C. M. Barendrecht

Primitive roots in number fields

Bachelor thesis

August 2, 2018

Thesis supervisor: prof.dr. P. Stevenhagen

Universiteit LeidenMathematisch Instituut

Contents

1 The rational Artin conjecture 41.1 Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2 The splitting of prime ideals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 The problem reformulated . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.4 The density of primitive roots in Q . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2 The Artin conjecture for number fields 102.1 Formulating the conjecture for number fields . . . . . . . . . . . . . . . . . . . . 102.2 The Frobenius element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.3 Applications of the Frobenius element . . . . . . . . . . . . . . . . . . . . . . . . 14

3 The density of primitive roots in number fields 183.1 The main theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.2 Proof of the main theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203.3 Applications of the main theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

References 24

1

Introduction

Let a ∈ Z\0 be a non-zero integer, and let p - a be a prime number. Since a is coprime to p,its residue modulo p is a unit in the finite field Fp = Z/pZ. The unit group F∗p is a cyclic groupof order p− 1. Naturally, the question rises, when the residue of a is a generator of F∗p. If it is,a is said to be a primitive root modulo p.

Example. It is a commonly known fact that a real number x ∈ R is a rational number if and onlyif its decimal expansion is ultimately periodic; given the decimal expansion x =

∑∞n=−d an ·10−n

of x, there exist integers k, n0 ∈ Z≥0 such that an = an+k for all n > n0. If p 6= 2, 5 is a primenumber, there exist minimal k,m ∈ Z>0 such that 1

p =∑∞i=1m · 10−ki. Here, k denotes the

period of the decimal expansion of 1p . This series is a geometric series, hence, 1p = m

10k−1 . It

follows that k is the minimal integer for which 10k = 1 mod p, which is precisely the order of10 in the unit group F∗p. Thus, 10 is a primitive root modulo p if and only if the period of the

decimal expansion of 1p has length p− 1.

In 1927, Emil Artin conjectured that the collection of prime numbers p for which a given integera is a primitive root modulo p, denoted by P (a), possesses a density in the collection of primenumbers. This natural density of P (a) is given by

δa = limn→∞

#p ∈ P (a) | p ≤ n#p prime | p ≤ n

.

Artin discovered that a being a primitive root modulo p, is determined by the splitting behaviour

of the prime ideal (p) ⊂ Z in the splitting field Fl = ΩXl−a

Q = Q(ζl, l√a), where l is a prime num-

ber dividing p− 1. He conjectured the following [7]:

Artin conjecture. Let a ∈ Z\0 be a non-zero integer, not equal to ±1, that is not a perfectpower. Then there are infinitely many prime numbers p for which a is primitive root modulop. Moreover, the collection of these primes P (a) posessess a density inside the collection of allprime numbers. This density is independent of a and is given by

δa =∏

l prime

(1− 1

l(l − 1)

)≈ 0.37396.

It is important to note that the number l(l− 1) is in fact the degree of the field extension Fl/Q.Moreover, if we replace the condition that a is an integer with the condition that a is a rationalnumber, we obtain the following generalization: Let a ∈ Q∗ be a rational number, not equal to±1, then P (a) possesses a natural density δa in the collection of prime numbers, given by:

δa =∏

l prime

(1− 1

[Fl : Q]

). (1)

In 1957, numerical calculations made by Derrick and Emma Lehmer gave unexpected results [7].They sent their results to Artin, who explained why the density δ5, should be approximately5% higher than the conjectural value of 0.37396. In 1967, Christopher Hooley proved a restatedtheorem, under assumption of the Generalized Riemann Hypothesis [2].

In 1977, Hendrik W. Lenstra proved several generalizations of the conjecture, also under as-sumption of the Generalized Riemann Hypothesis [3]. One of these generalizations is given by

2

replacing the field Q with an arbitrary number field K. This thesis is dedicated to provingthis generalization with minimal prior knowledge of algebraic number theory. Also an explicitformula will be given for the corresponding density δa, which relates this density to the densityof primes which split completely in a given finite field extension Fn/K, where Fn = K(ζn, n

√a).

3

1 The rational Artin conjecture

Before the Artin conjecture be formulated in arbitrary number fields, we will first consider thefield K = Q, and a non-zero rational number a ∈ Q∗. In this chapter, the concept of idealfactorisation and the splitting of prime ideals in number fields will be introduced. Moreover, anequivalence will be proved between the property of a being a primitive root modulo a prime pand the splitting behaviour of the ideal (p) in the fields Fl = Q(ζl, l

√a), where l | p− 1 is a prime

number. Finally, several results with respect to the density δa will be stated.

1.1 Conditions

Let a ∈ Q∗ be a non-zero rational number. The rational numbers allow unique prime factorisa-tion. Hence, a can be written uniquely as a = b

c , where b and c are two coprime integers, withc > 0. Let p be a prime number such that ordp(b) = ordp(c) = 0. The residues of b and c modulop are units in Fp, since p is coprime to b and c. The residue of a modulo p is therefore definedas c−1b and is a unit in Fp as well. The residue of a modulo p is denoted by a, and the order ofp in a is defined as ordp(a) = ordp(b) − ordp(c). We wish to classify the conditions for which ais a primitive root modulo p. The condition that a generates F∗p is equivalent to the conditionthat the group 〈a〉 has index 1 in F∗p:

〈a〉 = F∗p ⇔ [F∗p : 〈a〉] = 1

Hence, a is not a primitive root modulo p if and only if there exists a prime number l such thatl | [F∗p : 〈a〉]. The index of a subgroup of a finite group is always a divisor of the group order.For any odd prime number p, the group order of F∗p is divisible by 2. This observation allows usto impose several restrictions on a. This is illustrated in the following example.

Example 1.1. Let a ∈ Q∗ be a perfect square, and let p be a prime number with ordp(a) = 0.Then a is a primitive root modulo p, if and only if p = 2. Let p 6= 2 be an odd prime number.Since F∗p is a cyclic group of finite order p − 1, the collection of squares, F∗2p , is a subgroup ofindex 2. It follows from the multiplicativity of the index that 2 | [F∗p : 〈a〉], and hence a is not aprimitive root modulo p. The other implication is trivial.Furthermore, the integer −1 is an element of order 2 in Q∗. Hence, if p 6= 2 is an odd primenumber, then the equality 〈−1〉 = 1,−1 holds. It follows that −1 is a primitive root modulop if and only if p ∈ 2, 3.

Example 1.1 is in fact a direct result from the following lemma, which relates the divisibility ofthe index [F∗p : 〈a〉] to the splitting behaviour of certain polynomials in Fp[X]:

Lemma 1.2. Let a ∈ Q∗ be a rational number and p, l ∈ Z be two distinct prime numbers, suchthat ordp(a) = 0. Then the following are equivalent:

(1) l | [F∗p : 〈a〉].

(2) l | #F∗p and a ≡ yl mod p for certain y ∈ F∗p.

(3) F∗p contains an element of order l and a ≡ yl mod p for certain y ∈ F∗p.

(4) The polynomial X l − a splits in l distinct factors in Fp[X].

4

Proof. The proof of the equivalence of (1) and (2) relies on the fact that the unit group F∗p is acyclic group of finite order. Since it is, all subgroups of F∗p are cyclic, and for every d | #F∗p, there

exists a unique subgroup Hd ⊂ F∗p of index d, given by Hd = F∗dp = xd | x ∈ F∗p. Moreover,for every m | #F∗p and d | m, there is a natural inclusion Hm ⊂ Hd ⊂ F∗p. Hence if l | [F∗p : 〈a〉]we naturally have that l | F∗p, and by the above we have that the group 〈a〉 is contained in the

subgroup Hl of index l. Thus there exists an y ∈ F∗p such that a ≡ yl mod p. To proof the

converse we note that since l | F ∗p , and a ≡ yl mod p for certain y ∈ F∗p, the inclusion 〈a〉 ⊂ F∗lpholds. This inclusion now gives us that l | [F∗p : 〈a〉]. This proves the equivalence of (1) and (2).

The theorem of Cauchy states that for every prime number l | #F∗p there exists an element ζ ∈ F∗pof order l. Moreover, the order of every element in a finite group is a divisor of the group order.This proves the equivalence of (2) and (3).

(3) ⇒ (4): Since Fp is a field, the polynomial X l − a has at most l distinct roots in Fp. SinceF∗p contains an element ζ of order l, the polynomial X l − a splits into l distinct linear factors as

X l − a =∏lk=1(X − ζky) in Fp[X]. To show that these factors are distinct, it suffices to show

that X l − a has no common roots with its derivative. Since ddX (X l − a) = lX l−1 has 0 as only

root, we can conclude that all the roots of X l − a are distinct.

(4)⇒ (3): Clearly, there exists an y ∈ F∗p such that a ≡ yl mod p. The map ϕ : F∗p → F∗p, x 7→ xl

is a homomorphism of groups. From the first isomorphism theorem, it follows that all cosets ofker(ϕ) have the same cardinality. The polynomial X l − a splits in l distinct factors in Fp[X].Hence, all cosets have size l. Therefore, the kernel of ϕ is a subgroup of F∗p of order l. Thisproves the final implication.

It follows directly from Lemma 1.2 that a is a primitive root modulo p, if and only if there existsno prime number l | F∗p, for which the polynomial fl = X l−a ∈ Fp[X] splits completely in Fp[X].

Condition (4) is equivalent to the condition that the splitting field ΩXl−a

Fpof fl has degree 1 over

Fp. The splitting field of fl over Fp, is strongly related to the splitting field of fl over Q. Inorder to fully comprehend this relation, several objects have to be introduced.

1.2 The splitting of prime ideals

In order to examine the splitting behaviour of X l− a in Fp[X] more closely, several objects haveto be introduced.

Definition 1.3. Let K be a number field. The integral closure of Z in K is called the ring ofintegers of K. It is denoted by OK , and is given by

OK = α ∈ K | there exists a monic polynomial f ∈ Z[X] such that f(α) = 0.

The ring of integers OK of a number field K satisfies several useful properties. These propertiesare listed in the following proposition.

Proposition 1.4. Let K be a number field of degree n over Q. The ring of integers OK of Ksatisfies the following properties.

5

1. The ring of integers is a Dedekind domain: Every non-zero ideal I ⊂ OK can be decomposeduniquely as

I =∏I⊂p

pep ,

where p denote distinct prime ideals and ep ≥ 1 is an integer.

2. K has an integral basis: There exist ω1, ..., ωn ∈ OK such that

OK =

n⊕i=1

Z · ωi, K = Frac(OK) =

n⊕i=1

Q · ωi.

Every element x ∈ K can be written as x = αβ , with α, β ∈ OK , and β 6= 0.

These properties are certainly not trivial. The proof of the proposition will not be given in thisarticle however. For this we refer to [6]. The collection of non-zero prime ideals of OK is denotedby PK . Throughout this article, a prime p of K refers to an element of PK . A prime p is saidto divide an ideal I, if the inclusion I ⊂ p holds.

Every non-zero rational number q ∈ Q∗, permits a unique prime decomposition q =∏p∈PQ

pnp ,with np ∈ Z. Proposition 1.4 allows a similar decomposition in general number fields. It fallsbeyond the scope of this thesis to formally define this. However, for x ∈ K∗, we can definethe collection of prime ideals of OK that divide x. By Proposition 1.4, there exist algebraicintegers α, β ∈ OK\0, such that x = α

β . As the ideals generated by α and β allow a unique

factorisation, the divisor set DK(x) of x in K is defined as:

DK(x) = p ∈ PK : ordp(α)− ordp(β) 6= 0.

The divisor set DK(x) of an algebraic number in a number field K is always a finite set.

Let p be a prime number. Since OK is a Dedekind domain, the ideal pOK can be decomposeduniquely as

pOK =∏

p|pOK

pe(p/p). (2)

The prime ideals p of OK in this decomposition are called the primes extending p or the exten-sions of p in K. The index e(p/p) is called the ramification index of p in K. Note that p ∩ Zis a prime ideal of Z containing (p). Hence, p ∩ Z = (p) and therefore this index is dependsonly on the prime p. If there exists a prime p ∈ PK with e(p/p) > 1, then p is ramified in K.The prime numbers that ramify in K are precisely those prime numbers that divide the absolutediscriminant of K. This statement is given without proof.

Lemma 1.5. Let K/Q be a field extension of degree n. Let G = σ1, ..., σn denote the collectionof embeddings of K into C. Let ω1, ..., ωn be any integral basis of K. A prime number p isramified in K if and only if p divides the absolute discriminant ∆K of K, given by ∆K =(det(σi(ωj))i,j)

2. Moreover, if f ∈ Z[X] is an irreducible polynomial, and K = ΩfQ is its splittingfield over Q, then DQ(∆K) ⊂ DQ(∆(f)).

Since ∆K ∈ Z\0, there are at most finitely many prime numbers l that ramify in K. Leta ∈ Q∗ be a rational number, let l be a prime number, and let p ∈ PQ be a prime ideal of Q. Itfollows that if p ramifies in Fl, then p ∈ DQ(l) ∪DQ(a).

6

Let K be a number field of degree n over Q, and let p be a prime number. From the Chineseremainder theorem, it follows that

OK/pOK ∼=

∏p|pOK

OKpe(p/p). (3)

Since the ideals p are prime ideals in a number ring, the rings kp = OKp are finite fieldsof characteristic p. The field kp is called the residue class field of p. The degree of the fieldextension kp/Fp is called the residue class degree and is denoted by f(p/p). From Proposition1.4, it follows that

OK+pOK+=

n⊕i=1

Z · ωipZ · ωi∼= (Z/pZ)n (4)

From equation (3) we obtaint the equality |OK/pOK | =

∏p|p p

e(p/p)f(p/p). On the other hand,

we have by equality (4) that |OK/pOK | = pn, hence we conclude that

pn =

k∏p|p

pe(p/p)f(p/p) = p∑

p|p e(p/p)f(p/p).

And therefore,

[K : Q] =∑p|p

e(p/p)f(p/p). (5)

This gives rise to the definition of the splitting of prime ideals in number fields; let K/Q be afield extension of degree n with corresponding ring of integers OK . A prime ideal (p) is said tosplit completely in K if (p) has n distinct extensions in OK . Naturally, if p splits completelyin OK , then f(p/p) = 1 for all primes p of K extending p. Hence, kp = Fp for all primes ofK extending p. The prime p is said to be totally ramified in K, if there exists a prime p of Kextending p, with e(p/p) = n.

Example 1.6. Let l be a prime number. We will show that l is totally ramified in the extensionQ(ζl)/Q. It is a commonly known fact that is the ring of integers of Q(ζl) is given by Z[ζl]. Letτ : Z[ζl] → Fl be a homomorphism. τ is determined uniquely by its action on ζl, and since τis a homomorphism, τ(ζl) must be a root of the l-th cyclic polynomial Φl(X) modulo l. Thispolynomial splits as Φl(X) = (X − 1)l−1 over Fl, hence τ(ζl) = 1. Consequently, we have thatIm(τ) = Fl, and by the first isomorphism theorem p = ker τ is a prime ideal of Q(ζl), extendingl. In particular, this is the only prime of Q(ζl) extending l. It follows from equation (5) thate(p/l) = l− 1, and hence l is totally ramified in Q(ζl). Finally, since ∆(Φl(X)) = ±ll−2, we notethat l is the only prime that ramifies in Q(ζl).

1.3 The problem reformulated

Lemma 1.2 shows that the property of being a primitive root modulo p is closely related to thesplitting behaviour of fl in Fp[X]. The splitting behaviour of the polynomial fl is closely relatedto the splitting behaviour of the ideal (p) ⊂ OFl

. This relation is formalized in the followingtheorem.

7

Theorem 1.7. Let a ∈ Q∗ be a rational number, let l be a prime number and let Fl denote thesplitting field of fl = X l − a ∈ Q[X]. Let (p) ∈ PQ\(DQ(l) ∪DQ(a)) be a prime of Q such thatordp(a) = 0, and ordp(l) = 0. The following are equivalent.

(1) The polynomial fl splits completely in Fp[X].

(2) The ideal (p) splits completely in OFl.

Proof. In order to show that (p) splits completely in OFl, we need to show that (p) has [Fl : Q]

extensions in Fl. The polynomial fl is irreducible in Q[X] if and only if it has no rational roots[5]. Hence, we assume without loss of generality, that [Fl : Q] = l(l − 1). The ring OFl

containsthe subring R = Z[ζl, l

√a]. Consider the collection C of surjective homomorphisms τ : R → Fp.

It follows from the first isomorphism theorem that ker(τ) must be a prime ideal of R, for allτ ∈ C. Moreover, since Fp is a field, a homomorphism τ : R → Fp is surjective, if and only ifτ is non-trivial. Hence, if C is non-empty, then there are exactly l(l − 1) homomorphisms in C,since there are l − 1 distinct images of ζl, and exactly l distinct images of l

√a. Every τ ∈ C

corresponds to a unique prime ideal pτ extending p. Hence, there are l(l − 1) prime ideals p ofR that extend (p). None of these prime ideals are ramified in OFl

, and since there are at most[Fl : Q] = l(l − 1) primes of OFl

extending p, each prime p of R can be extended uniquely to aprime q of OFl

. Hence, (p) splits completely in Fl if C is non-empty, and C is non-empty if andonly if the polynomial fl splits completely in Fp[X], by Lemma 1.2.

On the other hand, if the prime (p) splits completely in OFl, it follows from equality (5) that

OFl/pOFl

∼= (Fp)l(l−1). Therefore, there exists a non-trivial homomorphism τ : OFl→ Fp, and

since τ |R is a non-trivial homomorphism from R to Fp, it follows that C 6= ∅.

Theorem 1.7 gives the first concrete result with respect to the Artin conjecture:

Corollary 1.7.1. Let a ∈ Q∗ be a non-zero rational number, and let p /∈ DQ(a) be a primenumber. Then a is a primitive root modulo p if and only if for all l | p− 1, the ideal p does notsplit completely in Fl.

1.4 The density of primitive roots in Q

Corollary 1.7.1 offers an alternative proof for Example 1.1, using field extensions. Let a ∈ Q∗ bea perfect square, and let p be an odd prime number coprime to a. Since a is a perfect square inQ∗, the extension F2/Q is trivial. And since p has exactly 1 extension in Q, the ideal (p) splitscompletely in F2. We conclude that a is not a primitive root modulo p, unless p = 2.

In fact, this alternative proof of Example 1.1, gives rise to a general statement with respect tothe density δa of the collection P (a) of primes p for which a is a primitive root modulo p, in thecollection of all rational primes:

Corollary 1.7.2. Let a ∈ Q∗ be a non-zero rational number. If there exists a prime number l,for which the field extension Fl/Q is trivial, then δa = 0.

This corollary is in fact equivalent with Example 1.1, since ζ1 = 1, and ζ2 = −1 are the onlyroots of unity contained in Z. In arbitrary number fields, this is less trivial however. We concludethis chapter by stating a correction of the Artin conjecture, under assumption of the GeneralizedRiemann Hypothesis [2].

8

Theorem 1.8 (C. Hooley, 1967). Let a ∈ Z\±1 be an integer that is not a perfect square. Leth denote the largest integer for which a is a perfect h-th power. Let a1 denote the square-freepart of a. Let C(h) be given by:

C(h) =∏

l prime,l|h

(1− 1

l − 1

) ∏l prime,l-h

(1− 1

l(l − 1)

).

Assume that the Generalized Riemann Hypothesis holds. If a1 6≡ 1 mod 4, then δa = C(h),while if a1 ≡ 1 mod 4, we have:

δa = C(h)

1− µ(|a1|) ·∏l|h,l|a1

1

l − 2

∏l-h,l|a1

1

l2 − l − 1

,

where µ denotes the Mobius counting function.

A simple calculation shows that for a = 5, we have:

δ5 = C(1) ·(

1− 1 · 1

52 − 5− 1

)= C(1) ·

(1− 1

19

)=

20

19

∏l prime

(1− 1

l(l − 1)

).

Hence for a = 5, the density δ5 is approximately 5% greater than the Artin constant, which isthe numerical deviation that Derrick and Emma Lehmer estimated in 1957 [7]. Not only doesHooley’s theorem give a method to explicitly calculate the density δa of a given integer a, it alsoprecisely states the conditions under which the density δa vanishes:

Corollary 1.8.1. Assume the Generalized Riemann Hypothesis holds. Let a ∈ Q∗\±1 be arational unit. The density δa vanishes if and only if a is a perfect square in Q∗.

Proof. We note that the proof of Hooley’s theorem does not require a to be an integer, it stillholds if a is replaced by a rational number that is not a root of unity. Assume that a is not aperfect square. Since 2 - h, it follows that C(h) > 0. Hence if a 6= 1 mod 4, then δa > 0. Notethat the correction factor when a ≡ 1 mod 4 as described in Hooley’s theorem never vanishes.It follows that δa > 0. The other implication is given by Corollary 1.7.2.

9

2 The Artin conjecture for number fields

In this chapter the Artin conjecture will be formulated for arbitrary number fields K, and anon-trivial example of a number field K and algebraic number a ∈ K∗, for which the density δavanishes, will be examined. Further, the Frobenius element and Frobenius symbol of unramifiedprimes in Galois extensions will be defined. We will classify primes of K, by their respectiveFrobenius symbol in the fields Fn = K(ζn, n

√a). We will show that the primes p of K, not

contained in the divisor set of a or 2∆K , for which a is not a primitive root modulo p, are thoseprimes for which there exists a prime number l | #k∗p, such that the Frobenius symbol in Fl istrivial. This chapter is concluded by proving a lemma regarding the vanishing of the density δa.

2.1 Formulating the conjecture for number fields

All proofs in chapter 1 can be generalized to arbitrary number field extensions L/K, by replacingZ with the ring of integers OK and the prime number p with general prime ideals p ∈ PK . Sincethe residue class fields kp are finite fields, the collection PK can be provided with the naturalordering induced by the cardinality of the residue class fields; let A ⊂ PK be a subset of PK , thenatural density of A in PK is defined as:

δ = limn→∞

p ∈ A | #kp ≤ np ∈ PK | #kp ≤ n

,

whenever this limit exists. This allows us to formalize the Artin conjecture over general numberfields K.

Generalized Artin conjecture. Let K be a number field with ring of integers OK . Let a ∈K∗\µK be a unit of K. Let P (a) ⊂ PK\DK(a) denote the set of all prime ideals p ⊂ OK , forwhich a is a primitive root modulo p. Then P (a) possesses a natural density δa in PK .

Under assumption of the Generalized Riemann Hypothesis, it was shown in section 1.4 that whenK = Q, δa = 0 if and only if a ∈ Q∗2 ∪±1. In general number fields however, this need not bethe case. This is illustrated in the following examples.

Example 2.1. The results from Example 1.1 can be generalized to arbitrary number fields K.Let K be a number field with ring of integers OK and let ζ ∈ µK be a primitive n-th root ofunity. Let p ∈ PK\DK(n) be a prime of K that is not contained in the divisor set of n. Notethat char(kp) - n, and since the polynomial Xn−1 splits completely in K[X], it splits completelyin kp as well. We therefore conclude that n | k∗p, and since ζ has order n in k∗p, it will only bea primitive root modulo p if #k∗p = n. This holds for only finitely many primes p. In general,if K contains a primitive n-th root of unity and a ∈ K∗n, then the polynomial Xn − a splitscompletely in K[X], and therefore splits completely over all but finitely many residue class fieldskp.

From Example 2.1, we conclude that if ζn ∈ K, and a ∈ K∗n, then δa = 0. In general the densityδa vanishes if there exists a square-free integer n, such that the polynomial Xn − a is reduciblein K[X]. A less trivial example is given when considering the field Q(

√5):

Example 2.2. Consider the number field K = Q(√

5), with corresponding ring of integers

OK = Z[ 1+√5

2 ]. Let x ∈ K∗\µK , be a fixed algebraic number to be determined later, and let

10

a = x15. Let p ∈ PK\DK(a) be a prime of K that is not contained in the divisor set of a.Analogous to Lemma 1.2, it can be concluded that 3 | [k∗p : 〈a〉] ⇔ X3 − 1 ∈ kp[X] splits indistinct factors, and there exists a y ∈ K such that a ≡ y3 mod p. The second statement holdstrivially since a = x15. Hence, it suffices to show that the polynomial X3−1 splits completely indistinct factors in kp[X]. If char(kp) 6= 3, this holds if and only if X2 +X + 1 splits completelyin kp[X] and since there are only finitely many primes extending 3, there are only finitely manyprimes p for which char(kp) = 3. Note that by the quadratic formula, the roots of this polynomial

are given by y = −1±√−3

2 .Hence, if char(kp) 6= 2, 3, the following can be concluded:

3 | [k∗p : 〈a〉] if and only if− 3 is a square in kp.

The same deductive strategy can be used to determine the conditions for which 5 | [k∗p : 〈a〉].Assume that char(kp) 6= 5. Since a = (x3)5 mod p, it can be concluded that 5 | [k∗p : 〈a〉] ifand only if X5 − 1 splits completely in kp[X]. This is equivalent with the statement that the

splitting field ΩX5−1

kphas degree one over kp. In order to examine this condition more closely,

we note that the extension ΩX5−1

Q /Q is cyclic of degree 4. Therefore, ΩX5−1

Q has a subfield L of

degree 2 over Q, given by L = Q(√

5) = K. Hence, the extension ΩX5−1

K /K has degree 2 andis given by a polynomial f = X2 − τX + c ∈ K[X]. Since K is a real number field, we note

that Gal(Q(ζ5)/K) = id, σ, where σ is given by complex conjugation. It follows that fζ5K isgiven by f = (X − ζ5)(X − σ(ζ5)) = X2 − (ζ5 + ζ5)X + ζ5ζ5. We note that ζ5 = ζ−15 and thus

fζ5K = X2 − τX + 1, where τ = ζ5 + ζ−15 = −1+√5

2 .Assume that ordp(τ) = 0. From the above it can be concluded that X5 − 1 splits completelyin kp[X] if and only if X2 − τX + 1 splits completely in kp[X]. By the quadratic formula, the

roots of X2 − τX + 1 are given by y = τ±√τ2−42 . Thus, if char(kp) 6= 2, 5 and ordp(τ) = 0, the

following can be concluded:

5 | [k∗p : 〈a〉] if and only if τ2 − 4 is a square in kp.

Now let x = −3(τ2 − 4). Assume that char(kp) 6= 2. Since the polynomial X2 − 1 splits indistinct factors in kp[X], it follows that 2 | [k∗p : 〈a〉] if and only if there exists a y ∈ K such thata ≡ y2 mod p.

Assume that char(kp) 6= 2, 3, 5 and assume ordp(x) = ordp(τ) = 0. Assume that [k∗p : 〈a〉] = 1.Since 3, 5 - [k∗p : 〈a〉] it follows that −3 /∈ k∗2p and τ2− 4 /∈ k∗2p . We note that k∗p is a cyclic groupand therefore, the product of two non-squares is a square. It follows that −3(τ2 − 4) = x ∈ k∗2p .We conclude that there exists a y ∈ K such that a ≡ y2 mod p. And thus, 2 | [k∗p : 〈a〉]. Thisgives a contradiction. Hence, a is not a primitive root modulo p for all but finitely many primesq.

In both example 1.1, and 2.1, there exists a prime number l, such that for all but finitely manyprimes p, the index [k∗p : 〈a〉] is divisible by l. In Example 2.2 this is not the case. For everyprime p of K, the index [k∗p : 〈a〉] is divisible by either 2, 3 or 5, however for each of theseindividual prime numbers there are infinitely many primes of K for which the index [k∗p : 〈a〉]is not divisible by K. In Example 2.2, there exists no prime number l, such that the extensionFl/K is trivial, however the density δa still vanishes. This example shows that Corollary 1.7.2does not generalize directly to arbitrary number fields K.

The result of Example 2.2 is not just a coincidence. In fact, the field K and the element xwere chosen in such a way that the fields F2, F3 and F5 all have degree 2 over K, whereas the

11

compositum F30 = F2 · F3 · F5 is of degree 4 over K instead of 8. Due to this dependence offields, every prime p of K splits completely in either F2, F3 or F5. In order to fully comprehendthis, several concepts have to be introduced.

2.2 The Frobenius element

There appear to be no restrictions, other than the restriction imposed by equality (5), on thesplitting behaviour of a prime p ∈ PK in OL. However, if L/K is a Galois extension, there ex-ists a strong relation between the primes of L extending p. This is shown in the following theorem.

Theorem 2.3. Let L/K be a Galois extension of number fields with Galois group G, and letp ∈ PK be a prime of K. Then the following statements hold:

1. The group G acts transitively on the collection DL(p) of primes in L extending p.

2. All residue class fields of the extensions q of p in L are isomorphic.

3. The residue class degree fp = f(q/p) and the ramification index e(q/p) depend only on p.

4. If gp denotes the number of extensions of p in L, the following equality holds:

epfpgp = [L : K].

Proof. Let σ ∈ G be an automorphism of L that acts trivially on K. Let x ∈ OL be an algebraicinteger with minimum polynomial fxOK

. Since σ is an automorphism that acts trivially on OK ,we have that σ(x) is a root of fxOK

and is therefore contained in OL. Let q be a prime of Lextending p. Let z = xy ∈ σq be given. Naturally, we have that σ−1(xy) = σ−1(x)σ−1(y) ∈ q.Since q is a prime ideal, it follows that either σ−1(x) ∈ q or σ−1(y) ∈ q holds, and hence eitherx ∈ σq or y ∈ σq holds. We conclude that σq is a prime of L extending p.

Assume that q and q′ are two primes of L extending q that are in different G-orbits. By applyingthe Chinese remainder theorem on OL/pOL, we find that there exists an x ∈ q that is notcontained in σq′, for all σ ∈ G. The minimum polynomial of x over OK divides the polynomialf =

∏σ∈G(X − σ(x)) ∈ OK [X]. It follows that a0 =

∏σ∈G σ(x) ∈ OK . Moreover, we have that

a0 /∈ q′. It follows that a0 ∈ q ∩ OK = p = q′ ∩ OK . This gives a contradiction. Hence, G actstransitively on DL(p).

Let q and q′ be two distinct primes of L extending p. By Theorem 2.3.1, there exists an automor-phism σ ∈ G such that σq = q′. The map σ : kq → kq′ , x mod q 7→ σ(x) mod q′, is thereforea well-defined homomorphism. Its inverse is given by σ−1 : kq′ → kq, x mod q′ 7→ σ−1(x)mod q. Therefore, σ is an isomorphism. This proves the second statement. The third and fourthstatement are a direct result from the second statement and equation (5).

Let L/K be a Galois extension of number fields with Galois group G, and let p ∈ PK be a primeof K that is unramified in L. Theorem 2.3 shows that G acts transitively on the collection ofprimes extending p. Hence, let q ∈ DL(p) be a prime of L extending p. Then q has a stabilizergroup Gq/p in G. This group is called the decomposition group of q over p. Since every automor-phism σ ∈ Gq/p acts trivially on q, the reduction map r : Gq/p → Gal(kq/kp) is well-defined. Infact, there is a strong relation between the decomposition group Gq/p and the Galois group of

12

kq over kp:

Lemma 2.4. Let L/K be a Galois extension of number fields with Galois group G. Let p be aprime of K that is unramified in L, and let q be a prime of L extending p. Let Gq/p denote thedecomposition group of q in G. Then the map

rq : Gq/p → Gal(kq/kp),

(σ : x 7→ σ(x)) 7→ (σ : x mod q 7→ σ(x) mod q),

is a well-defined isomorphism of groups.

Proof. Let σ, τ ∈ Gq/p, and x ∈ q be given. Since σ(x) ∈ q, the automorphism rq(σ) is well-defined. And since rq(στ) = rq(σ)rq(τ), the map rq is a well-defined homomorphism. SinceG acts transitively on the set DL(p) of primes extending p, there is a bijective correspondencebetween set G/Gq/p and DL(p). It therefore follows that #Gq/p = fp = #Gal(kq/kp). Ittherefore suffices to show that rq is surjective, in order to complete the proof.

There exists an α ∈ k∗q, such that kq = kp(α). Since every σ ∈ Im(rq) is determined by its action

on α, it suffices that Im(rq) acts transitively on the conjugates β of α in kq. By the Chineseremainder theorem, there exists an α ∈ OL, such that α ≡ α mod q, and α ∈ q′ for all primesq′ 6= q of L extending p. Let σ ∈ G\Gq/p be an automorphism of L. Since G acts transitively onthe collection of primes extending p there exists a prime q′ extending p, such that σq′ = q. Sinceα ∈ q′, it follows that σ(α) ∈ q for all σ ∈ G\Gq/p. The characteristic polynomial of α in K[X]is given by f =

∏σ∈G(X − σ(α)) =

∏σ∈Gq/p

(X − σ(α))∏σ∈G\Gq/p

(X − σ(α)). Its reduction

modulo q is therefore given by f = X(#G−fp)∏σ∈Gq/p

(X − σ(α)). The minimum polynomial

of α is a divisor of its characteristic polynomial. Hence, the minimum polynomial of α in kp[X]is an irreducible polynomial of degree fp that divides

∏σ∈Gq/p

(X − σ(α)). It follows that this

polynomial is the minimum polynomial of α over kp, and that Im(rq) acts transitively on thecollection of conjugates of α.

Since Gal(kq/kp) is a cyclic group generated by Frobq : x 7→ x#kp , it follows from Lemma 2.4that Gq/p is a cyclic group as well. Therefore, there exists an automorphism σ ∈ Gq/p such thatrq(σ) = Frobq. This automorphism is called the Frobenius element of q over p and is denotedby Frq. The Frobenius element has various useful properties:

Proposition 2.5. Let K ⊂ L ⊂ M be a tower of number field extensions such that L/K, andM/K are Galois. Let p be a prime of K that is unramified in M . Let q be a prime of Mextending p. Then q′ = q ∩ OL is a prime of L extending p, and Frq|L = Frq′ .

Proof. It is a trivial exercise to show that q′ is a prime of L, extending p. We note that kp ⊂ kq′ ⊂kq is a tower of finite Galois extensions. Since the Frobenius element Frq acts on kq by x 7→ x#kp ,and kq′ ⊂ kq, it has the same action on kq′ . Hence, the Frobenius automorphism of kq′ over kpis given by Frobq′(x) = Frq(x) mod q′. On the other hand, it is given by Frobq′(x) = Frq′(x)mod q′. Since M/L is Galois, we have that Gal(L/K) = Gal(M/K)|L. In particular we havethat Gq′/p = Gq/p|L, and hence Frq|L ∈ Gq′/p. It follows from the uniqueness of the Frobeniuselement that Frq|L = Frq′ .

13

2.3 Applications of the Frobenius element

Let L/K be a Galois extension of number fields, and let p ∈ PK be a prime of K that is unram-ified in L. If q and q′ are two primes of L extending p, then there exists a σ ∈ Gal(L/K) suchthat q′ = σq. Consequently, we have that Gq′/p = σ−1Gq/pσ. Since these primes are isomorphic,we conclude that the Frobenius element Frq′ of q′ over p is given by Frq′ = σFrqσ

−1. This givesrise to the definition of the Frobenius symbol :

Definition 2.6. Let L/K be a Galois extension of number fields, and let p ∈ PK be a prime ofK that is unramified in L. The Frobenius symbol (p, L/K) of p in L, is the union of all Frobeniuselements lying above p in L:

(p, L/K) =⋃q|p

Frq.

In particular, the Frobenius symbol is a conjugacy class.

Theorem 2.3 and Lemma 2.4, allow us to identify the splitting behaviour of a prime in a numberfield, with the order of the elements in its corresponding Frobenius symbol:

Lemma 2.7. Let L/K be a Galois extension of number fields, and let p be a prime of K that isunramified in L. Then, ord(σ) = fp, for all σ ∈ (p, L/K).

Proof. Since the Frobenius symbol is a conjugacy class all automorphisms in the Frobeniussymbol have the same order. Let σ ∈ (p, L/K) be given. There exists a prime q ∈ PL extendingp with σ = Frq. By Lemma 2.4 the decomposition group Gq/p is cyclic of order fp. Hence,ord(σ) = fp.

Lemma 2.7 has a direct result in the case that p splits completely:

Corollary 2.7.1. Let L/K be a Galois extension of number fields, and let p be a prime of Kthat is unramified in L. The following are equivalent:

(1) The prime p splits completely in L.

(2) The Frobenius symbol (p, L/K) of p in L, is the collection of the trivial element, (p, L/K) =idL.

(3) There exists a prime q in L extending p, for which the corresponding Frobenius elementFrq is the trivial automorphism.

Proof. A prime p splits completely in L, if and only if fp = 1, and the only element of order 1 isthe trivial element.

We wish to relate the property of a being a primitive root modulo a prime p, to the Frobeniussymbol of p in the splitting fields Fl. However, there exists primes of K for which a is a primitiveroot, but still have a trivial Frobenius symbol in certain fields Fl.

Example 2.8. Consider the field K = Q(ζ3), and let a = 10. Let p = (ζ3 − 1) be the primeextending 3. Since 3 is totally ramified in K, it follows that kp = F3. Since 10 ≡ 1 mod 9, itcan be shown that the prime p splits completely in F3 = 3

√10. However, 3 can never divide the

indexe [k∗p : 〈10〉], since #k∗p = 2. This gives a contradiction.

14

Fortunately, there are at most finitely many primes for which this may occur, hence this willhave no effect on the density δa. The following lemma restates the conditions.

Lemma 2.9. Let K be a number field and let p be a prime of K not contained in the divisor setof a or 2∆K , and let l be a prime number. Then,

l | [k∗p : 〈a〉] if and only if p splits completely in Fl.

Proof. The proof of Lemma 2.9 relies on the same argument as Theorem 1.7.Let p /∈ DK(a) ∪ DK(2∆K) be a prime of K and let l be a prime number. Consider the ringR = OK [ζl, l

√a], and the collection C of homomorphisms τ : R → kp, with the condition that

x 7→ x mod p for x ∈ OK . Analogous to Theorem 1.7, every map τ ∈ C corresponds to a uniqueprime q ∈ DFl

(p) extending p. Hence, p splits completely in Fl if, and only if #C = [Fl : K].

Assume that l | [k∗p : 〈a〉]. Then k∗p contains an element ζ of order l, and a is contained in the

unique subgroup of l-th powers k∗lp . Since k∗p contains l−1 elements of order l, there are precisely

[K(ζl) : K] embeddings of ζl in kp. And since X l − a has one root in k∗p, all of its roots arecontained in k∗p, and hence there are [Fl : K(ζl)] distinct embeddings of l

√a in kp. It follows that

#C = [Fl : K].

For the converse, we note that since K(ζl)/K is a totally ramified extension of degree l − 1 > 1for l - 2∆K , a prime q that splits completely in Fl/K can never divide l. Therefore, p doesnot divide l, and hence the polynomial X l − 1 is seperable in kp[X]. And since there exists ahomomorphism τ : R → kp, the group k∗p contains an element of order l. Since p /∈ DK(a), wehave that a ∈ k∗p, and hence τ( l

√a) ∈ k∗p. It follows that l | [k∗p : 〈a〉].

Lemma 2.9 shows that a is not a primitive root modulo a prime p /∈ DK(a) ∪DK(2∆K), if andonly if there exists a prime number l, such that p splits completely in the field Fl = K(ζl, l

√a).

By Corollary 2.7.1, this occurs if and only if the Frobenius symbol (p, L/K) is the collection ofthe identity automorphism. Combining these observations, it can be proven δa vanishes, if itsatisfies a certain condition.

Lemma 2.10. Let a ∈ K∗\µK be a non-zero algebraic number. If there exists a square-freeinteger n ∈ Z>0, with the property that:

Gal(Fn/K) =⋃l|n

Gal(Fn/Fl),

then there are at most finitely many primes p of K for which a is a primitive root modulo p.

Proof. Let n ∈ Z>0 be a square-free integer and assume Gal(Fn/K) =⋃l|n Gal(Fn/Fl). Let

p ∈ PK\(DK(a)∪DK(2∆K) be a prime of K, that is not contained in the divisor set of a or 2∆K .From Lemma 2.9 and Corollary 2.7.1, it follows that a is not a primitive root modulo p if and onlyif there exists a prime number l such that (p, Fl/K) = idFl

. Let σ ∈ (p, Fn/K) ⊂ Gal(Fn/K),per assumption there exists a prime number l | n, such that σ|Fl

= idFl. From Proposition 2.5 it

follows that (p, Fl/K) = (p, Fn/K)|Fl. Since idFl

∈ (p, Fl/K), it follows that (p, Fl/K) = idFl.

Hence, a is not a primitive root modulo p.

15

Given a square-free integer n, one might wonder whether there exists a field K and an algebraicnumber a ∈ K∗ such that Gal(Fn/K) =

⋃l|n Gal(Fn/Fl). In fact, if we exclude the pairs (K, a)

for which there exists a prime number l such that Fl = K, we can show that the smallest square-free integer n, for which there exists a pair (K, a) satisfying Lemma 2.10, is the integer n = 30.This is illustrated in the following examples.

Example 2.11. Let n be a square-free integer and assume that the equality Gal(Fn/K) =⋃l|n Gal(Fn/Fl) holds.

If n is the product of one prime number, we have that n = l. Hence Gal(Fn/K) = Gal(Fn/Fl) =idFl

. It follows that Fl = K, hence this is the trivial example where a ∈ K∗l and ζl ∈ K.

If n is the product of two prime numbers, then Gal(Fn/K) is the union of two subgroups. Sinceno group is the union of two proper subgroups, it follows that there exists a prime number l | nsuch that Gal(Fn/Fl) = Gal(Fn/K). Hence we arrive at the trivial example Fl = K.

Hence, the first non-trivial example is given when n is the product of three prime numbers.

Example 2.12. Let n = l1l2l3 be the product of three prime numbers. We wish to construct afield K and an algebraic number a ∈ K∗, such that the equality Gal(Fn/K) = ∪l|nGal(Fn/Fl)holds. From the above, we conclude that there exists an automorphism σ ∈ Gal(Fl1l2/K),with the property that σ /∈ Gal(Fl1l2/Fl1) ∪ Gal(Fl1l2/Fl2). If Fl3 is not contained in Fl1l2 ,then this automorphism can be extended to an automorphism σ ∈ Gal(Fn/K) such that σ /∈Gal(Fn/Fl3). Since its not contained in the Galois groups of Fn/Fl1 or Fn/Fl2 either, we arriveat a contradiction. We therefore must have that Fl3 ⊂ Fl1l2 . We are looking for prime numbersl1, l2 and l3, such that if a prime does not p split completely in Fl1 and Fl2 , then it must splitcompletely in Fl3 . We have the following diagram.

Fl1l2

Fl1 Fl2 Fl3

K

This diagram has great similarities with the diagram of the Klein four-group. In fact, it followsdirectly from Lemma 2.10, that δa vanishes if Gal(Fn/K) is isomorphic to the Klein four-group.Hence, the question rises if this diagram can be realized as a V4-diagram.

To this end, we let n = l1l2l3 be the product of three distinct odd prime numbers. All extensionsFl/K must be of degree 2 over K. If a /∈ K∗l, then K( l

√a)/K is an extension of degree l > 2.

Hence we choose ∈ K∗l1l2l3 . It follows that Fn = K(ζl1l2l3 , and the Galois group Gal(Fn/K)must be a subgroup of (Z/l1Z)∗× (Z/l2Z)∗× (Z/l3Z)∗, which is the Galois group of Q(ζl1l2l3)/Q.Since Fli/K must have degree 2 for all i, we conclude that Gal(Fl1l2l3/K) must be containedin the subgroup C2 × C2 × C2 of (Z/l1Z)∗ × (Z/l2Z)∗ × (Z/l3Z)∗, which exists since the primenumbers are odd. In particular, we are looking for the subgroup H, where each σ ∈ H, has theproperty that if σ|Fl1

= −1, and σ|Fl2= −1, then we must have that σ|Fl3

= 1. We note thatthis group H is precisely the group given by

H = (g1, g2, g3) ∈ (Z/l1Z)∗ × (Z/l2Z)∗ × (Z/l3Z)∗ | gi = ±1, g1g2g3 = 1.

This group is indeed isomorphic to the Klein four-group and has the property that Fl1 , Fl2 and

16

Fl3 all have degree 2 over the field K = Q(ζl1l2l3)H . Hence we arrive at a non-trivial examplewhere the density δa vanishes. Moreover, if L is any number field such that l1, l2, l3 - ∆L, thenthe extensions Q(ζl1l2l3)/Q and L(ζl1l2l3)/L, have the same Galois group. Hence, other examplewhere the Galois group Gal(Fl1l2l3/K) is isomorphic to the Klein four-group are given when Qis replaced by such a field L.

A similar construction exists when l3 = 2, in this case we let a ∈ K∗l1l2 . Again, we choose thefield K such that Fl1 and Fl2 are quadratic extensions. Since they are quadratic, there existalgebraic numbers α and β, such that Fl1 = K(α), Fl2 = K(β), and we have that α2, β2 ∈ K.Hence, if we let a = (αβ)l1l2 , then the field F2 is given by F2 = K(αβ). Since αβ /∈ K, this is asubfield of Fl1l2 of degree 2, and the Galois group of F2l1l2 is isomporhic to the Klein four-group.

17

3 The density of primitive roots in number fields

In this chapter two explicit expressions for the density δa corresponding an algebraic number a ∈K∗\µK will be conjectured. Using the Chebotarev Density Theorem [4], the theorem of GeorgeCooke and Peter Weinberger [1], and the theorem of Hendrik W. Lenstra [3], a conditional prooffor these expressions is given, under the assumption of the Generalized Riemann Hypothesis.To this end, the concept of linear disjointness of number fields is introduced. This chapter isconcluded by proving two corollaries of the main theorem, stating that the density δa vanishes ifand only if a certain square-free integer n(a,K) satisfies Lemma 2.10, where n(a,K) is an integerdepending only on a and K.

3.1 The main theorem

Theorem 3.1. Let K be a number field and let a ∈ K∗\µK be a non-zero algebraic number.Let PK denote the collection of prime ideals of OK , and let P (a) ⊂ PK denote the collectionof primes p of K for which a is a primitive root in kp. Under assumption of the GeneralizedRiemann Hypothesis, P (a) possesses a natural density δa in PK . Moreover, there exists a square-free integer d ∈ Z, such that the density is given by

δa =∑n|d

µ(n)

[Fn : K]·∏

l prime,l-d

(1− 1

l(l − 1)

). (6)

Here, µ(n) denotes the Mobius counting function.

In order to prove the theorem, we first determine for every prime number l, the primes p of K, forwhich the index [k∗p : 〈a〉] is divisible by l. As shown in section 2.3, for p /∈ DK(a) ∪DK(2∆K),this occurs if and only if p splits completely in Fl. In section 2.2, it was shown that this occurs ifand only if the Frobenius symbol (p, Fl/K) of p in Fl is the trivial element. By Proposition 2.5,this can be generalized to arbitrary, square-free n ∈ N. Let n ∈ N be a square-free integer, andlet σ ∈ Gal(Fn/K) be an automorphism. Let p /∈ DK(a)∪DK(2∆K) be a prime of K such thatσ ∈ (p, Fn/K). If there there exists a prime number l | n, for which the restriction σ|Fl

= idFl

holds, then a is not a primitive root modulo p. Hence if a is a primitive root modulo p, then σmust act non-trivially on all subfields Fl. Hence we define the collection

Cn = σ ∈ Gal(Fn/K) : σ|Fl6= idFl

, for all l | n.

If the Frobenius symbol (p, Fn/K) of a prime p is not contained in Cn, then a is not a primitiveroot modulo p. This proves fruitful in combination with the Chebotarev Density Theorem, whichstates the following [4]:

Theorem 3.2 (N. Chebotarev, 1926). Let L/K be a finite Galois extension of number fields.Let C ⊂ G = Gal(L/K) be a conjugacy class of Gal(L/K). The collection

p ∈ PK | p is unramified in L and (p, L/K) ⊂ C

has density #C#G in PK .

Since Cn is a conjugacy class, it can be concluded from this theorem that the collection Mn ofprimes p of K, that do not split completely in any subfield Fl of Fn, has density δa,n = #Cn

[Fn:K]

18

in PK . If m ∈ N is a divisor of n, then the splitting field Fm is a subfield of Fn. Everyσ ∈ Cm has [Fn : Fm] extensions in Gal(Fm/K). From the injectivity of the inclusion mapi : Cn → Cm × Gal(Fn/Fm), it is therefore concluded that #Cn ≤ #Cm · [Fn : Fm]. Thefollowing inequality therefore holds:

δa,n =#Cn

[Fn : K]≤ #Cm · [Fn : Fm]

[Fn : Fm] · [Fm : Fl]= δa,m.

Let P denote the collection of products of increasing prime numbers,

P =

∏p≤n

p

∣∣∣∣∣∣ n ∈ N, p prime

,

that is totally ordered by divisibility. Then, by the monotone convergence theorem, we concludethat the limit

limi∈P

δa,i

exists. Naively, one might conclude from the above that this limit is δa. Proving this is rathernon-trivial however. In 1975, George Cooke and Peter Weinberger proved that this equality doeshold under assumption of the General Riemann Hypothesis [1].

Theorem 3.3 (G. Cooke, P. Weinberger, 1975). Under assumption of the Generalized RiemannHypothesis, we have that

δa = limi∈P

δa,i.

In order to give an explicit expression for the density, it therefore suffices to find an explicit valuefor #Cn. This can be given using the principle of inclusion-exclusion:

Proposition 3.4. Let n ∈ N be a square-free integer. The density δa,n is given by

δa,n =∑d|n

µ(d)

[Fd : K].

Proof. Let σ ∈ Gal(Fn/K) be given. We note that σ /∈ Cn if and only if there exists a square-freeinteger d | n for which σ|Fd

= idFd. In particular, for d | n, define

Dd = σ ∈ Gal(Fn/K) : σ|Fl= idFl

, for all l | d.

Then Cn = Gal(Fn/K)\(⋃

l|nDl

). By the principle of inclusion and exclusion, #Cn is given by

#Cn =∑d|n

µ(d)#Dd.

Since Dd = σ ∈ Gal(Fn/K) : σ|Fl= idFl

, for all l | d = Gal(Fn/Fd), we conclude that

δa,n =#Cn

[Fn : K]=∑d|n

µ(d) · [Fn : Fd]

[Fn : K]=∑d|n

µ(d)

[Fd : K].

19

By taking the limit n ∈ P in Proposition 3.4, we obtain

limn∈P

δa,n = limn∈P

∑d|n

µ(d)

[Fd : K]=

∞∑n=1

µ(n)

[Fn : K]. (7)

Under assumption of the Generalized Riemann Hypothesis, this limit is equal to the density δa.

3.2 Proof of the main theorem

When comparing Theorem 3.1 with Artin’s original conjecture (1) for square-free a ∈ Q∗, it caneasily be seen that the equations differ by replacing the finite product

∏l|d(1− (l2 − l)−1

)with

the inclusion-exclusion sum

ε =∑n|d

µ(n)

[Fn : K].

Here d is a square-free integer, depending only on the choice of K and a ∈ K∗. In the previ-ous section it was concluded that ε is in fact the density of primes p ∈ PK , that do not splitcompletely in any subfield Fl ⊂ Fd. For almost all prime numbers l, the density of primes thatdo not split completely in Fl is given by δa,l = [Fl : K]−1 = (l2 − l)−1. Artin’s original con-jecture proved incorrect, since it did not take the ’linear dependence’ of the splitting fields intoaccount. As shown in Example 2.2, for the given field K and algebraic number a, a prime p ofK will split completely in either F2, F3 or F5. This is due to the fact that these fields are notlinearly disjoint over K. In order to formalize this, the concept of linear disjointness is introduced:

Definition 3.5. Let K be a number field. Let S = Lii∈I be a collection of number fields thatare Galois over K. Let L =

∏i∈I Li denote the compositum of these fields. The collection S is

linearly disjoint over K, if the natural injection

ι : Gal(L/K)→∏i∈I

Gal(Li/K),

σ 7→ (σ|Li)i∈I

is surjective.

This definition is equivalent with the condition that ι is a well-defined isomorphism of groups.If I is a finite collection, this definition is also equivalent with the condition that

[L : K] =∏i∈I

[Li : K]. (8)

In particular, two fields L and M are linearly disjoint over K if and only if they have intersec-tion K. This is illustrated in the following example.

Example 3.6. Let L/K and M/K be two finite Galois extensions of K. We will show that thefields L and M are linearly disjoint over K, if and only if they satisfy the equality, L∩M = K. Tothis end, we will show that there is a natural isomorphism ϕ : Gal(LM/L)

∼−→ Gal(M/M ∩ L),given by the restriction σ 7→ σ|M . Since M is a Galois extension, this map is well-defined.Let σ ∈ Gal(LM/L) be given and assume that σ|M = idM . Since σ|L = idL, it follows thatσ = idLM . Hence, ϕ is injective. Let τ ∈ Gal(M/M ∩ L) be given. Since an automorphism τ ∈Gal(LM/L∩M) is defined by its action on the fields M and L, we can define the automorphism

20

τ , by τ(x) = τ(x) for all x ∈ M and τ(x) = x, whenever x ∈ L. Since τ |L∩M = idL∩M ,this automorphism is well-defined, and contained in Gal(LM/L). Moreover, it satisfies thatτ |M = τ(x). Hence, ϕ is surjective. This is illustrated in the following diagram.

LM

L

M

L ∩M

K

||

||

We therefore have a natural isomorphism

Gal(LM/K)∼−→ Gal(M/L ∩M)×Gal(L/K), (9)

σ 7→ (σ|M , σ|L).

It follows directly from equation (8), and the definition of linear disjointness that, L and M arelinearly disjoint over K if and only if the L ∩M = K holds.

It is important to note that this alternative definition of linear disjointness does not generalizewhen S is a collection of cardinality greater than 2; the fields F2, F3 and F5 as described inExample 2.2 are pairwise linearly disjoint over Q(

√5). The collection F2, F3, F5 is not linearly

disjoint however, as the map ι is not surjective.

Let d,m be two square-free integers such that the fields Fd and Fm are linearly disjoint. Thenthe image of Cdm under the restriction map ι as defined in Definition 3.5 is the direct productof Cd and Cm. We therefore obtain the equality

δa,dm =#Cdm

[Fdm : K]=

#Cd ·#Cl[Fd : K][Fm : K]

= δa,d · δa,m.

If the collection Fl | l prime is linearly disjoint over Q, then equality (1) would hold. Forexample, if a = 2, and d is a square-free integer we obtain the equality

δ2,d =∏

l prime,l|d

δ2,l =∏

l prime,l|d

(1− 1

[Fl : Q]

).

This is not the case if a ≡ 1 mod 4 however. Galois theory shows that if p is an odd primenumber, then the field Q(ζp) has a subfield L of degree 2 given by L = Q(

√p∗), where p∗ =

(−1)(p−1)/2)p. [5, p. 54]. Hence, if p ≡ 1 mod 4, we have that p∗ = p and therefore, L = F2. Itfollows that F2p = Fp and hence the fields Fp and F2 are not linearly disjoint. In order to proveTheorem 3.1 in arbitrary number fields K, we need to determine a finite collection I of primenumbers, such that the collection Fll/∈I ∪

∏p∈I Fp is linearly disjoint over K. Over Q, such

a collection is given by I = 2 ∪DQ(a). the only eligble primes are the primes 2 and the primenumbers l for which ordl(a) 6= 0.For a number field K, a similar collection exists. In 1977, Hendrik W. Lenstra, analyzed theconditions for which the field Fl is linearly disjoint of all fields Fd, where d is a square-free integercoprime to d, [3]. This is summarized in the following theorem.

21

Theorem 3.7 (H. W. Lenstra, 1977). Let l be a prime number satisfying the following conditions:

1. l does not divide 2 ·∆K ,

2. there exists no x ∈ K∗ such that a = xl,

3. the divisor sets DK(l) and DK(a) are disjoint.

Further, let d ∈ Z be a square-free integer, coprime to l. Then the fields Fl and Fd are linearlydisjoint over K. Moreover, the field extension Fl/K has degree l(l − 1).

Proof. Let l be a prime number satisfying all the listed conditions. It follows from the firstcondition that the l-th cyclotomic polynomial Φl(X) is irreducible in K[X]. We therefore havethat [K(ζl) : K] = l − 1. From the second condition, it follows that the polynomial X l − a isirreducible in K(ζl)[X], [5, p. 72]. We conclude that [Fl : K] = l(l − 1).

Let d ∈ Z>0 be a square-free integer coprime to l. In order to prove that Fl and Fd are linearlydisjoint over K, it suffices to show that [Fdl : Fd] = l(l − 1). The extension K(ζl)/K is totallyramified of degree l − 1 > 1. The only primes that ramify in Fd are the primes contained inthe union DK(a) ∪DK(d). Since l is coprime to d, and the intersection DK(a) ∩DK(l) = ∅ isempty, the extension Fd/K is unramified over l. Hence, it follows that K(ζl) ∩ Fd = K, andtherefore the equality [Fd(ζl) : Fd] = (l − 1) holds. Fd is a Galois extension, hence if l

√a ∈ Fd,

then ζl ∈ Fd. We conclude that l√a /∈ Fd. Since l is a prime number, it therefore follows from

the second condition that the polynomial X l−a is irreducible in Fd(ζl)[X] as well. We concludethat [Fd(ζl, l

√a) : Fd(ζl)] = l, and hence the fields Fd and Fl are linearly disjoint over K.

There are only finitely many prime numbers l that violate a condition of Theorem 3.7. Let ddenote the product of these primes. For all other primes p coprime to d, we have that the fieldsFd and Fp are linearly disjoint over K. This observation allows us to prove Theorem 3.1.

Proof of Theorem 3.1. Let K be a number field and let a ∈ K∗\µK be a non-zero algebraicnumber that is not a root of unity. Let g ∈ N denote the largest integer for which the polynomialXg − a has a root in K. For all primes pi ∈ DK(a) in the divisor set of a, let pi = char(kpi

)denote the characteristic of the residue class field kpi

, and let P =∏i pi be the product of all

these prime numbers. Finally, let d be the square-free integer given by d = rad(2∆KgP ) ∈ Z,where rad denotes the radical function.

Let l be a prime number that is coprime to d. By Theorem 3.7, the fields Fd and Fl are linearlydisjoint over K. Let ι : Gal(Fdl/K) 7→ Gal(Fd/K)×Gal(Fl/K) be the restriction map as givenin Definition 3.5. From the equality ι(Cdl) = Cd × Cl, it follows that #Cdl = #Cd · #Cl, andtherefore the equality

δa,dl =#Cdl

[Fdl : K]=

#Cd ·#Cl[Fd : K][Fl : K]

(10)

holds. Since l is a prime number, the equality #Cl = [Fl : K] − 1 holds. Therefore, equation(10) reduces to

δa,dl =#Cd

[Fd : K]

([Fl : K]− 1

[Fl : K]

)= δa,d ·

(1− 1

[Fl : K]

). (11)

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Let m ∈ P be a product of prime numbers, such that d | m. By repeating the given argumentwe obtain

δa,m = δa,d∏

l prime,l|m/d

(1− 1

[Fl : K]

).

For all l - d, we have that [Fl : K] = l(l − 1) by Theorem 3.7. Theorem 3.3 finally gives

δa =∑n|d

µ(n)

[Fn : K]·∏

l prime,l-d

(1− 1

l(l − 1)

).

3.3 Applications of the main theorem

We will conclude this thesis by proving several corollaries of the main theorem.

Corollary 3.1.1. Let K be a number field and a ∈ K∗\µK be a non-zero algebraic number thatis not a root of unity. Then the following are equivalent:

1. δa = 0.

2. There exists a square-free n ∈ Z>0, for which the equality Gal(Fn/K) =⋃l|n Gal(Fn/Fl)

holds.

Proof. If δa = 0, then it follows from Theorem 3.1, that there exists a square-free integer n ∈ Zfor which δa,n = 0, and since δa,n = #Cn

[Fn:K] , we conclude that

σ ∈ Gal(Fn/K) : σ|Fl6= idFl

, for all l | n = ∅.

Hence, for all σ ∈ Gal(Fn/K) there exists a prime number l | n for which σ|Fl= idFl

, henceσ ∈ Gal(Fn/Fl). We conclude that Gal(Fn/K) =

⋃l|n Gal(Fn/Fl). The other implication was

proven in Lemma 2.10.

Corollary 3.1.2. Let K be a number field, and let a ∈ K∗\µK be an algebraic number that isnot a perfect power. Then δa does not vanish.

Proof. Let K be a number field, let a ∈ K∗ be an algebraic integer that is not a perfect power.We will show using induction on P, that Cn 6= ∅ for all n ∈ P.

We note that C2 6= ∅, hence the statement holds for n = 2. Let n ∈ P be given and assumethat Cn 6= ∅. Let l be the smallest prime number that does not divide n. We note that thedegree [Fn : K] is a divisor of

∏p|n[Fp : Q] =

∏p|n p(p − 1), where p ranges over all prime

numbers dividing n. Since a /∈ K∗l, we have that l | [Fl : K], and since l - n, we conclude thatl - [Fn : K], and therefore l | [Fl : Fl ∩ Fn]. From Example 3.6, we conclude that there is anatural isomorphism:

Gal(Fnl/Fn ∩ Fl)∼−→ Gal(Fl/Fn ∩ Fl)×Gal(Fn/Fn ∩ Fl),

σ 7→ (σ|Fl, σ|Fn

).

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Hence, there exists an automorphism σ ∈ Gal(Fnl/Fn ∩ Fl) of order l, such that σ|Fl6= idFl

,and σ|Fn

= idFn. This automorphism can be extended to an automorphism σ ∈ Gal(Fnl/K).

By the induction hypothesis, there exists a τ ∈ Cn, which can be extended to an automorphismτ ∈ Gal(Fnl/K), with order coprime to l. The composition φ = στ , is an automorphism of Fnl,for which the restriction φ|Fn

= τ holds. Since the order of τ and σ are coprime, it follows thatφ|Fl

6= idFl. Hence, this automorphism acts non-trivially on all subfields Fd ⊂ Fl and on Fl. It

therefore acts non-trivially on all subfields Fd ⊂ Fnl. We conclude that φ ∈ Cnl. It follows thatCn 6= ∅, for all n ∈ P.

To finalize the proof, we note that by the proof of Theorem 3.1, there exists an m ∈ P, suchthat δa = δa,m ·

∏l-m(1 − (l2 − l)−1), and by the Chebotarev Density theorem, it follows that

δa.m 6= 0.

References

[1] George Cooke and Peter J. Weinberger. On the construction of division chains in algebraicnumber rings,with applications to SL2. In: Communications in Algebra 3.6 (1975), pp. 481–524. doi: 10.1080/00927877508822057.

[2] Christopher Hooley. On Artin’s conjecture. In: Journal fur die reine und angewandte Math-ematik 225 (1967), pp. 209–220.

[3] H. W. Lenstra. On Artin’s Conjecture and Euclid’s Algorithm in Global Fields. In: Inven-tiones mathematicae 42 (1977), pp. 201–224.

[4] H. W. Lenstra & Peter Stevenhagen. Chebotarev and his density theorem. In: The Mathe-matical Intelligencer 18.2 (1996).

[5] Peter Stevenhagen. Algebra 3. 2012. url: websites.math.leidenuniv.nl/algebra/.[6] Peter Stevenhagen. Number Rings. Oct. 13, 2017. url: websites.math.leidenuniv.nl/

algebra/.[7] Peter Stevenhagen. The correction factor in Artin’s primitive root conjecture. In: Journal de

Theorie des Nombres de Bordeaux (2003).

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