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On primitive roots for Carlitz modules
Journal of Number Theory 100 (2003) 88–103
http://www.elsevier.com/locate/jnt
On primitive roots for Carlitz modules Wei-Chen Yaoa,* and Jing Yub a
Department of Mathematics and Computer Science Education, Taipei Municipal Teachers College, No. 1, Aikuo West Road, Taipei 100, Taiwan, ROC b Department of Mathematics, National Tsing-Hua University, No. 101, Sec. 2, Kuang Fu Road, Hsinchu 30043, Taiwan, ROC Received 28 November 2001 Communicated by D. Goss
Abstract Let A be the polynomial ring over a finite field. We prove that for every element a of a global A-field of finite A-characteristic the set of places P for which a is a primitive root under the Carlitz action possesses a Dirichlet density. We also give a criterion for this density to be positive. This is an analogue of Bilharz’ version of the primitive roots conjecture of Artin, with Gm replaced by the Carlitz module. r 2003 Elsevier Science (USA). All rights reserved. MSC: primary 11T55 Keywords: Primitive root; Carlitz module
1. Introduction Let Fq be the finite field of q elements, Fq ðtÞ the rational function field in one variable, and put A ¼ Fq ½t: We are interested in the arithmetic of Fq ðtÞ with elements of the ring A playing the role of integers, and the Carlitz module C (cf. [2]) playing the role of the multiplicative group Gm : Here the Carlitz module is the action C of A on the additive (rather than multiplicative) group given by Ct ðxÞ ¼ iðtÞx þ xq ; where i is a specified morphism from A into the global coefficient field. Given an element a of the coefficient field K: We want to investigate the set of places P of K; for which a
*Corresponding author. E-mail addresses: [email protected] (W.-C. Yao), [email protected] (J. Yu). 0022-314X/03/$ - see front matter r 2003 Elsevier Science (USA). All rights reserved. doi:10.1016/S0022-314X(02)00118-X
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modulo P generates the residue fields KP as finite A-modules (under the Carlitz action). In other words, we study the set of places P where a is a ‘‘primitive root’’ mod P: In the case K ¼ Fq ðtÞ and i is the inclusion A+Fq ðtÞ; this problem has been worked out by Hsu [7], where an analogue of Artin’s (1927) primitive roots conjecture is proved. In this paper, we aim at examining the case of finite Acharacteristic, i.e. the case i has a nontrivial kernel P into the coefficient field K which can be arbitrary. The kernel P of the A-field K is called the A-characteristic. Thus, what we are looking for is the parallel of the case of Gm over finite Zcharacteristic function fields. Note that the finite Z-characteristic case is actually where Bilharz [1] (cf. also [11]) is able to give the first affirmative answer to the primitive roots problem. Let CðKÞ denote the additive group of K together with the A-module structure from the Carlitz action. Given aAK and let Ma denote the set of places P of K where a is a primitive root mod P: The main results of this paper are: existence of the Dirichlet density for Ma ; and a criterion for the positivity of this density dðMa Þ: In particular, we can prove that if a has a pole at some place P of order prime to q; then dðMa Þ40: On the other hand, if CðKÞ does not contain too many A-torsion points, then the positivity criterion can be simplified to that dðMa Þ40 if and only if a is not a ‘‘Pth power’’ whenever CðKÞ contains nontrivial P-torsion for monic irreducible PAA (cf. Theorem 5.7). In Section 3, an analogue of a theorem of Romanoff for the polynomial ring A is proved. This is needed in Section 4 in order to apply the analytic tool based on Ceˇbotarev’s density theorem and Weil’s Riemann hypothesis. Existence of the Dirichlet density is worked out in Section 4. Positivity of the density is treated in Section 5, where our method follows to some extent the idea from Lenstra [8].
2. Notations and preliminaries Definition 2.1. Let L be an A-field, that is L is a field together with a structure homomorphism i : A-L: If i is injective, the A-characteristic of L is defined to be N: If i is not injective, the A-characteristic of L is defined to be ker i: Write Aþ ðresp: Pþ ) for the set of all monic polynomials (resp. monic irreducible polynomials) in A: Let PAPþ : Then A=ðPÞ is a finite field which has qdegðPÞ elements. For convenience, denote A=ðPÞ by FP : Let L be an A-field with Acharacteristic ðPÞ via the map i : A-FP +L: Put tðxÞ ¼ xq and denote by Lftg the twisted polynomial ring with the commutation rule t a ¼ aq t for aAL: Recall that the Carlitz module defined over L is the action given by the Fq -linear ring homomorphism C : A-Lftg; Q/CQ
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such that Ct ðxÞ ¼ iðtÞx þ tðxÞ: Then C gives an A-module structure on ðL; þÞ: In particular, if K is an algebraic function field (of one variable) over FP ; then one obtains an A-module structure on ðK; þÞ from the action C: For the sake of simplicity, denote this A-module by CðKÞ: Let K be an algebraic function field over FP with constant field kK containing FP and let K 0 be a finite extension of K: Given any place P of K; put: OP ¼ the valuation ring of P; KP ¼ OP =P ¼ the residue field at P; degðPÞ ¼ ½KP : kK ; eðP0 j PÞ ¼ the ramification index of P0 over P; vP ¼ the normalized valuation corresponding to P; mK ¼ ½kK : FP and q ¼ qmK deg P ¼ #kK : Let PK be the set of all places of K: Given PAPK ; one also regards KP as A-field of A-characteristic P via A-FP +KP : The structure of CðKP Þ can be determined from the Carlitz module theory: Theorem 2.1. CðKP Þ is isomorphic to A=ðPmK deg P 1Þ as an A-module. Proof. See [4, Theorem 3.6.3].
&
It is important to know the Carlitz action of the polynomial PAA: Theorem 2.2. CP ¼ tdeg P ; i.e., CP ðxÞ ¼ xq Proof. See [4, Theorem 3.6.2].
deg P
:
&
Definition 2.2. Given aAOP ; we define aAK % P to be the residue class of a modulo the maximal ideal of OP : We say that aAK is a primitive root modulo a place P if aAOP and a% generates the residue field KP as an A-module. An immediate consequence of Theorem 2.1 is the following: Lemma 2.3. Given aAK and PAPK : Assume that vP ðaÞ ¼ 0: Then, a is a primitive root modulo P if and only if there is no irreducible polynomial PAPþ satisfying the following conditions: P j ðPmK deg P 1Þ and
CðPmK deg P 1Þ=P ðaÞ 0 ðmod PÞ:
Next, we observe that the constants are precisely the A torsion points:
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Proposition 2.4. Let aAK: Then a is an A-torsion points in CðKÞ if and only if aAkK : Proof. Assume that aAkK : From the definition of CQ ; CQ ðxÞ is a polynomial over FP : Hence CQ ðaÞAkK : Since kK is a finite field and A has infinitely many elements, there exist two different polynomials Q1 ; Q2 AA such that CQ1 ðaÞ ¼ CQ2 ðaÞ: This implies CQ1 Q2 ðaÞ ¼ 0: Conversely, assume that aAK is an A-torsion point, i.e. there is a Qa0AA such that CQ ðaÞ ¼ 0: Since CQ ðxÞ is a polynomial over FP ; a lies in the algebraic closure of FP : Hence aAkK : & Let K ac be a fixed algebraic closure of K: Given QAAþ ; we set LQ ¼ faAK ac j CQ ðaÞ ¼ 0g: If P[Q; then LQ is isomorphic to A=ðQÞ as A-module [5]. In particular, it is an elementary abelian p-group with cardinality jQj ¼ qdeg Q : The extension KðLQ Þ=K; obtained by adjoining all Q torsion points of the Carlitz module, is always a constant field extension by Proposition 2.4. Lemma 2.5. Let QAAþ such that P[Q; and let f ðQÞ be the multiplicative order of PmK modulo Q. Then ½KðLQ Þ : K ¼ f ðQÞ: Proof. First we observe that kKðLQ Þ ¼ kK ðLQ Þ and ½kKðLQ Þ : kK ¼ ½KðLQ Þ : K: From Q j ðPmK f ðQÞ 1Þ and Theorem 2.2 we obtain aq
f ðQÞ
¼ CPmK f ðQÞ ðaÞ ¼ a
C
mK ½kKðL Þ :kK Q P 1
for all aALQ : This f ðQÞ j ½KðLQ Þ : K: &
implies
ðaÞ ¼ aq QjP
½kKðL Þ :kK Q
½kKðL Þ :kK Q
½KðLQ Þ:K
¼ aq
and
therefore
for all aALQ : Hence ½KðLQ Þ : K j f ðQÞ: On the other hand, aq for all aAkKðLQ Þ : By Theorem 2.2
¼a
a¼0
mK ½kKðLQ Þ :kK
1
also
Corollary 2.6. The torsion A-submodule of CðKÞ is isomorphic to A=ðPmK 1Þ: Proof. By Theorem 2.1, kK is a submodule of a cyclic A-module and therefore is cyclic itself. By Theorem 2.2 and counting ones sees PmK 1 is its annihilator. &
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Remark. Let Q be a square-free polynomial in Aþ relatively prime to P: From Lemma 2.5, one deduces that f ðQÞ equals to lcmf f ðPÞ j PAPþ and P j Qg: It also follows that the degree ½KðLQ Þ : K is always prime to the characteristic p: Lemma 2.7. Given PAPK : Then P j ðPmK deg P 1Þ if and only if P splits completely in KðLP Þ: Proof. Since KðLP Þ=K is a constant field extension, P splits completely in KðLP Þ if deg P and only if LP CKP (i.e. aq ¼ a for all aALP ). By Theorem 2.2, this is equivalent to CPmK deg P 1 ðaÞ ¼ 0 for all aALP : From the definition of LP ; the condition amounts precisely to P j ðPmK deg P 1Þ: & Given PAPþ ; we also set CP 1 ðaÞ ¼ fxAkac j CP ðxÞ ¼ ag; and EP ¼ KðLP ; CP 1 ðaÞÞ: Clearly EP are Galois extensions of K: If QAAþ is squarefree, we define EQ as the compositum of the fields fEP : P j Qg: It is not difficult to see that EQ1 EQ2 ¼ KðLQ1 Q2 ; CQ 11 Q2 ðaÞÞ for all Q1 ; Q2 such that ðQ1 ; Q2 Þ ¼ 1: Moreover, we have GalðEQ =KðLQ ÞÞ+LQ ; so that EQ =KðLQ Þ are always elementary abelian p-extensions. Proposition 2.8. Given aAK and PAPK such that vP ðaÞ ¼ 0: Then a is a primitive root mod P if and only if P does not split completely in any of the fields EP for all PaPAPþ : Proof. It is easy to see that CP ðxÞ is a monic separable polynomial for all PaPAPþ : If a is a root of CP ðxÞ a; then every root of CP ðxÞ a has the form a þ l for some lALP : Hence, we may write EP ¼ KðLP ÞðaÞ: Suppose that P splits completely in EP for some PaPAPþ and let P0 be a place of KðLP Þ which lies above P: Then CP ðxÞ a ðmod P0 Þ is solvable in OP0 : Also P splits completely in KðLP Þ: This implies (Lemma 2.7) that P j ðPmK deg ðPÞ 1Þ and CP ðxÞ a ðmod PÞ is solvable in OP : According to Theorem 2.1, CPmK deg ðPÞ 1 ðbÞ 0 ðmod PÞ for all bAOP and therefore CðPmK deg P 1Þ=P ðaÞ CPmK deg P 1=P ðCP ðbÞÞ
CPmK deg P 1 ðbÞ 0 ðmod PÞ: Thus a is not a primitive root mod P; by Lemma 2.3.
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Conversely, if a is not a primitive root modulo P; by Lemma 2.3, there is PAPþ such that P j ðPmK deg P 1Þ and CP ðxÞ a ðmod PÞ solvable in OP : By Lemma 2.7, the condition P j ðPmK deg P 1Þ implies P splits completely in KðLP Þ: Given P0 APKðLP Þ above P: We also have CP ðxÞ a ðmod P0 Þ solvable in OP0 : Therefore P splits completely in EP : This completes the proof. & Given aAK; we are interested in the Dirichlet density of the set Ma ¼ fPAPK j a is a primitive root modulo Pg: If aAkK ; Ma is necessary a finite set. Hence from now on we assume that aAK\kK : Let Q be a square-free polynomial in Aþ : Let hðQÞ denote the degree ½EQ : KðLQ Þ: Our first main theorem to be proved in Section 4 is Theorem 2.9. The set Ma has a Dirichlet density dðMa Þ given by dðMa Þ ¼
X QAAþ P[Q
mðQÞ ; hðQÞf ðQÞ
where m is the Mo¨bius function for A:
3. Analogue of Romanoff’s theorem We start this section with an analogue of Merterns’ theorem for A whose proof is very close to that of the classical case, hence is omitted. Lemma 3.1. Let jQj ¼ qdeg Q for QAA: If d-N; then Y PAPþ deg Ppd
1 e g 1 þO 2 ; 1 ¼ jPj d d
where g denotes Euler’s constant. We recall that the zeta function of A is zðsÞ ¼
Y 1 1 ¼ 1 1 s 1 q 1 PAPþ jPjs
for RðsÞ41:
We have the following easy consequence from Lemma 3.1.
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Lemma 3.2. Let RAAþ be any polynomial of degree n. X mðQÞ2 pz 1 ð2Þeg logq n þ Oð1Þ: jQj QAA þ
QjR
The following theorem is an analogue for A of a theorem of Romanoff [9]. It can be easily deduced from Lemma 3.2 as in the classical case. Theorem 3.3. The series X QAAþ P[Q Q square-free
1 jQjf ðQÞ
converges.
4. Existence of the density In this section, we will prove Theorem 2.9. We begin with Definition 4.1. Given a: Let Ma be the set of places of K which do not split completely in any of the fields EP for all PaPAPþ : Also let Ia be the finite set of places of K where a has poles. On the basis of Proposition 2.8 we see that Ma differs from Ma by at most a finite set. In order to study the Dirichlet density of Ma ; we recall the following theorem from [3] (cf. also [1]). Given fKj gjAJ ; a countable family of Galois extensions of K: Let kj be the algebraic closure of kK in Kj : Set rj ¼ ½Kj : K; cj ¼ ½kj : kK ; and q ¼ #kK : Let gj be the genus of Kj : For any finite set of indices I ¼ f j1 ; j2 ; y; js g; we also define KI to be the compositum Kj1 ?Kjs ; and put mðIÞ ¼ ð 1Þ#ðIÞ : Using the generalized Riemann hypothesis proved by Weil and Ceˇbotarev’s density theorem, one establishes the following important. Theorem 4.1 (Bilharz [1]; Clark and Kuwata [3]). Suppose that the following conditions hold for the family fKj gjAJ : P 1 (1) jAJ rj oN; P 1 (2) cj =2 oN; and jAJ cj q
(3) there exists a constant c such that gj pccrj for all jAJ: j
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Let M be the set of places in K that do not split completely in any Kj for all jAJ: Then the Dirichlet density of M exists and is given by
dðMÞ ¼
X mðIÞ : ½KI : K I
We will apply Theorem 4.1 to the family F ¼ fEP gPAPþ : We first estimates the degree ½EQ : KðLQ Þ ¼ hðQÞ: Lemma 4.2. Given aAK\kK and QAAþ square-free with P[Q: If aeCQ ðKÞ; then there is a constant 0oca p1 such that ca jQjphðQÞpjQj: Furthermore, if a has pole at place P of K with vP ðaÞ prime to the characteristic p, then hðQÞ ¼ jQj: Proof. Let a be a root of CQ ðxÞ a: Let Ia ¼ fPAPK j vP ðaÞo0g: Given PAIa ; we have vP0 ðajQj Þ ¼ vP0 ðaÞ for every P0 APEQ lying above P: It follows that jQjvP0 ðaÞ ¼ vP ðaÞeðP0 j PÞ and therefore vP0 ðaÞ 1 pjQj ¼ eðP0 jPÞphðQÞpjQj jQj jvP ðaÞj vP ðaÞ for all PAIa : Here jvP ðaÞj is the absolute value of vP ðaÞ: We then take ca ¼ maxfjvP1ðaÞjgPAIa : If a has pole at place P of K with vP ðaÞ prime to p; then the ramification index over places above P has to be jQj: Hence hðQÞ ¼ jQj: & Remark. As a consequence of Theorem 3.3 and Lemma 4.2, we have X QAAþ P[Q Q square-free
1 oN: hðQÞf ðQÞ
Also one observes that the extension EQ =K is always unramified outside the set of places P where the given a is not regular. We will estimate the genus of the function fields EQ ; using the following classical inequality of Castelnuovo: Lemma 4.3. Let F =k be a function field with constant field k. Suppose there are given two subfields F1 =k and F2 =k satisfying (1) F ¼ F1 F2 ; (2) ½F : Fi ¼ ni ; Fi =k has genus gi for i ¼ 1; 2:
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Then the genus g of F =k is bounded by gpn1 g1 þ n2 g2 þ ðn1 1Þðn2 1Þ: Proof. See [10, Theorem III.10.3].
&
Theorem 4.4. Let QAAþ be square-free with P[Q: Let gEQ be the genus of EQ : Then gEQ pc0a hðQÞ; where c0a is a constant depending only on a and K. Proof. Let a be a root of CQ ðxÞ ¼ a; and F1 be the rational function field kEQ ðaÞ which contains the rational function field kK ðaÞ: Let F2 be the constant field extension K kEQ of K: Then EQ ¼ F1 F2 : Let gK be the genus of K: By Lemma 4.3, we have gEQ p ½EQ : F2 gK þ ð½EQ : F1 1Þð½EQ : F2 1Þ p hðQÞgK þ ½K kEQ ðaÞ : kK ðaÞ kEQ ðaÞ hðQÞ p hðQÞgK þ ½K : kK ðaÞ hðQÞ ¼ c0a hðQÞ:
&
Proposition 4.5. X PAPþ PaP
1 oN: qf ðPÞ=2 f ðPÞ
Proof. Write the series as X PAPþ PaP qf ðPÞ=2 XjPj
1 qf ðPÞ=2 f ðPÞ
þ
X PAPþ PaP qf ðPÞ=2 ojPj
1 qf ðPÞ=2 f ðPÞ
:
The first sum converges by Theorem 3.3. For those PAPþ such that PaP and qf ðPÞ=2 ojPj: Let P1 ; P2 ; y; Pr be the monic irreducible factors of ðPmK Þf ðPÞ 1 such that jPi j4qf ðPÞ=2 : We have P1 ?Pr j ððPmK Þf ðPÞ 1Þ and therefore jðPmK Þf ðPÞ jXjP1 ?Pr j:
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This implies r ¼ 1: In other words, such a P is uniquely determined by f ðPÞ: Hence this series X PAPþ PaP f ðPÞ=2 ojPj q
N X 1 1 p ¼ log ð1 1=2 Þ: qf ðPÞ=2 f ðPÞ f ¼1 qf =2 f q
1
&
Proof of Theorem 2.9. Put F ¼ fEP gPAPþ : Since EP =KðLP Þ is an abelian pextension, and KðLP Þ=K is a constant field extension of degree f ðPÞ; we deduce that f ðPÞpcP pp f ðPÞ: Condition (1) of Theorem 4.1 is verified by Theorem 3.3 and Lemma 4.2, condition (2) by Proposition 4.5, and condition (3) by Theorem 4.4. Hence this theorem is proved. &
5. Positivity of the density Given nonconstant aAK: If aACP ðKÞ for some monic irreducible P j ðPmK 1Þ (i.e. hðPÞ ¼ f ðPÞ ¼ 1), then CðPmK deg P 1Þ=P ðaÞ ¼ 0 whenever aAOP (by Theorem 2.1) and certainly a cannot be a primitive root modulo P for any place PAPK : However, if aeCP ðKÞ for all monic irreducible P j ðPmK 1Þ; it may still happen (Example 2 at the end), though not frequently (see Theorems 5.6 and 5.7), that Ma has zero density. The purpose of this section is to give a necessary and sufficient condition for Ma to have positive density. The first step is to control those square-free Q with hðQÞojQj: Lemma 5.1. There is a square-free NAAþ such that P[N and the following hold for all square-free polynomial Q with ðP N; QÞ ¼ 1: (1) hðQÞ ¼ jQj; (2) hðQNÞ ¼ hðQÞhðNÞ: Proof. It is not difficult to see that hðQ1 Q2 ÞphðQ1 ÞhðQ2 Þ for all Q1 ; Q2 AAþ such that ðQ1 ; Q2 Þ ¼ 1: Let p be the characteristic. By Lemma 4.2, one can find the smallest integer nX0 such that hðQÞXjQj pn for all QAAþ : Choose any NAAþ with hðNÞ ¼ jNj pn : Given square-free polynomial Q with ðPN; QÞ ¼ 1; if hðQÞojQj; or hðQNÞohðQÞhðNÞ then hðQNÞphðQÞhðNÞojQNj contradicting to our pn ; choice of n: Also it is not possible that hðQNÞohðQÞhðNÞ: This completes the proof. & Lemma 5.2. Let N be a square-free polynomial satisfying the properties in Lemma 5.1. Then for all square-free Q in Aþ with ðPN; QÞ ¼ 1; the following are true: (1) EN -EQ ¼ EN -KðLQ Þ ¼ KðLN Þ-KðLQ Þ;
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(2) ½ENQ : K ¼ ½EN : KjQj½EN ðLQ Þ : EN ; (3) if Q ¼ Q1 Q2 ; then ðEN -EQ1 Þ ðEN -EQ2 Þ ¼ EN -EQ : Proof. (1) Since hðNQÞ ¼ hðNÞhðQÞ; we have EQ ðLN Þ and EN ðLQ Þ are linearly disjoint over KðLQ Þ KðLN Þ: This implies that EN -EQ must be a constant field extension of KðLN Þ-KðLQ Þ with degree prime to the characteristic p: Hence EN -EQ ¼ EN -KðLQ Þ ¼ KðLN Þ-KðLQ Þ: (2) Observe that KðLNQ Þ and EN are linearly disjoint over KðLN Þ: Hence ½EN ðLQ Þ : KðLNQ Þ ¼ hðNÞ: By Lemma 5.1(2), we deduce that ½ENQ : EN ðLQ Þ ¼ jQj: Thus ½ENQ : K ¼ jQj ½EN ðLQ Þ : EN ½EN : K: (3) Let G be the cyclic group GalðKðLQ Þ=KÞ: Define the subgroups H1 ; H2 ; H of G by Hi ¼ GalðKðLQ Þ=KðLQi ÞÞ; i ¼ 1; 2; H ¼ GalðKðLQ Þ=EN -KðLQ ÞÞ: Since Q is square-free, we have ðQ1 ; Q2 Þ ¼ 1: Hence H1 -H2 ¼ 1; therefore #H1 and #H2 are relatively prime. It follows that also the index of H in H H1 is relatively prime to the index of H in H H2 : Consequently H H1 -H H2 ¼ H: This implies ðEN -KðLQ1 ÞÞ ðEN -KðLQ2 ÞÞ ¼ EN -KðLQ Þ: By (1), we have ðEN -EQ1 Þ ðEN -EQ2 Þ ¼ EN -EQ : & Given a square-free polynomial QAAþ ; P[Q; we define SQ ¼ fsAGalðEQ =KÞ j sjEP aidEP for all PAPþ ; P j Qg and we put sQ ¼
#SQ #SQ ¼ : #GalðEQ =KÞ ½EQ : K
Proposition 5.3. We have sQ ¼
X RjQ
mðRÞ : ½ER : K
Proof. For R j Q; put DR ¼ fsAGalðEQ =KÞ j sjEL ¼ idEL for all L j Rg: It is not difficult to see that #SQ ¼
X RjQ
mðRÞ #DR :
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Since DR ¼ GalðEQ =ER Þ; we obtain P X mðRÞ R j Q mðRÞ #GalðEQ =ER Þ ¼ : sQ ¼ R j Q ½E : K ½EQ : K R
99
&
Remark. If Q1 j Q2 ; then sQ1 XsQ2 X0: It follows that the sequence fsQ g has a limit if Q ranges over all square-free monic polynomials not divided by P; ordered by divisibility. Combining Proposition 5.3 and Theorem 2.9, we have lim sQ ¼ dðMa Þ: Q
Our proof of positivity relies on the following: Lemma 5.4 (Heilbronn). Let f; c be two functions defined on square-free polynomials in Aþ such that 0pfðQÞp1 and cðQÞAN for all QAAþ : Moreover, we assume that fðQ1 Q2 Þ ¼ fðQ1 ÞfðQ2 Þ and cðQ1 Q2 Þ ¼ lcmðcðQ1 Þ; cðQ2 ÞÞ for Q1 ; Q2 AAþ with ðQ1 ; Q2 Þ ¼ 1: Then we have X mðQÞfðQÞ Y fðPÞ X 1 : cðQÞ cðPÞ QjQ PjQ QAAþ
PAPþ
The proof of this elementary lemma can be found in [6]. We arrive finally at the following criterion: Theorem 5.5. Let aAK\kK and let N0 AAþ be the product of all distinct irreducible factors of PmK 1: Then dðMa Þ40 if and only if SN0 a|: Proof. It is clear that if dðMa Þ40; all SQ a|: In particular SN0 a|: Conversely if SN0 a|; we claim that SQ a| for all QAAþ such that Q ¼ N0 R for some R relatively prime to N0 P: Hence, for every square-free polynomial QAAþ with P[Q; we may write Q ¼ Q1 Q2 with Q1 j N0 and ðQ2 ; N0 PÞ ¼ 1: Then sQ XsN0 Q2 40: Consider monic irreducible P j N0 : Then KðLP Þ ¼ K by Lemma 2.5 so that EP is an elementary abelian p-extension of K: Thus EN0 =K is elementary abelian pextension. Given Q ¼ N0 R for some R relatively prime to N0 P: We let s0 AGalðKðLR Þ=KÞ be the automorphism such that s0 jkK ðLR Þ is the generator of GalðkK ðLR Þ=kK Þ: Then s0 jKðLP Þ aid for all P dividing R; because kK ðLP ÞakK : Since EN0 -KðLR Þ ¼ K; one can first extend s0 to EN0 KðLR Þ by letting it coincide with an element in SN0 on EN0 : Then any extension of s0 from EN0 KðLR Þ to EQ will give an element of SQ : Hence we always have SQ a|: To complete the proof, let N be defined as in Lemma 5.2 and let QAAþ be squarefree such that ðPN; QÞ ¼ 1: For tASN ; define
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SQ ðtÞ ¼ fsAGalðENQ =KÞjsjEN ¼ t
sjEP aidEP for all P j Qg;
and
sQ ðtÞ ¼ #SQ ðtÞ=½EN : K
and
sðtÞ ¼ lim sQ ðtÞ Q
the limit being taken over all square-free monic polynomials relatively prime to PN; ordered by divisibility. Clearly, we have SNQ ¼
a
SQ ðtÞ;
tASN
sNQ ¼
X
sQ ðtÞ;
tASN
dðMa Þ ¼
X
sðtÞ:
tASN
We will show that sðtÞ40 for every tASN : Since SN is not empty, the desired inequality, dðMa Þ40; follows from the last inequality displayed above. Put ( cðt; QÞ ¼
1 if tAGalðEN =ðEN -EQ ÞÞ; 0 otherwise:
By Lemma 5.2(3), cðt; QÞ ¼ cðt; Q1 Þcðt; Q2 Þ if Q ¼ Q1 Q2 : Hence sQ ðtÞ ¼
X mðRÞcðt; RÞ RjQ
½ENQ : K
:
Following Lemma 5.2(2) sQ ðtÞ ¼
X mðRÞcðt; RÞ jRj 1 1 : ½EN : K R j Q ½EN ðLR Þ : EN
Put fðRÞ ¼ cðt; RÞjQj 1 ; cðRÞ ¼ ½EN ðLR Þ : EN : We have sQ ðtÞ ¼
X mðRÞfðRÞ 1 : ½EN : K R j Q cðRÞ
Since fðQ1 Q2 Þ ¼ fðQ1 ÞfðQ2 Þ and cðQ1 Q2 Þ ¼ lcmðcðQ1 Þ; cðQ2 ÞÞ for ðQ1 ; Q2 Þ ¼ 1; we have Y 1 fðPÞ 1 sQ ðtÞX ½EN : K PjQ cðPÞ PaP PAPþ
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by Lemma 5.4. The infinite product Y Y fðPÞ cðt; PÞ 1 1 ¼ cðPÞ jPj½EN ðLP Þ : EN P[PQ P[PN PAPþ
PAPþ
is convergent by Theorem 3.3. Hence Y 1 cðt; PÞ sðtÞX 1 40: ½EN : K P[PN jPj½EN ðLP Þ : EN
&
PAPþ
From our criterion, we can deduce Theorem 5.6. Let P be a place of K. Given any aAK which has pole at P with vP ðaÞ prime to the characteristic p. Then dðMa Þ40: Proof. By Lemma 4.2, our assumption gives hðQÞ ¼ jQj for all QAAþ : In particular, GalðEN0 =KÞ is the direct product of nontrivial groups GalðEP =KÞ with P j N0 : Hence SN0 a|: Applying Theorem 5.5 we obtain immediately dðMa Þ40: & On the other hand, when the A-module CðKÞ does not contain too many torsion points, we also have the following simpler necessary and sufficient condition for the positivity of the density. Theorem 5.7. Assume that the number of PAPþ such that PaP and LP CK; is less than p þ 1: Given aAK\kK : Then dðMa Þ40 if and only if aeCP ðKÞ for all monic irreducible P dividing PmK 1: Proof. If aACP ðKÞ for some monic irreducible P j ðPmK 1Þ; by Theorem 2.1 it is not difficult to see that dðMa Þ vanishes. Conversely, let N0 be the product of all distinct monic irreducible factors of PmK 1: If aeCP ðKÞ for P j N0 ; then EP =K is a proper extension, hence SP a| for all irreducible monic P j N0 : In view of Theorem 5.5, it is enough to prove SN0 a|: By Corollary 2.6, for monic irreducible P; LP CK if and only if P divides N0 : It follows that the extensions EP =K in question are always elementary abelian pextensions. We now choose a family of proper extensions EP0 =K; with P ranges through monic irreducibles dividing N0 ; EP0 CEP for each P; and EP0 =K is of degree p: To show that SN0 a|; it suffices to find sAGalðEN0 =KÞ whose restriction to each EP0 ; P j N0 irreducible, is not the identity. The number n0 of distinct fields EP0 in the family is no greater than the number of irreducible factors of N0 ; which is less than p þ 1 by our assumption. Let E 0 be the compositum of all the fields EP0 ; with P j N0 irreducible. Then GalðE 0 =KÞ ¼ V is a finite dimensional Fp -vector space. The subfield EP0 corresponds to the subspace GalðE 0 =EP0 Þ which is of codimension one in V : If V has dimension one, then extending any nonidentity automorphism in
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GalðE 0 =KÞ to EN0 gives element in SN0 : If V has dimension 41; then there are Xp þ 1 subspaces of V having codimension one. Since n0 op þ 1; we can find sAV such that seGalðE 0 =EP0 Þ for all irreducible P j N0 : It follows that this s can be extended to an element in SN0 : Thus, we have shown that SN0 is always nonempty which completes our proof. & Example 1. Let p ¼ q ¼ 2; and suppose K is an A field of A characteristic P satisfying 2pdeg Pp3 and kK ¼ FP : The monic irreducible factors of PmK 1 are t and t þ 1: Given nonconstant aAK: We have dðMa Þ40 if and only if aeCt ðKÞ,Ctþ1 ðKÞ: Note that elements in Ct ðKÞ,Ctþ1 ðKÞ are considered as the ‘‘squares’’ in our A-module CðKÞ; since jtj ¼ jt þ 1j ¼ 2: Finally, we close with an example which shows that the hypothesis CðKÞ does not contain ‘‘too many’’ A torsion points in Theorem 5.7 is indeed necessary: Example 2. Let p ¼ q ¼ 2 and let P ¼ t4 þ t þ 1: The monic irreducible factors of P 1 ¼ N0 are t; t þ 1; and t2 þ t þ 1: We will exhibit a function field K together with aAK such that SN0 ¼ | (hence dðMa Þ ¼ 0), although aeCP ðKÞ for all monic irreducible P dividing PmK 1: For convenience, we denote iðtÞ by a which is primitive in FP : We may write Ct ðxÞ ¼ ax þ x2 ; Ctþ1 ðxÞ ¼ a4 x þ x2 ; and Ct2 þtþ1 ðxÞ ¼ a10 x þ a10 x2 þ x4 : By an easy computation, also that Ct2 þtþ1 ðxÞ ¼ C1 3C2 ðxÞ where C1 ðxÞ ¼ a3 x þ x2 and C2 ðxÞ ¼ a7 x þ x2 : Let k ¼ FP ðyÞ be a rational function field over FP and let a ¼ y2 þ a7 y ¼ C2 ðyÞ: Introducing: x: a root of Ct ðxÞ ¼ a; r: a root of Ctþ1 ðxÞ ¼ a; d: a root of x2 þ a3 x ¼ a11 ðy2 þ a3 yÞ; d: a root of C1 ðxÞ ¼ y: (In fact, d ¼ a2 x þ a14 r þ d:) After elementary computations basing on Artin–Schreier theory one can show that kðx; rÞ=k; kðx; dÞ=k and kðd; rÞ=k are all extensions of type (2,2). Moreover kðdÞgkðx; rÞ: We take K ¼ kðdÞ: It follows Et ¼ KðxÞ ¼ kðd; xÞ; Etþ1 ¼ KðrÞ ¼ kðd; rÞ; and Et2 þtþ1 ¼ KðdÞ ¼ kðd; dÞCEt Etþ1 : We have aeCP ðKÞ for all monic irreducible P dividing N0 ¼ PmK 1: We contend that both Et2 þtþ1 Etþ1 and Et2 þtþ1 Et are extensions of type (2, 2) over K: This implies that SN0 ¼ |: To show that Et2 þtþ1 Et is an extension of type (2, 2) over K; it is enough to check xekðd; dÞ: Suppose this is not the case. Since d ¼ a2 x þ a14 r þ d; we obtain rAEt2 þtþ1 : Thus kðx; rÞCEt2 þtþ1 : Consequently kðx; rÞ ¼ Et2 þtþ1 which contradicts the fact that kðdÞgkðx; rÞ: By a similar argument, we can also deduce that Et2 þtþ1 Et is an extension of type (2, 2) over K:
References [1] H. Bilharz, Primdivisoren mit vorgegebener Primitivwu¨rzel, Math. Ann. 114 (1937) 476–492. [2] L. Carlitz, A class of polynomials, Trans. Amer. Math. Soc. 43 (1938) 167–182.
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[3] D.A. Clark, M. Kuwata, Generalized Artin’s conjecture for primitive roots and cyclicity mod P of Elliptic curves over function fields, Canad. Math. Bull. 38 (2) (1995) 167–173. [4] D. Goss, Basic Structures of Function field Arithmetic, Springer, Berlin, 1996. [5] D. Hayes, Explicit class field theory for rational function fields, Trans. Amer. Math. Soc. 189 (1974) 77–91. [6] H. Heilbronn, On an inequality in the elementary theory of numbers, Proc. Cambridge Philos. Soc. 33 (1937) 207–209. [7] C.-N. Hsu, On Artin’s conjecture for Carlitz modules, Compositio Math. 106 (1997) 247–266. [8] H.W. Lenstra, On Artin’s conjecture and Euclidean algorithm in global fields, Invent. Math. 42 (1977) 201–224. [9] N.P. Romanoff, Uber einiger Sa¨tze der Additive ZahlenTheorie, Math. Ann. 109 (1934) 668–678. [10] H. Stichtenoth, Algebraic Function Fields and Codes, Springer, Berlin, 1994. [11] Jing Yu, On a theorem of Bilharz, preprint.
http://www.elsevier.com/locate/jnt
On primitive roots for Carlitz modules Wei-Chen Yaoa,* and Jing Yub a
Department of Mathematics and Computer Science Education, Taipei Municipal Teachers College, No. 1, Aikuo West Road, Taipei 100, Taiwan, ROC b Department of Mathematics, National Tsing-Hua University, No. 101, Sec. 2, Kuang Fu Road, Hsinchu 30043, Taiwan, ROC Received 28 November 2001 Communicated by D. Goss
Abstract Let A be the polynomial ring over a finite field. We prove that for every element a of a global A-field of finite A-characteristic the set of places P for which a is a primitive root under the Carlitz action possesses a Dirichlet density. We also give a criterion for this density to be positive. This is an analogue of Bilharz’ version of the primitive roots conjecture of Artin, with Gm replaced by the Carlitz module. r 2003 Elsevier Science (USA). All rights reserved. MSC: primary 11T55 Keywords: Primitive root; Carlitz module
1. Introduction Let Fq be the finite field of q elements, Fq ðtÞ the rational function field in one variable, and put A ¼ Fq ½t: We are interested in the arithmetic of Fq ðtÞ with elements of the ring A playing the role of integers, and the Carlitz module C (cf. [2]) playing the role of the multiplicative group Gm : Here the Carlitz module is the action C of A on the additive (rather than multiplicative) group given by Ct ðxÞ ¼ iðtÞx þ xq ; where i is a specified morphism from A into the global coefficient field. Given an element a of the coefficient field K: We want to investigate the set of places P of K; for which a
*Corresponding author. E-mail addresses: [email protected] (W.-C. Yao), [email protected] (J. Yu). 0022-314X/03/$ - see front matter r 2003 Elsevier Science (USA). All rights reserved. doi:10.1016/S0022-314X(02)00118-X
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modulo P generates the residue fields KP as finite A-modules (under the Carlitz action). In other words, we study the set of places P where a is a ‘‘primitive root’’ mod P: In the case K ¼ Fq ðtÞ and i is the inclusion A+Fq ðtÞ; this problem has been worked out by Hsu [7], where an analogue of Artin’s (1927) primitive roots conjecture is proved. In this paper, we aim at examining the case of finite Acharacteristic, i.e. the case i has a nontrivial kernel P into the coefficient field K which can be arbitrary. The kernel P of the A-field K is called the A-characteristic. Thus, what we are looking for is the parallel of the case of Gm over finite Zcharacteristic function fields. Note that the finite Z-characteristic case is actually where Bilharz [1] (cf. also [11]) is able to give the first affirmative answer to the primitive roots problem. Let CðKÞ denote the additive group of K together with the A-module structure from the Carlitz action. Given aAK and let Ma denote the set of places P of K where a is a primitive root mod P: The main results of this paper are: existence of the Dirichlet density for Ma ; and a criterion for the positivity of this density dðMa Þ: In particular, we can prove that if a has a pole at some place P of order prime to q; then dðMa Þ40: On the other hand, if CðKÞ does not contain too many A-torsion points, then the positivity criterion can be simplified to that dðMa Þ40 if and only if a is not a ‘‘Pth power’’ whenever CðKÞ contains nontrivial P-torsion for monic irreducible PAA (cf. Theorem 5.7). In Section 3, an analogue of a theorem of Romanoff for the polynomial ring A is proved. This is needed in Section 4 in order to apply the analytic tool based on Ceˇbotarev’s density theorem and Weil’s Riemann hypothesis. Existence of the Dirichlet density is worked out in Section 4. Positivity of the density is treated in Section 5, where our method follows to some extent the idea from Lenstra [8].
2. Notations and preliminaries Definition 2.1. Let L be an A-field, that is L is a field together with a structure homomorphism i : A-L: If i is injective, the A-characteristic of L is defined to be N: If i is not injective, the A-characteristic of L is defined to be ker i: Write Aþ ðresp: Pþ ) for the set of all monic polynomials (resp. monic irreducible polynomials) in A: Let PAPþ : Then A=ðPÞ is a finite field which has qdegðPÞ elements. For convenience, denote A=ðPÞ by FP : Let L be an A-field with Acharacteristic ðPÞ via the map i : A-FP +L: Put tðxÞ ¼ xq and denote by Lftg the twisted polynomial ring with the commutation rule t a ¼ aq t for aAL: Recall that the Carlitz module defined over L is the action given by the Fq -linear ring homomorphism C : A-Lftg; Q/CQ
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such that Ct ðxÞ ¼ iðtÞx þ tðxÞ: Then C gives an A-module structure on ðL; þÞ: In particular, if K is an algebraic function field (of one variable) over FP ; then one obtains an A-module structure on ðK; þÞ from the action C: For the sake of simplicity, denote this A-module by CðKÞ: Let K be an algebraic function field over FP with constant field kK containing FP and let K 0 be a finite extension of K: Given any place P of K; put: OP ¼ the valuation ring of P; KP ¼ OP =P ¼ the residue field at P; degðPÞ ¼ ½KP : kK ; eðP0 j PÞ ¼ the ramification index of P0 over P; vP ¼ the normalized valuation corresponding to P; mK ¼ ½kK : FP and q ¼ qmK deg P ¼ #kK : Let PK be the set of all places of K: Given PAPK ; one also regards KP as A-field of A-characteristic P via A-FP +KP : The structure of CðKP Þ can be determined from the Carlitz module theory: Theorem 2.1. CðKP Þ is isomorphic to A=ðPmK deg P 1Þ as an A-module. Proof. See [4, Theorem 3.6.3].
&
It is important to know the Carlitz action of the polynomial PAA: Theorem 2.2. CP ¼ tdeg P ; i.e., CP ðxÞ ¼ xq Proof. See [4, Theorem 3.6.2].
deg P
:
&
Definition 2.2. Given aAOP ; we define aAK % P to be the residue class of a modulo the maximal ideal of OP : We say that aAK is a primitive root modulo a place P if aAOP and a% generates the residue field KP as an A-module. An immediate consequence of Theorem 2.1 is the following: Lemma 2.3. Given aAK and PAPK : Assume that vP ðaÞ ¼ 0: Then, a is a primitive root modulo P if and only if there is no irreducible polynomial PAPþ satisfying the following conditions: P j ðPmK deg P 1Þ and
CðPmK deg P 1Þ=P ðaÞ 0 ðmod PÞ:
Next, we observe that the constants are precisely the A torsion points:
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Proposition 2.4. Let aAK: Then a is an A-torsion points in CðKÞ if and only if aAkK : Proof. Assume that aAkK : From the definition of CQ ; CQ ðxÞ is a polynomial over FP : Hence CQ ðaÞAkK : Since kK is a finite field and A has infinitely many elements, there exist two different polynomials Q1 ; Q2 AA such that CQ1 ðaÞ ¼ CQ2 ðaÞ: This implies CQ1 Q2 ðaÞ ¼ 0: Conversely, assume that aAK is an A-torsion point, i.e. there is a Qa0AA such that CQ ðaÞ ¼ 0: Since CQ ðxÞ is a polynomial over FP ; a lies in the algebraic closure of FP : Hence aAkK : & Let K ac be a fixed algebraic closure of K: Given QAAþ ; we set LQ ¼ faAK ac j CQ ðaÞ ¼ 0g: If P[Q; then LQ is isomorphic to A=ðQÞ as A-module [5]. In particular, it is an elementary abelian p-group with cardinality jQj ¼ qdeg Q : The extension KðLQ Þ=K; obtained by adjoining all Q torsion points of the Carlitz module, is always a constant field extension by Proposition 2.4. Lemma 2.5. Let QAAþ such that P[Q; and let f ðQÞ be the multiplicative order of PmK modulo Q. Then ½KðLQ Þ : K ¼ f ðQÞ: Proof. First we observe that kKðLQ Þ ¼ kK ðLQ Þ and ½kKðLQ Þ : kK ¼ ½KðLQ Þ : K: From Q j ðPmK f ðQÞ 1Þ and Theorem 2.2 we obtain aq
f ðQÞ
¼ CPmK f ðQÞ ðaÞ ¼ a
C
mK ½kKðL Þ :kK Q P 1
for all aALQ : This f ðQÞ j ½KðLQ Þ : K: &
implies
ðaÞ ¼ aq QjP
½kKðL Þ :kK Q
½kKðL Þ :kK Q
½KðLQ Þ:K
¼ aq
and
therefore
for all aALQ : Hence ½KðLQ Þ : K j f ðQÞ: On the other hand, aq for all aAkKðLQ Þ : By Theorem 2.2
¼a
a¼0
mK ½kKðLQ Þ :kK
1
also
Corollary 2.6. The torsion A-submodule of CðKÞ is isomorphic to A=ðPmK 1Þ: Proof. By Theorem 2.1, kK is a submodule of a cyclic A-module and therefore is cyclic itself. By Theorem 2.2 and counting ones sees PmK 1 is its annihilator. &
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Remark. Let Q be a square-free polynomial in Aþ relatively prime to P: From Lemma 2.5, one deduces that f ðQÞ equals to lcmf f ðPÞ j PAPþ and P j Qg: It also follows that the degree ½KðLQ Þ : K is always prime to the characteristic p: Lemma 2.7. Given PAPK : Then P j ðPmK deg P 1Þ if and only if P splits completely in KðLP Þ: Proof. Since KðLP Þ=K is a constant field extension, P splits completely in KðLP Þ if deg P and only if LP CKP (i.e. aq ¼ a for all aALP ). By Theorem 2.2, this is equivalent to CPmK deg P 1 ðaÞ ¼ 0 for all aALP : From the definition of LP ; the condition amounts precisely to P j ðPmK deg P 1Þ: & Given PAPþ ; we also set CP 1 ðaÞ ¼ fxAkac j CP ðxÞ ¼ ag; and EP ¼ KðLP ; CP 1 ðaÞÞ: Clearly EP are Galois extensions of K: If QAAþ is squarefree, we define EQ as the compositum of the fields fEP : P j Qg: It is not difficult to see that EQ1 EQ2 ¼ KðLQ1 Q2 ; CQ 11 Q2 ðaÞÞ for all Q1 ; Q2 such that ðQ1 ; Q2 Þ ¼ 1: Moreover, we have GalðEQ =KðLQ ÞÞ+LQ ; so that EQ =KðLQ Þ are always elementary abelian p-extensions. Proposition 2.8. Given aAK and PAPK such that vP ðaÞ ¼ 0: Then a is a primitive root mod P if and only if P does not split completely in any of the fields EP for all PaPAPþ : Proof. It is easy to see that CP ðxÞ is a monic separable polynomial for all PaPAPþ : If a is a root of CP ðxÞ a; then every root of CP ðxÞ a has the form a þ l for some lALP : Hence, we may write EP ¼ KðLP ÞðaÞ: Suppose that P splits completely in EP for some PaPAPþ and let P0 be a place of KðLP Þ which lies above P: Then CP ðxÞ a ðmod P0 Þ is solvable in OP0 : Also P splits completely in KðLP Þ: This implies (Lemma 2.7) that P j ðPmK deg ðPÞ 1Þ and CP ðxÞ a ðmod PÞ is solvable in OP : According to Theorem 2.1, CPmK deg ðPÞ 1 ðbÞ 0 ðmod PÞ for all bAOP and therefore CðPmK deg P 1Þ=P ðaÞ CPmK deg P 1=P ðCP ðbÞÞ
CPmK deg P 1 ðbÞ 0 ðmod PÞ: Thus a is not a primitive root mod P; by Lemma 2.3.
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Conversely, if a is not a primitive root modulo P; by Lemma 2.3, there is PAPþ such that P j ðPmK deg P 1Þ and CP ðxÞ a ðmod PÞ solvable in OP : By Lemma 2.7, the condition P j ðPmK deg P 1Þ implies P splits completely in KðLP Þ: Given P0 APKðLP Þ above P: We also have CP ðxÞ a ðmod P0 Þ solvable in OP0 : Therefore P splits completely in EP : This completes the proof. & Given aAK; we are interested in the Dirichlet density of the set Ma ¼ fPAPK j a is a primitive root modulo Pg: If aAkK ; Ma is necessary a finite set. Hence from now on we assume that aAK\kK : Let Q be a square-free polynomial in Aþ : Let hðQÞ denote the degree ½EQ : KðLQ Þ: Our first main theorem to be proved in Section 4 is Theorem 2.9. The set Ma has a Dirichlet density dðMa Þ given by dðMa Þ ¼
X QAAþ P[Q
mðQÞ ; hðQÞf ðQÞ
where m is the Mo¨bius function for A:
3. Analogue of Romanoff’s theorem We start this section with an analogue of Merterns’ theorem for A whose proof is very close to that of the classical case, hence is omitted. Lemma 3.1. Let jQj ¼ qdeg Q for QAA: If d-N; then Y PAPþ deg Ppd
1 e g 1 þO 2 ; 1 ¼ jPj d d
where g denotes Euler’s constant. We recall that the zeta function of A is zðsÞ ¼
Y 1 1 ¼ 1 1 s 1 q 1 PAPþ jPjs
for RðsÞ41:
We have the following easy consequence from Lemma 3.1.
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Lemma 3.2. Let RAAþ be any polynomial of degree n. X mðQÞ2 pz 1 ð2Þeg logq n þ Oð1Þ: jQj QAA þ
QjR
The following theorem is an analogue for A of a theorem of Romanoff [9]. It can be easily deduced from Lemma 3.2 as in the classical case. Theorem 3.3. The series X QAAþ P[Q Q square-free
1 jQjf ðQÞ
converges.
4. Existence of the density In this section, we will prove Theorem 2.9. We begin with Definition 4.1. Given a: Let Ma be the set of places of K which do not split completely in any of the fields EP for all PaPAPþ : Also let Ia be the finite set of places of K where a has poles. On the basis of Proposition 2.8 we see that Ma differs from Ma by at most a finite set. In order to study the Dirichlet density of Ma ; we recall the following theorem from [3] (cf. also [1]). Given fKj gjAJ ; a countable family of Galois extensions of K: Let kj be the algebraic closure of kK in Kj : Set rj ¼ ½Kj : K; cj ¼ ½kj : kK ; and q ¼ #kK : Let gj be the genus of Kj : For any finite set of indices I ¼ f j1 ; j2 ; y; js g; we also define KI to be the compositum Kj1 ?Kjs ; and put mðIÞ ¼ ð 1Þ#ðIÞ : Using the generalized Riemann hypothesis proved by Weil and Ceˇbotarev’s density theorem, one establishes the following important. Theorem 4.1 (Bilharz [1]; Clark and Kuwata [3]). Suppose that the following conditions hold for the family fKj gjAJ : P 1 (1) jAJ rj oN; P 1 (2) cj =2 oN; and jAJ cj q
(3) there exists a constant c such that gj pccrj for all jAJ: j
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Let M be the set of places in K that do not split completely in any Kj for all jAJ: Then the Dirichlet density of M exists and is given by
dðMÞ ¼
X mðIÞ : ½KI : K I
We will apply Theorem 4.1 to the family F ¼ fEP gPAPþ : We first estimates the degree ½EQ : KðLQ Þ ¼ hðQÞ: Lemma 4.2. Given aAK\kK and QAAþ square-free with P[Q: If aeCQ ðKÞ; then there is a constant 0oca p1 such that ca jQjphðQÞpjQj: Furthermore, if a has pole at place P of K with vP ðaÞ prime to the characteristic p, then hðQÞ ¼ jQj: Proof. Let a be a root of CQ ðxÞ a: Let Ia ¼ fPAPK j vP ðaÞo0g: Given PAIa ; we have vP0 ðajQj Þ ¼ vP0 ðaÞ for every P0 APEQ lying above P: It follows that jQjvP0 ðaÞ ¼ vP ðaÞeðP0 j PÞ and therefore vP0 ðaÞ 1 pjQj ¼ eðP0 jPÞphðQÞpjQj jQj jvP ðaÞj vP ðaÞ for all PAIa : Here jvP ðaÞj is the absolute value of vP ðaÞ: We then take ca ¼ maxfjvP1ðaÞjgPAIa : If a has pole at place P of K with vP ðaÞ prime to p; then the ramification index over places above P has to be jQj: Hence hðQÞ ¼ jQj: & Remark. As a consequence of Theorem 3.3 and Lemma 4.2, we have X QAAþ P[Q Q square-free
1 oN: hðQÞf ðQÞ
Also one observes that the extension EQ =K is always unramified outside the set of places P where the given a is not regular. We will estimate the genus of the function fields EQ ; using the following classical inequality of Castelnuovo: Lemma 4.3. Let F =k be a function field with constant field k. Suppose there are given two subfields F1 =k and F2 =k satisfying (1) F ¼ F1 F2 ; (2) ½F : Fi ¼ ni ; Fi =k has genus gi for i ¼ 1; 2:
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Then the genus g of F =k is bounded by gpn1 g1 þ n2 g2 þ ðn1 1Þðn2 1Þ: Proof. See [10, Theorem III.10.3].
&
Theorem 4.4. Let QAAþ be square-free with P[Q: Let gEQ be the genus of EQ : Then gEQ pc0a hðQÞ; where c0a is a constant depending only on a and K. Proof. Let a be a root of CQ ðxÞ ¼ a; and F1 be the rational function field kEQ ðaÞ which contains the rational function field kK ðaÞ: Let F2 be the constant field extension K kEQ of K: Then EQ ¼ F1 F2 : Let gK be the genus of K: By Lemma 4.3, we have gEQ p ½EQ : F2 gK þ ð½EQ : F1 1Þð½EQ : F2 1Þ p hðQÞgK þ ½K kEQ ðaÞ : kK ðaÞ kEQ ðaÞ hðQÞ p hðQÞgK þ ½K : kK ðaÞ hðQÞ ¼ c0a hðQÞ:
&
Proposition 4.5. X PAPþ PaP
1 oN: qf ðPÞ=2 f ðPÞ
Proof. Write the series as X PAPþ PaP qf ðPÞ=2 XjPj
1 qf ðPÞ=2 f ðPÞ
þ
X PAPþ PaP qf ðPÞ=2 ojPj
1 qf ðPÞ=2 f ðPÞ
:
The first sum converges by Theorem 3.3. For those PAPþ such that PaP and qf ðPÞ=2 ojPj: Let P1 ; P2 ; y; Pr be the monic irreducible factors of ðPmK Þf ðPÞ 1 such that jPi j4qf ðPÞ=2 : We have P1 ?Pr j ððPmK Þf ðPÞ 1Þ and therefore jðPmK Þf ðPÞ jXjP1 ?Pr j:
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This implies r ¼ 1: In other words, such a P is uniquely determined by f ðPÞ: Hence this series X PAPþ PaP f ðPÞ=2 ojPj q
N X 1 1 p ¼ log ð1 1=2 Þ: qf ðPÞ=2 f ðPÞ f ¼1 qf =2 f q
1
&
Proof of Theorem 2.9. Put F ¼ fEP gPAPþ : Since EP =KðLP Þ is an abelian pextension, and KðLP Þ=K is a constant field extension of degree f ðPÞ; we deduce that f ðPÞpcP pp f ðPÞ: Condition (1) of Theorem 4.1 is verified by Theorem 3.3 and Lemma 4.2, condition (2) by Proposition 4.5, and condition (3) by Theorem 4.4. Hence this theorem is proved. &
5. Positivity of the density Given nonconstant aAK: If aACP ðKÞ for some monic irreducible P j ðPmK 1Þ (i.e. hðPÞ ¼ f ðPÞ ¼ 1), then CðPmK deg P 1Þ=P ðaÞ ¼ 0 whenever aAOP (by Theorem 2.1) and certainly a cannot be a primitive root modulo P for any place PAPK : However, if aeCP ðKÞ for all monic irreducible P j ðPmK 1Þ; it may still happen (Example 2 at the end), though not frequently (see Theorems 5.6 and 5.7), that Ma has zero density. The purpose of this section is to give a necessary and sufficient condition for Ma to have positive density. The first step is to control those square-free Q with hðQÞojQj: Lemma 5.1. There is a square-free NAAþ such that P[N and the following hold for all square-free polynomial Q with ðP N; QÞ ¼ 1: (1) hðQÞ ¼ jQj; (2) hðQNÞ ¼ hðQÞhðNÞ: Proof. It is not difficult to see that hðQ1 Q2 ÞphðQ1 ÞhðQ2 Þ for all Q1 ; Q2 AAþ such that ðQ1 ; Q2 Þ ¼ 1: Let p be the characteristic. By Lemma 4.2, one can find the smallest integer nX0 such that hðQÞXjQj pn for all QAAþ : Choose any NAAþ with hðNÞ ¼ jNj pn : Given square-free polynomial Q with ðPN; QÞ ¼ 1; if hðQÞojQj; or hðQNÞohðQÞhðNÞ then hðQNÞphðQÞhðNÞojQNj contradicting to our pn ; choice of n: Also it is not possible that hðQNÞohðQÞhðNÞ: This completes the proof. & Lemma 5.2. Let N be a square-free polynomial satisfying the properties in Lemma 5.1. Then for all square-free Q in Aþ with ðPN; QÞ ¼ 1; the following are true: (1) EN -EQ ¼ EN -KðLQ Þ ¼ KðLN Þ-KðLQ Þ;
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(2) ½ENQ : K ¼ ½EN : KjQj½EN ðLQ Þ : EN ; (3) if Q ¼ Q1 Q2 ; then ðEN -EQ1 Þ ðEN -EQ2 Þ ¼ EN -EQ : Proof. (1) Since hðNQÞ ¼ hðNÞhðQÞ; we have EQ ðLN Þ and EN ðLQ Þ are linearly disjoint over KðLQ Þ KðLN Þ: This implies that EN -EQ must be a constant field extension of KðLN Þ-KðLQ Þ with degree prime to the characteristic p: Hence EN -EQ ¼ EN -KðLQ Þ ¼ KðLN Þ-KðLQ Þ: (2) Observe that KðLNQ Þ and EN are linearly disjoint over KðLN Þ: Hence ½EN ðLQ Þ : KðLNQ Þ ¼ hðNÞ: By Lemma 5.1(2), we deduce that ½ENQ : EN ðLQ Þ ¼ jQj: Thus ½ENQ : K ¼ jQj ½EN ðLQ Þ : EN ½EN : K: (3) Let G be the cyclic group GalðKðLQ Þ=KÞ: Define the subgroups H1 ; H2 ; H of G by Hi ¼ GalðKðLQ Þ=KðLQi ÞÞ; i ¼ 1; 2; H ¼ GalðKðLQ Þ=EN -KðLQ ÞÞ: Since Q is square-free, we have ðQ1 ; Q2 Þ ¼ 1: Hence H1 -H2 ¼ 1; therefore #H1 and #H2 are relatively prime. It follows that also the index of H in H H1 is relatively prime to the index of H in H H2 : Consequently H H1 -H H2 ¼ H: This implies ðEN -KðLQ1 ÞÞ ðEN -KðLQ2 ÞÞ ¼ EN -KðLQ Þ: By (1), we have ðEN -EQ1 Þ ðEN -EQ2 Þ ¼ EN -EQ : & Given a square-free polynomial QAAþ ; P[Q; we define SQ ¼ fsAGalðEQ =KÞ j sjEP aidEP for all PAPþ ; P j Qg and we put sQ ¼
#SQ #SQ ¼ : #GalðEQ =KÞ ½EQ : K
Proposition 5.3. We have sQ ¼
X RjQ
mðRÞ : ½ER : K
Proof. For R j Q; put DR ¼ fsAGalðEQ =KÞ j sjEL ¼ idEL for all L j Rg: It is not difficult to see that #SQ ¼
X RjQ
mðRÞ #DR :
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Since DR ¼ GalðEQ =ER Þ; we obtain P X mðRÞ R j Q mðRÞ #GalðEQ =ER Þ ¼ : sQ ¼ R j Q ½E : K ½EQ : K R
99
&
Remark. If Q1 j Q2 ; then sQ1 XsQ2 X0: It follows that the sequence fsQ g has a limit if Q ranges over all square-free monic polynomials not divided by P; ordered by divisibility. Combining Proposition 5.3 and Theorem 2.9, we have lim sQ ¼ dðMa Þ: Q
Our proof of positivity relies on the following: Lemma 5.4 (Heilbronn). Let f; c be two functions defined on square-free polynomials in Aþ such that 0pfðQÞp1 and cðQÞAN for all QAAþ : Moreover, we assume that fðQ1 Q2 Þ ¼ fðQ1 ÞfðQ2 Þ and cðQ1 Q2 Þ ¼ lcmðcðQ1 Þ; cðQ2 ÞÞ for Q1 ; Q2 AAþ with ðQ1 ; Q2 Þ ¼ 1: Then we have X mðQÞfðQÞ Y fðPÞ X 1 : cðQÞ cðPÞ QjQ PjQ QAAþ
PAPþ
The proof of this elementary lemma can be found in [6]. We arrive finally at the following criterion: Theorem 5.5. Let aAK\kK and let N0 AAþ be the product of all distinct irreducible factors of PmK 1: Then dðMa Þ40 if and only if SN0 a|: Proof. It is clear that if dðMa Þ40; all SQ a|: In particular SN0 a|: Conversely if SN0 a|; we claim that SQ a| for all QAAþ such that Q ¼ N0 R for some R relatively prime to N0 P: Hence, for every square-free polynomial QAAþ with P[Q; we may write Q ¼ Q1 Q2 with Q1 j N0 and ðQ2 ; N0 PÞ ¼ 1: Then sQ XsN0 Q2 40: Consider monic irreducible P j N0 : Then KðLP Þ ¼ K by Lemma 2.5 so that EP is an elementary abelian p-extension of K: Thus EN0 =K is elementary abelian pextension. Given Q ¼ N0 R for some R relatively prime to N0 P: We let s0 AGalðKðLR Þ=KÞ be the automorphism such that s0 jkK ðLR Þ is the generator of GalðkK ðLR Þ=kK Þ: Then s0 jKðLP Þ aid for all P dividing R; because kK ðLP ÞakK : Since EN0 -KðLR Þ ¼ K; one can first extend s0 to EN0 KðLR Þ by letting it coincide with an element in SN0 on EN0 : Then any extension of s0 from EN0 KðLR Þ to EQ will give an element of SQ : Hence we always have SQ a|: To complete the proof, let N be defined as in Lemma 5.2 and let QAAþ be squarefree such that ðPN; QÞ ¼ 1: For tASN ; define
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SQ ðtÞ ¼ fsAGalðENQ =KÞjsjEN ¼ t
sjEP aidEP for all P j Qg;
and
sQ ðtÞ ¼ #SQ ðtÞ=½EN : K
and
sðtÞ ¼ lim sQ ðtÞ Q
the limit being taken over all square-free monic polynomials relatively prime to PN; ordered by divisibility. Clearly, we have SNQ ¼
a
SQ ðtÞ;
tASN
sNQ ¼
X
sQ ðtÞ;
tASN
dðMa Þ ¼
X
sðtÞ:
tASN
We will show that sðtÞ40 for every tASN : Since SN is not empty, the desired inequality, dðMa Þ40; follows from the last inequality displayed above. Put ( cðt; QÞ ¼
1 if tAGalðEN =ðEN -EQ ÞÞ; 0 otherwise:
By Lemma 5.2(3), cðt; QÞ ¼ cðt; Q1 Þcðt; Q2 Þ if Q ¼ Q1 Q2 : Hence sQ ðtÞ ¼
X mðRÞcðt; RÞ RjQ
½ENQ : K
:
Following Lemma 5.2(2) sQ ðtÞ ¼
X mðRÞcðt; RÞ jRj 1 1 : ½EN : K R j Q ½EN ðLR Þ : EN
Put fðRÞ ¼ cðt; RÞjQj 1 ; cðRÞ ¼ ½EN ðLR Þ : EN : We have sQ ðtÞ ¼
X mðRÞfðRÞ 1 : ½EN : K R j Q cðRÞ
Since fðQ1 Q2 Þ ¼ fðQ1 ÞfðQ2 Þ and cðQ1 Q2 Þ ¼ lcmðcðQ1 Þ; cðQ2 ÞÞ for ðQ1 ; Q2 Þ ¼ 1; we have Y 1 fðPÞ 1 sQ ðtÞX ½EN : K PjQ cðPÞ PaP PAPþ
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by Lemma 5.4. The infinite product Y Y fðPÞ cðt; PÞ 1 1 ¼ cðPÞ jPj½EN ðLP Þ : EN P[PQ P[PN PAPþ
PAPþ
is convergent by Theorem 3.3. Hence Y 1 cðt; PÞ sðtÞX 1 40: ½EN : K P[PN jPj½EN ðLP Þ : EN
&
PAPþ
From our criterion, we can deduce Theorem 5.6. Let P be a place of K. Given any aAK which has pole at P with vP ðaÞ prime to the characteristic p. Then dðMa Þ40: Proof. By Lemma 4.2, our assumption gives hðQÞ ¼ jQj for all QAAþ : In particular, GalðEN0 =KÞ is the direct product of nontrivial groups GalðEP =KÞ with P j N0 : Hence SN0 a|: Applying Theorem 5.5 we obtain immediately dðMa Þ40: & On the other hand, when the A-module CðKÞ does not contain too many torsion points, we also have the following simpler necessary and sufficient condition for the positivity of the density. Theorem 5.7. Assume that the number of PAPþ such that PaP and LP CK; is less than p þ 1: Given aAK\kK : Then dðMa Þ40 if and only if aeCP ðKÞ for all monic irreducible P dividing PmK 1: Proof. If aACP ðKÞ for some monic irreducible P j ðPmK 1Þ; by Theorem 2.1 it is not difficult to see that dðMa Þ vanishes. Conversely, let N0 be the product of all distinct monic irreducible factors of PmK 1: If aeCP ðKÞ for P j N0 ; then EP =K is a proper extension, hence SP a| for all irreducible monic P j N0 : In view of Theorem 5.5, it is enough to prove SN0 a|: By Corollary 2.6, for monic irreducible P; LP CK if and only if P divides N0 : It follows that the extensions EP =K in question are always elementary abelian pextensions. We now choose a family of proper extensions EP0 =K; with P ranges through monic irreducibles dividing N0 ; EP0 CEP for each P; and EP0 =K is of degree p: To show that SN0 a|; it suffices to find sAGalðEN0 =KÞ whose restriction to each EP0 ; P j N0 irreducible, is not the identity. The number n0 of distinct fields EP0 in the family is no greater than the number of irreducible factors of N0 ; which is less than p þ 1 by our assumption. Let E 0 be the compositum of all the fields EP0 ; with P j N0 irreducible. Then GalðE 0 =KÞ ¼ V is a finite dimensional Fp -vector space. The subfield EP0 corresponds to the subspace GalðE 0 =EP0 Þ which is of codimension one in V : If V has dimension one, then extending any nonidentity automorphism in
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GalðE 0 =KÞ to EN0 gives element in SN0 : If V has dimension 41; then there are Xp þ 1 subspaces of V having codimension one. Since n0 op þ 1; we can find sAV such that seGalðE 0 =EP0 Þ for all irreducible P j N0 : It follows that this s can be extended to an element in SN0 : Thus, we have shown that SN0 is always nonempty which completes our proof. & Example 1. Let p ¼ q ¼ 2; and suppose K is an A field of A characteristic P satisfying 2pdeg Pp3 and kK ¼ FP : The monic irreducible factors of PmK 1 are t and t þ 1: Given nonconstant aAK: We have dðMa Þ40 if and only if aeCt ðKÞ,Ctþ1 ðKÞ: Note that elements in Ct ðKÞ,Ctþ1 ðKÞ are considered as the ‘‘squares’’ in our A-module CðKÞ; since jtj ¼ jt þ 1j ¼ 2: Finally, we close with an example which shows that the hypothesis CðKÞ does not contain ‘‘too many’’ A torsion points in Theorem 5.7 is indeed necessary: Example 2. Let p ¼ q ¼ 2 and let P ¼ t4 þ t þ 1: The monic irreducible factors of P 1 ¼ N0 are t; t þ 1; and t2 þ t þ 1: We will exhibit a function field K together with aAK such that SN0 ¼ | (hence dðMa Þ ¼ 0), although aeCP ðKÞ for all monic irreducible P dividing PmK 1: For convenience, we denote iðtÞ by a which is primitive in FP : We may write Ct ðxÞ ¼ ax þ x2 ; Ctþ1 ðxÞ ¼ a4 x þ x2 ; and Ct2 þtþ1 ðxÞ ¼ a10 x þ a10 x2 þ x4 : By an easy computation, also that Ct2 þtþ1 ðxÞ ¼ C1 3C2 ðxÞ where C1 ðxÞ ¼ a3 x þ x2 and C2 ðxÞ ¼ a7 x þ x2 : Let k ¼ FP ðyÞ be a rational function field over FP and let a ¼ y2 þ a7 y ¼ C2 ðyÞ: Introducing: x: a root of Ct ðxÞ ¼ a; r: a root of Ctþ1 ðxÞ ¼ a; d: a root of x2 þ a3 x ¼ a11 ðy2 þ a3 yÞ; d: a root of C1 ðxÞ ¼ y: (In fact, d ¼ a2 x þ a14 r þ d:) After elementary computations basing on Artin–Schreier theory one can show that kðx; rÞ=k; kðx; dÞ=k and kðd; rÞ=k are all extensions of type (2,2). Moreover kðdÞgkðx; rÞ: We take K ¼ kðdÞ: It follows Et ¼ KðxÞ ¼ kðd; xÞ; Etþ1 ¼ KðrÞ ¼ kðd; rÞ; and Et2 þtþ1 ¼ KðdÞ ¼ kðd; dÞCEt Etþ1 : We have aeCP ðKÞ for all monic irreducible P dividing N0 ¼ PmK 1: We contend that both Et2 þtþ1 Etþ1 and Et2 þtþ1 Et are extensions of type (2, 2) over K: This implies that SN0 ¼ |: To show that Et2 þtþ1 Et is an extension of type (2, 2) over K; it is enough to check xekðd; dÞ: Suppose this is not the case. Since d ¼ a2 x þ a14 r þ d; we obtain rAEt2 þtþ1 : Thus kðx; rÞCEt2 þtþ1 : Consequently kðx; rÞ ¼ Et2 þtþ1 which contradicts the fact that kðdÞgkðx; rÞ: By a similar argument, we can also deduce that Et2 þtþ1 Et is an extension of type (2, 2) over K:
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