On Artin's Conjecture for Primitive Roots

On Artin’s Conjecture for Primitive RootsbyFrancesco Pappalardi ∗Department of Mathematics and StatisticsA thesis submitted in partial fulfillmentof the requirements of the degree ofDoctor of Philosophy at McGill UniversityFebruary 1993∗ c○Francesco Pappalardi - (1993)

ACKNOWLEDGMENTSThere are many people who I would like to thank at this point:My Master thesis Supervisor Marco Fontana for having trusted me and supportedmy choice to come to Canada. Professor Paulo Ribenboim for having always advisedme on my choices and his precious mathematical teaching.Professors Jal Choksi, Ian Connell, Hershy Kisilevsky, John Labute, V. Seshadriand Georg Schmidt, whose help and politeness made my time at McGill pleasant.Raffaella Bruno who corrected the english of this thesis. None of my problemswere ever unsolvable for her.Valerie McConnell, Elaine Tremblay for never having denied a smile and theirvaluable help.My friends Mark Fels and Djordje ˘Cubrić, the first having shown me the humanside of being a graduate student, and the second , with whom I shared an office wasa true gentleman and a solid office-mate. It will not be easy to find others like them.Masato Kuwata, who has helped me considerably with the computers and DamienRoy who helped me with the theory of Kummer’s Extensions.My wife Claudia and my daughter Silvia, the first for support and encouragementduring these difficult years, and the second for having waited that extra weeknecessary to finish this thesis.Last but certainly not least I would like to thank my supervisor Ram Murty. Heis an exceptional man, who placed his trust in me and guided me to many of theproblems which lead to this thesis. His teaching did not cover only Mathematics buthe has also been a precious counsellor on any issue that I have brought to him. Ihope to be able to pass on what I have learned from Professor Murty to others in thefuture.

ABSTRACTVarious generalizations of the Artin’s Conjecture for primitive roots are considered.It is proven that for at least half of the primes p, the first log p primes generatea primitive root. A uniform version of the Chebotarev Density Theorem for the fieldQ(ζ l , 2 1/l ) valid for the range l < log x is proven. A uniform asymptotic formula forthe number of primes up to x for which there exists a primitive root less than s isestablished. Lower bounds for the exponent of the class group of imaginary quadraticfields valid for density one sets of discriminants are determined.RESUMÉNous considérons différentes généralisations de la conjecture d’Artin pour lesracines primitives. Nous démontrons que pour au moins la moitié des nombres premiersp, les premiers log p nombres premiers engendrent une racine primitive. Nousdémontrons une version uniforme du Théorème de Densité de Chebotarev pour lecorps Q(ζ l , 2 1/l ) pour l’intervalle l < log x. On établit une formule asymptotiqueuniforme pour les nombres de premiers plus petits que x tels qu’ il existe une racineprimitive plus petite que s. Nous déterminons des minorants pour l’exposant dugroupe de classe des corps quadratiques imaginaires valides pour ensembles de discriminantsde densité 1.

ContentsINTRODUCTION 31 ON HOOLEY’S THEOREM 91.1 A generalization of Hooley’s Theorem . . . . . . . . . . . . . . . . . . 91.2 Computation of the Densities . . . . . . . . . . . . . . . . . . . . . . 161.3 The Main Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222 ON THE ARTIN L-FUNCTIONS OF Q(ζ l , 2 1/l ) 242.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242.2 Artin L-functions of L/Q . . . . . . . . . . . . . . . . . . . . . . . . . 272.3 On the non-Abelian L-function of Q(ζ l , 2 1/l ) . . . . . . . . . . . . . . 312.4 An Application to Chebotarev Density Theorem . . . . . . . . . . . . 403 ON THE NUMBER OF PRIMES GENERATING F ∗ p 433.1 Extending Hooley’s Method . . . . . . . . . . . . . . . . . . . . . . . 433.2 Relaxation of the Hypothesis and Improvements . . . . . . . . . . . . 483.3 A Density One Result . . . . . . . . . . . . . . . . . . . . . . . . . . 554 MORE ON PRIMITIVE ROOTS 604.1 Another Generalization of Hooley’s Theorem . . . . . . . . . . . . . . 614.2 Calculation of the Densities . . . . . . . . . . . . . . . . . . . . . . . 674.3 An Application to the Least Prime Primitive Root . . . . . . . . . . . 801

APPENDIX A: ON DIVISORS OF p − 1 83APPENDIX B: ON THE EXPONENT OF THE IDEAL CLASS GROUPOF Q(−d) 90APPENDIX C: OPEN QUESTIONS AND FUTURE RESEARCH 99Variants of the Bombieri-Vinogradov Theorem . . . . . . . . . . . . . . . . 99The Lang-Trotter Conjecture for Abelian Varieties . . . . . . . . . . . . . 101REFERENCES 1042

INTRODUCTIONThe famous Artin Conjecture for primitive roots states that any integer a ≠ ±1 whichis not a perfect square is a primitive root for infinitely many primes. More precisely,if N a (x) is the set of such primes up to x, thenN a (x) ∼ A(a)π(x)where A(a) ≠ 0.Artin also gave an explicit formula for A(a) and his intuition was based on thefollowing heuristic argument (see [1]):For any prime p less than x, let P a (q) be the probability that the prime q divides theindex [F ∗ p : 〈a〉]; then,by considering such instances independent, we haveA(a) = ∏ q(1 − P a (q)).In order to have q|[F ∗ p : 〈a〉], the two necessary and sufficient conditionsp ≡ 1(modq) and a (p−1)/q ≡ 1(modp) (1)must be satisfied.Now consider the field L q= Q(ζ q , a 1/q ), let p be a rational prime that splitscompletely in L q and let P be a prime over p. The residue field at P has p elements,therefore (1) holds. Conversely if (1) holds for p, then p splits completely in L q .The Chebotarev Density Theorem indeed states that the probability that p splitscompletely in a normal extension K, equals 1/[K : Q] and therefore the probabilityP a (q) is 1/q(q − 1) andA(a) = ∏ q(1 −)1.q(q − 1)3

Later, calculations made by D. H. Lehmer and E. Lehmer (see [35]) suggested thatin some cases the expression of A(a) was not correct and the factors of the productexpansion of A(a) corresponding to the prime divisors of a had to be replaced byother expressions.In 1965, C. Hooley (see [26]) used the linear sieve to prove that if the validity ofthe Generalized Riemann Hypothesis is assumed for the Dedekind zeta function ofthe fields L q then the Artin’s Conjecture is true, with the corrections indicated byLehmer.The main tool used by Hooley is the effective version of the Chebotarev densityTheorem valid under the assumption of the Riemann Hypothesis for the Dedekindzeta function of K. That is#{p ≤ x | p splits completely in K} = 1n Kli(x) + O(x 1/2 (log x + log d 1/n KK )),where n K = [K : Q] and d K is the discriminant.This version of the Chebotarev Density Theorem has been for a long time theonly effective one available until 1977 when J. C. Lagarias and A. M. Odlyzko proveda version of the Theorem valid with the condition (see [29]):√log xn K≫ max { d 1/n KK , log d K}.For a Kummer’s extension of the type L q , this is equivalent to q < log 1/6 x.Such a discovery, unfortunately, does not allow one to eliminate the use of theRiemann Hypothesis on the proof of the Theorem of Hooley, however it gives auniform result for q < log 1/6 x.In 1984 R. Gupta and R. Murty (see [15]) published the first result in which thevalidity of the Artin Conjecture is established for at least one value of a. Indeed,4

they constructed a set of 13 numbers for which at least one is primitive root for anumber of primes p up to x which is ≫Heath-Brown (see [21]) to a set of 3 elements.x . This result was later sharpened bylog 2 xThe idea of Gupta and Murty also allowed them to deal with the analogousstatement of the Artin Conjecture for rational points on Elliptic Curves (see. [17]).This is the Lang-Trotter Conjecture. From this they were led to consider a high-rankversion of the Artin Conjecture.Given a 1 , . . . , a r ∈ Z, we say that a 1 , . . . , a r are multiplicatively independentif, whenever there are integers n 1 , . . . , n r such thatwe have n 1 = n 2 = · · · = n r = 0.It makes sense to ask ifa n 11 · · · a nrr = 1,〈a 1 , . . . , a r mod p〉 = F ∗ p (2)for infinitely many primes p and to speculate whether the density of such primes canbe calculated. It is necessary to express the condition for a prime q to divide theindex of the group generated by a 1 , . . . , a r in terms of splitting conditions on somefields. The natural generalization of Artin’s original idea is in:Theorem 1 Let 〈a 1 , . . . , a r 〉 be the subgroup of F ∗ p generated by the multiplicativelyindependent a 1 , . . . , a r . For any prime qq ∣ ∣[F∗p : 〈a 1 , . . . , a r 〉] ⇐⇒ p splits completely in Q ( ζ q , a 1/q )1 , . . . , a 1/qr .✷This result and the consequent application of the Chebotarev Density Theoremsuggests that the density of primes for which (2) holds equals()∏ 11 −. (3)q r (q − 1)q5

In Chapter 1 we prove that if the Generalized Riemann Hypothesis holds for thefields in Theorem 1, thenN a1 ,...,a r(x) = # { p ≤ x | 〈a 1 , . . . , a r 〉 = F ∗ p}∼ Aa1 ,...,a rπ(x). (4)where the constant A a1 ,...,a requals the product in (3), up to finitely many factors.The complete formulas, with the analogous corrections of those suggested by Lehmerfor the Artin Conjecture, are worked out in Section 2.1 by the use of some propertiesof Kummer’s extension.The proof follows the original one of Hooley but now the estimate for the numberof primes for which there is a large prime divisor of the index is made using a Lemmadue to C.R. Matthews which is an application of the pigeon-hole principle.The new parameter given by the rank, suggests to take r as a function of x andtry to adapt the proof to obtain a result independent of the Riemann Hypothesis.This is done in Section 3.1 and the conclusion is that for a positive density of primesp, F ∗ p can be generated by about log p multiplicatively independent integers.The main obstacle comes from those primes for which the index[F ∗ p : 〈a 1 , . . . , a r 〉]has some prime divisor in the interval [log 1/6 x, log 2 x].The range [log 1/6 x, log x] is dealt by using a version of Chebotarev Density Theoremfor the field Q(ζ l , 2 1/l ) valid for a range of l up to log x/(log log x) 2 which is provenin Chapter 2. Such a proof uses properties of the single non-Abelian L-function ofQ(ζ l , 2 1/l ), and is of course stronger than the one of Lagarias and Odlyzko of [29].This establishes a Conjecture of H. Zassenhaus of 1969 (see [4]).In Section 3.3, we work out the bound of r ≥ log 2 p for a set of density one ofprimes p for which F ∗ p is generated by r elements. Such a result is stronger than6

the one that was known as a consequence of the work of Burgess and Elliot (seeProposition 1.11) and uses the Large Sieve Inequality.The Lemma of Matthews used in the proof of the asymptotic formula in (4) allowsone to conclude that for almost all primes p the index[F ∗ p : 〈a 1 , . . . , a r 〉] ≥ pr/(r+1)log p .In Appendix A, we improve such a lower bound to[F ∗ p : 〈a 1 , . . . , a r 〉] ≥ p r/(r+1) exp{log δ p}. (5)This is done by deducing an upper bound for the number of primes p for which p − 1has a divisor in the range (x h , x h exp{log δ p}) which is due to Murty and Erdös (see[14]) and proven here with the uniformity conditions that allow estimates of the type(5) uniform with respect to r.Next we take into consideration the problem of determining an asymptotic formulafor the number of primes for which two given numbers (or more in general s givennumbers) are simultaneously primitive roots. An heuristic argument similar to Artin’ssuggests a densityδ =∏q prime(1 − 2q − 1 ),q 2 (q − 1)and again this is proven to be the right one up to finitely many factors. Completeformulas are worked out in the case where the given numbers are primes. We laterdiscovered that a general version of this statement has been proven by K. R. Matthewsin his Ph.D. Thesis (see [36]). However, our proof is different and by the use of aTauberian Theorem, we get a better uniform error term.This result has, as an application, a uniform asymptotic formula for the numberof primes for which the least prime primitive root is less than a parameter y. Such aformula has applications to the problem of the distribution of least primitive roots.7

In Appendix B, we consider the problem of the exponent of the class group ofimaginary quadratic fields.If e(d) is the exponent of the ideal classgroup of the imaginary quadratic fieldsQ( √ −d), the Iwasawa Conjecture states thatlim e(d) = +∞.d→+∞In 1972, D.W. Boyd and H. Kisilevsky (see [3]) proved that if the Extended RiemannHypothesis holds for certain Dirichlets L-functions, then the Iwasawa Conjecture istrue.The proof consists on noticing a link between the least prime p for which −d is aquadratic residue and e(d) (this is p e(d) ≫ d) and then use the Riemann Hypothesisto prove that p ≪ log 2 d. This argument establishes the bounde(d) ≫log dlog log d . (6)We prove unconditionally that (6) holds for a set of discriminants of density one,by calculating uniform asymptotic formulas for the number of integers (resp. squarefreeintegers) d < x for which the least prime p with ( )−dp = 1 is smaller than s.8

1 ON HOOLEY’S THEOREM1.1 A generalization of Hooley’s TheoremSuppose a 1 , . . . , a r are multiplicatively independent integers and let Γ be the subgroupof Q × generated by a 1 , . . . , a r . For all but finitely many primes p, it makes sense toconsider the reduction of Γ modulo p which we indicate by Γ p which can be viewedas a subgroup of F ∗ p.In the case r = 1, Hooley has shown that if the generalized Riemann Hypothesisholds for the Dedekind zeta function of the fields Q(ζ l , a 1/l1 ), with l prime, then theset of primes p for which F ∗ p = Γ p has non zero density (see [26]).We will consider the following generalization first introduced by R. Gupta and R.Murty in [15].Theorem 1.1 Let Γ be as above, n m = [Q(ζ m , a 1/m1 , . . . , a 1/m ) : Q] and letδ Γ =∞∑m=1µ(m)n mif the Generalized Riemann Hypothesis holds for the Dedekind zeta function of thefields Q(ζ l , a 1/l1 ), l prime, thenRemark: a) Note thatN Γ (x) = #{p ≤ x|F ∗ xp = Γ p } ∼ δ Γlog x .n m ≥ [Q(ζ m , a 1/m1 ) : Q] ≫ φ(m)m, (1)therefore δ Γ is a convergent series and thus a well defined number. We will prove inthe second section that δ Γ ≠ 0.r9

) Theorem 1.1 can also be proven on the weaker assumption that there exists a ∈ Γwith the property that all the Dedekind zeta functions of the fields Q(ζ l , a 1/l ) (l largeprime) have no zeroes in the regionσ > 1 − 1r + 1 .Proof: Let us assume r > 1. The first steps of the proof follow the original ideaof Hooley who considered the following functions:N Γ (x, y) = #{p ≤ x | ∀l, l ≤ y, l ̸ | [F ∗ p : Γ p ]},M Γ (x, y, z) = #{p ≤ x | ∃l, y ≤ l ≤ z, l| [F ∗ p : Γ p ]},M Γ (x, z) = #{p ≤ x | ∃l, l ≥ z, l| [F ∗ p : Γ p ]},where y = 16r+8 log log x and z = x1/(r+1) log x.Clearly,N Γ (x, y) ≥ N Γ (x) ≥ N Γ (x, y) − M Γ (x, y, z) − M Γ (x, z), (2)and establishing the following:the Theorem would be proven.a) N Γ (x, y) = δ Γxlog x + o( xlog x );b) M Γ (x, y, z) = o( xlog x );c) M Γ (x, z) = o( xlog x ),In his original work, Hooley used the GRH to treat both the main term N Γ (x, y)and the term M Γ (x, y, z). In this proof we will show that a choice of y = 1 log log x6r+8enables to remove the GRH from the treatment of the main term.element for subsequent applications.This is a key10

a) By the inclusion-exclusion formula,N Γ (x, y) =∗∑µ(m)π m (x)mwhere µ is the Möbius function, the upper ∗ means that the sum is extended to allthe integers m whose prime divisors are distinct and less than y (note that this forcesm ≤ ∏ q

Now, let d m be the discriminant of L m and n m its degree. The Hensel inequality(see. page. 259 of [42]) states thattherefored 1/nmm≤ ∏log |d m | ≤ n m∑q|d mlog q, (4)q|d mq ≤ ma 1 . . . a r ≤ n m ≤ log d msince indeed in any field log d ≥ n. We can also prove the following.Corollary 1.3 If m ≤ (log x) 13r+4 thenfor some absolute positive constant A.π m (x) = li(x)n m+ O(x exp −A(log x) 1/3 )Proof of Corollary 1.3: The inequality assumed for m and the Hensel inequalityin (4), imply (n m ≤ m r+1 ):c n 1/2m log d m ≤ c n 3/2m∑log q ≤ m 3r+42 ≤ (log x) 1/2 .q|d mHence, Lemma 1.2 givesπ m (x) − li(x)n m⎛ ⎛ ( ) ⎞⎞1/2log x= O ⎝x exp ⎝−A⎠⎠( ())= O x exp −A(log x) 1 2 − r+12(3r+4)= O ( x exp ( −A(log x) 1/3)) .✷n mThe choice made for y allows us to apply Corollary 1.3 to all the m ≤ e 2y =(log x) 13r+4 . Using the estimate (1) for the degree nm , we get:N Γ (x, y) =∗∑m( )1µ(m) li(x) + O(x exp −A(log x) 1/3 ) =n m12

=∞∑µ(m)n mm=1∞∑=m=1⎛li(x) + O ⎝ ∑ m>yµ(m)n m⎞1mφ(m) li(x) ⎠ + O(e y x exp −A(log x) 1/3 )xlog x + o(li(x)) + O((log x) x exp(−A(log x)1/3 ))( )x x= δ Γlog x + o .log xc) To deal with the last term, we will make use of the following result due toMatthews (see [37]):Lemma 1.4#{p | |Γ p | ≤ t} = O(t 1+1/r ∑ ilog a i )where the constants involved in the O symbol do not depend on t nor r, nor on{a 1 , . . . , a r }.Proof of Lemma 1.4: Consider the set S = {a n 11 · . . . · a nrr| 0 ≤ n i ≤ t 1/r }. Asa 1 , . . . , a r are multiplicatively independent, the number of elements of S exceeds([t 1/r ] + 1) r > t.If p is prime such that |Γ p | ≤ t, then two distinct elements of S are congruent(modp). Hence, p divides the numerator N ofa m 11 · · · a mrr − 1for some m 1 , m 2 , . . . , m r satisfying |m 1 | ≤ t 1/r , 1 ≤ i ≤ r.For a fixed choice of m 1 , m 2 , . . . , m r , the number of such primes is bounded byr∑log N ≤ t 1/r log a ii=1Taking in account the number of possibilities for m 1 , m 2 , . . . , m r , the total number ofprimes p cannot exceedr∑O(t 1+1/r log a i ).i=113

This completes the proof of the Lemma.✷Now note thatM Γ (x, z) ≤ #{p ≤ x | ∃ l ≥ z, l| p − 1|Γ p |{≤ # p ≤ x | |Γ p | ≤ x }z}and applying Lemma 1.4 (no dependence on r is required here), we getM Γ (x, z) = O( ) ( )x(1−1/(r+1))(1+1/r) x= o .(log x) 1+1/r log xb) For the middle term we assume the GRH which allows to state the followingversion of the Chebotarev Density Theorem (a proof can be found in [26] or also in[30]):#{p ≤ x | p splits completely in Q(ζ l , a 1/l1 )} =1l(l − 1) li(x) + O(x1/2 log xl) (5)Now, as in the main term, l|[F ∗ p : Γ p ] if and only if p splits completely in theKummer extension Q(ζ l , a 1/l1 , . . . , a 1/l ) and thus, in particular, p splits completely inQ(ζ l , a 1/l1 ). From this we get:rM Γ (x, y, z) ≤ #{p ≤ x | ∃ l, y ≤ l ≤ z, p splits completely in Q(ζ l , a 1/l1 )}≤ ∑ ()1l(l − 1) li(x) + O(x1/2 log xl) .As ∑ l≥y1l(l−1)y≤l≤zis the tail of a convergent sequence and∑l 1 this yields to an estimate of the type:M Γ (x, y, z) ≪ 1 y li(x) + O(x 1 2 + 1r+1 log x) (6)14

which is o(log x/x), and this completes the proof for r > 1.For completeness we add the the proof of the remain case r = 1. Estimate (6)has no meaning anymore. We have thatM Γ (x, y, x 1/2 log x) ≤ M Γ (x, y, x 1/2 / log 3 x) + M Γ (x, x 1/2 / log 3 x, x 1/2 log x).The first term is treated as the general case and leads to the coresponding estimateof (6) that in this case is o ( )xlog x . For the second term we proceed as Hooley and wenote that z = x 1/2 log x andM Γ (x, x 1/2 / log 3 x, x 1/2 log x) ≪≪xlog x∑x 1/2 / log 3 x

1.2 Computation of the DensitiesThe density δ Γ can always be expressed as an Euler product. Doing so one can provethat the density is not zero. In this section we will calculate δ Γ in the case whena i = p i is an odd prime for any i ≥ 1, we will also be able to prove that in thisparticular caselim δ Γ = 1.r→∞The first step is to calculate the degrees of L m over Q.Theorem 1.5 Let p 1 , . . . , p r be odd primes, m a square-free integer and letn m = [Q(ζ m , p 1/m1 , . . . , p 1/m ) : Q].Suppose (m, p 1 · · · p r ) = p i1 · · · p it , then n m = φ(m)mr , where2 α⎧0⎪⎨m is odd or t = 0α = t if p i1 ≡ p i2 ≡ · · · ≡ p it ≡ 1 (mod 4)r⎪⎩t − 1otherwise.Proof: Fix m > 1, we may assume without loss of generality that p 1 · · · p t =(p 1 · · · p r , m), we let K = Q(ζ m ), A = K(p 1/m1 , . . . , p 1/mt ) and for any 1 ≤ i ≤ r − t,let B i = A(p 1/mt+1 , . . . , p 1/mt+i ). We have thatn m = [B r−t : Q] = [B r−t : A][A : K][K : Q]and clearly [K : Q] = φ(m).Step 1): We claim that [B r−t : A] = m r−t .Since the polynomial x m − p t+1 splits completely in B 1 = A(p 1/mt+1 ), we know that[B 1 : A] = m d. Let q|d be a prime, then [A(p1/q t+1) : A] = 1 or q. If it was q, we wouldhave q = [A(p 1/qt+1) : A]|[B 1 : A] = m , which is a contradiction since m is square-free.d16

Therefore p 1/qt+1 ∈ A, which implies that p t+1 ramifies in A/Q, but, from Kummer’sTheory, we know that the only primes that ramify in A are p 1 , . . . , p t and those thatdivide m, and since (p t+1 , m) = 1, we conclude that d = 1. Now, by induction, wehave that[B r−t : A] = [B r−t : B r−t−1 ][B r−t−1 : A] = [B r−t : B r−t−1 ]m r−t−1 ,and again, [B r−t : B r−t−1 ] = m d[B r−t : A] = m r−t .and since (p r, m) = 1, we conclude that d = 1. HenceStep 2) Let A i = K(p 1/m1 , . . . , p 1/mi ), then A i+1 = A i (p 1/mi+1 ), and for the samereason as above, [A i+1 : A i ] = m . We claim that e = 1 or 2.eLet q|e be a prime divisor and consider A i (p 1/qi+1), since m is square-free, we have thatp 1/qi+1 ∈ A i . If p 1/qi+1 ∈ K, then we would have a cyclic extension of prime degree (overQ) Q(p 1/qi+1) ⊂ K and this is only possible when q = 2. Therefore we may assumethat p 1/qi+1 ∉ K, having extensions:K ⊆ K(p 1/qi+1) ⊆ A i .Note that Gal(A i /K) is the direct product of cyclic groups and a general subgroup oforder q has as fixed field K((p s1 · · · p sk ) 1/q ), with 1 ≤ s 1 ≤ · · · ≤ s k ≤ i−1. Therefore,K(p 1 qi+1) = K((p s1 · · · p sk ) 1/q ) and from Lemma 3 in page 160 of Cassels and Fröhlich[7], we have that there exists 0 ≤ i ≤ q − 1 such thatand again this implies that q = 2.() 1/qp i+1∈ K,(p s1 · · · p sk ) iTherefore, if m is odd, [A i+1 : A i ] = m for every i, and thus [A t : K] = m t .From the Theory of Cyclotomic Fields, we know that the general quadratic subfieldof K has the form Q −1(√ ( ) )D D , where D is a positive divisor of m. We gatherthat if p i ≡ 1 (mod 4), 1 ≤ i ≤ t, then ( −1p i)= 1, hence√pi ∈ K.17

Step 3) If p 1 ≡ p 2 ≡ · · · ≡ p t ≡ 1 (mod 4),then let ζ m be a primitive m-th root of unity, then Gal(A 1 /K) is generated by σ :p 1/m1 ↦→ ζ 2 mp 1/m1 , (Note that σ( √ p 1 ) = (σ(p 1/m1 )) m/2 = (ζ 2 m) m/2 p (1/m)(m/2)1 = √ p 1 ) andhence, |Gal(A 1 /K)| = [A 1 : K] = m 2 .Similarly Gal(A i+1 /A i ) is generated by σ : p 1/mi+1 ↦→ ζ 2 mp 1/mi+1 , therefore [A i+1 : A i ] = m 2and [A : K] = mt2 t .Step 4) If it exists 1 ≤ i ≤ t such that p i ≡ 3 (mod 4),then we can suppose without loss of generality that p 1 ≡ 3 (mod 4). Let us considerA i = K(p 1/m1 ). We have that [A 1 : K] = m (If not, we would have K( √ p 1 ) = K,but this only happens when p 1 ≡ 1 (mod 4), which is a contradiction). Now consideri > 1, and A i = A i−1 (p 1/mi ). We claim that [A i : A i−1 ] = m 2 . Indeed either p i ≡ 1(mod 4) or p i ≡ 3 (mod 4); in the first case √ p 1 ∈ K, in the second case √ p 1 p i ∈ K.In any case, Gal(A i /A i−1 ) is always generated by σ : p 1/mi ↦→ ζmp 2 1/mi . Finally we get[A i : A i−1 ] = m 2and [A : K] =mt2 t−1 .This concludes the proof of the Theorem.✷Corollary 1.6 With the same notation of Theorem 1.5, we haven m ≥ m r φ(m)/2 min(r,ν(m)−1)(where ν(m) is the number of distinct prime divisors of m), furthermore such a lowerbound is the best possible.✷We are now ready to express the density as an Euler product. The case r = 1 hasbeen dealt with by C. Hooley in [26]. He proved that:Lemma 1.7 Let p be a prime, n m = [Q(ζ m , p 1/m ) : Q] and letA =∏ ()11 −l(l − 1)l prime18

e the Artin’s constant, then we have:∞∑m=1⎧µ(m)⎪⎨=n m⎪⎩A if p ≢ 1 (mod 4),A ( )1 + 1p 2 −p−1 if p ≡ 1 (mod 4).Proof: If p ≢ 1 (mod 4), then n m = mφ(m) for every m and the result followsfrom the definition of the Artin’s constant. We can therefore assume that p ≡ 1(mod 4), having:∞∑m=1µ(m)n m= Σ o + Σ e ,where Σ o is the sum extended to the odd values of m and Σ e to the even values.Clearly Σ o = 2A and Σ e = − 1 2 Σ′ e, with2A +−1p(p − 1)∑Σ ′ e =m=1(m,2p)=1∞∑m=1(m,2p)=1µ(m)mφ(m) + 2µ(m)mφ(m) = 2A +Finally Σ o + Σ e = A ( )1 + 1p 2 −p−1 .✷∞ ∑m=1p|m,modd−1p(p − 1)µ(m)mφ(m) =2A(1 −1p(p−1)The general case is similar but a little more complicated:) = 2A −2Ap 2 − p − 1 .Theorem 1.8 Let p 1 , . . . , p r be odd primes, n m = [Q(ζ m , p 1/m1 , . . . , p 1/m ) : Q], letα i = p r i (p i − 1) − 1 and define the r-dimensional incomplete Artin’s constant to be:A(r) =∏l odd prime(1 −)1,l r (l − 1)rthen:∞∑m=1µ(m)n m= A(r){1 − 1 [ r∏(1 −2 r+1 i=1( −1p i) 1α i)+r∏i=1(1 − 1 α i) ]} .19

Proof: As in the case r = 1, note that if m is odd, then n m = m r φ(m), thus wecan write:∞∑m=1µ(m)n m= A(r) − Σwhere Σ is the sum extended to the even values of m.Let P = p 1 · · · p r and ˜P = ∏ ri=1, pi ≡1(4) p i , if m is an odd positive integer and Q =(m, P ) then, by Theorem 1.5, we have⎧⎪⎨n 2m =⎪⎩2 r m r φ(m)2 ν(Q) if Q| ˜P2 r m r φ(m)2 ν(Q)−1 otherwise.For any Q|P , let S(Q) = {m ∈ N| (m, P ) = Q}. We have that N = ⋃ Q|P S(Q), andthe union is disjoint. Therefore,Σ = ∑ Q|P∑m∈S(Q)µ(2m)n 2m.Now divide the set of divisors of P into two sets; the divisors of ˜P , and its complement.It follows thatΣ = ∑ Q| ˜P12 r+1 ⎧⎪⎨⎪ ⎩∑Q| ˜P∑m∈S(Q)2 ν(Q) ∑µ(2m)2 ν(Q)2 r m r φ(m) + ∑ Q|PQ|/ ˜Pm∈S(Q)The sum over m ∈ S(Q) is easy to evaluate,∑m∈S(Q)µ(2m)m r φ(m) = −(−1)ν(Q) Q r φ(Q)Substituting we get:Σ = −A(r)2 r+1 r∏i=1∑(m,2P )=1(1 − 1α i + 1∑m∈S(Q)µ(2m)2 ν(Q)−12 r m r φ(m)⎫µ(2m)m r φ(m) + ∑ 2 ν(Q) ∑ µ(2m)⎪⎬mQ|P m∈S(Q)r φ(m) ⎪⎭ .µ(m)r∏(m r φ(m) = −(−1)ν(Q) Q r φ(Q) A(r) 1 − 1 ) −1.i=1α i + 1⎛) −1⎜∑⎝20Q| ˜P(−2) ν(Q)Q r φ(Q) + ∑ Q|P=⎞(−2) ν(Q) ⎟Q r ⎠ =φ(Q)

−A(r)2 r+1 r∏i=1⎛−A(r)⎜2 r+1 ⎝⎛( ) αi + 1⎜α⎝ ir∏i=1p i ≡1(4)The claim is therefore deduced.✷r∏i=1p i ≡1(4)(1 − 2 )+α i + 1r∏i=1(1 − 1 ) r∏ (1 + 1 )+α i i=1α ip i ≡3(4)⎞(1 − 2 )⎟α i + 1⎠ =r∏i=1⎞(1 − 1 )⎟α⎠ .iCorollary 1.9 Let {a i } i>1 be a sequence of odd primes and let δ r be the density ofthe set of primes p for which F ∗ p is generated by a 1 , . . . , a r , thenlim δ r = 1.✷r→∞Remark:The method just exposed can be easily extended to any set of r multiplicativelyindependent numbers which are pairwise coprime. The first step of the inductionin the general form is in [26]. It is also conceivable that for any infinite sequence ofmultiplicatively independent integers (that is a sequence of integers such that a i < a i+1and for any r, a 1 , . . . , a r are multiplicatively independent), one has that lim r→∞ δ r =1. Not being able to provide a proof of this property here, we will include it in thehypothesis when ever needed.21

1.3 The Main ProblemSuppose f(p) is a monotone function of p that tends to infinity with p and let {a n } n∈Nbe an infinite sequence of multiplicatively independent integers. LetΓ f,p = 〈a i(mod p)| 1 ≤ i ≤ f(p)〉.Question:Does a function exist f such that, Γ f,p = F ∗ p for almost all primes p ?Using Theorem 1.1, we can prove:Theorem 1.10 Let {a i } i∈Nsuch thatbe a sequence of multiplicatively independent integerslim δ Γ = 1r→∞(We noticed in the last section that when the a i ’s are all primes this is true) supposethe Generalized Riemann Hypothesis holds for the Dedekind function of the fieldQ(ζ l , a 1/l1 ), l prime, then for any monotone function f(p) that tends to infinity, wehave that#{p ≤ | Γ f,p = F ∗ p} ∼ π(x).Proof: Let us fix r ∈ N. For all but finitely many primes p, we have:Γ f,p ⊃ Γ p = 〈a 1 , . . . , a r 〉.ThereforeA = #{p ≤ x | Γ f,p = F ∗ p} ≥ #{p ≤ x | Γ p = F ∗ p} + O(1).From Theorem 1.1, we get:Alimx→∞ xlog x22≥ δ Γ .

Now let r tend to infinity and prove the statement.✷Our intention in the following Chapters is to prove statements of the type ofTheorem 1.10, restricting our assumptions only on the rate of growth of f. Forexample it is not difficult to prove:Proposition 1.11 Let f(p) = log 2+ɛ p then there exists a sequence of multiplicativelyindependent integers such that, for almost all primes p, Γ f,p = F ∗ p.Proof: This is a consequence of a Theorem of Burgess and Elliott (see [5]) onthe average of the least primitive root. They proved that if g(p) is the least primitiveroot, then for large x,π(x) −1 ∑ g(p) ≪ log 2 x(log log x) 2 .p≤xIf U is the number of primes up to x for which g(p) ≥ f(p), we get that:U log 2+ɛ x ≪∑g(p) + o ( π(x) log 2+ɛ x ) ≪ π(x) log 2 x(log log x) 2xlog x ≤p≤xwhich is equivalent to saying that for almost all primes g(p) ≤ f(p).Now let a i = p i be the i-th prime number, since g(p) ≤ log 2+ɛ p, every prime thatdivide g(p) is also less than f(p), therefore Γ f,p contains a primitive root for almostall primes p.✷23

2 ON THE ARTIN L-FUNCTIONS OF Q(ζ l , 2 1/l )2.1 IntroductionLet L = Q(ζ l , 2 1/l ) and G = Gal(L/Q). Ifτ : L −→ L and ν : L −→ L2 1/l ↦→ ζ l 2 1/l 2 1/l ↦→ 2 1/lζ l ↦→ ζ l ζ l ↦→ ζ g l(where g is a primitive root modulo p), then G is generated by τ and ν, more precisely,G =< τ, ν | τ l = ν l−1 = 1, ν −1 τν = τ g∗ >is a presentation (here g ∗ is any integer such that gg ∗ ≡ 1 mod p).Hence G is the semidirect product of a cyclic group of order l by a cyclic group oforder l − 1. Note also that τ generates the Galois group of L/Q(ζ l ) and the subgroupgenerated by ν has as fixed field, the non-Galois field K = Q(2 1/l ).For any t = 1, . . . , l − 1, the mapχ t : G → G, τ ↦→ 1, ν ↦→ e 2πitl−1is clearly a character and a quick computation shows that G has l conjugate classesand the remaining character of G can be calculated via the orthogonality relations.That is⎧(l − 1)⎪⎨If a = b = 0χ l (τ a ν b ) = 0 If b ≠ 0⎪⎩−1 If b = 0, and a ≠ 0.Note also that χ l is induced by any non-trivial character of the normal subgroupgenerated by τ and if φ t : ν ↦→ e 2πit/(l−1) , t < l − 1 is a character of the subgroup 〈ν〉,24

then ind G 〈ν〉φ t = χ t + χ l .Hence χ 1 , . . . , χ l is a complete list of the irreducible characters of G.Let us now take a step back and describe the concept of non-Abelian Artin L-function. Let E/F be a Galois extension and ρ a representation of Gal(E/F ), wedefine the Artin L-function of ρ to beL(s, ρ, E/F ) = ∏ ℘L ℘ (s)where, if ℘ does not ramify, the Artin symbol σ ℘ is the conjugacy class inGal(E/F ) determined by the Frobenius automorphism of the residue field of anyprime of E over ℘ (note also that, if ℘ does not ramify in E, then σ ℘ = {1} if andonly if ℘ splits completely in E) and L ℘ (s) is the characteristic polynomial of σ ℘evaluated at N(℘) −s , i.e.L ℘ (s) = det(I − N(℘) −s ρ(σ ℘ )) −1and, if ℘ is ramified, L ℘ (s) is the characteristic polynomial of the Frobenius elementat ℘ acting on the subspace fixed by the inertia group I ℘ evaluated at N(℘) −s .Simple arguments on the bounds of the eigenvalues of the representation showthat L(s, ρ, E/F ) converges absolutely for R(s) > 1. Since the determinant of amatrix is the product of its eigenvalues, we also have that:log L(s, ρ, E/F ) = ∑ ℘, mtr(ρ(σ ℘ ) m )m℘ msSometimes, we might indicate L(s, ρ, E/F ) by L(s, χ, E/F ) where χ is the characterof ρ.We describe here the basic properties of L-functions. For a more complete picture,see [33] Chapter XII.25

PROPERTIES:A) If Z F (s) is the usual Dedekind zeta function of the field F , thenZ F (s) = L(s, 1, E/F );B) If χ 1 , χ 2 are two characters of Gal(E/F ), thenL(s, χ 1 + χ 2 , E/F ) = L(s, χ 1 , E/F ) + L(s, χ 2 , E/F );C) If E ′ ⊃ E ⊃ F , where E ′ /F is also Galois, then any character of Gal(E/F ) canbe viewed as a character of Gal(E ′ /F ) (by composing Gal(E ′ /F ) − → Gal(E/F )C), and we haveL(s, χ, E/F ) = L(s, χ, E ′ /F );χ−→D) If E ⊃ E ′ ⊃ F then Gal(E/E ′ ) ⊂ Gal(E/F ), therefore any character χ ofGal(E/E ′ ) induces a character Ind(χ) of Gal(E/F ) and one has:L(s, Ind(χ), E/F ) = L(s, χ, E/E ′ );E) If E/F is an abelian extension then for every character χ, L(s, χ, E/F ) has anextension to an entire function and verifies a functional equation;F) If χreg is the character of the regular representation of Gal(E/F ), thenL(s, χreg, E/F ) = Z E (s).(This is a consequence of the fact that the regular representation is induced by thetrivial character on the trivial identity subgroup which is the Galois group of L/Ltherefore, D) and A) give this claim);G) The Brauer Theorem for characters, states that any character is equal to asum with integer coefficients of characters induced from elementary subgroups (see26

[46] Chapter X). By properties B), C), and D) this implies that any Artin L-functioncan be written as product of powers with integer exponents of entire functions, andtherefore, any such a function is certainly meromorphic. Artin had actually conjecturedthat these functions are always entire whenever χ does not contain the trivialcharacter;H) Whenever the Galois group of an extension has the property that every characteris induced by the character of an abelian subgroups (such characters are calledmonomials) then the Artin Conjecture holds for such an extension. This is the caseof nilpotent extensions (as well as supersolvable extensions).2.2 Artin L-functions of L/QThe Galois group G of L/Q is certainly supersolvable. Thus all the Artin L-functionsof G are entire and by the properties F), D) and A), we have the following factorization:Z L (s) = ζ(s)On the other hand, if K = Q(2 1/l ),( l−2)∏L(s, χ t , L/Q) L(s, χ l , L/Q) l−1 .t=1l−2 ∏Z L (s) = L(s, χ reg , L/K) = Z K (s) L(s, φ t , L/K) =t=1t=1l−2 ∏Z K (s) L(s, χ t , L/Q)L(s, χ l , L/Q),the last identity being obtained noticing that Ind(φ t ) = χ t + χ l and applying propertiesB) and D). Putting the two together, we getZ K (s) =Z L (s)( ∏l−2t=1 L(s, χ t , L/Q) ) L(s, χ l , L/Q) l−2= ζ(s) ( ∏ l−2t=1 L(s, χ t , L/Q) ) L(s, χ l , L/Q) l−1( ∏l−2t=1 L(s, χ t , L/Q) ) L(s, χ l , L/Q) l−2 = ζ(s)L(s, χ l , L/Q).27

Therefore the zeroes of L(s, χ l , L/Q) are in particular zeroes of Z K (s) andL(1, χ l , L/Q) = Res s=1 Z K (s) ≠ 0The identity also allows us to compute the functional equation for L(s, χ l , L/Q).It is indeed a classical result that if K is any number field, and( ) s r1F K (s) = A s Γ Γ(s)r 2Z K (s) (1)2(where A = 2 −r 2d 1/2K π −nK/2 , r 1 and r 2 are respectively the number of real and complexembeddings of K, d K is the absolute value of the discriminant of K and n K its degreeover Q) then F K (s) = F K (1 − s).In our case d K = 2 l−1 l l , n K = l, r 1 = 1, r 2 = (l − 1)/2, andZ K (s) = ζ(s)L(s, χ l , L/Q),so we get:( ) (l/2)s ( l sF K (s) = Γ Γ(s)π 2)(l−1)/2 ζ(s)L(s, χ l , L/Q).Using the fact that the value of π s/2 Γ ( )s2 ζ(s) does not change if we substitute swith 1 − s, we get that if( ) (l/2)slG(s) = π s/2 Γ(s) l−12 L(s, χl , L/Q) (2)πthen G(s) = G(1 − s), which is the functional equation.An asymmetric functional equation can also be deduced using the formula:which isΓ(s)Γ(1 − s) =πsin πs ;L(1 − s, χ l , L/Q) = ll(s−1/2)π (l−1)s (sin πs)(l−1)/2 Γ(s) l−1 L(s, χ l , L/Q) (3)28

These results can be used to determine all the zeroes of L(s, χ l , L/Q) outside of thecritical strip. Indeed L(s, χ l , L/Q) has just a zero of order l−12for s = 0, −1, −2, . . .and is non-vanishing elsewhere (outside the critical strip).We conclude this Section with a classical general result that we will use later (thisresult can be found in [31], in that version, though, are missing all the uniformityconditions which are necessary for subsequent applications).Lemma 2.1 Let K be a number field, n = [K : Q], d the absolute value of thediscriminant and let Z K (s) be its Dedekind zeta function.absolute numerical constant c 1 such that in the regionZ K (s) has no zeroes.σ ≥ 1 −c 1log d(t + 2) n , t ≥ 0There exists a positiveProof of Lemma 2.1: We will follow the classical proof for the Riemann Zetafunction (See. [6] §13). Let H K (s) = 1s(s − 1)F 2 K(s) where F K (s) has been definedin (1). H K (s) is an integral function of order 1, verifies H K (1 − s) = H K (s) = H K (¯s)and admits the following Weierstrass product expansion:H K (s) = e a+bs ∏ (1 − s )e s/ρ (4)ρ ρwhere the product is extended to all the non-trivial zeroes of Z K (s).Taking the logarithmic derivative and using the functional equation, we getH K′ (s) = b + ∑ ( 1H K ρ s − ρ + 1 )= −b − ∑ (1ρρ 1 − s − ¯ρ + 1¯ρ)= − H′ K(1 − ¯s).H KSince, if ρ is a root then also 1 − ¯ρ is, we deduce that(R b + 1 (∑ 1¯ρ)) 12 ρ + = 0,ρ29

thereforeR( ) H′K(s) = ∑ H K ρR( ) 1.s − ρSubstituting inside the real part of the logarithmic derivative of (1), we have theidentity:( ) (∑ 1 1R = Rρ s − ρ s + 1s − 1 + log A + r 1 Γ ′ (s/2)2 Γ(s/2) + r Γ ′ )(s)2Γ(s) + Z′ K(s). (5)Z K (s)Now consider this expression for s = σ, σ + it, σ + 2it,log(A) ≪ log d and sinceR( ) 1= σ − βs − ρ |s − ρ| > 0, 21 < σ ≤ 2, t ≥ 0. Sincethere exist three absolute positive constants c 2 , c 3 , c 4 such that if we take t = γ to bethe ordinate of the zero ρ = β + iγ, then−R Z′ K(σ)Z K (σ) < 1σ − 1 + c 2 log(d);−R Z′ K(σ + it)Z K (σ + it) < c 3 log(d(t + 2) n ) − 1σ − β ;−R Z′ K(σ + 2it)Z K (σ + 2it) < c 4 log(d(t + 2) n )because of the Stirling formula for the Gamma function. Finally the standard inequalityimplies3[A choice of σ = 1 +]− Z′ K(σ)Z K (σ)+ 4[]−R Z′ K(σ + it)Z K (σ + it)+[]−R Z′ K(σ + 2it)Z K (σ + 2it)4σ − β < 3σ − 1 + c 5 log(d(t + 2) n ).δlog(d(t+2) n )yields, for an opportune δβ < 1 −which is equivalent to the statement.✷c 1log(d(t + 2) n )30≥ 0

Proof of the Lemma: If N K (T ) and N(T ) are respectively the number ofzeroes of Z K (s) and ζ(s) in the region in question, then we have that N(T, χ l ) =N K (T ) − N(T ) and since, from the classical theory, we know thatN(T ) = T 2π log T 2π − T 2π+ O(log T ).It is enough to show thatN K (T ) = l T 2π log T 2π − l T 2π + ( log dK2π)T + O(log d K T l ).Also, in the same way as in the classical result, we can write( ∫H ′ )K4πN K (T ) = I (s)dsR H k(6)where H K (s) is the function defined during the proof of the lemma in the last Sectionand R is the rectangle, described counterclockwise, having as vertices:5/2 − iT, 5/2 + iT, −3/2 + iT, −3/2 − iT .Since H K (s) = 1 2 s(s − 1)F K, by the residue theorem, we can write that (6) is equalto( ∫IR( 1s + 1s − 1 + F K′ ) )( ∫F ′ )K(s) ds = 4π + I (s)dsF K R F KIf L denotes the line from 5/2 to 5/2+iT and then to 1 +iT , then using the functional2equation, we quickly get that (7) equals:∫F ′ ∫K4π + 4I (s)ds = 4π + 4T log A + 4IL F K L((7)r 1 /2 Γ′ (s/2)Γ(s/2) + r Γ ′ )(s)2Γ(s) + Z′ K(s)dsZ K (s)and, by the Stirling formula we know that( ∫Γ ′ )( )(s/2)1IL Γ(s/2) ds = 2I log Γ4 + iT = T log T 2 2 − T − π ( ) 14 − O ,Tand( ∫ILΓ ′ )( )( )(s)1 1Γ(s) ds = I log Γ2 + iT = T log T − T + O .T32

This yieldsN K (T ) = l T 2π log T 2π − l T 2π + ( log dK2π)T + 8 − r 18( ) ( l ∫+ O + 4IT LZ K(s)′ )Z K (s) ds .The problem amounts to proving that( ∫Z ′ )IK(s)L Z K (s) ds = O(log d K T l ). (8)Since Z K (s) is real on the reals, we have that ∆ L arg(Z K (s)) = arg Z K ( 1 2in the classical case:+ iT ). AsLemma 2.4 For all T > 0, we have that∑ρ11 + (T − γ) 2 = O(log d K(T + 2) l )where ρ runs over all the non-trivial zeroes of Z K (s).Proof of Lemma 2.4: From the same argument used in the proof of Lemma 2.1,we know that for 1 < σ ≤ 2 and t > 0, there exists an absolute positive constant c 0such that−R Z′ K(s)Z K (s) < c 0 log(d K (t + 2) l ) − ∑ ρSince for a choice of s = 2 + iT , we haveZ ′ K(2 + iT )∣Z K∣ ≪ log d K,we obtainFinally note that∑ρR 1σ − β =and prove the Lemma.✷R 1σ − β < c 1 log(d K (T + 2) l ).R 1σ − β .2 − β(2 − β) 2 + (T − γ) ≫ 12 1 + (T − γ) 2As in the classical case we have the following implications:33

Corollary 2.5 For T ≥ 2, we have thata) N(T + 1) − N(T − 1) = O(log d K T n );b)∑ 1(T − γ) = O(log d KT n ).✷2ρ,|γ−T |>1Now we are ready to prove (8). Take the identity (5) for s = σ + iT and s = 2 + iT ,subtract and get:Z K(σ ′ + iT )Z K (σ + iT ) = Z′ K(2 + iT )Z K (2 + iT ) + O(log T l ) + ∑ ρO(log d K T l ) +∑ρ,|T −γ| 1,1∣σ + iT − ρ − 12 + iT − ρ∣ ≤ 3|γ − T | ,andFinally,( ∫IL∑ρ,|T −γ|

Lemma 2.6 Let x be an integer and 2 ≤ T < x, thenψ(x, χ l ) = − ∑|γ|≤Tx ρ ( )xlρ + O T (log lxT )2 + l 2 log 2 l . (10)where the sum is extended over all those zeroes ρ whose imaginary part γ is in absolutevalue less or equal than TProof of Lemma 2.6: As in the classical case of the zeta function, ψ(x, χ l ) isthe sum of the coefficients of the logarithmic derivative, more precisely, if c > 1 andT is large, then the Lemma in §17 of [6] gives that if:then⎛(⎜∞∑( xψ(x, χ l ) = J(x, T )+O ⎝ Λ l (n)nJ(x, T ) = 1 ∫ [ ] c+iT− L′ (s) xs2πi c−iT L(s) s ds,n=1n≠x) c 1)T | log(x/n)|⎞+ cΛ l(x) ⎟⎠+O(l log x) (11)TIf we choose c = 1 + 1/ log x (x c = ex) and we treat the four ranges separately:(n ≤ 3 4 x, n ≥ 5 4 x) ( 3 4 x < n < x − 2) (x − 2 ≤ n ≤ x + 2) (x + 2 < n < 5 4 x).For the values of n, the first range, we have that |log(x/n)| ≫ 1 therefore thecontribution of these terms in the sum in(11) is≪ x Twhere we just noticed that for c > 1[ ]− L′ (c)≪ x l log l, (12)L(c) T( )− L′ (c) Z′L(c) = − K (c)Z K (c) − ζ′ (c)= 1ζ(c) c − 1 − 1c − 1 + O(log d K) = O(l log l),and that d K = 2 l−1 l l .35

For the values of n, the second range, set [x] − n = r, and note that 1 ≤ r ≤ x/4therefore( ( )x xlog = log ≥n)[x] − r(∣ logWe gather that the contribution of these terms in (11) is≪ x T∑1≤r≤x/4Λ l (x − r)r −1 ≪xl log xT1 − r )∣ ∣∣∣∣≥ r[x] 2x .∑1≤r≤x/4r −1 ≪ xl log2 xT(13)Analogously, for the values of n in the fourth range, set n − [x] = r ′ (now 2 ≤r ′ < x + 1) and thus4∣( )∣ (∣∣∣ x ∣∣∣log = − log 1 − n − x )≫ r nn xand the contribution of these terms in (11) is≪ xl log2 x. (14)TFinally for the (at most five) values of n in the third range, we have a contributionwhich is ≪ l log x. Putting this together with the estimates in (12), (13) and (14),we get:ψ(x, χ l ) = J(x, T ) + O( )xlT (log2 x + log l) . (15)Now we replace the vertical segment of integration by the other three sides of therectangle with verticesc − iT, c + iT, −U + iT, −U − iTwhere U is a large half integer (i.e. U = m/2, m odd integer). If T ≠ γ for any zeroρ = β + iγ of L(s) is the critical strip, the residue Theorem givesψ(x, χ l ) = − ∑ x ρρ − Res ( xsL ′ )(s)− l − 1 U∑ x −ms = 0 s L(s) 2|γ|

since by the functional equation in (2), the integrand in J(x, T ) has simple poles ats = ρ, poles of order (l − 1)/2 at s = −m (m ≥ 1), and an extra pole of order 2 ats = 0.The residue at s = 0 can be estimated as follows: Since L(s) = Z k (s)/ζ(s), andsince x s /s = 1/s + log x + . . ., we haveRes( xss = 0 sL ′ )(s)= Res ( 1L(s) s = 0 sZ K(s)′ )+ O(log x). (17)Z K (s)From the functional equation in (1) and the Weierstrass product expansion in (4) weget that(Res 1 Z ′ )K(s)= b − log A + 1 − (l/2) Γ′ (1)s = 0 s Z K (s)Γ(1)= b + O(l log l). (18)If we substitute s = 2 in (4), use the functional equation again and note thatZ ′ K(2)/Z K (2) ≪ l, we deduce thatb = ∑ ρ( 12 − ρ + 1 )+ O(l log l). (19)ρFor the terms of this series with |γ| ≥ 1, we have∑∣|γ|≥112 − ρ + 1 ρ∣ = 2 ∑|γ|≥11|ρ(2 − ρ)| ≪ ∑ ρ12≪ l log l. (20)|2 − ρ|The last sum being estimated as O(log d k ) using Lemma 2.4 with t = 0. The sameestimate applies to∑|γ|

and being the number of zeroes in question ≪ log d K by Lemma 2.3 with T = 1, wehave∑|γ|

provided that circles of radius 1/3 around the trivial zeroes s = −m of L(s) areexcluded.Using (23) and (25) we have∫ c±iT−U±iT≪ l log2 lTT∫ c−∞x σ dσ ≪ xl log2 lTT log x ,while (25) gives∫ −U+iT−U−iT≪ l log2 lUU∫ T−Tx U dt ≪ T l log2 lUUx U = o(1), (for U → ∞).Inserting these estimates in (16) we get the wanted claim.✷Proof of Theorem 2.2: The zero-free region proved for Z K (s) in Lemma 2.1holds also for L(s), therefore, if ρ = β + iγ is a zero of L(s) with γ < T < x we havethatβ < 1 −where c is an absolute positive constant.cl log lTWe gather that|x ρ | = x β < x exp(−c log xl log lT). (26)The sum ∑ γ

We minimize it by choosing T such thatand we get that (27) isl log 2 lT = log x,⎛ √ ⎞log x≪ xl exp ⎝−c/2 ⎠lwhich by partial summation is equivalent to statement.✷Remark: If we assume the strong Hypothesis that for any prime l, the Dedekindzeta function Z K (s) has the zero-free regionσ > 1 −clog T , T ≥ 0then, using exactly the same method we would be able to prove that uniformly forl < x( √ )π(x, χ l ) ≪ xl exp −c log x . (28)2.4 An Application to Chebotarev Density TheoremIn this Section we apply Theorem 2.2 to the Chebotarev Density Theorem, obtainingfor the special case of Q(ζ l , 2 1/l ) a stronger result than Lemma 1.2. This will be usedlater in Theorem 3.1, which is actually a motivation for such a result.Theorem 2.7 There exists a constant B such that uniformly for all l withwe havel

Proof Let χ G be the character of the regular representation. That is:We have that⎧⎪⎨ |G| = l(l − 1) if x = 1,χ G (x) =⎪⎩ 0 otherwise.1 ∑χ G (σ p ) = #{p ≤ x | σ p is trivial} = #{p ≤ x | p splits completely in L}.l(l − 1)p≤xOn the other hand, χ G = χ 1 +. . .+χ l−1 +(l −1)χ l is the canonical decompositionof the regular character, therefore:1 ∑χ G (σ p ) =l(l − 1)p≤x1 ∑l(l − 1)p≤xl−1 ∑i=1χ i (σ p ) + 1 l∑χ l (σ p ).p≤xThe orthogonality relations for the characters of the subgroup H < G give:⎧1 ∑l−1⎪⎨χ i (h) =l − 1i=1⎪⎩1 if h = 1,0 otherwise.Therefore, the first l − 1 characters count the number of primes up to x such thattheir Artin symbol is trivial modulo H. These are the primes that split completelyin the Cyclotomic field Q(ζ l ) (whose Galois group is isomorphic to H). Finally, ifπ(x, l, 1) = {p ≤ x | p ≡ 1 mod l}, thenP (x, l) = 1 l (π(x, l, 1) + π(x, χ l)) .The Siegel-Walfisz Theorem (see [6] in §22) states that given any positive constantC, if l ≤ (log x) C , one hasπ(x, l, 1) = #{p ≤ x | p ≡ 1 mod l } = 1l − 1 li(x) + O(x exp(−A √log x))for some constant A = A(C), uniformly in l.41

This result with C = 2 and Theorem 2.2 gives the wanted claim.✷We conclude with the following statement whose proof is a consequence of theRemark at the end of the last Section.Theorem 2.8 Assuming the strong Hypothesis that for any prime l and for any nontrivialzero β + iγ of the Dedekind zeta function Z K (s), β < 1 −positive constant C, uniformly for l ≤ (log x) C , we haveP (x, l) =for some constant A = A(C).✷( (1√ ))l(l − 1) li(x) + O x exp −A log x ,c , then given anylog γ42

3 ON THE NUMBER OF PRIMES GENERAT-ING F ∗ p3.1 Extending Hooley’s MethodIn this Section we will extend the ideas illustrated in Chapter 1 proving the followingTheorem 3.1 Suppose Γ p is the subgroup of F ∗ p generated by the classes of the firstlog p primes, letN(x) = #{p ≤ x | Γ p = F ∗ p}thenN(x)limx→∞ xlog x≥ 1 − log 2.̸Proof: If we assume p ≥ x 1/2 , then, for every x,Γ p ⊇ Γ r,p = 〈p 1 , . . . , p r 〉, with r = [ 1 log x]2andN(x) ≥ Ñ(x) = #{p ≤ x | Γ r,p = F ∗ p}. (1)Now, as in the standard Hooley’s case, we define for given η 1 and η 2 :Ñ(x, η 1 ) = #{p ≤ x | ∀l, l ≤ η 1 , l |[F ∗ p : Γ r,p ]};M(x, η 1 , η 2 ) = #{p ≤ x | ∃l, η 1 ≤ l ≤ η 2 , l|[F ∗ p : Γ r,p ]};M(x, η 2 ) = #{p ≤ x | ∃l, l ≥ η 2 , l|[F ∗ p : Γ r,p ]},and clearlyÑ(x) ≥ Ñ(x, ξ 1) − M(x, ξ 1 , ξ 2 ) − M(x, ξ 2 , ξ 3 ) − M(x, ξ 3 ) (2)where ξ 1 = 1/4 √ log log x, ξ 2 =log xB(log log x) 2fixed positive number to be chosen later.and ξ 3 = (log 2 x)(log log x) 2 and B is a43

• The last term of (2) can be treated as in Theorem 1.1 using Lemma 1.4. Wehave thatM(x, ξ 3 ) ≤ #{p ≤ x | |Γ r,p | ≤}x(log 2 x)(log log x) 2and since ∑ i log p i = O(p r ) = O(r log r), Lemma 1.4 givesM(x, ξ 3 )≪≪() 1/rxx(log 2 x)(log log x) 2 (log 2 r log rx)(log log x) 2x 1log x log log x . (3)• To handle the third term of (2), we will make use of the already quoted Siegel-Walfisz Theorem, which states that given any positive constant C, then if l ≤(log x) C , one hasπ(x, l, 1) = #{p ≤ x | p ≡ 1 mod l } = 1l − 1 li(x) + O(x exp(−A √log x)) (4)for some constant A = A(C), uniformly in l.This result yields to,M(x, ξ 2 , ξ 3 ) ≤ #{p ≤ x | ∃l, ξ 2 < l < ξ 3 , p ≡ 1 mod l } ≤ ∑=where we have chosen C = 3 say.∑( 1√ )ξ 2

Using this result in (5), we getM(x, ξ 2 , ξ 3 ) ≤ li(x)(log( )) (log ξ3√ )+ O((log ξ 2 ) −1 ) + O ξ 3 x exp −c log xlog ξ 2≤x (()) ( )1 + log log log x/ log log xxlog 2 + log+ olog x1 − (log B + 2 log log log x)/ log log x log x= x( ) xlog x log 2 + o (7)log x• Theorem 2.7 in Chapter 2 is the ingredient to the estimate of the second termof (2). Indeed, l < ξ 2 yields toM(x, ξ 1 , ξ 2 , )∑≤ #{p ≤ x | p splits completely in Q(ζ l , 2 1/l )}ξ 1

Finally, the Hensel inequality (see (4) of Section 1.1) givesc n 1/2m log d m ≤ m 3 2 t+1 (log m + ∑ √log p i ) ≤ m 2t ≤ e 2ξ1t ≤ log x.i≤tTherefore the Chebotarev Density Theorem (see Lemma 1.2 of Section 1.1)gives=∞∑m=1N 0 (x, ξ 1 ) =µ(m)n m⎛⎛ ⎛∗∑µ(m) ⎝ 1√ ⎞⎞⎞log xli(x) + O ⎝x exp ⎝−A ⎠⎠⎠m n m n m⎛xlog x + O ⎝ ∑m>ξ 11⎞ ( ( √ ))m li(x) ⎠ + O e ξ 1 log xx exp −At+1 (log x) 1/2( )x 1= δ Γlog x + O x, (9)ξ 1 log xwhere δ Γ is as in Section 1.1 and where we used the fact that in this range ofm, n m ≤ e 2tξ 1≤ √ log x.Putting together the estimates (3), (7), (8) and (9) and using (1) and (2), we getN(x) ≥ N 0 (x, ξ 1 ) − M(x, ξ 1 , ξ 2 ) − M(x, ξ 2 , ξ 3 ) − M(x, ξ 3 )( )x x≥ δ Γlog x + o − log 2x ( ) xlog x log x + o .log xTherefore, by Corollary 1.9N(x)limx/ log x ≥ lim (δ Γ − log 2 + o(1)) ≥ 1 − log 2x→∞x→∞which is the wanted claim.✷The estimate in (8) is the real obstacle to achieve an asymptotic formula for N(x).Such estimate is connected with the range of validity of the Chebotarev DensityTheorem of the field K = Q(ζ l , 2 1/l ). As we have seen in Chapter 2, such a rangedepends on the determination of zero-free regions for L(s, χ l ) and thus of the Dedekindzeta function Z K (s). Indeed we have46

Theorem 3.2 Assuming in the strong Hypothesis that for any prime l and for anynon-trivial zero β + iγ of the Dedekind zeta function Z K (s), β < 1 −almost all primes p, F ∗ p is generated by the first [2 log p] primes.c , then forlog TProof. Using exactly the same notation of Theorem 3.1, we now choose ξ 1 =1/4 √ log log x, ξ 2 = log 2 x and ξ 3 = (log 2 x)(log log x) 2 .The estimate of the main and last terms in (2) is the same, for the third term,again we use the Siegel-Walfisz Theorem (4) and Merten’s formula (6),M(x, ξ 2 , ξ 3 ) ≤ ∑ ( ( (1√ )))ξ 2

Finally note that using this approach improvements, in terms of determining asignificantly smaller set of generators of F ∗ p for almost all p, are not possible. Thechoice r = log p is in fact imposed by the statement of Lemma 1.4. The next Sectionis devoted to analyzing this aspect in a more detailed manner.3.2 Relaxation of the Hypothesis and ImprovementsOur first intention is to prove a version of Theorem 3.1 in which the number ofgenerators is optimal with respect to the method used. As we have already remarked,the choice of the minimal number of generators of F ∗ p for a positive proportion of p’sis imposed by Lemma 1.4. PreciselyTheorem 3.3 a) Let f be a (monotone) function of p with f(p) → +∞ for p → ∞and let Γ p be the subgroup of F ∗ p generated by the classes of the firstlog pf(p)log log pprimes, then for a set of primes of density greater than 1 − log 2, we have Γ p = F ∗ p.b) Let α be a real number with 0 < α < e − 2 and let Γ ′ p be the subgroup of F ∗ pgenerated by the classes of the firstlog pα log log pprimes, then for a set of primes of density greater than 1 − log(2 + α), it follows thatΓ ′ p = F ∗ p.Proof: a) The proof starts in the same way as in Theorem 3.1, where we assumedp ≥ x 1/2 and noticed that, for every x,Γ p ⊇ Γ r,p =< p 1 , . . . , p r >, with r =48[ f(x)2]log xlog log x

and#{p ≤ x | Γ p = F ∗ p} ≥ Ñ(x, ξ 1) − M(x, ξ 1 , ξ 2 ) − M(x, ξ 2 , ξ 3 ) − M(x, ξ 3 ) (10)where the notations, ξ 1 and ξ 2 are the same as in Theorem 3.1 and ξ 3 will be chosenlater.The estimate of the main term and the second term are exactly the same, whilethis time the estimate of the last term of (10) using Lemma 1.4 is the following:M(x, ξ 3 ) ≤ #{p ≤ x | |Γ r,p | ≤ x }≪ x ( ) 1/rxr log rξ 3 ξ 3 ξ 3≪x ξ 3(log x) 2/f(x) f(x) log x ≪x(log x)1+ɛ(x)ξ 3(11)where we have put ɛ = ɛ(x) = 2/f(x) + (log f)/ log log x and assumed that f(x) ≪log log x, say.If we now choose ξ 3 = (log x) 2+ɛ log log x, we get that (11) is≪xlog x log log x .Finally we deal with the third term similarly as we did in Theorem 3.1, using theSiegel-Walfisz Theorem and the Merten’s Formula:M(x, ξ 2 , ξ 3 ) ≤ ∑ ( 1√ )ξ 2

) In this case we need to be a little more careful with the definition of r. Wecan assume that p ≥ x 1−ɛ(x) where ɛ(x) is a given function which is o(therefore we defineand the rest is as in a).The last term is{p ≤ x | |Γ ′ r,p| ≤ x }ξ 3M(x, ξ 3 ) ≤ #[]log xr = (1 − ɛ(x))α log log xlog log x), andlog x≪ x ξ 3( xξ 3) 1/rr log r ≪ x ξ 3(log x) 1+α/(1−ɛ(x)) .and choosing ξ 3 = (log x) 2+α/(1−ɛ(x)) log log x, we would make it o(x/ log x). Finallythe estimate for the third term is:M(x, ξ 2 , ξ 3 ) ≤ li(x)≤=(log(xlog x log 2 +and this completes the proof.✷( ) log ξ3log ξ 2+ O((log ξ 2 ) −1 ))α1 − ɛ(x)+ o( )xxlog x log(2 + α) + o log xNow we turn our attention to another aspect.)( x)log x(√ )+ O ξ 3 x exp −c log xNote that neither the proof ofTheorem 3.1 nor the one of Theorem 3.3 use in any way the fact that each Γ p isgenerated by the first [log p] primes except for the fact that the sum of their logarithmsis ≪ log p log log p and that lim r→∞ δ Γ = 1. The statement remains true if we considera sequence a 1 , a 2 . . . of multiplicatively independent integers such that for any r,r∑i=1log a 1 ≪ r log r and lim r→∞δ Γ = 1.It is conceivable to ask if a choice of a 1 , . . . , a r exists for which we could prove astronger Theorem. That would amount to having a better estimate for the sum of50

the logarithms. For this purpose one could setΥ(r) = min{ r∑log a i | a 1 , . . . , a r ,i=0multiplicatively independent r−tuple}.The following holds:Proposition 3.4Υ(r) = r log r + O(r).Proof: For any multiplicatively independent a 1 , . . . , a r , we can assumea 1 ≥ 1, . . . , a r ≥ r, thereforer∑r∑log a i ≥ log i = log r! = r log r − r + O(log r)i=1i=1the last identity, by the Stirling formula. ThusΥ(r) ≥ r log r + O(r).The choice a 1 = 2, . . . , a r= p r , the r th prime, and the Prime Number Theorem,prove thatΥ(r) ≤r∑i=0√√log p r = p r + O(p r exp −A log p r ) = r log r + O(r exp −A log r).✷Although many of the results that we will state can be extended to any sequence ofmultiplicatively independent integers, from now on we will only consider the sequenceof the prime numbers.It is now clear that the problem amounts to estimating the number of those primesp ≤ x for which [F ∗ p : Γ r ] has a prime divisor in the range (log 1−ɛ 1(x) x, log 2+ɛ 2(x) x)where ɛ i (x) = o(1).We note that it is enough to restrict our attention to thoseprimes for which the prime divisor is in (log x, log 2 x), since by the argument wealready used more than once (the Siegel-Walfisz Theorem and the Mertens Formula),51

that the number of primes p for which p − 1 has a prime divisor in (log 1−ɛ1(x) x, log x)or in (log 2 x, log 2+ɛ2(x) x) is o(π(x)).We can also assume that [F ∗ p : Γ r ] has exactly one prime divisor in (log x, log 2 x).Indeed, if p is a prime in the set under consideration for which this is not the case,we would have∣ ∣∣[F ∗∃l 1 , l 2 , l 1 l 2 p : Γ r ] ∣ =⇒ |Γ p | ≤ p − 1 ≤ xl 1 l 2 log 2 xand an application of Lemma 1.4 shows that the number of such primes is o(π(x)).Finally for the same reason we can assume no divisors of [F ∗ p : Γ r ] are > log 2 x.Putting these remarks together, we have the followingProposition 3.5 With the same notation of Theorem 3.3, for almost all primes pup to x eitherF ∗ p = Γ ror the index [F ∗ p : Γ r ] has exactly one prime divisor in the range (log x, log 2 x) and nodivisor > log 2 x, i.e.N(x) =wherex( ) xlog x − A(x) + o ,log xA(x) = { p ≤ x | ∃!l ∈ ( log x, log 2 x ) , l|[F ∗ p : Γ r ], and [F ∗ p : Γ r ] ≤ log 2 x } .✷This fact will be used in Section 3.3. We conclude the Section mentioning howthis argument can be extended to the case of any r → ∞, in particular the followingholds:Theorem 3.6 Let r be a function of p that tends to ∞, then for almost all primesp, either Γ r = F ∗ p or∃!l, l|[F ∗ p : Γ r ] with l ∈ ( log p, rp 1/r log p ) ,and [F ∗ p : Γ r ] < rp 1/r log p.✷52

Corollary 3.7 Let α > 0 be fixed. For almost all primes p either F ∗ p is generated bythe classes of the first 1 αlog plog log pprimes, or∃!l, l|[F ∗ p : Γ p ] with l ∈ ( log p, log 2+α p ) ,and [F ∗ p : Γ p ] < log 2+α p.✷As we have seen in Proposition 3.5, the problem is now to determine upper boundsfor the quantity:A(x) = { p ≤ x | ∃!l ∈ ( log x, log 2 x ) , l|[F ∗ p : Γ r ], and [F ∗ p : Γ r ] ≤ log 2 x }where we can suppose r ≫ log x. We already noticed that the Siegel-Walfisz Theoremand the Merten’s Formula, give:A(x) ≤ log 2x ( ) xlog x + o .log xThe idea that allows one to improve this upper bound is coming from the Brun’sSieve, more precisely we will use the following result:Lemma 3.8 Let B n (x) be the number of primes up to x for which p−1 is not divisibleby any of the primes in the interval (log x, log n x), then we haveB n (x) ∼ 1 n π(x).Proof: It is an application of the version of the Brun’s Sieve that is on Theorem2.5 ′ at page 83 of [18] to the set:A = {p − 1 | p is primes, p ≤ x}.Hypothesis (R 0 ) and (R 1 (k, a)) are easily satisfied, the latter using the Bombieri-VinogradovTheorem.✷The application to our problem with the following:53

Corollary 3.9 With the same notation as above, for any r,A(x) ≤ 1 ( )x x2 log x + o .log xProof: We have thatA(x) ≤ { p ≤ x | ∃l ∈ ( log x, log 2 x ) , l|[F ∗ p : Γ r ] } = π(x) − B 2 (x) ∼ 1 2 π(x).✷We conclude withTheorem 3.10 a) Let f be a (monotone) function of p with f(p) → +∞ for p → ∞and let Γ p be the subgroup of F ∗ p generated by the classes of the firstlog pf(p)log log pprimes, then for at least half of the primes p, we have Γ p = F ∗ p.b) Let α be a real number with 0 < α and let Γ ′ p be the subgroup of F ∗ p generatedby the classes of the firstlog pα log log pprimes, then for a set of primes of density greater than 12+α , it results Γ′ p = F ∗ p.✷Proof of b): The proof goes as the one in Theorem 3.3, except that to estimatethe third term we make use of Lemma 3.8 with n = 2 + α.✷54

3.3 A Density One ResultWe will discuss in this section a method to find an estimate for the size of the set{H m,r (x) = p ≤ x | |Γ r | = p − 1 }mwhere m is a given integer greater than one and r a function of p that tends to infinity.The idea is that Γ r is a subgroup of the cyclic group F ∗ p and therefore is itselfcyclic. For any integer m, m ≡ 1( mod p), the subgroup of m-th powers is a subgroupof F ∗ p of order (p−1)/m and since a finite cyclic group has a unique subgroup for everydivisor of its order, we deduce that Γ r is the group of m-th powers mod p. Since agroup is made out of m-th powers if and only if it is generated by m-th powers, thisimplies:H m,r (x) = {p ≤ x | p ≡ 1(modm) and p i is an m-th power (modp) ∀i = 1, . . . , r} .If n m (p) is the least prime which is not congruent to an m-th power (modp), thenwe can also write:H m,r (x) = {p ≤ x | p ≡ 1(modm) and n m (p) > p r } .As r grows, the possibility that all the p i ’s are m-th powers becomes less probable.The idea is to find the minimum r such that H m,r (x) is o( 1 π(x)). We will do thismmaking use of the large sieve, the proof of which can be found in [6] or [2], that is:Lemma 3.11 (The Large Sieve)Let N be a set of integers contained in the interval {1, . . . , z} and for any primep ≤ x, let Ω p = {h(modp) | ∀n ∈ N , n ≢ h(modp)} andL = ∑ µ 2 (q) ∏q≤x p|q55|Ω p |p − |Ω p | ,

then|N | ≤ z + 3x2 .✷LIn our case, let N = {n ≤ z | ∀q|n, q < p r } and note that if p ∈ H m,r (x), thenΩ p ⊃ {h(modp) | h is not an m-th power (modp)}therefore, for such p’s, |Ω p | ≥ p − 1 − (p − 1)/m andL ≥∑p∈H m,r(x)|Ω p |p − |Ω p | ≥ m − 1 |H m,r (x)|.2Applying the Large Sieve with z = x 2 , we get:Theorem 3.12 Let Ψ(s, t) = #{n ≤ s‖ ∀q|n, q < t}, thenH m,r (x) ≤8x 2(m − 1)Ψ(x 2 , p r ) .✷Estimating the function Ψ(z, y) is a classical problem in Number Theory. In 1985,D. Hensley proved (see [22]) the following:Lemma 3.13 Let u = log zlog y⎧⎪⎨⎪⎩and let ρ(u) be the function determined by:ρ(u) = 1 if 0 ≤ u ≤ 1;uρ ′ (u) = −ρ(u − 1) if u > 1,then, for 1 + log log z ≤ log y ≤ (log log z) 2 , and ɛ > 0,Ψ(z, y) ≫ zρ(u) exp(−u exp(−(log y) (3/5−ɛ) )).✷In our case, this gives:56

Corollary 3.14 Let r be a function of x such that log p r ∈ [1+log log x 2 , (log log x 2 ) 2 ]and let u = 2 log x/ log p r , thenH m,r (x) ≪ m −1 1ρ(u) exp ( u exp ( −u exp ( − log (3/5−ɛ) p r))).✷An asymptotic formula for ρ(u) was found by de Bruijn in [10] and is the following:Lemma 3.15 Let u > 0, then{ (ρ(u) = exp−ulog u + log log u − 1 +In our case we get the following:(log log u) − 1log u( ))}(log log u)2+ Olog 2 .✷uTheorem 3.16 If p r ≥ log 2 x thenH m,r (x) ≪ 1 mexp {xlog x2 log log x} = o(π(x)).Proof: From Corollary 3.14 and the asymptotic formula of Lemma 3.15, we canwrite the estimate:H m,r (x) ≪ 1 ( ) log log um exp(u(log u + log 2 u − 1 + O)).log uwhere u = 2 log xlog p r. Now, take p r ≥ log 2 x, and note thatThereforeu = log x ; log u = log log 2 x 2 x − log 3 x; log 2 u = log 3 x + log ( 1 − log 3 x )log 2 xand log u ≍ log 2 x; log 2 u ≤ log 3 x − log 3 xlog 2 x ≍ log 3 x.mH m,r (x) ≪ expexp{log x{ ( log xloglog 2 x 2 x − log 3 x + (log 3 x − log ( ))}3 xlog 2 x ) − 1 + O log3 x≪log 2 x(1 − 1 ( ))}log 2 x + O log3 xlog 2 2 x57≪exp {xlog x2 log log x} = o(π(x)).✷

Remark:The choice of p r = log 2 x is not optimal in Theorem 3.16. A simple but long calculationshows that if p r = ( )log x 2,e then the asymptotic formula on Lemma 3.15 givesthe estimate(1ρ(u) ≪ x 1+O ( log 2 3 xlog 3 2)),xwhich is useless to our purpose. However, if we fix δ < 1 and set p r = ( )log x 2,e thenδthe same calculation gives(1ρ(u) ≪ x 1− 1−δlog 2 x − 1−δ2 +O( log 2log 2 3 x2 x log 3 2)),xwhich is a valid estimate and is the optimal one.We are now ready to prove:Theorem 3.17 Let Γ r = 〈p 1 , . . . , p r 〉 be the subgroup of F ∗ p generated by all theprimes up to p r = log 2 p (r ∼ log2 p), then for almost all primes p,log log pΓ r = F ∗ pProof: We want to estimate the size of the setS = { p ≤ x | [ F ∗ p : Γ r]> 1}.The index [ F ∗ p : Γ r]can be at most x since it is a divisor of p − 1.Since we may suppose p > x 1−ɛ , we can take p r = log 2 p > (1 − ɛ) 2 log 2 x andapply Theorem 3.16, to S,Remarks:(x∑x∑)1S = H m,r (x) =m=1m=1m exp {58xlog x2 log log x} = o(π(x)).✷

a) This is an improvement with respect to the result of Burgess and Elliot of [5]deduced in Proposition 1.11 where for almost all primes ≤ x, the size of p r (leastprimitive root) was proven to be ≥ log 2 x(log log x) 4 ;b) Improvements to this result using this approach do not stay in the possibility toapply a stronger version of Lemma 3.13 (which exists in the literature, see for examplethe work of A. Hildebrand in [23] or Canfield, Erdös and Pomerance in [8]). It is theasymptotic formula of De Bruijn for the function ρ(u) that forces a choice p r of sizeclose to log 2 x.59

4 MORE ON PRIMITIVE ROOTSThis Chapter is devoted to the problem of finding a uniform asymptotic formula forthe number of primes p up to x such that s distinct numbers (which for simplicitywe suppose to be prime) are all at the same time primitive roots (modp). It turnsout that, under the assumption of the G.R.H., there is always a positive density ofprimes with such a property.Heuristically, the probability that a prime l divides one the indexes [F ∗ p : 〈p 1 〉]or [F ∗ p : 〈p 2 〉] is the probability that p splits completely in the fields Q(ζ l , p 1/l1 )and Q(ζ l , p 1/l2 ), minus the probability that p splits completely in the compositumQ(ζ l , p 1/l1 , p 1/l2 ). By the Chebotarev density Theorem, we get that the probabilitythat l does not divide both the indexes is()11 −[Q(ζ l , p 1/l1 ) : Q] + 1[Q(ζ l , p 1/l2 ) : Q] − 1.[Q(ζ l , p 1/l1 , p 1/l2 ) : Q]Multiplying for all primes l, we get the formula:δ =∏l prime(1 − 2l − 1 ).l 2 (l − 1)A natural generalization of this argument to the case of r distinct primes with theapplication of an inclusion exclusion principle, yields to:δ =∏l prime(1 − 1 [ (1 − 1 − 1 ) s ]).l − 1 lWe will prove that this heuristic argument is correct with the assumption of theGeneralized Riemann Hypothesis and some adjustments of the same type of thosenoticed by Lehmer in the case of the Artin Conjecture for primitive roots. This willbe applied in Section 4.3 to the problem of determining the least prime primitive rootmodp for almost all primes p.60

4.1 Another Generalization of Hooley’s TheoremWe present in this section a very natural generalization of Hooley’s Theorem forprimitive roots.Theorem 4.1 Let P = {p 1 , . . . , p s } be a set of odd primes, L(d 1 , . . . , d s ) be thecompositum field:L(d 1 , . . . , d s ) =n d1 ,...,d s= [L(d 1 , . . . , d s ) : Q], and letDefineδ P =∞∑d 1 =1,...,d s=1s∏i=1Q(ζ di , p 1/d ii ),µ(d 1 ) · · · µ(d s )n d1 ,...,d s.N P (x) = #{p ≤ x | ∀i = 1, . . . , s p i is a primitive root(mod p)}.Then, if the Generalized Riemann Hypothesis holds for the fields L(d 1 , . . . , d s ), wehave that( )x x log log xs∑N P (x) = δ Plog x + O log 2 x cs 0 log p ii=1for some absolute constant c 0 , uniformly respect to s = |P| and p 1 , . . . , p s .Remark: It is not straightforward to see that δ P is well defined nor that it is nonzero. Indeed, the series defining the density, converges absolutely. We will assume itfor the moment and prove it in the next section in Corollary 4.12.Proof: We will follow the same approach of Hooley who first noticed that inorder p 1 , . . . , p s be all primitive roots for the same prime p, one has to have that∀l prime, l ̸ |[F ∗ p : 〈p i 〉], ∀i = 1, . . . , s,thereforeN P (x) = N P (x, y) + M P (x, y)61

whereN P (x, y) = # { p ≤ x | ∀l prime, l < y and ∀i = 1, . . . , s l ̸ |[F ∗ p : 〈p i 〉] }andM P (x, y) = # { p ≤ x | ∃l prime, l > y with l|[F ∗ p : 〈p i 〉], for some i = 1, . . . , s } .We choose y = 1 log x for a reason that will become clear later.6sx log log xs∑Step 1): M P (x, y) ≪log 2 xs log p i .i=1Clearlys∑M P (x, y) ≤ M {pi }(x, y),i=1therefore it is enough to show that, uniformly with respect to P,M {q} (x, y) ≪x log log xlog 2 s log q.xThis was proven already by Hooley in his original paper [26], and we will report ithere just for completeness.Note thatM {q} (x, y) ≤ |A| + |B| + |C|,whereA = { p ∈ M {q} (x, y)| ∃l|[F ∗ p : 〈q〉], l > x 1/2 log x } ;B = { p ∈ M {q} (x, y)| ∃l|[F ∗ p : 〈q〉], x1/2 < l ≤ log 2 x x1/2 log x } ;C = { }p ∈ M {q} (x, y)| ∃l|[F ∗ p : 〈q〉], y ≤ l ≤ x1/2log 2 x .|A| can be estimated as follows:l|[F ∗ p : 〈q〉] implies thatq p−1l ≡ 1 (mod p).62

Therefore, since l > x 1/2 log x and p ≤ x, any prime in A must divide the positiveproduct∏m≤x 1/2 log −1 x(q m − 1).Now note that the number of divisors of a natural number N is O(log N), therefore|A| ≪∑m≤x 1/2 log −1 x|B| can be estimated as follows:m log q ≪xlog 2 log q.xRetaining only the condition l|p − 1 for the primes p ∈ B, we get|B| ≤∑x 1/2log 2 x

If we use this formula we get:|C| =∑y≤l≤ x1/2log 2 x1l(l − 1) li(x) + O(x1/2 log x · l · q).P (x, l) =( 1Oy∑y≤l≤ x1/2log 2 x[)x+ Olog x]1l(l − 1) li(x) + O(x1/2 log x · l · q)( xlog 2 x log q ).Taking into account that y = 1 log x, we get6s( ) x|C| = Olog 2 x s log q ,which is the desired estimate.Step 2): We can now turn our attention to N P (x, y). We claim thatN P (x, y) =∗∑a 1· · ·∗∑a sµ(a 1 ) · · · µ(a s )P (x, a 1 , . . . , a s ) (1)where the ∗ over the sums means that the sums are extended to those values of a i forwhich all its prime divisors are less than y (note that this implies a i < e 2y = x 1/3s ),andP (x, a 1 , . . . , a s ) = # { p ≤ x | a i∣ ∣∣[F ∗p : 〈p i 〉] ∀i = 1, . . . , s } .This claim can be proven by induction on s: If s = 1 then we have the standardinclusion-exclusion principle:N {p1 }(x, y) = π(x) − ∑ P (x, l) +l

and the recursive relation holds;P t (x, a t+1 , . . . , a s ) =∗∑a tµ(a t )P t−1 (x, a t , . . . , a s ).Hence inductivelyN P (x, y) = P s (x) =∗∑a sµ(a s ) . . .∗∑a 1µ(a 1 )P 0 (x, a 1 , . . . , a s ).Now note that the condition a i∣ ∣∣[F ∗p : 〈p i 〉] is equivalent top splits completely in Q(ζ ai , p 1/a ii ),and that a prime splits completely in a set of fields if and only if it splits completelyin their compositum.Therefore a i∣ ∣∣[F ∗p : 〈p i 〉] for all i = 1, . . . , s, if and only if p splits completely inWe gather thats∏i=1Q(ζ ai , p 1/a ii ) = Q(ζ [a1 ,...,a s], p 1/a 11 , . . . , p 1/ass ).P (x, a 1 , . . . , a s ) = # {p ≤ x | p splits completely in L(a 1 , . . . , a s )}and the Generalized Riemann Hypothesis allows us to write the Chebotarev Densityformula:P (x, a 1 , . . . , a s ) =(1li(x) + O(x 1/2 log x + log D ))a 1 ,...,a s.n a1 ,...,a sn a1 ,...,a swhere D a1 ,...,a sis the discriminant of L(a 1 , . . . , a s ). Recall that by the Hensel Inequality(See page 259 of [42]) we can writelog D a1 ,...,a sn a1 ,...,a s≤ log[a 1 , . . . , a s ] +s∑log p i ,i=165

since only p 1 , . . . , p s and the primes dividing [a 1 , . . . , a s ] ramify in L(a 1 , . . . , a s ). Nowsubstitute inside (1) and get:N P (x, y) =∗∑a 1· · ·∗∑a sµ(a 1 ) · · · µ(a s )⎛δ P li(x) + O ⎝{ li(x)∑n a1 ,...,a s+ O(x 1/2 (⎞µ 2 (a 1 ) · · · µ 2 (a s )⎠ li(x)+n a1 ,...,a slog x + log D ))}a 1 ,...,a s=n a1 ,...,a s(a 1 ,...,a s)∈S⎛(∑s∑O ⎝x 1/2 log x + log p 1 + log[a 1 , . . . , a s ]) ⎞ ⎠ (2)a 1

4.2 Calculation of the DensitiesWe can now give the expression for the density δ P , and as we did in Section 1.2, thefirst step is to calculate the dimension of the fieldsL = L a1 ,...,a s=s∏i=1Q ( )ζ ai , p 1/a i = Q(ζ[a1 ,...,a s], p 1/a 11 , . . . , p 1/ass ),for any s-tuple a 1 , . . . , a s of square-free integers. This is done in the following:Theorem 4.2 Let n = n a1 ,...,a sproduct of those p i such that p i |M and a i is even.factors of P , thenwith= [L : Q] , M = [a 1 , . . . , a s ] and suppose P is then = φ(M)a 1 · · · a s2 α ,⎧⎪⎨ t if ∀q|P, q ≡ 1 (mod 4),α =⎪⎩ t − 1 if ∃q|P, with q ≡ 3 (mod 4).Let t be the number of primeProof: The argument is similar to the one in the proof of Theorem 1.5. We cansuppose, with out loss of generality, that P = p 1 · · · p t , and defineC 0 = Q(ζ M ),C i = C i−1 (p 1/a ii ).Clearly L = C s and[L : Q] = [C s : C s−1 ] · · · [C 2 : C 1 ]φ(M).Step 1): For 1 ≤ i ≤ s, it results [C i : C i−1 ] = a i or a i /2 .Since x a 1− p i splits into linear factors over C i−1 , we have that [C i : C i−1 ] = a i, if q|ddis a prime, then we have the fields:C i−1 ⊆ C i−1 (p 1/qi ) ⊆ C i .67

Now, either q = 1 or[C i−1 (p 1/qi ) : C i−1 ] |[C i : C i−1 ] = a idand since a i is square-free, we deduce that p 1 qi∈ C i−1 . Either p i ∈ C 0 which implyq = 2 since the only subfields of a cyclotomic field of the type Q(p 1/qi ) are quadratic,or C 0 (p 1/qi ) is a Kummer extension of degree q of C 0 . Now by Galois Theory, we getthat such an extension has to be of the following type:C 0 (p 1/qi ) = C 0((ps1 · · · p sk ) 1/q) ,where 1 ≤ s 1 ≤ s 2 · · · ≤ s k ≤ i − 1 Finally, the Theory of Kummer extensions, (SeeLemma 3 at page 160 of Cassels and Fröhlich [7]) implies that there exists 0 ≤ i ≤ q−1such that () 1/qp i∈ C(p s1 · · · p sk ) i 0 ,which again implies q = 2.Step 2): [C i : C i−1 ] = a i for t < i ≤ s.In the case a i odd then clearly Step 1) implies Step 2), thus suppose a i is even and[C i : C i−1 ] = a i /2. In this case, we have that √ p i is in C i−1 because a i is squarefree([C i−1 ( √ p i ) : C i−1 ] |a i /2). This implies that p i ramifies in C i−1 , but since, bythe Kummer Theory, the only primes that ramify in C i−1 are p 1 , . . . , p i−1 and thosedividing M, we get a contradiction and conclude that [C i : C i−1 ] = a i .This also implies that [C s : C t ] = a t+1 · · · a s .Step 3): If every prime dividing P is ≡ 1( mod 4), then [C i : C i−1 ] = a i /2 for every1 ≤ i ≤ t.From the Theory of Cyclotomic fields we know that a generic quadratic subfield ofC 0 = Q(ζ M ) is of the following type:⎛√ (−1⎞)Q ⎝ D⎠ , where D |M ,D68

since ( )−1√q = 1 if and only if q ≡ 1(mod4), we deduce that qi ∈ C 0 and the Galoisgroup of C i over C i−1 is generated by the mapσ : p 1/a ii↦→ ζ 2 a ip 1/a ii(note that σ( √ p i ) = (σ(p 1/a ii )) a i/2 = √ p i ), which has clearly order a i /2.In this case we have [C t : C 0 ] = a 1 · · · a t /2 t .Step 4): If it exists q |P with q ≡ 3( mod 4) (we assume, without loss of generality,that q = p 1 ), then [C 1 : C 0 ] = a 1 and [C i : C i−1 ] = a i /2 for every 1 < i ≤ t.The assumption [C 1 : C 0 ] = a 1 /2 would imply again that √ p 1 ∈ C 0 . By the sameargument of Step 3), this implies that the Legendre symbol ( −1p 1)= 1; which is acontradiction. Therefore we are left to show the second part of the statement of thisStep.If 1 < i ≤ t then √ p i ∈ C 1 because, either p i ≡ 1(mod4) and thus √ p i ∈ C 0 ⊂ C 1 ,or p i ≡ 3(mod4) and √ p 1 p i ∈ C 0 hence √ p i = √ p 1 p i / √ p 1 ∈ C 1 . In both cases, theGalois group of C i over C i−1 is generated by the mapσ : p 1/a ii↦→ ζ 2 a ip 1/a iiwhich again has order a i /2.Finally, in this case we have [C t : C 0 ] = a 1 · · · a t /2 t−1 and this concludes the proof.✷Corollary 4.3 We have the following lower bound for the dimension n a1 ,...,a sfield L a1 ,...,a sover Q:of then a1 ,...,a s≥ φ([a 1, . . . , a s ])a 1 · · · a s2 s .✷We have now enough tools to prove the property we used during the proof ofTheorem 4.1.69

Proposition 4.4 Recall that S = {(a 1 , . . . , a s ) ∈ N s | ∃i with a i > y }.thatfor some absolute constant c 1 .∑(a 1 ,...,a s)∈Sµ 2 (a 1 ) · · · µ 2 (a s )≪ cs 1n a1 ,...,a syIt results(3)Before proving Proposition 4.4, we need the following technical Lemma:Lemma 4.5 Consider the multiplicative function d t (n) defined as the number of waysto write n as product of t natural numbers, and denote:⎛⎞σ(t) =∏ ⎝1 + 1 ∑ d t (l k )⎠ .l l kl primek≥1we have:σ(t) ≤ ζ(2) 2t .HenceProof: Note that d t (l 2k−1 ) ≤ d t (l 2k ), therefore1 + 1 l(dt (l)l+ d t(l 2 ) ()dt (l 2 )+ · · · ≤ 1 + 2 + d t(l 4 ))+ · · ·l 2 l 2 l 4(≤ 1 + d t(l 2 )+ d t(l 4 ) 2 ()+ · · · = 1 + 1 l 2 l 4 l + 1 ·) 2t 2 l + · · .4σ(t) ≤∏l prime(1 + 1 l 2 + 1 l 4 + · · ·) 2t= ζ(2) 2t .✷Proof of Proposition 4.4: First note that from Corollary 4.3, we get thatn a1 ,...,a s≥ φ([a 1, . . . , a s ])a 1 · · · a s2 s ,therefore (3) is≪ 2 s ∑ Sµ 2 (a 1 ) · · · µ 2 (a s )a 1 · · · a s φ([a 1 , . . . , a s ]) , (4)70

But the sum on (4) is completely symmetric on the a i ’s, therefore is≪ 2 s s ∑≪ 2 s ∑ µ 2 (a 1 ) · · · µ 2 (a s )sa 1 >y a 1 · · · a s φ([a 1 , . . . , a s ])(a 2 ,...,as)∈N s−1a 1 >yµ 2 (a)a∑1(a 2 ,...,a s)∈Na s−1 2 · · · a s φ([a 2 , . . . , a s ]) . (5)Using the multiplicative function defined above and the function γ defined asγ(n) = ∏ p|np,we can write that (5) is equal to:2 s s ∑ a>yµ 2 (a)a∞∑b=1d s−1 (b)φ(γ(ab)) = 2s s ∑ µ 2 (a)a>y φ(a)a∞∑b=1d s−1 (b)φ((a, b)).φ(γ(b))Since all functions inside the second sum are multiplicative, we can write (5) as2 s s ∑ a>yµ 2 (a)φ(a)a∏l primeIf we can prove the estimate:{1 +2 s sσ(s − 1) ∑ a>yφ((a, l))φ(l)(ds−1 (l)⎛µ 2 (a)⎝ ∏ ∑φ(a)al|a k≥0l+ d s−1(l 2 )})+ · · · ≪l 2⎞d s−1 (l k )⎠ .l k∑a>yf(a)a 2 ≪σ(s − 1)y(6)wheref(a) = aµ2 (a)φ(a)∏ ∑l|a k≥0d s−1 (l k )l k= µ 2 (a)( ) sa,φ(a)then the estimate for the function σ(t) of Lemma 4.5 would imply the claim.From the Theory of Dirichlet series, we know that (6) is≪∫ ∞yF (x)dx,x 371

where F (x) = ∑ n≤x f(n) is the average of f.If we could prove the asymptoticformula:for x that tends to infinity, then we would have that (6) is≪∫ ∞yF (x) ∼ σ(s − 1)x, (7)σ(s − 1)xx 3 dx =To prove (7), we make use of the Dirichlet series:H(z) =∞∑n=1σ(s − 1).yf(n)n z . (8)Since f(n) ≪ (log log n) s , we know that H(z) converges in the semi-plane R(z) > 1and we can write the Euler product expansion:whereH(z) = ∏ lK(z) = ∏ l⎛⎝1 + 1 l z⎛⎝1 + 1 l zl ∑l − 1k≥0l ∑l − 1k≥0d s−1l k⎞d s−1l k⎠ = ζ(z)K(z)⎞⎠(1 − 1 )l zconverges for R(z) > 0. This decomposition gives an analytic continuation for H(z)and therefore we can calculate the residue at z = 1 of H(z) which is going to belimz→1 (z − 1)ζ(z)K(z) = K(1) = ∏ l⎛⎛⎝1 − 1 l + 1 ⎝1 + ∑ lk≥1d s−1 (l k )l k⎞⎞⎠⎠ = σ(s − 1).The Ikehara Tauberian Theorem (see [33], page 311) implies the claim of (7) and theProposition results proven.✷Remark: The details used in the last part of the proof of Proposition 4.4 aremissing from the original proof of Hooley in the case s = 1. In that circumstancethe level of precision that we need here for the application on Section 4.3 was notrequired.72

If n d1 ,...,d sis the dimension over Q of ∏ si=1 Q(ζ di , p 1/d ii ) then we have seen thatwhenever ([d 1 , . . . , d s ], p 1 · · · p s ) = 1, it resultsn d1 ,...,d s= d 1 · · · d s φ([d 1 , . . . , d s ]),this leads us to consider the function:∆ s =∞∑a 1 =1· · ·∞∑a s=1µ(a 1 ) · · · µ(a s )a 1 · · · a s φ([a 1 , . . . , a s ]) .Our goal is to show that ∆ s ≠ 0 and that lim s→∞ ∆ s = 0, the best way to do thisis again to express ∆ s as an Euler product. This will also confirm the heuristicargument illustrated at the beginning of this Chapter.Proposition 4.6∆ s =∏l prime(1 − 1 [ (1 − 1 − 1 ) s ]).l − 1 lWe need two lemmas:Lemma 4.7 For any prime l, let it be given a function α l , and defineΓ t+1 (α l ) =∞∑a 1 =1. . .∞∑a t+1 =1(note that Γ s (1/(l − 1)) = ∆ s ) then:⎛⎝ µ(a 1) . . . µ(a t+1 )a 1 · · · a t+1( )l − 1Γ t+1 (α l ) = Γ t α ll − α l∏q prime(∏l|[a 1 ,...,a t+1 ])1 − α qq).α l⎞⎠ , (9)Proof: The right hand side of (9) is equal to:⎛⎛⎞∞∑ ∞∑· · · ⎝ µ(a 1) . . . µ(a t ) ∏α l⎠∞∑⎜a 1 =1 a t=1a 1 · · · a t⎝l|[a 1 ,...,a t]) x=1µ(x)x∏l∣ [a 1 ,...,a t ,x][a 1 ,...,a t ]⎞α l ⎟⎠ , (10)73

where we just renamed a t+1 to be x. Note that for any multiplicative function f(n)and H ∈ N, if we defineF (n) = µ(n)f([H, n]/H),then F (n) is again multiplicative. If we take H = [a 1 , . . . , a t ] and f(n) = ∏ l|n α l , weget that (10) is equal to⎛ ⎛⎛⎞∞∑ ∞∑· · · ⎝ µ(a 1) · · · µ(a t ) ∏α l⎠∏⎜a 1 =1 a t=1a 1 · · · a t⎝ 1 − 1 ⎜l|[a 1 ,...,a t] q primeq ⎝∣lSince l ∣ [H,q]H⎛∞∑ ∞∑· · · ⎝ µ(a 1) · · · µ(a t ) ∏a 1 =1 a t=1a 1 · · · a t l|[a 1 ,...,a t]∏∣ [a 1 ,...,a t ,q][a 1 ,...,a t ]if and only if l = q and q ̸ |H, we gather that (11) is equal to:α l⎞⎠∏l|[a 1 ,...,a t](1 − 1 )l∏q primeq̸|[a 1 ,...,a t ]Multiplying and dividing by the missing terms, we get the claim.✷⎞⎞α l ⎟⎟⎠⎠ . (11)(1 − α qq).Lemma 4.8 Define inductively the functions β i = β i (l) in the following :then:β 1 = 1l − 1 and β n = β n−1l − 1l − β n−1,∆ s =∏(s∏l prime n=11 − β n(l)lProof: By induction on s, the case s = 1 being the definition of the Artinconstant. Lemma 4.7 implies that:∆ s = Γ s (β 1 ) = Γ s−1 (β 2 ) ∏Γ s−2 (β 3 ) ∏l prime(Γ 1 (β s ) ∏l prime1 − β 1(l)ll prime) (().1 − β 1(l)l1 − β 2(l)l.[ s−1( ∏1 − β )]i(l).i=1l74))==

FinallyΓ 1 (β s ) =and the claim follows.✷∞∑a=1µ(a)a∏β s (l) =∏ (1 − β )s(l),l|al primelFirst Proof of Proposition 4.6: Note that by the inductive definition of theβ i ’s, we have that(1 − β i(l)land, more in general(Finally1 − β i(l)l∏l prime) () (1 − β i+1 (l)1 − β i+1(l)l(1 −)=(1 − β i (l)(1 −(1 − 1 ) 2))l(1 − 1 ) k)) ( ( (= 1 − β i (l) 1 − 1 − 1 ) k+1)).ll∆ s =∏ [(1 − β ) (1(l)· · · 1 − β ) (s−1(l)1 − β )]s(l)=l primelll[(1 − β ) (1(l)· · · 1 − β ) (( (s−2(l)1 − β s−1 (l) 1 − 1 − 1 ) 2))]lll∏l prime[(1 − β 1(l)l) (.1 − β 2 (l)Substitute β 1 (l) = 1/(l − 1), and get the claim.✷(1 −Second Proof of Proposition 4.6: If we definef(m) =∑a 1 ,...,as[a 1 ,...,as]=mµ(a 1 ) · · · µ(a s )a 1 · · · a s,(1 − 1 ) s−1))].lthen it results∆ s =∞∑m=1µ 2 (m)f(m)φ(m)75

since the lowest common multiple of square-free integers is itself square-free.We now claim that f(m) is multiplicative, which imply:∆ s =∏l primeIndeed, if m and m ′ are coprime integers, then the map(1 + f(l) ). (12)l − 1(a 1 , . . . , a s ), (b 1 , . . . , b s ) ↦−→ (a 1 b 1 , . . . , a s b s )is a bijection from the set of s-tuples of integers with lowest common multiple m crossthe set of r-tuples of integers with lowest common multiple m ′ to the set of s-tuplesof integers with lowest common multiple mm ′ , whose inverse map is given by:We gather that(c 1 , . . . , c s ) ↦−→ ((c 1 , m), . . . , (c s , m)), ((c 1 , m ′ ), . . . , (c s , m ′ )).f(m)f(m ′ ) =∑a 1 ,...,as[a 1 ,...,as]=m∑ µ(a 1 b 1 ) · · · µ(a s b s )=b 1a,...,bs 1 b 1 · · · a s b s[b 1 ,...,bs]=m ′∑ µ(c 1 ) · · · µ(c s )= f(mm ′ ).c 1 ,...,cs c 1 · · · c s[c 1 ,...,cs]=mm ′Finally, if [a 1 , . . . , a s ] = p, then each a i can be equal to 1 or to p, and each possibilityis possible except a i = 1∀i = 1, . . . , s. Hence⎛ ⎞⎛ ⎞f(p) = s −1p + ⎜ s ⎟ 1 ⎝ ⎠2 p + · · · + ⎜ s ⎟2 ⎝ ⎠ (−1)kk p kSubstitute in (12) and get the claim.✷(+ · · · + (−1)s = 1 − 1 s− 1.p p) sCorollary 4.9 With the same notation of Theorem 4.1, we have that:( )∆ s = O 2 −s 1.log s76

Proof: For any fixed N > 0, we have that:where f(l) =2 s ∆ s ≤ ∏ l

Furthermore, for any subset I of [s] def= {1, . . . s}, we denote by P I the product ofelements in I and by Ĩ the subset of I of those i’s for which p i ≡ 1(mod4). Now call[a] I the set of s-tuples of integers for which a i is even for all i ∈ I and a i is odd forall i ∉ I. It is clear that {[a] I } I∈[s] is a partition of N s , thereforeδ P = ∑ aµ(a)n a= ∑ ∑ µ(a)nI⊆[s] a∈[a] I a= ∑I⊆[s](−1) |I|2 |I| ∑a odd2 β Iµ(a)aφ([a])where, if Q = (P I , [a]),⎧⎪⎨ ν(Q)β I =⎪⎩ ν(Q) − 1if Q |PĨotherwisethe possibility Q = 1 belonging to the first case. We gather that∑a odd2 β µ(a)aφ([a]) = ∑ J⊆I∑a oddP J =(P I ,[a])⎧= 1 ⎪⎨ ∑2 ν(P J ) ∑2 ⎪⎩ J⊆Ĩ2 β µ(a)aφ([a]) = ∑ 2 ν(P J ) ∑ µ(a)J⊆Ĩ a oddaφ([a]) + ∑ 2 ν(P J )−1 ∑ µ(a)J⊈Ĩa oddaφ([a])P J =(P I ,[a])J⊆IP J =(P I ,[a])a oddP J =(P I ,[a])Now note that for J ⊆ I, the condition P Jµ(a)aφ([a]) + ∑ 2 ν(P J ) ∑J⊆Ia oddP J =(P I ,[a])⎫µ(a)⎪⎬.aφ([a]) ⎪⎭= (P I , [a]) is equivalent to P J |[a] and([a], P I−J ) = 1. As we did during the second proof of Proposition 4.6, we can write∑a oddP J =(P I ,[a])µ(a)aφ([a]) =∞ ∑m=1m odd,P J |m,(m,P I−J )=1µ 2 (m)φ(m)∑a odd[a]=mµ(a)a , (13)again the function inside is multiplicative, thus we can write that (13) is equal to= 1φ(P J )∏∞∑m=1m odd,P J |m,(m,P I−J )=1l|P J[(1 − 1 lµ 2 (m)φ(m)) s− 1] ∞ ∑2 s ∆ s∏∏l|mm=1m odd, (P I ,m)=1p|P Jα p78[(1 − 1 ) s ]− 1lµ 2 (m)φ(m)∏ 1.p|P I1 + α p∏l|m[(1 − 1 ) s ]− 1 =l

Putting everything together,⎧ ⎫δ P = 2 s ∑ (−1) |I| ⎨ ∑∆ s 2 J ∏ ∏ 1α2I⊆[s]|I|+1 ⎩p + ∑ 2 J ∏ ∏ 1 ⎬α pJ⊆I p|P J p|P I1 + α pJ⊆Ĩ p|P J p|P I1 + α p⎭ .Finally∑I⊆[s](−1) |I|2 |I| ∑and similarly2 J ∏ ∏ 1α pJ⊆I p|P J p|P I1 + α p= ∑I⊆[s](−1) |I| ∏ 1 + 2α p= 1 s∏ 12 |I| p|P I1 + α p 2 s i=11 + α p1∑ (−1) |I| ∑2 J ∏ ∏ 1α2I⊆[s]|I| p = ∑ (−1) |I| ∏(1 + 2αJ⊆Ĩ p|P J p|P I1 + α p 2I⊆[s]|I| p ) ∏ 1=p|PĨp|P I1 + α p12 s s∏i=1p 1 ≡1 mod 4therefore the claim.✷11 + α ps∏i=1p 1 ≡3 mod 4(2 − 11 + α p)=We conclude this Section with two Corollaries:( ( ) )s∏ −1 αp1 −i=1p 1 + α pCorollary 4.11 Under tha same assumptions of Theorem 4.1, if every prime in Pis congruent to 1(mod4) then the density of primes for which all elements of P areprimitive roots, is∏(1 + α l ) .✷l∉PCorollary 4.12 For any set of odd primes P, δ P is a well defined number.✷79

4.3 An Application to the Least Prime Primitive RootIn this last section we apply Theorem 4.1 to the classical problem of the study ofthe function G(p) defined as the least prime primitive root (modp). More precisely,by the use of the inclusion exclusion principle, we determine a uniform asymptoticformula for the number of primes p ≤ x such that G(p) < r.Theorem 4.13 With the same notation and hypothesis of Theorem 4.1, let q n be then-th odd prime,T r (x) = #{p ≤ x | ∃i ≤ r | q i is a prime primitive root (mod p)}, ,and∑ˆδ r = (−1) |P| δ PP⊆{p 1 ,...,p r}(δ ∅ = 1). We have that( )T r (x) = (1 − ˆδ x x log log xr )log x + O log 2 x Cr 2for some absolute constant C 2 , uniformly respect to r.Proof: Let S r be the number of primes p up to x such that none of the first rprimes is a primitive root modp.As a straightforward application of the inclusion exclusion principle, we get that∑S r = (−1) |P| N P (x)P⊆{p 1 ,...,p r}where as in Theorem 4.1,N P = # {p ≤ x |∀q ∈ P, q is a primitive root (modq)}and N ∅ (x) = π(x).80

Applying Theorem 4.1, we get for a suitable positive constant c 0 ,S r =∑P⊆{p 1 ,...,p r}(−1) |P| (δ P(x x log log xlog x + O log 2 x( )x x log log xˆδ rlog x + O log 2 x Crwhere we have taken C = 2c 0 , say. Finally, noticing that T r = π(x) − S r , we get theclaim.✷c|P| o))=Corollary 4.14 Let f(x) be any monotone function of x that tends to infinity. Supposealso that f(x) = o(log log x), then, if the generalized Riemann Hypothesis holds,we have that G(p) ≤ f(x) for all primes with the exception of a set of primes of sizeˆδ f(x)(x x log log xlog x + O log 2 xCf(x) 2Proof:It is enough to notice that by the assumption made on f, the error termis o(π(x)).✷The problem now amounts to estimating the behaviour of the function ˆδ r . Computercalculations suggest that ˆδ r = O ( )1log r , but we do not hazard in any preciseclaim.We are not even able to present a direct proof of the fact that).lim ˆδ r = 0,r→∞which, of course would imply that G(p) < f(p) for almost all p (f → ∞), under theRiemann Hypothesis.The latter assertion has been proven by L. Murata in [38] and is equivalent, underthe Riemann Hypothesis, to ˆδ r = o(1).81

Upper and Lower estimates for ˆδ r would allow to determine (under the RiemannHypothesis) Ω ± -type of estimates for the size of set of primes p for which G(p) < f(p).We do feel that such a problem is not too difficult and we are planning to addressthese questions in the near future.Finally we would like to mention that in principle this approach could be extenedto the analogous problem for the function g(p), the least primitive root (modp). Wefound out just recently that a general form of Theorem 4.1 has been found by K. R.Matthews in [36]. The asymptotic formula found by Matthews is not uniform andprovides a weeker error term than Theorem 4.1, however the proof can be adjusted toyield to an analogous result of Corollary 4.14 for g(p). The expression for the densityin this case would be much more complicated and even computer calculations seemat the moment very hard to perform.82

APPENDIX A: ON DIVISORS OF p − 1We recall that Lemma 1.4 states that for any given sequence of multiplicativelyindependent integers, the number of primes for which the group generated by thefirst r elements of the sequence is smaller than t isO ( t 1+1/r r log r ) ,uniformly with respect to r.(Note that we are using the statement of Proposition 3.4 according to which the sumof the logarithms of the first r elements of a sequence of multiplicatively independentnumbers in asymptotic to r log r). A consequence of this is that, if r is fixed, thenfor almost all primes,|Γ r | ≥ p rr+1log p .Indeed, if we take t = x r/(r+1) / log x, in Lemma 1.4, we get that the number of primesfor which |Γ r | < p r/(r+1) / log p ≤ t is O(π(x)/ log 1/r x), therefore for almost all primeswe have the desired inequality. It is natural to ask what would be an estimate of |Γ r |uniform respect to r? Using the same method of the fixed r case, we get that, if f(p)is any divergent function, then|Γ r | ≥(pf(p) log p r log r) rr+1for almost all primes, uniformly respect to r. We need of course to ensure certaingrowing conditions to be met.The goal of this section is to improve the preceding results making use of thefollowing:83

Theorem A. 1 It exist, β and δ positive such that, for all h ∈ [log −β x, 1 − log −β x],and y = x h , one has uniformly on h:# { p ≤ x | ∃u|p − 1, with u ∈ [y, y exp log δ x] } = owhere the constant implied by the o symbol is absolute.( x)log xBefore starting the proof of the Theorem we need some preliminary lemmas:Lemma A. 2 (Erdös) Let Ω(n) be the number of prime divisors counted with multiplicityof a natural number n than the normal order of Ω(p − 1) is log log p; moreprecisely, for every ɛ > 0, it exists η = η(ɛ) such that the number of p up to x forwhich Ω(p − 1) > ɛ log log p isProof: See [12].✷o( ) x.log xLemma A. 3 (de Bruijn) Let Ψ(x, y) be the number of natural numbers up to xwhose greatest prime divisor is less than y, thenProof: See [11].✷{ }log xΨ(x, y) ≪ x exp −c 1 .log yLemma A. 4 (Hardy-Ramanujan) For any 0 < ɛ < 1 there exists τ > 0 suchthat the number of integers n up to x such that Ω(n) < ɛ log log x isProof: See [19].✷O( ) xlog τ .x84

Lemma A. 5 Murty (Weak Brun’s sieve) For any natural number m < x, denoteby N(x, m) the number of solutions ofp − 1 = qmwhere p and q are prime numbers ≤ x. Then for some absolute constant B > 0, wehaveProof: See [43].✷N(x, m) ≤We are now ready to prove Theorem A.1:Bx(log log x/m)2φ(m) log 2 (x/m) .Proof: Let S = { p ≤ x | ∃u|p − 1, with u ∈ [y, y exp log δ x] } . Without loss, we canassume that p ≥x , and for a suitable δ to be chosen later, p ∈ S means thatlog 2 xp − 1 = uv with u ∈ [ y, y exp log δ x ] and v ∈[ xy exp logδ x, x y]If Ω(u) > 2 log log x and Ω(v) > 2 log log x then Ω(p − 1) > 4 log log x, the number3 3 3of p ∈ S for which this holds is certainly less thenand for Lemma A.2, this is o(π(x)).#{p ≤ x | Ω(p − 1) > 4 }3 log log xRemark: A stronger statement than Lemma A.2 can be found in [39]. Usingsuch a statement, our proof would yield to |S| ≪ x/ log α x. For the purpose of theapplication that will follow, our assumption is enough.On the other hand, for a fixed u, the number of v’s for which the maximum primedivisor is less then z is, by Lemma A.3,O( xu exp {−c 1log(x/u)log z85})..

Fix ɛ > 0, let β > 0 be a number to be chosen later and put log z = log 1−β−ɛ x.We notice that u < y exp log δ x = x h exp log δ x and thus we get that the number weare estimating is≪ x {u exp log x − log y − log δ }x−c 1log 1−β−ɛ x≪ x {u exp (1 − h) log x − log δ }x−c 2log 1−β−ɛ ≪ x xu exp {−c 3 log ɛ x} .(Note that this put the constraint 1 − β > δ.)Therefore, the number of p ∈ S for which this holds is≪ ∑ u′ x u exp {−c 3 log ɛ x} ≪ x exp {−c 3 log ɛ x} ∑ u′ 1 u≪ x exp {−c 4 log ɛ x}(here the dash on the sum sign means that the sum is extended to all the valuesof u for p ∈ S). A similar argument shows that h > log −β x implies that we canalso exclude the possibility that the maximum prime divisor of u is smaller thanexp ( log 1−β−ɛ x ) . Therefore we can assume thatp − 1 = u 1 v 1 q, with u 1 and v 1 in the desired range, q > exp ( log 1−β−ɛ x ) and Ω(u 1 )or Ω(v 1 ) is less than 2 log log x.3isFrom Lemma A.5, we get that for fixed u 1 and v 1 , the number of possible solutions≪x(log logxu 1 v 1) 2u 1 v 1 log 2 (x/u 1 v 1 )≪x(log log x)2u 1 v 1 log 2 (x/u 1 v 1 ) .As u 1 v 1 ≤ x exp(− log 1−β−ɛ x) and (log log x) 2 ≪ log ɛ x, the number is≪xu 1 v 1 log 2−2β−3ɛ x .As applications of Lemma A.4 we know that#{n ≤ x | Ω(n) < 2 } ( ) x3 log log x = Olog τ x86

for some τ > 0. Partial summation implies that, ifS(t) = # { n ≤ t | Ω(n) < 2 3 log log t} , then∑Ω(n)

Hence |Γ r | ≥ p rr+1log pNow setand note thatthere exists δ such that if{for almost all p’s.r log log x− < 1 − 1r+1 log x log β xT =y = x rr+1log x = x r log log x− r+1 log xp ≤ x | ∃l|p − 1, l ∈then |T | = o(π(x)). Finally, since{p ≤ x | |Γ r | ∈we get that for almost all primes p,for x large enough. Theorem A.2 gives that[ xrr+1log x , x rr+1 exp(log δ x )]}[ xrr+1log x , x rr+1 exp(log δ x )]} ⊂ T ,|Γ r | ≥ x rr+1 exp(log δ x) ≥ p rr+1 exp(log δ p).✷The case in which r grows with p can be treated in an analogous fashion. Theonly care is to consider the version of Lemma 1.4 which is uniform with respect to r.In particular:Theorem A. 7 There exist β and δ such that if r ≤ (log β p) − 1, then for almost allprimes p,|Γ r | ≥ p rr+1 exp(log δ p).Proof: The uniform version of Lemma 1.4 states thatif we take t = x r+1r, we getr log 2 x{# p ≤ x | |Γ r | < p rr+1#{p ≤ x | |Γ r | < t} = O ( t 1+ 1 r r log r),}1r log 2 ≪p88xlog 2 x log 2 r xr − 1 r log r = o(π(x)).

Now setand note that h =x rr+1r log 2 x = x rr+12 log log x− − log rlog x log xr 2 log log x− − log r < 1 − 1r+1 log x log x log β xif r ≤ (log β p) − 1 and x is large enough.Therefore Theorem A.1 gives#{p ≤ x |which clearly implies the claim.✷= x h = yp r}r+1r log 2 p ≤ |Γ r| ≤ p rr+1 exp(log δ p) = o(π(x))Remark: If l|[F ∗ p : |Γ r |], then Theorem A.7 puts the constraintfor almost all primes p.l < p − 1Γ r< p − 1r+1 exp(− log δ p)Unfortunately the position r = (log p) β − 1 and the constraint β + δ < 1 remarkedduring the proof of Theorem A.1, implies thatl < exp ( log 1−β p − log δ p ) ( ) 1≤ exp2 log1−β p .Such a bound is too high to make possible the use of any of our techniques for therange of r’s under consideration.89

APPENDIX B: ON THE EXPONENT OF THEIDEAL CLASS GROUP OF Q( √ −d)Let d be a positive square-free integer and let m(d) denote the exponent of theclass group of Q( √ −d), i.e. m(d) is the least positive integer m, such that x m = 1for every x in the class group.In 1972 D.W. Boyd and H. Kisilevsky (see [3]) proved that if the Extended RiemannHypothesis holds, then for any η > 0, and d sufficiently large,m(d) >log d(2 + η) log log d(1)which of course implies that m(d) → ∞ as d → ∞.The goal of this note is to establish unconditional inequalities of the type (1) fordensity-one sets of values of d. Before doing this, let us review the method used byBoyd and Kisilevsky to prove (1).First they noticed that if α is an integer of Q( √ −d) which is not in Z, thenN(α) ≥ d/4 and that if p is a rational prime that splits in Q( √ −d) and ϖ is a primeideal above p, then ϖ m(d) is a principal ideal (α) thusN(ϖ) m(d) = p m(d) = N(α) m(d) ≥ (d/4) m(d) .In conclusion, ( ) −d= 1 =⇒ p ≥ (d/4) 1/m(d) . (2)pThen they proved thatIf the Extended Riemann Hypothesis holds then, for any integer d, there exists aprime less then log 2+η d for which −d is a quadratic residue and this gives (1).90

Now, let us take p = 3 and ask how often is a square-free d a quadratic residue(mod3)? This happens when d ≡ 1 mod 3, and the density of such d’s is certainlypositiveFor a positive proportion of square-free integers d,m(d) ≥In general we will be able to prove thatlog d/4log 3 .Theorem B. 1 For any d < x there exists a prime < log d for which d is a quadraticresidue with at the most O ( x 1−A(log log x)−1) exceptions.This is an consequence of Theorem B.3 below and by (2) impliesCorollary B. 2 For all discriminant d < x, we have thatm(d) >with at most O ( x 1−A(log log x)−1) exceptions.i.e.log d/4log log dFor an integer n, let M(n) be the least prime for which n is quadratic residue,{( ) }nM(n) = min p∣ p is prime and = 1 .pLet K(x, s) (respectively K 1 (x, s)) be the set of numbers (resp. square-free numbers)up to x such that M(n) > s. We have thatTheorem B. 3 Let k(x, s) = |K(x, s)| and k 1 (x, s) = |K 1 (x, s)|, thena) k(x, s) = x ∏(1 + 1 2 π(s) p − 2 ) ( e θ(s) log 3 )s+ O;p 2 2 π(s)p≤sb) k 1 (x, s) = 6 π 2 x2 π(s) ∏p≤s(1 + 1 )p + 191+ O( x 1/2 e θ(s)2 π(s) log s).

uniformly with respect to s (where as usual π(s) and θ(s) are respectively the numberof primes up to s and the sum of the logarithms of the primes up to s).Proof: Let us define P to be the product of all primes up to s.b) In order for a square-free number n ≤ x to be in K 1 (x, s), one must have)= 0 or −1 for all primes p up to s. For any divisor Q of P , let AQ be the set of( npn ∈ K 1 (x, s) such that( np)= 0 for any p |Q and( )n P= −1 for any pp ∣Q ,ClearlyK 1 (x, s) =where the union is disjoint. Note also that|A Q | = #{= ∑ ∗ #n ≤ x Q{◦⋃Q|PA Q (3)( ) ( )}n Q ∣ (n, Q) = 1, n square-free, P= − for any pp p ∣Qn ≤ x Q | (n, Q) = 1, n square-free, n ≡ g i(modq i ), i = 1, . . . , twhere we have put P Q = q 1 · · · q t and ∑ ∗ means that the sum is extended to all thet-tuples (g 1 , . . . , g t ), g i being a congruence class modq i such that ( g iq i)= −( Qq i).By the Chinese remainder Theorem, for each t-tuple (g 1 , . . . , g t ), there exists aunique congruence class M = M(g 1 , . . . , g t )(mod P ) such thatQn ≡ g i (modq i ), ∀i = 1, . . . , t ⇐⇒ n ≡ M(mod P Q)}(4)therefore (4) equals∑ ∗#{n ≤ x Q(mod∣ (n, Q) = 1, n square-free, n ≡ M P )}. (5)QNow we need the following two Lemmas:92

Lemma B. 4 Let R 1 , R 2 , R 3 be positive integers with (R 1 , R 3 ) = (R 2 , R 3 ) = 1 anddefineB R1 ,R 2 ,R 3(y) = # {n ≤ y | (n, R 1 ) = 1, n ≡ R 2 (modR 3 )}then, uniformly with respect to R 1 , R 2 , R 3 < y, we haveB R1 ,R 2 ,R 3(y) = y ϕ(R 1)R 1 R 3+ O (ϑ(R 1 )) ,where ϑ(R 1 ) is the number of square-free divisors of R 1 .Lemma B. 5 Let Q 1 , Q 2 , Q 3 be positive integers with (Q 1 , Q 2 ) = (Q 2 , Q 3 ) = 1 anddefineC Q1 ,Q 2 ,Q 3(z) = # {n ≤ z | n square-free, (n, Q 1 ) = 1, n ≡ Q 3 (modQ 2 )} ,then, uniformly respect to Q 1 , Q 2 , Q 3 < z, we haveC Q1 ,Q 2 ,Q 3(z) = 6 π 2 z ϕ(Q 1)Q 1 Q 2(∏1 − 1 ) −1+ O ( z 1/2 ϑ(Qp|Q 1 Q 2p 2 1 ) ) .Remark: Lemma B.4 and B.5 are due respectively Cohen (See [9]) and to Landau(See p. 633-636 of [30]). Their version is slightly less general though the proof issimilar. One might think that a stronger version of Lemma B.4, say valid on a rangeof R 1 of the same order of the range given by the Brun’s Sieve, would yield a bettererror term in Theorem B.3. On the contrary, it will become clear how this is notinfluential to the main goal of our discussion.Proof of Lemma B.4: We have thatB R1 ,R 2 ,R 3(y) = ∑ d|R 1µ(d)# {n ≤ y | d|n, and n ≡ R 2 (modR 3 )}= ∑ {µ(d)# n ≤ y }d|R 1d | n ≡ R 2d ∗ (modR 3 )= ∑ [ ] yµ(d)d|R 1dR 393

where d ∗ is the unique congruence class mod R 3 defined by dd ∗ ≡ 1 mod R 3 (Such aclass exists since we have assumed that (R 1 , R 3 ) = and d|R 3 ). FinallyB R1 ,R 2 ,R 3(y) = ∑ ( )yµ(d) + O (1)d|R 1dR 3= y ϕ(R 1)R 1 R 3+ O (ϑ(R 1 )) .✷Proof of Lemma B.5: This is based on the identityµ 2 (n) = ∑ d 2 |nµ(d),We have thatC Q1 ,Q 2 ,Q 3(z) ===∑n≤z(n,Q 1 )=1n≡Q 3 ( mod Q 2 )∑d≤z 1/2(d,Q 1 )=(d,Q 2 )=1∑d≤z 1/2(d,Q 1 )=(d,Q 2 )=1µ 2 (n) =µ(d)∑d 2 δ≤z(d,Q 1 )=(δ,Q 1 )=1d 2 δ≡Q 3 ( mod Q 2 )∑δ≤ zd 2(δ,Q 1 )=1δ≡Q 3 d ∗2 ( mod Q 2 )1 =µ(d)µ(d)B Q1 ,Q 3 d ∗2 ,Q 2(z/d 2 ) (6)where the condition (Q 2 , Q 3 ) = 1 implies (d, Q 2 ) = 1 and d ∗ has the same meaningas in the proof of Lemma B.4. Now apply Lemma B.4 and get that (6) equals( )∑ z ϕ(Q 1 )µ(d) + O(ϑ(Qd 2 1 )) =Q 1 Q 2= zϕ(Q 1)Q 1 Q 2and since clearly ϕ(Q 1)Q 1d≤z 1/2(d,Q 1 Q 2 )=1∞ ∑d=1(d,Q 1 Q 2 )=1µ(d)d 2< ϑ(Q 1 ) and∞∑d=1(d,Q 1 Q 2 )=1⎛⎞+ O ⎝ ∑ zϕ(Q 1 )⎠ + O ( z 1/2 ϑ(Qd 2 1 ) )Qd>z 1/2 1 Q 2µ(d)d 2 = 6 π 2 ∏p|Q 1 Q 2(1 − 1 p 2 ) −1,94

the claim is deduced.✷Now we can apply Lemma B.5 to (4) with Q 1 = Q, Q 2 = P/Q, Q 3 = M andz = x/Q. Note that the number of summands in (5) is ϕ( P )/ϑ( P ), thereforeQ Q|A Q | = ϕ( P )⎧⎨Q 6 xϑ( P ) ⎩π Q 2 Q= 6 π 2 x2 π(s) ∏p|P(ϕ(Q)P1 + 1 p(∏1 − 1 ) ⎛−1 ( ) ⎫1/2+ O ⎝ x ⎬ϑ(Q)⎞⎠pp|P2 Q⎭) −1 (ϑ(Q) x1/2Q + O ϑ 2 (Q)2 π(s) Q 1/2 ϕ(Q)e θ(s) )log swhere we just noticed that ϑ(P ) = 2 π(s) and ϕ(P ) ≪ eθ(s) . Now use (3) and getlog sk 1 (x, s) = ∑ |A Q |Q|P= 6 π 2 x2 π(s) ∏p|P(= 6 π 2 x2 π(s) ∏p|P(1 + 1 ) −1 ∑ ϑ(Q)p QQ|P) −1 ( ∏p|P1 + 1 pThe last identity follows since ∑ Q|Pproof of b).✷ϑ 2 (Q)Q 1/2 ϕ(Q)1 + 2 p,(7)⎛⎞+ O ⎝ x1/2 e θ(s) ∑ ϑ 2 (Q)⎠ (8)2 π(s) log s QQ|P1/2 ϕ(Q)) ( x1/2e θ(s) )+ O2 π(s) log sconverges as s → ∞. This concludes thea) This is simpler than b). For any Q|P , define A Q to be the set of n ∈ K(x, s)such that ( np)Again= 0 for any p |Q andk(x, s) = ∑ Q|P( )n P= −1 for any pp ∣Q .|A Q | (9)and now|A Q | =∗∑#{n ≤ x Q | n ≡ g i(modq i ), i = 1, . . . , t= ∑ {∗ # n ≤ x Q ∣ n ≡ M(mod P }Q ) .}95

where the g i , i = 1, . . . , t and M = M(g 1 , . . . , g t ) are defined as above. Now applyLemma B.4 with R 1 = Q, R 2 = M, R 3 = P/Q and y = x Q|A Q | = ϕ( P ) { } (Q x ϕ(Q)ϑ( P ) Q P + O(ϑ(Q)) = 2 −π(s)QFinally by (9),⎛k(x, s) = 2 −π(s) ⎝x ∏ (1 − 1 ) (1 + 2 )p≤sp pWhich is the claim of a).✷x ϕ(P )P+ O⎛and getϑ(Q)Q⎝ eθ(s)log s(∏p≤s⎛= 2 −π(s) ⎝x ∏ (1 + 1p≤sp − 2 )+ O ( e θ(s) log 3 s )⎞ ⎠p 2Proof of Theorem B.1: We want to estimate+ O ( eθ(s)log s1 + 4p − 1ϑ 2 ))(Q).ϕ(Q)) ⎞ ⎠⎞# {d ≤ x | M(d) > log d} (10)Note that, since the contribution for d < x 1/2 is O(x 1/2 ), we have that (10) equals# { d : x 1/2 ≤ d ≤ x | M(d) > log d } + O(x 1/2 )≤ #{d ≤ x | M(d) > 1 }2 log x + O(x 1/2 ). (11)Now apply Theorem 3 a), with s = 1 log x and get that (11) is ≪ then22 −π( 1 2 log x) ( x log log x + e θ( 1 2 log x)) (log log x) 3)≪ x exp(−A log x/ log log x)where we took A < 1 log 2, say, and this proves the claim.✷2Remark: Note that although in Theorem B.1 we consider discriminants of imaginaryquadratic fields which are by definition squarefree numbers, statement b) ofTheorem B.3, does not give anything more than statement a). This is due to the factthat square-free numbers have non-zero density.96⎠

Theorem B.3 b) can be improved using a version of Lemma B.5 in which the errorterm depends on Q 2 . This has been done by K. Prachar in [45] for the case Q 1 = 1,and his proof can adapted to proof the following:Lemma B. 6 With the same notations of Lemma B.5 we have that, uniformly respectto the parameters,C Q1 ,Q 2 ,Q 3(z) = 6 π 2 z ϕ(Q 1)Q 1 Q 2for any ɛ > 0.✷(∏1 − 1 ) −1+ O ( (z 1/2 Q −1/4+ɛp|Q 1 Q 2p 2 2 + Q 1/2+ɛ2 )ϑ(Q 1 ) ) ,Corollary B. 7 With the same notation as above, we have thatk 1 (x, s) = 6 π 2x2 π(s) ∏p≤s(1 + 1 )p + 1+ O((x 1/2)e + e θ(s) )θ(s)(1/4−ɛ) eθ(s)(1/2+ɛ) .2 π(s) log sProof: It is similar to the proof of Theorem B.1 b), but in this case we haveand therefore⎛⎛+O ⎝O(( x1/2⎝ x1/2 ∑P −1/4+ɛ= 6 π 2 x2 π(s) ∏p≤s|A Q | = 6 π 2 x2 π(s) ∏p|P(1 + 1 pQ 1/2 P −1/4+ɛ Q 1/4−ɛ + P 1/2+ɛQ 1/2+ɛ )Q|P(k 1 (x, s) = ∑ |A Q |Q|P= 6 π 2 x2 π(s) ∏p≤s(1 + 1p + 1) −1ϑ(Q)Q +ϑ 2 (Q) e θ(s) )2 π(s) ϕ(Q) log s⎞⎞ϑ 2 (Q)Q 1/2−1/4+ɛ ϕ(Q) + P 1/2+ɛ ∑ ϑ 2 (Q)⎠eθ(s) ⎠QQ|P1/2+ɛ ϕ(Q) 2 π(s) log s1 + 1 ) ((x 1/2)+ Op + 1 e + e θ(s) )θ(s)(1/4−ɛ) eθ(s)(1/2+ɛ) .2 π(s) log s97)+

The last identity because both the series ∑ Q|Pconverge as s → ∞.✷ϑ 2 (Q)Q 1/2+ɛ ϕ(Q) and ∑ Q|PRemark A general form of Theorem B.1 can also be proven.ϑ 2 (Q)Q 1/2−1/4+ɛ ϕ(Q)It is a uniformasymptotic formula for k m−1 (x, s), the number of m-free numbers d up to x for whichM(d) < s. The m-free version of Lemma B.6 is also in [45]. Finally, the results ofPrachar have been improved by Hooley in [24] and the use of this last one would givea further improvement of Theorem B.3.98

APPENDIX C: OPEN QUESTIONS ANDFUTURE RESEARCHVariants of the Bombieri-Vinogradov TheoremA form of the famous Bombieri-Vinogradov Theorem for primes in arithmeticprogression states thatFor any real number A > 0, it exists a B > 0 such that∣∑1 ∣∣∣∣ ψ(x, m, 1) −∣ φ(m) x ≪where ψ(x, m, 1) = ∑m≤ x1/2log A xp≤xp≡1 mod mlog p.xlog B xThis important result provides us with an estimate on average of the error term forthe prime number Theorem for primes in arithmetic progressions which is as strongas the one that could be deduced useing the Extended Riemann Hypothesis for theDirichlet L-functions of all the characters modm.Such a Theorem can be interpreted as an estimate on average of the error term ofthe Chebotarev Density Theorem for cyclotomic fields. More precisely, let us considerthe following statement:For every integer m, let us suppose K m is a given finite Galois extension of Qand let n(m) = [K m : Q]. Further, setWe have thatψ(x, K m ) =∑m≤ x1/2log A xLet us note the following facts:∑p≤xp splits completely in Km∣ ψ(x, K m) − 1∣ ∣∣∣∣n(m) x ≪log p.xlog B x(1)99

• If the Generalized Riemann Hypothesis holds for all the (non-abelian) ArtinL-functions of K m then the statement is true.• If, for any m, K m is the cyclotomic field Q(ζ m ), then the statement is a consequenceof the famous Bombieri-Vinogradov Theorem.• If, for any m, K m is the Galois Extension Q(ζ m , a 1/m ) and the statement istrue, then the Artin Conjecture for primitive roots is true for the number a.The last fact has been noticed by R. Murty in his thesis and he gave a resultwhich is in the spirit of this approach.We can refer to (1) as the general non-abelian Bombieri-Vinogradov Theorem andask for which families K m it holdsA proof of the general statements is certainly a very difficult problem, and to ourknowledge, the Theorem of R. Murty and K. Murty in [41] is the only significantcontribution toward this direction and it states that:If π K (x, q) is the number of primes p up to x such that p splits completely in agiven fixed Galois extension K of Q and p ≡ 1(modq) (i.e. p splits completely inK(ζ q )), then for any A > 0, there exists B = B(A) such that(where α = minK ∩ Q(ζ q ) = Q.∑q≤x α (log x) −B ∣ ∣∣∣∣π K (x, q) −2[K:Q] , 1 2)1[K(ζ q ) : Q] li(x) ∣ ∣∣∣∣≪xlog A xand the sum is extended to all the values of q for whichIn general, one could try to settle for something less and restrict the sum in (1) tom ≤ log C x for some fixed positive integer C. We would get a weaker statement butwith still quite a few interesting arithmetical consequences. Fox example, if we prove100

the statement with C = 2 (C = 1 is actually Theorem 2.7), then it can be proventhe following substantial improovement of Theorem 3.1:For amost all primes p, thelog plog log p primes generate F∗ p.Such a problem admits an analogous situation where we substitute the ArtinL-function with the L-series attached to modular forms.The Lang-Trotter Conjecture for Abelian VarietiesIn 1977 J.-P. Serre (see [47]) has proven the following result:Let E be an elliptic curve defined over Q and let K n = Q(E[n]) where by E[n] wedenote the set of n-points of E (i.e. Q ∈ E such that [n]Q = 0). Let us putδ =∞∑n=1µ(n)[K n : Q] ,where µ(n) denotes the µ function of Möbius. If the Generalized Riemann Hypothesisholds for K n , then#{p ≤ x | Ē(F ∗ p) is cyclic} ∼ δxlog xThis result has been reconsidered by R. Murty and R. Gupta. In 1990 (see [16])without any unproved hypothesis they have characterized elliptic curves for whichE(F ∗ p) is cyclic for infinitely many values of p.Gupta and Murty considered as well a similar problem to Serre’s Theorem, namelyThe Lang-Trotter Conjecture (see [34]):Let E be an elliptic curve defined over Q and let P be a rational point of E withinfinite order. We denote by N(x, P ) the number of primes p up to x such that〈P 〉 = E(F ∗ p), thenxN(x, P ) ∼ δ E (P )log x101

where δ E (P ) can be expressed in terms of the decomposition of primes in the extensionsQ(E[n], n −1 P ) over Q.Both this result and the statement of Serre’s Theorem are analogous to the Artin’sConjecture for primitive roots.Many of these conjectures admit a very natural generalization to the case ofabelian varieties. The problem can be stated as follows:Let A be an abelian variety defined over Q and let P ∈ A be a rational point(with infinite order). For all (but finitely many) prime numbers p, it makes sense toconsider the reduction of A modulo p that we can denote by A(F ∗ p).A(F ∗ p) is a finite group and we can indicate with ¯P ∈ A(F ∗ p) the reduction of Pmodulo p. Various questions can be formulated, for example:• Under which conditions A(F ∗ p) is cyclic (or more particularly 〈 ¯P 〉 = A(F ∗ p)) forinfinitely many p?• What is the distribution of the prime numbers with this property ?• Is it possible to write a formula for the density of such sets of primes?In the case dim A = 1, (i.e. A is an elliptic curve), then the Lang-Trotter conjecturetoghether with the Theorem of Serre and the contribution of Gupta and Murty,provide with a precise indication on what should be the answer to these questions.In the case dim A > 1, there are not, at the moment in the literature conjecturesthat give any answer to this question, nevertheless it is natural to suspect that manyof the arguments that worked in the case of elliptic curves, extend to the general caseand the first problem is as usual to express, for any prime number l, the condition[ l ∣ A[F∗p] : 〈P 〉 ]102

in terms of particular decompositions of p in algebraic extensions K(l, P ).Similarly as in the case of elliptic curves in which it has been necessary to distinguishbetween Complex Multiplication curves and curves without Complex Multiplication(see [17]), it is natural to expect that the properties under considerationdepend heavily on the structure of the ring End Q A.103

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