 1 (mod Pa).

(c) If Γr < k = Γ0, then{(1 + βΓrγ

Γr + . . .+ βΓ0−1γΓ0−1)p

b ≡ 1 (mod Pa−1)

pbβΓ0γΓ0 +

(pb

p

)(βΓ0γ

Γ0)p ≡ 1− (1 + βΓrγΓr + . . .+ βΓ0−1γ

Γ0−1)pb

(mod Pa).

40 Boaz Cohen

(d) If Γr = Γ0 < k, thenpbβΓ0γ

Γ0 +(pb

p

)(βΓ0γ

Γ0)p ≡ 0 (mod Pa−(k−Γ0))

pbβΓ0+1γΓ0+1 ≡ 1− (1 + βΓ0γ

Γ0)pb

(mod Pa−(k−Γ0)+1)...

pbβkγk ≡ 1− (1 + βΓ0γ

Γ0 + . . .+ βk−1γk−1)p

b(mod Pa).

(e) If Γr < Γ0 < k, then

(1 + βΓrγΓr + . . .+ βΓ0−1γ

Γ0−1)pb ≡ 1 (mod Pa−1−(k−Γ0))

pbβΓ0γΓ0 +

(pb

p

)(βΓ0γ

Γ0)p ≡ 1− (1 + βΓrγΓr + . . .+ βΓ0−1γ

Γ0−1)pb

(mod Pa−(k−Γ0))

pbβΓ0+1γΓ0+1 ≡ 1− (1 + βΓrγ

Γr + . . .+ βΓ0γΓ0)p

b(mod Pa−(k−Γ0)+1)

...

pbβkγk ≡ 1− (1 + βΓrγ

Γr + . . .+ βk−1γk−1)p

b(mod Pa).

Proof. Clearly Γr 6 Γ0 and since p− 1 | e, it follows that both Γ0 and Γr arepositive integers. By our assumptions a > E. If a = E, then Proposition 4.4(4)implies that 0 6 k < Γr. This remark will be referred to as “Remark 1”. Weshall break our proof into 5 parts, depending upon the value of k.

Part 1: Suppose that 0 6 k < Γr. In this case Theorem 4.6 implies thatthe incongruent solutions of xp


x ≡ 1 + θk+1γk+1 + θk+2γ

k+2 + . . .+ θa−2γa−2 + θa−1γ

a−1 (mod Pa),

where θk+1, . . . , θa−2, θa−1 ∈M are arbitrary, as required in assertion (A).

Part 2: Suppose that Γr = k 6 Γ0. By our assumptions we have

mΓr < a− eb 6 mΓr+1.

In Part 2 we shall establish item (a) of this theorem and the “edge case”Γr = k < Γ0 of item (b).

Our first goal is to prove using induction on a, that under our assumptionsthe incongruent solutions of xp


x ≡ 1 + βΓrγΓr + θΓr+1γ

Γr+1 + . . .+ θa−2γa−2 + θa−1γ

a−1 (mod Pa),

where θΓr+1, . . . , θa−2, θa−1 ∈M are arbitrary and βΓr ∈M satisfies

(1 + βΓrγΓr)p

b ≡ 1 (mod Pa).


Since Γr = k, it follows by Remark 1 that a − 1 > E. Moreover, part (5) ofProposition 4.4 implies that E 6 mΓr + 1 + eb. So the smallest a satisfyinga > E and mΓr < a− eb 6 mΓr+1 is a = mΓr + 1 + eb.

So suppose that mΓr = (a−1)−eb. As Γr−1 > 1−1 = 0 and (a−1)−eb =mΓr = m(Γr−1)+1, it follows that

mΓr−1 < (a− 1)− eb = m(Γr−1)+1.

Since Γr − 1 < Γr and a − 1 > E, Part 1 of this proof implies that theincongruent solutions of xp


x ≡ 1 + θΓrγΓr + θΓr+1γ

Γr+1 + . . .+ θa−2γa−2 (mod Pa−1),

where θΓr , θΓr+1, . . . , θa−2 ∈ M are arbitrary. Notice that Γr 6 (a − 1) − 1 =a−2 by Proposition 4.4(2). By applying Proposition 4.3(a) with α = 1+θΓrγ

Γr ,k = Γr and n = a− 2, we get

(1 + θΓrγΓr + θΓr+1γ

Γr+1 + . . .+ θa−2γa−2)p

b ≡ (1 + θΓrγΓr)p

b

(mod Pa).

Denote θΓr = βΓr ∈M satisfying

(1 + βΓrγΓr)p

b ≡ 1 (mod Pa).

Then it follows by Proposition 2.3 that the incongruent solutions of the con-gruence xp



Γr+1 + . . .+ θa−2γa−2 + θa−1γ

a−1 (mod Pa),


(1 + βΓrγΓr)p

b ≡ 1 (mod Pa),

as required. The proof for the induction basis a = mΓr +1+eb is now complete.Let now a > mΓr + 1 + eb and suppose that our assertion holds for a ∈

{mΓr + 1 + e,mΓr + 2 + e, . . . , a− 1}. Our assumption that mΓr + 1 < a− eb 6mΓr+1 implies that

mΓr < (a− 1)− eb < mΓr+1,

and by Remark 1 we may assume that a−1 > E. Hence it follows, by applyinginduction for a− 1, that the incongruent solutions of xp



Γr+1 + . . .+ θa−2γa−2 (mod Pa−1),

where θΓr+1, . . . , θa−2 ∈M are arbitrary and βΓr ∈M satisfies

(1 + βΓrγΓr)p

b ≡ 1 (mod Pa−1).

42 Boaz Cohen

By applying part (a) of Proposition 4.3 with α = 1 + βΓrγΓr , k = Γr and

n = a− 2, we obtain

(1 + βΓrγΓr + θΓr+1γ

Γr+1 + . . .+ θa−2γa−2)p

b ≡ (1 + βΓrγΓr)p

b

(mod Pa).

Therefore, by Proposition 2.3, the incongruent solutions of xpb ≡ 1 (mod Pa)

are


Γr+1 + . . .+ θa−2γa−2 + θa−1γ

a−1 (mod Pa),


(1 + βΓrγΓr)p

b ≡ 1 (mod Pa),

as required. The inductive proof of our claim is now complete. Notice thatthe result indeed establishes the “edge case” Γr = k < Γ0 of item (b).

In order to see how our result establishes item (a) of the theorem, notethat if Γr = k = Γ0, then βΓr = βΓ0 and the condition for βΓr becomes:(1+βΓ0γ

Γ0)pb ≡ 1(modPa). We shall prove now that this condition is equivalent

to the condition

pbβΓ0γΓ0 +

(pb

p

)(βΓ0γ

Γ0)p ≡ 0 (mod Pa),

thus establishing item (a).In order to do so, observe that parts (d) and (e) of Proposition 4.2 imply

that mΓ0 = Γ0, mΓ0+1 = Γ0 + 1 and that TΓ0(u) achieves its minimal integralvalues at u ∈ {0, 1}. Thus, since mΓ0 < a − eb 6 mΓ0+1, it follows thata−eb = Γ0+1 and by part (b) of Proposition 4.3 the condition (1+βΓ0γ

Γ0)pb ≡

1 (mod Pa) is equivalent to

1 +

(pb

1

)βΓ0γ

Γ0 +

(pb

p

)(βΓ0γ

Γ0)p ≡ 1 (mod Pa),

that is

pbβΓ0γΓ0 +

(pb

p

)(βΓ0γ

Γ0)p ≡ 0 (mod Pa),

as desired.

Part 3: Suppose that Γr 6 k < Γ0. Our goal is to show that the incon-gruent solutions of xp


x ≡ 1 + βΓrγΓr + . . .+ βkγ

k + θk+1γk+1 + . . .+ θa−1γ

a−1 (mod Pa),


where θk+1, . . . , θa−1 ∈M are arbitrary and βΓr , . . . , βk ∈M satisfy


Γr+1 + . . .+ βkγk)p

b ≡ 1 (mod Pa).

This result, whose proof relies on Part 2, will establish item (b) of the theorem.We shall prove our claim by induction on k. In the “edge case” k = Γr, our

assertion holds by Part 2. So let Γr < k < Γ0 and suppose that our assertionholds for k ∈ {Γr,Γr + 1, . . . , k− 1}. We shall continue our proof by inductionon a.

By our assumptions mk < a− eb 6 mk+1. Hence

mk 6 (a− 1)− eb < mk+1

and by Remark 1 we may assume that a−1 > E. By part (5) of Proposition 4.4the basis of our induction is a = mk + 1 + eb.

So suppose, first, that (a − 1) − eb = mk. Since k > Γr > 1, we havek − 1 > 0 and (a− 1)− eb = m(k−1)+1. Hence

mk−1 < (a− 1)− eb = m(k−1)+1

and by our assumption concerning k − 1, the incongruent solutions of xpb ≡

1 (mod Pa−1) are

x ≡ 1 + βΓrγΓr + . . .+ βk−1γ

k−1 + θkγk + . . .+ θa−2γ

a−2 (mod Pa−1),

where θk, θk+1, . . . , θa−2 ∈M are arbitrary and βΓr , βΓr+1 , . . . βk−1 ∈M satisfy

(1 + βΓrγΓr + . . .+ βk−1γ

k−1)pb ≡ 1 (mod Pa−1).

By applying Proposition 4.3(a) with α = 1 + βΓrγΓr + . . . + βk−1γ

k−1 + θkγk

and n = a− 2, we obtain

(1 + βΓrγΓr + . . .+ βk−1γ

k−1 + θkγk + . . .+ θa−2γ

a−2)pb ≡

(1 + βΓrγΓr + . . .+ βk−1γ

k−1 + θkγk)p

b(mod Pa).

Denote θk = βk ∈M and suppose that the βi’s satisfy

(1 + βΓrγΓr + . . .+ βkγ

k)pb ≡ 1 (mod Pa).

Then if follows by Proposition 2.3 that the incongruent solutions of the con-gruence xp


x ≡ 1 + βΓrγΓr + . . .+ βkγ

k + θk+1γk+1 + . . .+ θa−1γ

a−1 (mod Pa),

where θk+1, . . . , θa−1 ∈M are arbitrary and βΓr , βΓr+1, . . . βk ∈M satisfy


Γr+1 + . . .+ βkγk)p

b ≡ 1 (mod Pa),

44 Boaz Cohen

as desired.Notice that βΓr , βΓr+1, . . . , βk−1 satisfying the last congruence, satisfy also

the previous condition

(1 + βΓrγΓr + . . .+ βk−1γ

k−1)pb ≡ 1 (mod Pa−1).

Indeed, denote

B = 1 + βΓrγΓr + βΓr+1γ

Γr+1 + . . .+ βk−1γk−1.

Then

(B + βkγk)p

b

= Bpb +

pb∑j=1

(pb

j

)Bpb−jγkjβjk ≡ 1 (mod Pa−1),

and as shown in the proof of Proposition 4.3,(pb

j

)Bpb−jγkjβjk ≡ 0 (mod Pe(b−u)+kj),

where 1 6 j 6 pb and u is the non-negative integer satisfying pu‖j. Sincej > pu, we get

e(b−u)+kj > e(b−u)+kpu = (kpu−eu)+eb = Tk(u)+eb > mk+eb = a−1.

This inequality implies that Bpb ≡ 1 (mod Pa−1), as claimed. The proof for theinduction basis a = mk + 1 + eb is now complete.

Let now a > mk + 1 + eb and suppose that our assertion holds for a ∈{mk + 1 + eb,mk + 2 + eb, . . . , a− 1}. Since mk < (a− 1)− eb, our assumptionthat mk < a − eb 6 mk+1 implies that mk < (a − 1) − eb < mk+1, and byRemark 1 we may assume that a − 1 > E. Hence it follows, by applyinginduction for a− 1, that the incongruent solutions of xp


x ≡ 1 + βΓrγΓr + . . .+ βkγ

k + θk+1γk+1 + . . .+ θa−2γ

a−2 (mod Pa−1),

where θk+1, θk+2, . . . , θa−2 ∈M are arbitrary and βΓr , βΓr+1, . . . βk ∈M satisfy

(1 + βΓrγΓr + . . .+ βkγ

k)pb ≡ 1 (mod Pa−1).

By applying Proposition 4.3(a) with α = 1+βΓrγΓr + . . .+βkγ

k and n = a−2,we obtain

(1 + βΓrγΓr + · · ·+ βkγ

k + θk+1γk+1 + · · ·+ θa−2γ

a−2)pb ≡

(1 + βΓrγΓr + . . .+ βkγ

k)pb

(mod Pa).


Therefore, by Proposition 2.3, the incongruent solutions of xpb ≡ 1 (mod Pa)

are

x ≡ 1 + βΓrγΓr + . . .+ βkγ

k + θk+1γk+1 + . . .+ θa−1γ

a−1 (mod Pa),

where θk+1, θk+2, . . . , θa−1 ∈M are arbitrary and βΓr , βΓr+1, . . . βk ∈M satisfy

(1 + βΓrγΓr + . . .+ βkγ

k)pb ≡ 1 (mod Pa),

as required.

Part 4: Suppose that Γr < k = Γ0. Our goal is to show that the incon-gruent solutions of xp


x ≡ 1 + βΓrγΓr + . . .+ βΓ0γ

Γ0 + θΓ0+1γΓ0+1 + . . .+ θa−1γ

a−1 (mod Pa),

where θΓ0+1, . . . , θa−1 ∈ M are arbitrary and βΓr , . . . , βΓ0 ∈ M satisfy thefollowing system of congruences:{

(1 + βΓrγΓr + . . .+ βΓ0−1γ

Γ0−1)pb ≡ 1 (mod Pa−1)

pbβΓ0γΓ0 +

(pb

p

)(βΓ0γ

Γ0)p ≡ 1− (1 + βΓrγΓr + . . .+ βΓ0−1γ

Γ0−1)pb

(mod Pa).

This result will establish item (c) of the theorem.Notice that parts (d) and (e) of Proposition 4.2 imply that mΓ0 = Γ0 and

mΓ0+1 = Γ0 + 1. Since mΓ0 < a− eb 6 mΓ0+1, it follows that a− eb = Γ0 + 1.Thus mΓ0−1 < (a − 1) − eb = m(Γ0−1)+1 and by Remark 1 we may assumethat a − 1 > E. Hence it follows by Part 3 applied to k = Γ0 − 1, that theincongruent solutions of xp


x ≡ 1 + βΓrγΓr + . . .+ βΓ0−1γ

Γ0−1 + θΓ0γΓ0 + . . .+ θa−2γ

a−2 (mod Pa−1),

where θΓ0 , . . . , θa−2 ∈ M are arbitrary and βΓr , . . . , βΓ0−1 ∈ M satisfy thefollowing congruence:

(1 + βΓrγΓr + . . .+ βΓ0−1γ

Γ0−1)pb ≡ 1 (mod Pa−1).

Set B = 1 + βΓrγΓr + . . .+ βΓ0−1γ

Γ0−1. Clearly

B + θΓ0γΓ0 + . . .+ θa−2γ

a−2 ≡ B + θΓ0γΓ0 (mod PΓ0+1),

and since ep−1

= Γ0 < Γ0 + 1, it follows by Proposition 4.8 that

(B + θΓ0γΓ0 + . . .+ θa−2γ

a−2)pb ≡ (B + θΓ0γ

Γ0)pb

(mod PΓ0+1+eb).

But Γ0 + 1 + eb = a, so in order to be able to apply Proposition 2.3, weneed to consider the congruence (B + θΓ0γ

Γ0)pb ≡ 1 (mod Pa). By part (d) of

46 Boaz Cohen

Proposition 4.2, TΓ0(u) attains its integral minimum value for u ∈ {0, 1}. Thusit follows by Proposition 4.3(b) with α = B and k = Γ0, that (B + θΓ0γ

Γ0)pb ≡

1 (mod Pa) is equivalent to

Bpb +

(pb

1

)θΓ0γ

Γ0Bpb−1 +

(pb

p

)(θΓ0γ

Γ0)pBpb−p ≡ 1 (mod Pa).

This congruence may be a bit simplified. Observe that B ≡ 1 (mod P), soBpb−1 ≡ 1 (mod P). In addition, pbθΓ0γ

Γ0 ≡ 0 (mod PΓ0+eb), so pbθΓ0γΓ0 ≡

0 (mod Pa−1). Therefore(pb

1

)θΓ0γ

Γ0Bpb−1 ≡(pb

1

)θΓ0γ

Γ0 (mod Pa).

Similarly, by Proposition II.2.2,(pb

p

)(θΓ0γ

Γ0)pBpb−p ≡(pb

p

)(θΓ0γ

Γ0)p (mod Pe(b−1)+pΓ0+1).

Since e(b − 1) + pΓ0 + 1 = eb + (pΓ0 − e) + 1 = eb + Γ0 + 1 = a, the abovecongruence may be written as

pbθΓ0γΓ0 +

(pb

p

)(θΓ0γ

Γ0)p ≡ 1− Bpb (mod Pa).

Denote θΓ0 = βΓ0 ∈ M satisfying the above congruence. Then it follows byProposition 2.3 that the incongruent solutions of xp


x ≡ 1 + βΓrγΓr + . . .+ βΓ0γ

Γ0 + θΓ0+1γΓ0+1 + . . .+ θa−1γ

a−1 (mod Pa),

where θΓ0+1, . . . , θa−1 ∈ M are arbitrary and βΓr , . . . , βΓ0 ∈ M satisfy thefollowing system of congruences:{

(1 + βΓrγΓr + . . .+ βΓ0−1γ

Γ0−1)pb ≡ 1 (mod Pa−1)

pbβΓ0γΓ0 +

(pb

p

)(βΓ0γ

Γ0)p ≡ 1− (1 + βΓrγΓr + . . .+ βΓ0−1γ

Γ0−1)pb

(mod Pa),

as desired.

Part 5: Suppose that Γ0 6 k. Our goal is to show that the incongruentsolutions of xp


x ≡ 1 + βΓrγΓr + . . .+ βkγ

k + θk+1γk+1 + . . .+ θa−1γ

a−1 (mod Pa),

where θk+1, . . . , θa−1 ∈M are arbitrary and βΓr , . . . , βk ∈M satisfy one of thefollowing systems of congruences:

pbβΓ0γΓ0 +

(pb

p

)(βΓ0γ

Γ0)p ≡ 0 (mod Pa−(k−Γ0))

pbβΓ0+1γΓ0+1 ≡ 1− (1 + βΓ0γ

Γ0)pb

(mod Pa−(k−Γ0)+1)...


Γ0 + . . .+ βk−1γk−1)p

b(mod Pa)


if Γr = Γ0 6 k, and

(1 + βΓrγΓr + · · ·+ βΓ0−1γ

Γ0−1)pb ≡ 1 (mod Pa−1−(k−Γ0))

pbβΓ0γΓ0 +

(pb

p

)(βΓ0γ

Γ0)p ≡ 1− (1 + βΓrγΓr + . . .+ βΓ0−1γ

Γ0−1)pb


pbβΓ0+1γΓ0+1 ≡ 1− (1 + βΓrγ

Γr + . . .+ βΓ0γΓ0)p


...


Γr + . . .+ βk−1γk−1)p

b(mod Pa)

if Γr < Γ0 6 k. This result will establish items (d) and (e) of the theorem.These two systems of congruences will be referred to as the “required systemsof congruences”.

If k = Γ0, then either Γr = k = Γ0 and our claim holds by item (a), orΓr < k = Γ0 and our claim holds by item (c). So suppose that k > Γ0 and ourclaim holds for k ∈ {Γ0,Γ0 + 1, . . . , k − 1}.

Since k > Γ0, part (e) of Proposition 4.2 implies that mk = k and mk+1 =k + 1. Since mk < a − eb 6 mk+1, it follows that a − eb = k + 1. Thusa − 1 − eb = k, mk−1 < (a − 1) − eb = mk and by Remark 1 we may assumethat a − 1 > E. By the induction hypothesis, the incongruent solutions ofxp


x ≡ 1 + βΓrγΓr + . . .+ βk−1γ

k−1 + θkγk + . . .+ θa−2γ

a−2 (mod Pa−1),

where θk, . . . , θa−2 ∈ M are arbitrary and βΓr , . . . , βk−1 ∈ M satisfy the fol-lowing systems of congruences:

pbβΓ0γΓ0 +

(pb

p

)(βΓ0γ

Γ0)p ≡ 0 (mod Pa−1−(k−1−Γ0))

pbβΓ0+1γΓ0+1 ≡ 1− (1 + βΓ0γ

Γ0)pb

(mod Pa−1−(k−1−Γ0)+1),...

pbβk−1γk−1 ≡ 1− (1 + βΓ0γ

Γ0 + . . .+ βk−2γk−2)p

b(mod Pa−1)

if Γr = Γ0 6 k − 1, and

(1 + βΓrγΓr + . . .+ βΓ0−1γ

Γ0−1)pb ≡ 1 (mod P(a−1)−1−(k−1−Γ0))

pbβΓ0γΓ0 +

(pb

p

)(βΓ0γ

Γ0)p ≡ 1− (1 + βΓrγΓr + . . .+ βΓ0−1γ

Γ0−1)pb

(mod Pa−1−(k−1−Γ0))

pbβΓ0+1γΓ0+1 ≡ 1− (1 + βΓrγ

Γr + . . .+ βΓ0γΓ0)p

b(mod Pa−1−(k−1−Γ0)+1)

...

pbβk−1γk−1 ≡ 1− (1 + βΓrγ

Γr + . . .+ βk−2γk−2)p

b(mod Pa−1)

if Γr < Γ0 6 k− 1. These systems of congruences may be rewritten as follows:pbβΓ0γ

Γ0 +(pb

p

)(βΓ0γ

Γ0)p ≡ 0 (mod Pa−(k−Γ0)

pbβΓ0+1γΓ0+1 ≡ 1− (1 + βΓ0γ

Γ0)pb

(mod Pa−(k−Γ0)+1)...

pbβk−1γk−1 ≡ 1− (1 + βΓ0γ

Γ0 + . . .+ βk−2γk−2)p

b(mod Pa−1)

48 Boaz Cohen

if Γr = Γ0 6 k − 1, and

(1 + βΓrγΓr + . . .+ βΓ0−1γ

Γ0−1)pb ≡ 1 (mod P(a−1)−(k−Γ0))

pbβΓ0γΓ0 +

(pb

p

)(βΓ0γ

Γ0)p ≡ 1− (1 + βΓrγΓr + . . .+ βΓ0−1γ

Γ0−1)pb


pbβΓ0+1γΓ0+1 ≡ 1− (1 + βΓrγ

Γr + . . .+ βΓ0γΓ0)p


...

pbβk−1γk−1 ≡ 1− (1 + βΓrγ

Γr + . . .+ βk−2γk−2)p

b(mod Pa−1)

if Γr < Γ0 6 k − 1. These two systems of congruences will be referred to asthe “last systems of congruences”.

Set B = 1+βΓrγΓr +. . .+βk−1γ

k−1. We shall show now that the incongruentsolutions of xp


x ≡ 1 + βΓrγΓr + . . .+ βkγ

k + θk+1γk+1 + . . .+ θa−1γ

a−1 (mod Pa),

where θk+1, . . . , θa−2 ∈ M are arbitrary and βΓr , . . . βk−1 ∈ M satisfy one ofthe “required systems of congruences”.

Recall that the incongruent solutions of xpb ≡ 1 (mod Pa−1) are

x ≡ B + θkγk + . . .+ θa−2γ

a−2 (mod Pa−1),

where θk, . . . , θa−2 ∈M are arbitrary and βΓr , . . . , βk−1 satisfy one of the “lastsystems of congruences”. Clearly

B + θkγk + . . .+ θa−2γ

a−2 ≡ B + θkγk (mod Pk+1).

Since Γ0 = ep−1

6 k − 1, it follows by Proposition 4.8 that

(B + θkγk + . . .+ θa−2γ

a−2)pb ≡ (B + θkγ

k)pb

(mod Pk+1+eb).

But k + 1 + eb = a, so in order to be able to apply Proposition 2.3, we needto consider the congruence (B + θkγ

k)pb ≡ 1 (mod Pa). Since k > Γ0, it follows

by part (e) of Proposition 4.2 that Tk(u) attains its integral minimum valueonly for u = 0. Thus it follows by Proposition 4.3(b) with α = B, that thecongruence (B + θkγ

k)pb ≡ 1 (mod Pa) is equivalent to

Bpb +

(pb

1

)θkγ

kBpb−1 ≡ 1 (mod Pa).

Denote θk = βk ∈M. Since Bpb−1 ≡ 1 (mod P) and pbβkγk ≡ 0 (mod Pk+eb), it

follows that pbβkγkBpb−1 ≡ pbβkγ

k (mod Pa). Thus the condition for βk is:

pbβkγk ≡ 1− Bpb (mod Pa).


By Proposition 2.3, the the incongruent solutions of xpb ≡ 1 (mod Pa) are

x ≡ 1 + βΓrγΓr + . . .+ βkγ

k + θk+1γk+1 + . . .+ θa−1γ

a−1 (mod Pa),

where θk+1, . . . , θa−1 ∈ M are arbitrary, βΓr , . . . βk−1 ∈ M satisfy one of “thelast systems of congruences” and in addition, βk ∈M satisfies


Γr + . . .+ βk−1γk−1)p

b

.

By adding this congruence to each of the “last systems of congruences”, weobtain the “required system of congruences”, as claimed. The proof of Theo-rem 4.9 is complete.

Let us add a few remarks concerning Theorem 4.9. First note that eventhough Theorem 4.9 gives us the general structure of the solutions, these so-lutions depend upon the tuples (βΓr , . . . , βk) which are not given explicitly,but only as solutions of another set of congruences. Thus, in order to find thesolutions explicitly, we need to find these β’s first. If Γr < Γ0 6 k, then the β’sin first Γ0−Γr positions satisfy a non-linear congruence, in which case unfortu-nately, there is not much we can say about its solutions, unless we know moreabout the structure of the field K. For instance, as will be shown in part IV ofthis study, if K is a quadratic field or a cyclotomic field, the structure of thesolutions can be fully determined. However, it is important to notice that wehave at least one certain tuple, namely (βΓr , . . . , βΓ0−1) = (0, 0, . . . , 0), whichsatisfies trivially this non-linear congruence. Next, βΓ0 satisfies another non-linear congruence of degree p. This congruence can be always reduced to a

congruence modulo P. Indeed, observe that pbγΓ0 and(pb

p

)γpΓ0 are 0 modulo

Pa−1−(k−Γ0) because a− eb = k + 1. In addition

1− (1 + βΓrγΓr + . . .+ βΓ0−1γ

Γ0−1)pb ≡ 0 (mod Pa−1−(k−Γ0)).

Now, by Proposition I.2.1(d), given any α ∈ OK such that α ≡ 0 (mod Pn−1),we can always find α′ ∈ OK such that α ≡ γn−1α′ (mod Pn). Hence, byProposition I.2.1(c), a suitable reduction can be made to obtain a congruencemodulo P. In particular this means that βΓ0 can have at most p possiblevalues. Note that choosing βΓr = . . . = βΓ0 = 0 will surely solve the first twocongruences. Each of the remaining β’s satisfies a linear congruence, which inthe same way as above, can be reduced to a congruence modulo P. Unlike theother two congruences, this linear congruence is uniquely solvable for everyappropriate choice of βΓr , . . . , βΓ0 .

When p− 1 | e but e is not divisible by p, then r = 0. Therefore, it followsfrom the above remarks that in this case the β’s satisfy only one congruenceof degree p and a series of linear congruences, all of which, depend upon thevalue of βΓ0 . In this special case, the system gets a simpler form:

50 Boaz Cohen

Theorem 4.10. Assume Notation 4.1. Set ∆ = a− eb, Γ = ep−1

and supposethat Γ < ∆. Then the following statements hold.

(a) If p− 1 - e, then the incongruent solutions of xpb ≡ 1 (mod Pa) are

x ≡ 1 + θ∆γ∆ + . . .+ θa−1γ

a−1 (mod Pa),

where θ∆, . . . , θa−1 ∈M are arbitrary.

(b) If p− 1 | e but p - e, then the incongruent solutions of xpb ≡ 1 (mod Pa)

are

x ≡ 1 + βΓγΓ + . . .+ β∆−1γ

∆−1 + θ∆γ∆ + . . .+ θa−1γ

a−1 (mod Pa),

where θ∆, . . . , θa−1 ∈ M are arbitrary and βΓ ∈ M satisfies one of thefollowing congruences: either

βΓ ≡ 0 (mod P) or (γΓβΓ)p−1 ≡ −p (mod Pe+1).

Moreover, if Γ < ∆ − 1, then βΓ, . . . , β∆−1 ∈ M satisfy the followingsystem of congruences:

pbβΓ+1γΓ+1 ≡ 1− (1 + βΓγ

Γ)pb

(mod Pa−(∆−Γ)+2)

pbβΓ+2γΓ+2 ≡ 1− (1 + βΓγ

Γ + βΓ+1γΓ+1)p

b(mod Pa−(∆−Γ)+3)

...

pbβ∆−1γ∆−1 ≡ 1− (1 + βΓγ

Γ + . . .+ β∆−2γ∆−2)p

b(mod Pa).

Proof. (a) By our assumption ∆ > ep−1

, so by part (e) of Proposition 4.2we get that m∆ = ∆. Hence, k = ∆ − 1 is the unique integer such thatmk < a− eb 6 mk+1. In addition, note that 0 < Γ < a− eb, so 1 6 e 6 eb < aand hence 2 6 a and e < a. Thus max{2, e} = E 6 a. The assertion thenfollows from Theorem 4.6.

(b) As in part (a), we deduce that max{2, e} = E 6 a and k = ∆ − 1 isthe unique integer such that mk < a − eb 6 mk+1. In addition, p − 1 | e butp - e, so r = 0 and Γr = Γ0 = Γ. Hence, by parts (a) and (d) of Theorem 4.9we deduce that the incongruent solutions of xp


x ≡ 1 + βΓγΓ + . . .+ β∆−1γ

∆−1 + θ∆γ∆ + . . .+ θa−1γ

a−1 (mod Pa),

where θ∆, . . . , θa−1 ∈M are arbitrary, βΓ ∈M satisfies the congruence

(∗) pbβΓγΓ +

(pb

p

)(βΓγ

Γ)p ≡ 0 (mod Pa−(∆−Γ)+1)


and if Γ < ∆ − 1, then βΓ+1, . . . , β∆−1 ∈ M satisfy the following system ofcongruences:

pbβΓ+1γΓ+1 ≡ 1− (1 + βΓγ

Γ)pb

(mod Pa−(∆−Γ)+2)

pbβΓ+2γΓ+2 ≡ 1− (1 + βΓγ

Γ + βΓ+1γΓ+1)p

b(mod Pa−(∆−Γ)+3)

...

pbβ∆−1γ∆−1 ≡ 1− (1 + βΓγ

Γ + . . .+ β∆−2γ∆−2)p

b(mod Pa).

We claim that the congruence (∗) is equivalent to the following statement:either

βΓ ≡ 0 (mod P) or (γΓβΓ)p−1 ≡ −p (mod Pe+1).

To prove that statement, we shall deal separately with the cases p > 2 andp = 2.

Suppose that p > 2. First we prove that the following congruence holds:

(∗∗)(pb

p

)≡ pb−1 (mod Pa−(∆−Γ)+1).

Indeed, suppose that j is an integer satisfying 1 6 j 6 p−12

. Note that

(pb − j)(pb − p+ j) = pb(pb − p) + j(p− j) = p2b − pb+1 + j(p− j).

Since p > 2 and p− 1 | e, we have e > 2. Thus

1 =1

2+

1

2>

1

p− 1+

1

e,

and since a− eb = ∆ we obtain that

e(b+ 1) = eb+ e > eb+ e

(1

p− 1+

1

e

)= eb+ Γ + 1 = a− (∆− Γ) + 1.

In addition, 2eb > e(b+1), so p2b ≡ pb+1 ≡ 0(modPa−(∆−Γ)+1) and we concludethat (pb − j)(pb − p + j) ≡ j(p − j) (mod Pa−(∆−Γ)+1) for each 1 6 j 6 p−1

2.

Now,

(pb − 1)!

(pb − p)!= (pb − 1)(pb − 2) · · · (pb − p+ 1)

=

p−12∏j=1

(pb − j)(pb − p+ j)

≡

p−12∏j=1

j(p− j) = (p− 1)! (mod Pa−(∆−Γ)+1).

52 Boaz Cohen

Since (p− 1)! is invertible modulo Pa, it follows that(pb − 1

pb − p

)=

(pb − 1)!

(pb − p)! (p− 1)!≡ 1 (mod Pa−(∆−Γ)+1),

and hence(pb

p

)=

(pb − 1

pb − p

)pb

p=

(pb − 1

pb − p

)pb−1 ≡ pb−1 (mod Pa−(∆−Γ)+1),

as desired. Plugging (∗∗) in (∗) yields

pbβΓγΓ + pb−1(βΓγ

Γ)p ≡ 0 (mod Pa−(∆−Γ)+1)

pb−1γΓβΓ

(p+ (γΓβΓ)p−1

)≡ 0 (mod Pa−(∆−Γ)+1)

βΓ

(p+ (γΓβΓ)p−1

)≡ 0 (mod Pa−(∆−Γ)+1−e(b−1)−Γ)

βΓ

(p+ (γΓβΓ)p−1

)≡ 0 (mod Pe+1)

mβΓ ≡ 0 (mod P) or (γΓβΓ)p−1 ≡ −p (mod Pe+1).

Since each of our operations is reversible, the proof of the required equivalencefor p > 2 us complete.

Suppose, finally, that p = 2. Then(

2b

2

)= 2b−1(2b − 1), and plugging this

equality into (∗) gives

2bβΓγΓ + 2b−1(2b − 1)(βΓγ

Γ)2 ≡ 0 (mod Pa−(∆−Γ)+1)

2b−1γΓβΓ

(2 + (2b − 1)γΓβΓ

)≡ 0 (mod Pa−(∆−Γ)+1)

βΓ

(2 + (2b − 1)γΓβΓ

)≡ 0 (mod Pe+1)

mβΓ ≡ 0 (mod P) or (2b − 1)γΓβΓ ≡ −2 (mod Pe+1).

If b = 1, then the above right-hand congruence is γΓβΓ ≡ −2 (mod Pe+1). Thisis also the case if b > 2. Indeed, since eb > e+1, we have 2b−1 ≡ −1(modPe+1).Hence, (2b−1)γΓβΓ ≡ −2(modPe+1) becomes γΓβΓ ≡ 2(modPe+1). But since2 ≡ −2 (mod Pe+1), we obtain

γΓβΓ ≡ −2 (mod Pe+1).

Since each of our operations is reversible, the proof of the required equivalencefor p = 2 is complete. The proof of Theorem 4.10 is now complete.

The following example illustrates the result of Theorem 4.10.

Example 4.11. Suppose that K = Q(√

6). As one can verify, OK = Z+Z√

6.We want to solve the congruence

x9 ≡ 1 (mod P10), where P = (3,√

6).


In this case: a = 10, b = 2 and P2 = (3). Thus P is lying above p = 3, withramification index e = 2. Since NP = 3, we can choose M = {−1, 0, 1} to bea complete residue system modulo P. Here

∆ = a− eb = 6, Γ =e

p− 1= 1, ∆− 1 = 5

and we can choose the uniformizer γ =√

6. Since p − 1 | e, it follows bypart (b) of Theorem 4.10 that the solutions are of the form

x ≡ 1 +β1γ+β2γ2 +β3γ

3 +β4γ4 +β5γ

5 + θ6γ6 + θ7γ

7 + θ8γ8 + θ9γ

9 (mod P10),

where θ6, θ7, θ8 and θ9 are arbitrary elements of M and either β1 = β2 = β3 =β4 = β5 = 0 or β1 satisfies (

√6β1)2 ≡ −3 (mod P3) and β2, β3, β4, β5 satisfy

the following system of congruences9(√

6)2β2 ≡ 1− (1 +√

6β1)9 (mod P7)

9(√

6)3β3 ≡ 1− (1 +√

6β1 + 6β2)9 (mod P8)

9(√

6)4β4 ≡ 1− (1 +√

6β1 + 6β2 + 6√

6β3)9 (mod P9)

9(√

6)5β5 ≡ 1− (1 +√

6β1 + 6β2 + 6√

6β3 + 36β4)9 (mod P10).

We first solve:

(√

6β1)2 ≡ −3 (mod P3)6β2

1 ≡ −3 (mod P3)m

2β21 ≡ −1 (mod P)

−β21 ≡ −1 (mod P)

β21 ≡ 1 (mod P).

Hence β1 ≡ ±1(modP). Assigning β1 = 1 in the first congruence of the systemand noticing that

√6 ∈ P yields

54β2 ≡ 1− (1 +√

6)9 (mod P7)

54β2 ≡ 34560− 14121√

6 (mod P7)m

2β2 ≡ −1280− 523√

6 (mod P)

−β2 ≡ −1280− 523√

6 (mod P)β2 ≡ 1280 (mod P)β2 ≡ −1 (mod P)

As one can verify, plugging β1 = 1 and β2 = −1 into the second congruenceof the system gives β3 ≡ 0 (mod P). In the same way we can determine theother β’s. Continuing this process gives finally the tuple (1,−1, 0,−1,−1).

54 Boaz Cohen

Starting from β1 = −1 will produce the tuple (−1,−1, 0,−1, 1). Computing1 + β1γ + . . .+ β5γ

5 for all three tuples gives the following roots

x ≡ 1 + θ6γ6 + θ7γ

7 + θ8γ8 + θ9γ

9 (mod P10)

x ≡ 1 + γ − γ2 − γ4 − γ5 + θ6γ6 + θ7γ

7 + θ8γ8 + θ9γ

9 (mod P10)

x ≡ 1− γ − γ2 − γ4 + γ5 + θ6γ6 + θ7γ

7 + θ8γ8 + θ9γ

9 (mod P10).

where θi ∈ {−1, 0, 1}.

The results of Theorem 3.2, Theorem 4.6 and Theorem 4.9 help us tounderstand the structure of the tree corresponding to the pb-th roots of unitymodulo Pa. These theorems describe different levels of the tree. Theorem 3.2discusses the initial levels of the tree, namely the first e-th levels. In these levelsthe “growth” of the tree is determined by the number ∆ = [a/pb]. Theorem 4.6and Theorem 4.9 discuss further levels of the tree, namely when e < a. Thestructure of the tree for levels a > e depends upon whether p−1 - e or p−1 | e.In both cases, the number of branches in the initial and intermediate levelsgradually increases. This growth “stabilizes” at some point and becomes afixed pattern. This behavior is described in the following trees.

Figure 4a: A typical form of a tree when p− 1 - e


Figure 4b: A typical form of a tree when p− 1 | e

The following theorem asserts that this stabilization occurs when ep−1

< a−eb.Recall that we denote e

p−1by Γ0.

Theorem 4.12. Assume Notation 4.1. Set ∆ = a − eb and suppose thatΓ0 < ∆. Then the following statements hold.

(a) If p−1 - e, then the incongruent solutions modulo Pa of xpb ≡ 1(modPa)

are

x ≡ 1 + θ∆γ∆ + . . .+ θa−1γ

a−1,

where θ∆, . . . , θa−1 ∈M are arbitrary.

Consequently, the a-th level of the tree corresponding to these solutions has(NP)eb vertices. Moreover, only (NP)eb−1 of these vertices have children, andthe vertices with children are the solutions with θ∆ = 0 and have exactly NPchildren each.

(b) If p−1 | e, then the incongruent solutions modulo Pa of xpb ≡ 1(modPa)

are

x ≡ 1+βΓrγΓr+. . .+βΓ0γ

Γ0+βΓ0+1γΓ0+1+. . .+β∆−1γ

∆−1+θ∆γ∆+. . .+θa−1γ

a−1,

where βΓr , . . . , βΓ0 ∈M satisfy the systems in parts (d) or (e) of Theorem 4.9respectively by taking k = ∆ − 1, βΓ0+1, . . . , θ∆ ∈ M are determined uniquelyby the βΓr , . . . , βΓ0 and θ∆, . . . , θa−1 ∈M are arbitrary.

Consequently, the a-th level of the tree corresponding to these solutions hass(NP)eb vertices, where s is the number of the tuples (βΓr , . . . , βΓ0). Moreover,only s(NP)eb−1 of these vertices have children, and those with children are thesolutions with θ∆ = β∆ and have exactly NP children each.

56 Boaz Cohen

Proof. (a) Suppose that p − 1 - e. If Γ0 < ∆, then by the proof of Theo-rem 4.10(a) it follows that k = ∆ − 1. Therefore, by Theorem 4.6, the pb-throots of unity modulo Pa are

x ≡ 1 + θ∆γ∆ + θ∆+1γ

∆+1 + . . .+ θa−1γa−1,

as required. Since there are NP possibilities for each one of the θ’s, the totalnumber of solutions is (NP)a−∆ = (NP)eb. Now, the pb-th roots of unitymodulo Pa+1 are

x ≡ 1 + θ∆+1γ∆+1 + . . .+ θa−1γ

a−1 + θaγa.

Since

1 + θ∆γ∆ + . . .+ θa−1γ

a−1 ≡ 1 + θ∆+1γ∆+1 + . . .+ θa−1γ

a−1 + θaγa (mod Pa)

if and only if θ∆ = 0, it follows that the solutions modulo Pa+1 that are beinglifted from the modulo Pa solutions are exactly those with θ∆ = 0, and theirnumber is (NP)eb−1. Furthermore, each such vertex v has NP children of theform v + θaγ

a.(b) Suppose that p−1 | e. If Γ0 < ∆, then by the proof of Theorem 4.10(b)

it follows again that k = ∆−1. Therefore, by part (d) and (e) of Theorem 4.9,the incongruent solutions modulo Pa of xp


x ≡ 1+βΓrγΓr+. . .+βΓ0γ

Γ0+βΓ0+1γΓ0+1+. . .+β∆−1γ

∆−1+θ∆γ∆+. . .+θa−1γ

a−1,

as required. As mentioned after the proof of Theorem 4.9, each one of thecoefficients βΓ0+1, . . . , β∆−1 satisfies a linear congruence, which can be reducedto a congruence modulo P and which is uniquely solvable for every appropriatechoice of βΓr , . . . , βΓ0 . It follows that the number of solutions is s(NP)a−∆ =s(NP)eb. The rest of the claim follows in the same manner as in the previouscase.

Example 4.13. To illustrate the fixed pattern described in Theorem 4.12, letus return to Example 2.4 where we considered the congruence x3 ≡ 1 (mod Pa)for K = Q(

√3) and P = (

√3). In this example, p = 3, e = 2, b = 1 and

γ =√

3, so p − 1 | e, r = 0, Γ0 = 1 and ∆ = a − 2. In addition, in thesesettings M = {−1, 0, 1}, so NP = 3.

By Theorem 4.12, if Γ0 < ∆, that is, if 4 6 a, then the incongruentsolutions of x3 ≡ 1 (mod Pa) are

x ≡ 1 + β1γ + β2γ2 + . . .+ βa−3γ

a−3 + θa−2γa−2 + θa−1γ

a−1 (mod Pa).

Since in this case p - e, we may apply Theorem 4.10(b) which simplifies a bitthe system of congruences which determine the β’s. Accordingly,

β1 ≡ 0 (mod P) or (√

3β1)2 ≡ −3 (mod P3).


Since P2 = (3), the right-hand congruence is equivalent to the congruence3β2

1 ≡ −3 (mod 3√

3), that is β21 ≡ −1 (mod

√3). Trying successively the

elements of M, we find that none of them is a solution of β21 ≡ −1 (mod

√3).

It follows that β1 = 0, so the number of 1-tuples (β1) is s = 1. As one canverify, by plugging β1 = 0 into the rest of the congruences in Theorem 4.10(b)we obtain that β2 = . . . = βa−3 = 0. Therefore, if a > 4, the solutions ofx3 ≡ 1 (mod Pa) are

x ≡ 1 + θa−2γa−2 + θa−1γ

a−1 (mod Pa),

where θa−2, θa−1 ∈M are arbitrary. Consequently, if a > 4, then the a-th levelof the tree corresponding to these solutions has s(NP)eb = 9 vertices. Moreover,only (NP)eb−1 = 3 of these vertices have NP = 3 children. As can be seen,these results are reflected in the tree diagram presented in Example 2.4.

5 The general structure of the P-adic pb-th roots

of unity.

We turn now to discuss the results we have obtained from the perspective ofP-adic fields. Recall that each infinite “branch” of the tree corresponding tothe solutions of xp

b ≡ 1 (mod Pa), converges to a solution of xpb

= 1 in the P-adic field KP. As we shall see, when p−1 - e there is only one infinite “branch”and this branch converges to the trivial solution x = 1 (see Figure 4a). Whenp− 1 | e the tree may have several infinite branches (see Figure 4b). In orderto establish these claims we need the following proposition.

Proposition 5.1. Let K be a number field, P a prime ideal lying above therational prime p, a ∈ OP, M a complete residue system modulo P, γ ∈ P2 \ Pand n a positive integer. If a = α0 + α1γ + α2γ

2 + . . . is the P-adic expansionof a, where αi ∈M and if

a ≡ β0 + β1γ + β2γ2 + . . .+ βn−1γ

n−1 (mod Pn),

where βi ∈M, then αi = βi for every 0 6 i 6 n− 1

Proof. Set an = α0 + α1γ + . . .+ αn−1γn−1. Note that

|a− an|P = |αnγn + αn+1γn+1 + αn+2γ

n+2 + . . . |P= |γn|P|αn + αn+1γ + αn+2γ

2 + . . . |P6 |γn|P = p−n

Hence, by Proposition I.2.4 we get an ≡ a (mod Pn), so by the assumption

an ≡ β0 + β1γ + . . .+ βn−1γn−1 (mod Pn)

⇓an ≡ β0 + β1γ + . . .+ βn−1γ

n−1 (mod P).

58 Boaz Cohen

Therefore α0 ≡ β0 (mod P). But α0, β0 ∈M, so α0 = β0. It follows that

α1γ + α2γ2 + . . .+ αn−1γ

n−1 ≡ β1γ + β2γ2 + . . .+ βn−1γ

n−1 (mod Pn),

so

α1 + α2γ + . . .+ αn−1γn−2 ≡ β1 + β2γ + . . .+ βn−1γ

n−2 (mod Pn−1).

By reducing modulo P we deduce as above that α1 = β1. Repeating thatprocess gives eventually that αi = βi for every i, as required.

The following theorem characterizes the general structure of the P-adicpb-th roots of unity.

Theorem 5.2. Assume Notation 4.1.

(a) If p− 1 - e, then the only P-adic pb-th root of unity is x = 1.

(b) If p− 1 | e, then the P-adic pb-th roots of unity are

x = 1 + βΓrγΓr + . . .+ βΓ0γ

Γ0 + βΓ0+1γΓ0+1 + . . .

where the β’s are in M and satisfy the following conditions:

(∗)

(1 + βΓrγΓr + . . .+ βΓ0−1γ

Γ0−1)pb ≡ 1 (mod Peb+Γ0)

pbβΓ0γΓ0 +

(pb

p

)(βΓ0γ

Γ0)p ≡ 1− (1 + βΓrγΓr + . . .+ βΓ0−1γ

Γ0−1)pb

(mod Peb+Γ0+1)

pbβΓ0+1γΓ0+1 ≡ 1− (1 + βΓrγ

Γr + . . .+ βΓ0γΓ0)p

b(mod Peb+Γ0+2)

pbβΓ0+2γΓ0+2 ≡ 1− (1 + βΓrγ

Γr + . . .+ βΓ0+1γΓ0+1)p

b(mod Peb+Γ0+3)

...

if Γr < Γ0, and

(∗∗)

βΓ0 ≡ 0 (mod P) or (γΓ0βΓ0)p−1 ≡ −p (mod Pe+1)

pbβΓ0+1γΓ0+1 ≡ 1− (1 + βΓ0γ

Γ0)pb

(mod Peb+Γ0+2)

pbβΓ0+2γΓ0+2 ≡ 1− (1 + βΓ0γ

Γ0 + βΓ0+1γΓ0+1)p

b(mod Peb+Γ0+3)

pbβΓ0+3γΓ0+3 ≡ 1− (1 + βΓ0γ

Γ0 + βΓ0+1γΓ0+1 + βΓ0+2γ

Γ0+2)pb

(mod Peb+Γ0+4)...

if Γr = Γ0.

Proof. Suppose that a ∈ OP satisfies apb

= 1 and let

a = α0 + α1γ + α2γ2 + α3γ

3 + . . .

where αi ∈ M, be the P-adic expansion of a. For every positive integer adefine ∆a = a − eb. Since ap

b= 1, we have ap

b ≡ 1 (mod Pa) for every a. In


particular, fix a such that Γ0 + 1 < ∆a and max{2, e} 6 a. Note that mk = kfor every k > Γ0 by parts (d) and (e) of Proposition 4.2. Hence, since Γ0 < ∆a

we get that mk < ∆a 6 mk+1 for k = ∆a − 1.

(a) Suppose that p − 1 - e. We shall prove that in this case a = 1, that isα0 = 1 and αi = 0 for every i > 1.

Since Γ0 < ∆a, Theorem 4.10(a) implies that there are θ∆a , . . . , θa−1 ∈Msuch that

a ≡ 1 + θ∆aγ∆a + . . .+ θa−1γ

a−1 (mod Pa).

Hence, by Proposition 5.1 we get that α0 = 1 and αi = 0 for every 1 6 i < ∆a.Now, since a can be taken arbitrary large and since ∆a → ∞ when a → ∞,it follows that α0 = 1 and αi = 0 for every 1 6 i, as required. Thus a = 1.Since x = 1 is clearly a solution of xp

b= 1, we conclude that x = 1 is the only

solution of xpb

= 1.

(b) Suppose that p − 1 | e. We shall prove that in this case α0 = 1, αi = 0for every 1 6 i < Γr and (αΓr , αΓr+1, αΓr+2, . . .) satisfy the system (∗) or (∗∗)according to whether Γr < Γ0 or Γr = Γ0, respectively.

Suppose first that Γr < Γ0. Recall that Γ0 + 1 < ∆a and mk < ∆a 6 mk+1

for k = ∆a − 1. Thus Γ0 < k, so by part (e) of Theorem 4.9 there areθ∆a , . . . , θa−1 ∈M and βΓr , . . . , β∆a−1 ∈M such that

a ≡ 1 + βΓrγΓr + . . .+ β∆a−1γ

∆a−1 + θ∆aγ∆a + . . .+ θa−1γ

a−1 (mod Pa).

Hence, by Proposition 5.1 we get that α0 = 1, αi = 0 for 1 6 i < Γr andαi = βi for every Γr 6 i < ∆a. Moreover, the tuple (βΓr , . . . , β∆a−1) satisfiesthe following system of congruences:

(1 + βΓrγΓr + . . .+ βΓ0−1γ

Γ0−1)pb ≡ 1 (mod Pa−1−(k−Γ0))

pbβΓ0γΓ0 +

(pb

p

)(βΓ0γ

Γ0)p ≡ 1− (1 + βΓrγΓr + . . .+ βΓ0−1γ

Γ0−1)pb


pbβΓ0+1γΓ0+1 ≡ 1− (1 + βΓrγ

Γr + . . .+ βΓ0γΓ0)p


...


Γr + . . .+ βk−1γk−1)p

b(mod Pa),

where k = ∆a − 1. Plugging a− k = eb+ 1 gives us the system

(1 + βΓrγΓr + . . .+ βΓ0−1γ

Γ0−1)pb ≡ 1 (mod Peb+Γ0)

pbβΓ0γΓ0 +

(pb

p

)(βΓ0γ

Γ0)p ≡ 1− (1 + βΓrγΓr + . . .+ βΓ0−1γ

Γ0−1)pb

(mod Peb+Γ0+1)

pbβΓ0+1γΓ0+1 ≡ 1− (1 + βΓrγ

Γr + . . .+ βΓ0γΓ0)p

b(mod Peb+Γ0+2)

...

pbβ∆a−1γ∆a−1 ≡ 1− (1 + βΓrγ

Γr + . . .+ β∆a−2γ∆a−2)p

b(mod Peb+∆a).

60 Boaz Cohen

Now, since a can be taken arbitrary large and since ∆a → ∞ when a → ∞,it follows that α0 = 1, αi = 0 for 1 6 i < Γr and αi = βi for every Γr 6 i.Moreover, the β’s, and therefore the α’s, satisfy the desired system (∗).

Suppose now that Γr = Γ0. As above Γ0 < k, so by part (d) of Theorem 4.9there are θ∆a , . . . , θa−1 ∈M and βΓr , . . . , β∆a−1 ∈M such that

a ≡ 1 + βΓ0γΓ0 + . . .+ β∆a−1γ

∆a−1 + θ∆aγ∆a + . . .+ θa−1γ

a−1 (mod Pa).

As above, we have α0 = 1, αi = 0 for 1 6 i < Γ0 and αi = βi for everyΓ0 6 i < ∆a. Moreover, the tuple (βΓ0 , . . . , β∆a−1) satisfies the followingsystem of congruences:

pbβΓ0γΓ0 +

(pb

p

)(βΓ0γ

Γ0)p ≡ 0 (mod Pa−(k−Γ0))

pbβΓ0+1γΓ0+1 ≡ 1− (1 + βΓ0γ

Γ0)pb

(mod Pa−(k−Γ0)+1)...


Γ0 + . . .+ βk−1γk−1)p

b(mod Pa),

where k = ∆a−1. As was shown in the proof of Theorem 4.10, the congruence

pbβΓ0γΓ0 +

(pb

p

)(βΓ0γ

Γ0)p ≡ 0 (mod Pa−(k−Γ0))

is equivalent to the congruences

either βΓ0 ≡ 0 (mod P) or (γΓ0βΓ0)p−1 ≡ −p (mod Pe+1)

As above, taking a→∞ we get that ∆a →∞ and by plugging a− k = eb+ 1we get that the α’s satisfy the desired system (∗∗).

To complete the proof in the case p − 1 | e, we need to prove that ifβΓr , βΓr+1, . . . ∈M satisfy the system (∗) or (∗∗) and if

b = 1 + βΓrγΓr + . . .+ βΓ0γ

Γ0 + βΓ0+1γΓ0+1 + . . .

then b is a solution of xpb

= 1. To do so, for each positive integer n > Γr letus define

bn = 1 + βΓrγΓr + . . .+ βnγ

n.

Clearly, bn → b as n→∞. Now, choose an integer a such that Γ0 + 1 < ∆a.Hence as above k = ∆a − 1 and Γ0 < k, so according to whether Γr = Γ0

or Γr < Γ0, by parts (d) or (e) of Theorem 4.9, respectively, the solutions ofxp


x ≡ b∆a−1 + θ∆aγ∆a + . . .+ θa−1γ

a−1 (mod Pa),

where θ∆a , . . . , θa−1 ∈M are chosen arbitrarily. In particular, for θ∆a = · · · =θa−1 = 0, we obtain that b∆a−1 satisfies bp

b

∆a−1 ≡ 1 (mod Pa). Since ∆a → ∞when a → ∞, it follows by Proposition I.2.5 that bp

b

∆a−1 → 1 when a → ∞.

But since bpb

∆a−1 → bpb, we get bp

b= 1, as required.


To illustrate Theorem 5.2, consider the equation x9 = 1 in KP, whereK = Q(

√6) and P = (3,

√6). As we saw in Example 4.11, here p = 3 and

e = 2, so p−1 | e. According to the computation performed in Example 4.11 itfollows that this equation has 3 different solutions and that the P-expansionsof these roots are

x = 1

x = 1− γ − γ2 − γ4 + γ5 + . . .

x = 1 + γ − γ2 − γ4 − γ5 + . . .

where γ =√

6.In Theorem II.3.2 we were able to describe the number of pb-th roots of

unity in KP in the case where P‖(p). In the next theorem we shall provideresults regarding some ramified cases. We need first the following proposition(for a proof see [3, p. 80]):

Proposition 5.3. Let F be a finite field with q elements and let a ∈ F∗. Thenxn = a is solvable iff a(q−1)/d = 1, where d = gcd(n, q − 1). Moreover, if theequation is solvable, then there are exactly d solutions.

Theorem 5.4. Let K be a number field, b a positive integer, P be a primeideal lying above the rational prime p with ramification index e > 2 and let Ndenote the number of pb-th roots of unity in KP. Then the following statementshold:

(a) If p− 1 - e, then N = 1.

(b) If p− 1 | e but p - e, then

N =

{p if xp−1 ≡ −p (mod Pe+1) is solvable1 otherwise.

Proof. If p−1 - e, then by part (a) of Theorem 5.2, the only pb-th root of unityin KP is x = 1, as required.

Now suppose that p− 1 | e but p - e and assume Notation 4.1. Since p - e,we get that r = 0, so Γr = Γ0. Thus, by part (b) of Theorem 5.2 the pb-throots of unity in KP are of the form

x = 1 + βΓ0γΓ0 + βΓ0+1γ

Γ0+1 + . . .

where the β’s are in M and satisfy the following system of congruences:βΓ0 ≡ 0 (mod P) or (γΓ0βΓ0)

p−1 ≡ −p (mod Pe+1)

pbβΓ0+1γΓ0+1 ≡ 1− (1 + βΓ0γ

Γ0)pb

(mod Peb+Γ0+2)

pbβΓ0+2γΓ0+2 ≡ 1− (1 + βΓ0γ

Γ0 + βΓ0+1γΓ0+1)p

b(mod Peb+Γ0+3)

...

62 Boaz Cohen

Note that once we have picked up βΓ0 , the other β’s, namely βΓ0+1, βΓ0+2, . . .are determined uniquely modulo P. Since βΓ0 = 0 produces the solutionx = 1 + 0γΓ0 + 0γΓ0+1 + . . . = 1, the total number of pb-th roots of unity is 1plus the number of incongruent solutions modulo P of

(∗) (γΓ0βΓ0)p−1 ≡ −p (mod Pe+1).

Therefore, it suffices to prove that (∗) is solvable iff xp−1 ≡ −p (mod Pe+1) issolvable and if (∗) is solvable, then it has p− 1 incongruent solutions moduloP.

Clearly, if the (∗) is solvable, then xp−1 ≡ −p (mod Pe+1) is solvable. Con-versely, assume that the congruence xp−1 ≡ −p (mod Pe+1) is solvable and letx = ξ be a solution. Then

ξp−1 ≡ −p ≡ 0 (mod Pe),

and since Γ0 = ep−1

, we deduce by Proposition 3.1 that ξ ≡ γΓ0θΓ0 + . . . +

γe−1θe−1(modPe), where the θ’s are in M. Since Γ0 6 e, we deduce in particularthat ξ ≡ 0 (mod PΓ0), that is PΓ0 | (ξ). Thus gcd((γΓ0),Pe+1) = PΓ0 | (ξ) andit follows by Proposition I.2.1(c) that γΓ0βΓ0 ≡ ξ (mod Pe+1) is solvable, soalso (∗) is solvable.

To count the number of incongruent solutions modulo P of (∗), note thatsince (p− 1)Γ0 = e, we can write (∗) in the following equivalent form

γeβp−1Γ0≡ −p (mod Pe+1).

Since gcd((γe),Pe+1) = Pe | (−p), the congruence can be reduced to the formβp−1

Γ0≡ α (modP), where α ∈ OK. Moreover α 6≡ 0 (modP), because otherwise

βp−1Γ0≡ 0 (mod P), so −p ≡ γeβp−1

Γ0≡ 0 (mod Pe+1), a contradiction. Since

OK/P is a finite field and βp−1Γ0≡ α (mod P) is solvable, we deduce from

Proposition 5.3 that it has exactly gcd(p− 1,NP− 1) = p− 1 solutions. Theproof is therefore complete.

As we can see, the question whether the congruence xp−1 ≡ −p (mod Pe+1)is solvable is crucial for determining the structure of some pb-th roots of unity.Theorem 5.5 below gives a sufficient condition for solvability of this congruence.

Theorem 5.5. Let K be a number field and let P be a prime ideal lying abovethe prime number p with ramification index e. If K contains a non-trivial (thatis 6= 1) complex p-th roots of unity ζ, then x = ζ − 1 satisfies the congruencexp−1 ≡ −p (mod Pe+1).

Proof. First note that since ζ is a complex root of the monic polynomial xp−1,it follows that ζ ∈ OK. Set π = ζ − 1. Since ζ 6= 1, we obtain

0 =ζp − 1

ζ − 1=

(1 + π)p − 1

π=

1

π

p∑k=1

(p

k

)πk =

p∑k=1

(p

k

)πk−1,


and therefore

0 = p+ pπ

θ︷︸︸︷p−1∑k=2

(pk

)pπk−2 +πp−1.

Thus πp−1 = −p − pπθ and θ ∈ OK. Now, clearly ζp ≡ 1 (mod P), so ζ ≡1 (mod P) by Proposition I.3.2, that is π = ζ − 1 ≡ 0 (mod P). But Pe | (p)implies that pπ ≡ 0 (mod Pe+1). Hence

(ζ − 1)p−1 = πp−1 = −p− pπθ ≡ −p (mod Pe+1),

as claimed.

The condition in Theorem 5.5 above is sufficient, but not necessary. Toshow this, let us take

K1 = Q(√

3), P = (√

3), so OK = Z+ Z√

3 , (3) = P2.

K2 = Q(√

6), P = (√

6, 3), so OK = Z+ Z√

6 , (3) = P2.

First, both K1 and K2 do not contain a non-trivial third root of unity (namely

−12±√

32i). Now, in the case of K1, x2 ≡ −3 (mod P3) is not solvable: Indeed,

set x = a+ b√

3, a, b ∈ Z and assume that

(a+ b√

3)2 ≡ −3 (mod√

33)

a2 + 3b2 + 3 + 2ab√

3 ≡ 0 (mod 3√

3).

then

a2 + 3b2 + 3 + 2ab√

3

3√

3=

2ab

3+a2 + 3b2 + 3

9

√3 ∈ Z+ Z

√3,

and since 9 | a2 + 3b2 + 3, it follows that 3 | a2. Hence 3 | a, so 9 | 3b2 + 3, thatis 3b2 ≡ −3 (mod 9). But then b2 ≡ −1 (mod 3), which is impossible.InK2, on the other hand, the congruence x2 ≡ −3(mod P3) is solvable. Indeed,since P4 | (9), it follows that P3|(9) and (

√6)2 ≡ −3 (mod P3).

References

[1] B. Cohen, The structure of the n-th roots of unity in Residue rings ofprime ideals P over p in Algebraic Number Fields. Part I: n-th roots ofunity when p - n, International Mathematical Forum, 12 (2017), no. 9,439-455. https://doi.org/10.12988/imf.2017.7220

64 Boaz Cohen

[2] B. Cohen, The structure of the n-th roots of unity in Residue rings ofprime ideals P over p in Algebraic Number Fields. Part II: n-th roots ofunity when P‖(p), International Mathematical Forum, 12 (2017), no. 10,457-468 https://doi.org/10.12988/imf.2017.7221

[3] K. Ireland and M. Rosen, A Classical Inroduction to Modern NumberTheory, second edition, Springer, 1998

Received: December 29, 2017; Published: January 18, 2018

the structure of the n-th roots of unity · 16 boaz cohen the p-adic number eld and by o p the ring...

Documents