MODULES
WEI-CHEN YAO AND JING YU
1. Introduction
Let A = Fq[t] be the polynomial ring over a finite field. We [14]
proved that for every nonconstant 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 and gave the criterion for the density to be positive. In this paper, we generalize this result on “primitive roots” to general rank one Drinfeld Modules.
Throughout this paper, we work over a global function field k of characteristic p and its field of constant Fq , where q = ps. Let ∞ be a
fixed prime divisor of k and let A be the ring of the elements of k which are integral at all primes P of k other than ∞. We are interested in the rank one Drinfeld A-modules. Given an element a of the coefficient field K. We want to investigate the set of places P of K, for which a modulo P generates the residue field K(P) as finite A-modules(under a given Drinfeld A-module action). In other word, we study the set of places P where a is a primitive root modulo P for a give Drinfeld A-module.
We will discuss this question in the following two parts. In section 3 and Section 4, we will only discuss the case of constant Drinfeld modules of fnite A-characteristic. The idea is similar to the case of
1991 Mathematics Subject Classification. Primary 11T55.
Carlitz modules. In Section5, we will apply the results in Section 3, 4 to general rank one Drinfeld A-modules.
2. Notations and Preliminaries
Let I (resp. P) be the set of all ideals (resp. prime ideals) of A. For every ideal a ∈ I, let a = pr1
1 · · · prnn be prime decomposition of a. We
define the M¨obius function for I as follows:
µ(a) =
(
1, if ri = 1 for i = 1 · · · n,
0, otherwise.
Furthermore, we say that a is a square-free ideal if µ(a) = 1.
Definition 2.1. Let L be an A-field, that is L is a field together with a
structure homorphism ι : A → L. If ι is injective, the A-characteristic of L is defined to be ∞. If ι is not injective, the A-characteristic of L is defined to be ker ι.
Let P ∈ P. Then A/P is a finite field which has qdeg Pelements where
deg(P) = [A/P : Fq]. For convenience, denote A/P by FP. Let K be
an algebraic function field over FP with constant field FK containing
FP, K0 be a finite extension of K, and let Kac be an algebraic closure
of K. Given any place P of K, put :
OP = the valuation ring of P,
K(P) = the residue field at P,
deg(P) = [K(P) : FK],
PK = the set of all places of K.
mK = [FK : FP] and q = qmKdeg P = #FK.
Let a be an ideal of k. We denote by f (a) the multiplicative order of PmK modulo a(i.e. f (a) is the smallest positive integer r such that
a | (PmKr− 1)). Given P ∈ P
K, we always regard K(P) as A-field of
A-characteristic P via A → FP,→ K (P).
2.1. Drinfeld Modules. Let L be an A-field with A-characteristic P via the map ι : A → FP ,→ L. Put τ (x) = xq and denote by L{τ } the
twisted polynomial ring with commutation rule τ · a = aq· τ for a ∈ L.
Let f (τ ) =Pvi=0aiτi ∈ L{τ }. We define Df := a0. It is clear that the
mapping L{τ } → L, f 7→ a0, is a morphism of FP-algebras. We then
have the definition of Drinfeld Module.
Definition 2.2. Let φ : A → L{τ }, a 7→ φa be a homomorphism of
Fq-algebras. Then φ is a Drinfeld A-module over L if and only if
(a) D ◦ φ = ι;
(b) For some a ∈ A, φa6= ι(a)τ0.
In particular, if K is an algebraic function field of one variable over FP, then one obtains an A-module structure on (K, φ, +).
Further-more, if φa ∈ FK{τ }, then φ is called a constant Drinfeld A-module.
Let φ be a Drinfeld module over K, and let a ⊂ A be an ideal. As A is a Dedekind domain, a may be generated by at most two elements
{a1, a2} ⊂ a. Since K{τ } has a right division algorithm, there exists
a right greatest common divisor in K{τ }. It is the monic generator of the left ideal of K{τ } generated by a1, a2.
Definition 2.3. We set φa to be the monic generator of the left ideal
of k{τ } generated by φa1, φa2.
It is can be shown that deg φa = N(a) where N(a) is the norm of
Definition 2.4. (a) Let φ and φ0 be Drinfeld A-Modules over K. An
isogeny from φ to φ0 is a twisted polynomial ρ ∈ K{τ } such that ρφ x =
φ0
xρ for all x ∈ A.
(b) Two Drinfeld A-Modules φ and φ0 over K are isomorphic if there
is an element w 6= 0 ∈ K such that φx = w−1· φ0x· w for all x ∈ a.
It is clearly to see that the product of isogenies is again an isogeny [3]. Let a be an ideal of A, put Ia,φ = K{τ } · φa be an left ideal of K{τ }.
Clearly, Ia,φ is carried into itself by multiplication on the right by the
φx, x ∈ A. Therefore, for every x ∈ A there is a uniquely defined
φ0
x ∈ K{τ } such that
φaφx = φ0xφa.
One can check that the map φ0 : A → K{τ } defined by x 7→ φ0 x is a
Drinfeld A-module [4]. We denote φ0 by a ? φ.
Lemma 2.1. Let a and b be non-zero ideals in A. Then
φab = (b ? φ)aφb
and
a ? (b ? φ) = (ab) ? φ.
The proof can be found in [Theorem 3.10] [4].
Definition 2.5. For any non-zero ideal a ∈ I, let φ[a] = {x ∈ Kac |
φa(x) = 0}. If a ∈ A, then we set φ[a] = φ[(a)].
We can check that there exists a positive integer d such that deg φa(τ )
= −dn∞v∞(a) = d deg(a) where n∞is the degree of the divisor ∞ and
Definition 2.6. The integer d that we described above is the rank of
the Drinfeld module φ.
Remark. If a ∈ A is prime to the characteristic of K, we see that
φ[a] ' (A/(a))d.
In this paper, we only concern rank one Drinfeld modules. The following result can be found in [5, Corollary 5.9].
Theorem 2.2. Let φ be a rank one Drinfeld A-module. Then φP =
τdeg P, i.e. φ
P(x) = xq
deg P .
Theorem 2.3. φ(K(P)) is isomorphic to A/(PmKdeg P− 1) as an
A-module.
The proof can be found in [Proposition 2.1] [8]
Definition 2.7. Let φ be a rank one Drinfeld A-module. Given a ∈
OP, we denote by a to be the canonical image of a in φ(K(P)). We
say that a ∈ K is a primitive root modulo a place P for φ if a ∈ OP
and a generates φ(K(P)) as an A-module.
Given a ∈ K, we are interested in the Dirichlet density of the set Mφa,K = {P ∈ PK | a is a primitive root modulo P for φ}.
An immediate consequence of Theorem 2.3 is the following.
Lemma 2.4. Given a ∈ K and P ∈ PK. Assume that vP(a) = 0.
Then, P ∈ Mφa,K if and only if there is no prime ideal p of A satisfying the following conditions
p | PmKdeg P− 1 and φ
2.2. Some Analytic Results. We start this part with an analogue of Mertern’s Theorem for A. Let k be a function field over a finite field, denote by gk the genus of k. Let a ∈ I. Denote deg a(resp. N(a)) by
the degree (resp. norm) of a. We recall that the zeta function for A,
ζA, is defined by ζA(s) = X a∈I N(a)−s =Y p∈P (1 − 1 N(p)s) −1. Moreover, ζA(s) = (1 − q−n∞s)L k(q−s) (1 − q−s)(1 − q1−s)
where n∞is the degree of ∞ and Lk(u) ∈ Z[u] is a polynomial of degree
2gk. Lemma 2.5. If d → ∞, then Y p∈P deg p≤d (1 − 1 N(p)) = 1 − q−1 Lk(q−1)(1 − q−n∞) · e −γ d + O( 1 d2)
where γ denotes Euler’s constant. Proof. We claim that
X p∈P N(p)≤x 1 N(p) = ln logqx + γ + F (0) + ln Lk(q−1)(1 − q−n∞) (1 − q−1) + O( 1 logqx)
where F (0) is the value of the following function
F (δ) =X p∈P (ln(1 − 1 N(p)1+δ) + 1 N(p)1+δ) for δ ≥ 0.
Using Prime Theorem for function fields, it is not difficult to deduce X p∈P N(p)≤x 1 N(p) = ln logqx + γ + c + O( 1 logqx)
for some constant c. It is sufficient to show
F (0) = c + lnLk(q
−1)(1 − q−n∞)
For δ > 0, we have F (δ) =X p∈P 1 N(p)1+δ − ln ζA(1 + δ) = lim x→∞[− Z x q X p∈P N(p)≤t 1 N(p)d( 1 tδ)] − ln ζA(1 + δ) = lim x→∞[δ Z x q (γ + c + ln logqt + O( 1 logqt))t −1−δdt] − ln ζ A(1 + δ) = δ Z ∞ q (γ + c)t−1−δdt + δ Z ∞ q (ln logqt)t−1−δdt + δ Z ∞ q O( 1 logqt)t −1−δdt − ln ζ A(1 + δ) = γ + c qδ − γ − ln(δ ln q) − δ Z q 1 t−1−δln log qtdt + δ Z ∞ q O( 1 logqt)t −1−δdt − ln ζ A(1 + δ). Hence F (0) = lim δ→0( γ + c qδ −γ−ln(δ ln q)−ln ζA(1+δ)) = c+ln Lk(q−1)(1 − q−n∞) (1 − q−1) . In view of fact X p∈P N(p)>x [ln(1 − 1 N(p)) + 1 N(p)] = O( 1 x), we have ln( Y p∈P N(p)≤x (1 − 1 N(p))) = F (0) − X p∈P N(p)≤x 1 N(p) + O( 1 x) = −γ − ln logqx − lnLk(q −1)(1 − q−n∞) (1 − q−1) + O( 1 logqx). It follows that Y p∈P N(p)≤x (1 − 1 N(p)) = (1 − q−1 (1 − q−n∞)Lk(q−1) · e−γ logqx + O( 1 (logqx)2). ¤
Lemma 2.6. Let b be an ideal of A with degree n. X a|b µ(a)2 N(a) ≤ ζk(2) −1 1 Lk(q−1) · eγlog qn + O(1).
Proof. First we observe that
X a|b µ(a)2 N(a) = Y p|b deg p<logqn (1 + 1 N(p)) Y p|b deg p≥logqn (1 + 1 N(p)),
where p run through all prime ideal such that p | b. Let m(b) be the number of such prime ideals with degree ≥ logqn. Then m(b) ≤ n/ logqn. Hence (1) Y p|b deg p≥logqn (1 + 1 N(p)) ≤ (1 + 1 n) n/ logqn = 1 + O( 1 logqn).
On the other hand, Y p|b deg p<logqn (1 + 1 N(p)) = Y p|b deg p<logqn (1 − 1 N(p)2) Y p|b deg p<logqn (1 − 1 N(p)) −1.
It is not difficult to see that the first product can be written as
(2) Y p|b deg p<logqn (1 − 1 N(p)2) ≤ Y deg p<logqn (1 − 1 N(p)2) = ζA(2) −1+ O(1 n).
By Lemma 2.5, we also get Y p|b deg p<logqn (1 − 1 N(p)) −1 ≤ 1 − q−1 (1 − q−n∞)L k(q−1) · eγlogqn + O(1) ≤ 1 Lk(q−1) · eγlog qn + O(1). (3)
Multiply (1), (2) and (3), we get the desire inequality. ¤ The following Theorem is an analogue for A of a theorem of Ro-manoff [11]
Theorem 2.7. The series X P-a a∈I µ(a)2 N(a)f (a) converges.
Proof. For any positive integer n, let D(0) = 0 and
D(n) = X
f (a)≤n
µ(a)2
N(a).
Any a enter the above sum must be a square-free divisor of
A(n) =
n
Y
i=1
((PmK)i− 1).
Clearly, deg(A(n)) = mKdeg(P)n(n + 1)/2. By Lemma 2.6, there is a
constant c > 0 such that
D(n) ≤ X a|A(n) µ(a)2 N(a) ≤ ζA(2)−1 1 Lk(q−1) · eγlog q(mkdeg(P) n(n + 1) 2 ) + O(1) ≤ c logqn. We derive X P-a µ(a)2 N(a)f (a) = ∞ X n=1 1 n( X f (a)=n µ(a)2 N(a)) = ∞ X n=1 D(n) − D(n − 1) n . Since N X n=1 D(n) − D(n − 1) n = N X n=1 D(n) n(n + 1) + D(N) N , X P-a µ(a)2 N(a)f (a) ≤ c ∞ X n=1 logqn n(n + 1) < ∞. ¤ Apply Theorem 2.7, we have the following consequence.
Proposition 2.8. X p∈P p6=P 1 qf (p)2 f (p) < ∞
Proof. Write the series as
X p∈P p6=P qf (p)2 >N(p) 1 qf (p)2 f (p) + X p∈P p6=P qf (p)2 ≤N(p) 1 qf (p)2 f (p) .
The first sum is converges by Theorem 2.7. For those p ∈ P and qf (p)2 ≤ N(p). Let p1, p2, · · · , pr be the prime factors of (PmK)f (p)− 1
such that N(pi) > q f (p) 2 . We have p1p2· · · pr | ((PmK)f (p) − 1) and therefore N((PmK)f (p)) ≥ N(p 1· · · pr).
This implies r = 1. In other words, such p is unique determined by
f (p). Hence the series
X p∈P p6=P qf (p)2 ≤N(p) 1 qf (p)2 f (p) ≤ ∞ X f =1 1 qf2f = log(1 − 1 q1/2). ¤ 3. Dirichlet Density of Mφa,K
Let φ be a constant Drinfeld A-module. In this section, we will prove the existence of the Dirichlet density for Mφa,K. First of all, we observe that the constants are precisely the A-torsion.
Proposition 3.1. Let φ be a constant Drinfeld A-module and let a ∈
K. There a is an A-torsion element in φ(K) if and only if a ∈ FK.
The proof of this lemma is same as we showed in [14, Proposition 2.4]. Hence we omit it.
Corollary 3.2. Let φ be a rank one constant Drinfeld A-module over
K. The torsion A-submodule of φ(K) is isomorphic to A/(PmK − 1).
Lemma 3.3. Let φ be a rank one constant Drinfeld module and let a
be an ideal of A such that vP(a) = 0. Then [K(φ[a]) : K] = f (a).
Proof. First we observe that FK(φ[a]) = FK(φ[a]) and [FK(φ[a]) : FK] =
[K(φ[a]) : K]. From a | (PmKf (a)− 1) and Theorem 2.2 we obtain
αqf (a)
= φPmK f(a)(α) = α
for all α ∈ φ[a]. Hence [K(φ[a]) : K] | f (a). On the other hand,
αq[K(φ[a]):K]
= αq[FK(φ[a]):FK ] = α
for all α ∈ kK(φ[a]). By Theorem 2.2
φPmK [FK(φ[a]):FK ]−1(α) = αq
[FK(φ[a]):FK ]
− α = 0
for all α ∈ φ[a]. This implies a | PmK[FK(φ[a]):FK]− 1 and therefore also
f (a) | [K(φ[a]) : K]. ¤
Remark. Let a be a square-free ideal of A relatively prime to P. From Lemma 3.3, one deduces that f (a) equals to
lcm{f (p) | p is a prime ideal of k and p | a}.
It also follows that the degree [K(φ[a]) : K] is always prime to P. Furthermore, if P - a, b are relatively prime square-free ideals, then
K(φ[a]) · K(φ[b]) = K(φ[ab]).
Lemma 3.4. Given P ∈ PK and let φ be a constant Drinfeld module
over K. Then p | (PmKdeg P− 1) if and only if P splits completely in
Proof. Since K(φ[p])/K is a constant field extension, P splits
com-pletely in K(φ[p]) if and only if φ[p] ⊂ K(P)(i.e. αqdeg P
= α for all
α ∈ φ[p]). By Theorem 2.2, this equivalent to φPmK deg P−1(α) = 0 for all
α ∈ φ[p]. From the definition of φ[p], the condition amounts precisely
to p | (PmKdeg P− 1). ¤
Given a prime ideal p of K, we set
φ−1p (a) = {x ∈ Kac | φp(x) = a},
and Ep = K(φ[p], φ−1p (a)). Clearly Ea are Galois extension of K. If a
is a square-free ideal of A, we define Ea as the compositum of the field
{Ep : p | a}. It is not difficult to see that Ea1·Ea2 = K(φ[a1a2], φ −1
a1a2(a))
for all a1, a2 such that (a1, a2) = 1. Moreover we have
Gal(Ea/K(φ[a])) ,→ φ[a],
so that Ea/K(φ[a]) are always elementary abelian p−extensions.
Proposition 3.5. Given a ∈ K and P ∈ PK such that vP(a) = 0.
Then a is a primitive root modulo P if and only if P does not split completely in any of the fields Ep for all prime ideal of a such that
p 6= P.
Proof. It is easy to see that φp(x) is a monic separable polynomial for
all p 6= P ∈ P. If α is a root of φp(x)−a, then every root of φp(x)−a has
the form α+λ for some λ ∈ φ[p]. Hence we may write Ep = K(φ[p])(α).
Suppose that P splits completely in Ep for some p 6= P ∈ P and
let P0 be a place of K(φ[p]) which lies above P. Then φ
p(x) ≡ a
(mod P0) is solvable in O
P0. Also P splits completely in K(φ[p]). This
implies (Lemma 3.4) that p | (PmKdeg P− 1) and φ
p(x) ≡ a (mod P)
(mod P). According to Theorem 2.3, φPmK deg P−1(β) ≡ 0 (mod P) for
all β ∈ OP and therefore
φ(PmK deg P−1)p−1(a) ≡ φ(PmK deg P−1)p−1(φp(β))
≡ (p−1? φ)PmK deg P−1(β) ≡ 0 (mod P),
where p−1 ∈ I such that pp−1 is a principal ideal.
Thus a is not a primitive root modulo P, by Lemma 2.4.
Conversely, if a is not a primitive root modulo P, by Lemma 2.4, there is p ∈ P such that p | (PmKdeg P− 1) and φ
p(x) ≡ a (mod P)
solvable in OP. By Lemma 3.4, the condition p | (PmKdeg P− 1) implies
P splits completely in K(φ[p]). Given P0 ∈ P
K(φ[p]) above P. We
also have φp(x) ≡ a (mod P0) solvable in OP0. Therefore P splits
completely in Ep. This completes the proof. ¤
Let a ∈ I. Let h(a) denote the degree [Ea : K(φ[a])]. Recall that
N(a) is the norm of a for a ∈ I. Then we have the following
Lemma 3.6. Let φ be a constant rank one Drinfeld module over K.
Given a ∈ K\FK and a ∈ I square-free with P - a. If a /∈ φa(K),
then there is a constant 0 < ca ≤ 1 such that caN(a) ≤ h(a) ≤ N(a).
Furthermore if a has pole at place P of K with vP(a) prime to the
characteristic p, then h(a) = N(a).
Proof. Let α be a root of φa(x) − a and let Ia = {P ∈ PK | vP(a) < 0}.
Given P ∈ Ia, we have vP0(αdeg φa) = vP0(αN(a)) = vP0(a) for every
P0 ∈ P
EQ lying above P. It follows that N(a)vP0(α) = vP(a)e(P
0 | P) and therefore N(a) 1 |vP(a)| ≤ N(a)vP0(α) vP(a)
for all P ∈ Ia. Here |vP(a)| is the absolute value of vP(a). We then
take ca= max{|vP1(a)|}P∈Ia.
If a has a pole at place P of K with vP(a) prime to p, then the
ramification index over places above P has to be N(a). Hence h(a) =
N(a). ¤
We will estimate the genus of the function fields Ea.
Theorem 3.7. Let a be an ideal of k with P - a. Let gEa be its genus of Ea. Then gEa ≤ c0ah0(a), where h0(a) = [Ea : K · FEa] and c0a is a
constant depending only on a and K.
Proof. Let α be a root of φa(x) = a, and F1 be the rational function
field FEa(α) which contains the rational function field FK(a). Let F2
be the constant field extension K ·FEa of K. Then Ea = F1·F2. Let gK
be the genus of K. By Castelnuovo’s Inequality [11, Theorem III.10.3], we have
gEa ≤ [Ea : F2]gK+ ([Ea : F1] − 1)([Ea : F2] − 1) ≤ h0(a)gK+ [K · FEa(α) : FK(a) · FEa(α)]h
0(a)
≤ h0(a)gK+ [K : FK(a)]h0(a) = c0ah0(a).
¤ Given a ∈ K, recall that
Mφa,K = {P ∈ PK | a is a primitive root modulo P for φ}.
If a ∈ FK, Mφa,K is necessary a finite set for φ is a constant Drinfeld
module. Hence from now on we assume that a ∈ K\FK.
Theorem 3.8. Let φ be a constant Drinfeld module over K. The set Mφa,K has a Dirichlet density δ(Mφa,K) given by
δ(Mφa,K) =X
a∈A P-a
µ(a) h(a)f (a).
Before we prove Theorem 3.8, we introduce the following Definitions and Theorems.
Definition 3.1. Let Mφa,K be the set of places of K which do not split completely in any of the fields Ep for all p 6= P ∈ P. Also let Ia be the
finite set of places where a has poles.
On the basis of Proposition 3.5 we see that Mφa,K differs from Mφa,K by at most a finite set. In order to study the Dirichlet density of Mφa,K, we recall the following theorem from [2] (c.f. also [1]). Given
{Kj}j∈J, a countable family of Galois extensions of K. Let kj be the
algebraic closure of kK in Kj. Set rj = [Kj : K], cj = [kj : kK],
and q = #kK. Let gj be the genus of Kj. For any finite set of indices
I = {j1, j2, · · · , js}, we also define KIto be the compositum Kj1· · · Kjs,
and put µ(I) = (−1)#(I).
Using the generalized Riemann hypothesis proved by A. Weil and Cˇebotarev’s density theorem, one establishes the following important Theorem 3.9 (Bilharz [1], Clark and Kuwata [2]). Suppose that the
following conditions hold for the family {Kj}j∈J.
(1) Pj∈J 1
rj < ∞,
(2) Pj∈J 1
cjqcj /2 < ∞, and
(3) There exists a constant c such that gj ≤ crcjj for all j ∈ J.
Let M be the set of places in K that do not split completely in any Kj for all j ∈ J. Then the Dirichlet density of M exists and is given
by δ(M) =X I µ(I) [KI : K] .
Proof of Theorem 3.8. Put F = {Ep}p∈P. Since K(φ[p]) is a constant
field extension of degree f (P ) and Ep/K(φ[p]) is an abelian p-extension,
we deduce that f (p) ≤ cp ≤ p · f (p). The condition (1) of Theorem
3.9 is verified by Theorem 2.7 and Lemma 3.6, the condition (2) by Proposition 2.8, and the condition (3) by Theorem 3.7. Hence this Theorem is proved.
4. Positivity of Density
Lemma 4.1. There is a square-free ideal n ∈ I such that P - n and the
following hold for all square-free ideal a with (Pn, a) = 1: (1) h(a) = N(a), (2) h(an) = h(a)h(n).
Proof. It is not difficult to see that h(a1a2) ≤ h(a1)h(a2) for all a1, a2 ∈
I such that (a1, a2) = 1. Let p be the characteristic of k. By Lemma
3.6, one can find the smallest integer r ≥ 0 such that h(a) ≥ N(a)pr for
all a ∈ I. Choose any n ∈ I with h(n) = N(n)pr .
¤ Lemma 4.2. Let n be a square-free ideal of k satisfying the properties
in Lemma 4.1. Then for all square-free ideal a ∈ I with (Pn, a) = 1, the following are true:
(1) En∩ Ea = En∩ K(φ[a]) = K(φ[n]) ∩ K(φ[a]),
(2) [Ena : K] = [En : K]N(a)[En(φ[a]) : En],
(3) If a = a1a2, then (En∩ Ea1) · (En∩ Ea2) = En∩ Ea.
Proof. (1) Since h(na) = h(n)h(a), we have Ea(φ[n]) and En(φ[a]) are
be a constant field extension of K(φ[n]) ∩ K(φ[a]) with degree prime to the characteristic p. Hence En∩Ea = En∩K(φ[a]) = K(φ[n])∩K(φ[a]).
(2) Observe that K(φ[na]) and En are linearly disjoint over K(φ[n]).
Hence [En(φ[a]) : K(φ[na])] = h(n). By Lemma 4.1, we deduce that
[Ena: En(φ[a])] = N(a). Thus [Ena : K] = N(a)[En(φ[a]) : En][En: K].
(3) Let G be the cyclic group Gal(K(φ[a])/K). Define the subgroups
H1, H2, H of G by :
Hi = Gal(K(φ[a])/K(φ[ai])), i = 1, 2,
H = Gal(K(φ[a])/En∩ K(φ[a])).
Since a is square-free, we have (a1, a2) = 1. Hence H1 ∩ H2 = 1,
therefore #H1 and #H2 are relatively prime. It follows that also the
index of H in H · H1 are relatively prime to the index of H in H · H2.
Consequently H · H1∩ H · H2 = H. This implies (En∩ K(φ[a1])) · (En∩
K(φ[a2])) = En∩ K(φ[a]). By (1), we have (En∩ Ea1) · (En∩ Ea2) =
En∩ Ea. ¤
Given a ∈ I, P - a, we define
Sa = {σ ∈ Gal(Ea/K) : σ|Ep 6= idEp for all p ∈ P, p | a}
and we put sa = #Sa # Gal(Ea/K) = #Sa [Ea : K] . Proposition 4.3. We have sa = X b|a µ(b) [Eb : K] .
Proof. For b | a, put
It is not difficult to see that #Sa =
X
b|a
µ(b) · #Db.
Since Db = Gal(Ea/Eb), we obtain
sa = P b|aµ(b) · # Gal(Ea/Eb) [Ea : K] =X b|a µ(b) [Eb : K] . ¤ Remark. If a1 | a2, then sa1 ≥ sa2 ≥ 0. It follows that the sequence {sQ} has a limit if a range over all square-free ideals of A not divided
by P, ordered by divisibility. Combining Proposition 4.3 and Theorem 3.8, we have
lim
a sa = δ(M
φ a,K).
Our proof of positivity relies on the following
Lemma 4.4. Let ψ, ρ be two functions defined on square-free ideals in I such that 0 ≤ ψ(a) ≤ 1 and ρ(a) ∈ N for all a ∈ I. Moreover, we
assume that ψ(a1a2) = ψ(a1)ψ(a2) and ρ(a1a2) = lcm(ρ(a1), ρ(a2)) for
all a1, a2 ∈ I with (a1, a2) = 1. Then we have
X b|a b∈I µ(b)ψ(b) ρ(b) ≥ Y p|a p∈P (1 −ψ(p) ρ(p)).
The proof of this Lemma is similar to Heilbronn’s Theorem [6], hence we omit it.
Theorem 4.5. Let φ be a rank one constant Drinfeld module over K,
a ∈ K\FK, and let N0 ∈ I be the product of all distinct irreducible
factors of PmK − 1. Then δ(Mφ
a,K) > 0 if and only if SN0 6= ∅.
Proof. It is clear that if δ(Mφa,K) > 0, then Sa 6= ∅ for all square-free
Conversely if SN0 6= ∅, we claim that Sa 6= ∅ for all a ∈ I such that
a = N0b for some b relatively prime to N0P. Consequently we also have
Sa 6= ∅ for all a ∈ I such that (a, P) = 1.
Consider prime ideal p such that p | N0. Then K(φ[p]) = K by
Lemma 3.3 so that Ep is an elementary abelian p-extension of K. Thus
EN0/K is elementary abelian p-extension. Given a = N0b for some b
relatively prime to N0P. We let σ0 ∈ Gal(K(φ[b])/K) be the
auto-morphism such that σ0 |FK(φ[a]) is the generator of Gal(FK(φ[b])/FK).
Then σ0 |FK(φ[p])6= id for all p dividing b, because FK(φ[p]) 6= FK. Since
EN0∩K(φ[b]) = K, one can first extend σ0 to EN0. Then any extension
of σ0 from EN0 · K(φ[b]) to Ea will given an element of Sa. Hence we
always have Sa 6= ∅.
To complete the proof, let n be defined in Lemma 4.2 and let a ∈ I be square-free such that (Pn, a) = 1. For τ ∈ Sn, define
Sa(τ ) = {σ ∈ Gal(Ena/K) : σ|En = τ and σ|Ep 6= idEp for all p | a},
sa(τ ) =
#Sa(τ )
[En: K]
, and s(τ ) = lim
a sa(τ )
the limit being taken over all square-free ideals relatively prime to Pn, ordered by divisibility. Clearly, we have
Sna = qτ ∈SnSa(τ ), sna = X τ ∈Sn sa(τ ), δ(Mφa,K) = X τ ∈Sn s(τ ).
We claim that s(τ ) > 0 for every τ ∈ Sn. Since sn > 0, Sn is not
empty. This implies that δ(Mφa,K) > 0. Put
c(τ, a) =
(
1 if τ ∈ Gal(En/(En∩ Ea)),
By Lemma 4.2 (3), c(τ, a) = c(τ, a1)c(τ, a2) if a = a1a2. Hence sa(τ ) = X b|a µ(b)c(τ, b) [Ena : K] . Following Lemma 4.2 (2), sa(τ ) = 1 [En: K] X b|a µ(b)c(τ, b)N(b)−1 [En(φ[b]) : En] . Put ψ(b) = c(τ, b)N(b)−1, ρ(b) = [E n(φ[b]) : En]. We have sa(τ ) = 1 [En: K] X b|a µ(b)ψ(b) ρ(b) .
Since ψ(a1a2) = ψ(a1)ψ(a2) and ρ(a1a2) = lcm(ρ(a1), ρ(a2)) for (Q1, Q2) =
1, we have sa(τ ) ≥ 1 [En : K] Y p|a p6=P p∈P (1 −ψ(p) ρ(p))
be Lemma 4.4. The infinite product Y p-Pn p∈P (1 −ψ(p) ρ(p)) = Y p-Pn p∈P (1 − c(τ, p) N(p)[En(φ[p]) : En] ) > 0. ¤ Theorem 4.6. Let P be a place of K and φ be a rank one constant
Drinfeld module one over K. Given any a ∈ K which has pole at P with vP(a) prime to the characteristic p. Then δ(Mφa,K) > 0.
Proof. By Lemma 3.6, our assumption gives h(a) = N(a) for all a ∈ I.
In particular, Gal(EN0/K) is the direct product of non-trivial groups
Gal(EP/K) with P | N0. Hence SN0 6= ∅. Applying Theorem 4.5 we
obtain immediately δ(Mφa,K) > 0. ¤
On the other hand, when the A-module φ(K) does not contain too many torsions, we also have the following simpler necessary and suffi-cient condition for positivity of the density.
Theorem 4.7. Assume that the number of p ∈ P such that p 6= P and
φ[p] ⊂ K, is less than p + 1. Given a ∈ K\FK. Then δ(Mφa,K) > 0 if
and only if a /∈ φp(K) for all prime ideal p with p | (PmK − 1).
Proof. If a ∈ φ(K) for some irreducible ideal p | (PmK−1), by Theorem
2.3 it is not difficult to see that δ(Mφa,K) vanishes. Conversely, let N0
be the product of product of all distinct irreducible factors of PmK− 1.
If a /∈ φp(K) for p | N0, then Ep/K is a proper extension, hence Sp 6= ∅
for all irreducible p | N0. In view of Theorem 4.5, it is enough to prove
SN0 6= ∅.
By Corollary 3.2, for irreducible ideal p, φ[p] ⊂ K if and only if p divides N0. It follows that the extensions Ep/K in question are always
elementary abelian p-extensions. We now choose a family of proper extensions E0
p/K, with p ranges through irreducible ideals dividing
N0, Ep0 ⊂ Ep for each p, and Ep0/K is of degree p. To show that
SN0 6= ∅, it suffices to find σ ∈ Gal(EN0/K)whose restriction to each E0
p, p | N0 irreducible, is not identity. The number n0 of distinct
fields E0
p in the family is no greater than the number of irreducible
factors of N0, which is less than p + 1 by our assumption. Let E0
be the compositum of all the fields E0
p, with p | N0 irreducible. Then
Gal(E0/K) = V is a finite dimensional F
p-vector space. The subfield Ep0
corresponds to the subspace Gal(E0/E0
p) which is codimension one in V .
If V has dimension one, then extending any non-identity automorphism in Gal(E0/K) to E
N0 gives element in SN0. If V has dimension > 1, then
there are ≥ p + 1 subspace of V having codimension one. Since n0 <
p + 1, we can find σ ∈ V such that σ /∈ Gal(E0/E0
p) for all irreducible
Thus we have shown that SN0 is always nonempty which completes our
proof. ¤
5. General Case
For this section, we will generalize our result in section to every Drinfeld of rank one. In order to show our result, we need the following Theorem.
Theorem 5.1. Let φ be a rank one Drinfeld module over K. Then
there exist a finite extension K0 over K and a constant Drinfeld module
φ0 over K0, such that φ is isomorphic to φ0 over K0.
Proof. ¤
Before we prove our main result, we need the following Lemmas. Lemma 5.2. Let φ and φ0 be rank one Drinfeld modules over K such
that φ is isomorphic to φ0, (i.e. there is a w ∈ K such that φ x =
w−1· φ0
x · w for all x ∈ A). Given a ∈ K and P ∈ PK. Assume that
vP(a) 6= 0 and vP(w) 6= 0. Then, a is a primitive root modulo P for φ
if and only if wa is a primitive root modulo P for φ0.
Proof. Since φx = w−1· φ0x· w, φa ∈ w−1· Ia,φ0· w and φ0a ∈ w · Ia,φ· w−1
for every a ∈ I where Ia,φ(resp. Ia,φ0) be the left ideal generated by
φa(resp. φ0a). This Lemma is thereofore an immediate consequence of
Lemma 2.4. ¤
The following Lemma is a consequence of Lemma 2.4.
Lemma 5.3. Let φ be a rank one Drinfeld module over K and let K0
P0 ∈ P
K0 lies above P with relative degree one. Then P ∈ Mφa,K if and
only if P0 ∈ Mφ a,K0.
Let Kφ denote by the field K0 satisfying the property in Theorem
5.1 with [K0 : K] is smallest and let w
φ be an element Kφ× such that
wφ·φ·wφ−1is a constant Drinfeld A-module over Kφ. Our main Theorem
stated as follows.
Theorem 5.4. Let Kφ and wφ be defined as above and let φ0 = wφ·
φ · wφ−1 be a constant Drinfeld A-module over Kφ. Assume that the
number of p ∈ P such that p 6= P and φ[p] ⊂ Kφ, is less than p + 1.
Given a ∈ K with wφa ∈ Kφ\FKφ. Then δ(M
φ
a,K) > 0 if and only if
wa /∈ φ0
p(Kφ) for all prime ideal p with p | (PmKφ − 1).
Proof. By Lemma 5.3 and Lemma 5.2,
δ(Mφa,K) = δ(Mφa,Kφ) = δ(Mφw0φa,Kφ). Hence, this Theorem is a consequnce of Theorem 4.7.
¤
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Wei-Chen Yao, Department of Mathematics and Computer Science Education, Taipei Municipal Teachers College, No. 1, Aikuo West Road, Taipei, 100 Taiwan, R.O.C
Jing Yu, Department of Mathematics, National Tsing-Hua Univer-sity, No. 101, Sec. 2, Kuang Fu Road, Hsinchu, 30043, Taiwan, R.O.C
E-mail address: [email protected] E-mail address: [email protected]
