DOI: 10.1007/s00209-002-0451-5

Math. Z. 243, 201–215 (2003) Mathematische Zeitschrift

On primitive roots of tori: The case of function fields

Yen-Mei J. Chen1, Yoshiyuki Kitaoka2, Jing Yu3,4

1 Department of Math., Tamkang University, Tamshui, Taipei, Taiwan(e-mail: ymjchen@mail.tku.edu.tw)

2 Department of Math., Meijo University, Tenpaku, Nagoya, 468-8502, Japan(e-mail: kitaoka@meijo-u.ac.jp)

3 Institute of Math., Academia Sinica, Nankang, Taipei, Taiwan(e-mail: yu@math.sinica.edu.tw)and National Center for Theoretical Sciences, Hsinchu, Taiwan

Received: 16 October 2000 / Published online: 2 December 2002 – c© Springer-Verlag 2002

Abstract. We generalize Bilharz’s Theorem for Gm to all one-dimensionaltori over global function fields of finite constant field. As an application,we also derive an analogue, in the setting of function fields, of a theorem(Chen-Kitaoka-Yu, Roskam) on the distribution of fundamental units mod-ulo primes.

Mathematics Subject Classification (1991): 11R11, 11R45, 11R58

0. Introduction

Let k = Fr be a finite field with r elements, and let F be a global functionfield having k as field of constants. We are interested in dimension one tori T

defined over F . Given such a torus T/F together with a non-torsion rationalpoint P0 ∈ T(F ), we want to study the set MP0 consisting of prime divisorsv of F where T has a good reduction T and P0 modulo v generates theabelian group T(F (v)), where F (v) is the finite residue field of F at v. Inthe case T = Gm, it was claimed by Bilharz [2] that if P0 /∈ T(F )q for allprimes q dividing r − 1, then MP0 always has positive (Dirichlet) density.The purpose of this note is to extend Bilharz’s theorem to all dimension onetori.

Main Theorem. Let P0 ∈ T(F ) be a non-torsion rational point. Then theset MP0 has a positive density if and only if P0 /∈ T(F )q for all prime qdividing #Tor(T(F )).

Research partially supported by National Science Council, Rep. of China.

202 Y.-M.J. Chen et al.

Thus it is enough to consider only those tori T which are not isomorphicto Gm over F . Since the only automorphisms of Gm are±identity, any suchtorus is isomorphic to Gm over an unique separable quadratic extension Kof F . If r is odd, then such T can be described by an equation of the formx2 − my2 = 1 with m ∈ F � \ F �2. On the other hand if r is even thensuch T is described by an equation of the form x2 + xy + my2 = 1 withm ∈ F \ (F 2 +F ). It follows that points in T(F ) can always be put in one-to-one correspondence with elements of norm 1 in the separable quadraticextension K = F (α), with α2 = m if 2 � r and α2 + α = m if 2 | r:(x, y)←→ ε = x + yα

Given a prime divisor v of F , we set |v| = rdeg v = #(F (v)). It is notdifficult to see that, if we restrict ourselves to prime divisors of F where T

has a good reduction, for those v which are inert in K, the group T(F (v))is cyclic of order |v|+ 1, whereas for those v which splits in K, T(F (v)))is cyclic of order |v| − 1. What we are looking for are the prime divisorsof F for which a given point P0 = (x0, y0) under good reduction has themaximal possible order .

The main theorem is proved in Sect. 2, Theorem 2.4 and 2.5. In Sect. 3we use our results to establish an analogue of a theorem (Chen-Kitaoka-Yu[3], Roskam [10]) on the distribution of fundamental units modulo primesfor “real”quadratic function fields .

1. Algebraic preliminaries

We shall use multiplicative notations for the abelian groups T(F ). Given anon-torsion P0 = (x0, y0) ∈ T(F ), the corresponding norm 1 element inK is denoted by ε0. We say that the point P0 has a good reduction P0 at aprime divisor v of F if both x0 and y0 are regular at v. The set of all primedivisors of F where both T and P0 have good reduction will be denoted byR. Let I be the set consisting of 1 and all primes dividing rs + 1, for somes ∈ N, and also letN be the set of all square-free natural numbers divisibleonly by elements in I. Of particular interests to us are the following twosets of prime divisors whose union is MP0 :

M−P0

= {v ∈ R : v is inert in K, P0 ∈ T(F (v)) is of order |v|+ 1},M+

P0= {v ∈ R : v splits in K, P0 ∈ T(F (v)) is of order |v| − 1}.

On primitive roots of tori: The case of function fields 203

Notations: The Frobenius element of Gal(k/k) will be denoted by τ . Let νbe a rational prime in I.

µµµν = the group of ν-th roots of unity.

Kν = K(µµµν , ν√

ε0).

Kn =∏ν|n

Kν , for a square-free integern.

kn = the field of constants of Kn.

K1 = K.

K ′ = F (√

ε0 +1√ε0

), K ′′ = F (√

ε0 − 1√ε0

).

Gn = Gal(Kn/F ).dn = [Kn : K].

C−1 = {σ ∈ G1 : σ �= id}.

C−2 = {σ ∈ G2 : σ|K �= id, σ|K′ = id}.

C−ν = {σ ∈ Gν : σ|K �= id, σ|k(µµµν) �= id, and σ2 = id},

for odd ν ∈ I.C−

n = {σ ∈ Gn : σ|Kν ∈ C−ν for all ν | n}, for n ∈ N .

(v, E/F ) = the Artin symbol of the prime divisor v of E/F ,

where E is a finite Galois extenstion over F .

Let f(ν) = [k(µµµν) : k] which equals the order of the integer r moduloν. Note that if ν ∈ I and ν is odd, then f(ν) must be even. We definef(n) = lcmν|nf(ν) for a square-free integer n.

If a prime divisor v is inert in K/F , the residue field of K at v will bedenoted by K(v) whileOv denoting the associated valuation ring containedin K. If ε ∈ Ov, its image in K(v) is denoted by ε. If moreover the inert vlies in R, then T(F (v)) is isomorphic to the kernel in K(v)� of the normmap NK(v)/F (v) under the correspondence P ↔ ε. This leads to the basic

Lemma 1.1. Let v ∈ R be a prime divisor which is inert in K/F and ν anodd prime in I. Then the following conditions are equivalent :

(1) ν | (|v|+ 1) and P0

|v|+1ν = 1 in T(F (v)),

(2) ν | (|v|+ 1) and v splits completely in Kν/K,(3) (v, Kν/F ) ⊆ C−

ν .

Proof. Note first that (1) is equivalent to the condition ε0|v|+1

ν = 1 in K(v)�.The hypothesis |v| = rdeg v ≡ −1 (mod ν) implies f(ν) | 2 deg v. Know-ing that the degree of the constant field extension K(µµµν)/K divides f(ν),

204 Y.-M.J. Chen et al.

and v has a residue class of degree 2 in K/F , we see that f(ν) | 2 deg vmeans v splits completely in K(µµµν)/K. Thus v splits completely in Kν/Kif and only if xν = ε0 has a solution in K(v). The last condition implies

ε0|v|+1

ν = 1 in K(v)� so that (1) holds.Since K(v)� is cyclic of order |v|2 − 1, ε0

(v+1)/ν = 1, we havex(|v|−1)ν = ε0 has a solution in K(v). It follows xν = ε0 has a solution inK(v). Hence (1)⇔ (2). On the other hand, ν cannot divide |v| − 1, there-fore f(ν) � deg v, we have (v, F (µµµν)/F ) �= id, and (v, Kν/F )|kν �= id.Consequently (2)⇔ (3) follows from (v, Kν/K) = (v, Kν/F )2 = id. ��

Since P0 is a non-torsion point, y0 �= 0. We let R′ denote the subset ofR consisting of prime divisors v of F where y0 does not vanish. For ν = 2,Lemma 1.1 is replaced by

Lemma 1.2. Assume r is odd. Let v ∈ R′ be a prime divisor which is inertin K/F . Then the following conditions are equivalent:

(1) P0

|v|+12 = 1 in T(F (v)).

(2) v splits in K ′/F .(3) (v, K2/F ) ⊆ C−

2 .

Proof. The equivalence between (2) and (3) follows from the definitions.

It is clear that ε0|v|2−1

2 = 1 in K(v)� if and only if ε0 = η2 in K(v)�.Since T(F (v)) can be identified as the kernel of NK(v)/F (v) in K(v)�, we

see that P|v|+1

20 = 1 in T(F (v)) if and only if ε0 = η2 in K(v)� with

NK(v)/F (v)(η) = 1. On the other hand, NK(v)/F (v)(η) = 1 is equivalent toTrK(v)/F (v)(η) = η + 1

η . The last condition implies η + 1η ∈ F (v), hence

v splits in K ′/F . Conversely, if v splits in K ′/F , then ε0 = η2 in K(v)�

with η + 1η ∈ F (v). We contend that NK(v)/F (v)(η) has to be 1. Otherwise,

NK(v)/F (v)(η) = −1 and η − 1η ∈ F (v), therefore η ∈ F (v) holds. It

follows that ε0 = −1 ∈ F (v) which contradicts the assumption v ∈ R′. ��Corollary 1.3. Let v ∈ R′ be a prime divisor which is inert in K/F . Thenv ∈M−

P0if and only if (v, Kν/F ) �⊆ C−

ν for all prime ν ∈ I.

If a prime divisor v splits in K/F and w is one of the two prime divisors ofK lying above v, then K(w) = F (v). Suppose v ∈ R. There is one-to-onecorrespondence between T(F (v)) and K(w)� = F (v)�, with P0 ↔ ε0

′,where ε′

0 is the image of ε0 in K(w). We have

Lemma 1.4. Let v ∈ R be a prime divisor which splits in K/F and ν aprime �= char(k). Then the following conditions are equivalent :

(1) ν | (|v| − 1) and P0

|v|−1ν = 1 in T(F (v)),

On primitive roots of tori: The case of function fields 205

(2) v splits completely in Kν/F .

Proof. Since ν | (|v|−1), it follows that f(ν) | deg v and v splits completelyin K(µµµν)/F . Thus (2) is equivalent to the solvability of xν = ε0 over theresidue field F (v). Let w be any prime divisors of K lying above v. If ε0

′ isthe image of ε0 in K(w), then the solvability of xν = ε0 over F (v) amounts

to the solvability of xν = ε0′. Consequently, (2) ⇐⇒ (ε0

′)|v|−1

ν = 1 inK(w)�. On the other hand the one-to-one correspondence between T(F (v))and K(w)� gives (1)⇐⇒ (ε0

′)|v|−1

ν = 1. This completes the proof. ��Corollary 1.5. Let v ∈ R be a prime divisor which splits in K/F . Thenv ∈ M+

P0if and only if v does not split completely in any of the fields Kν ,

for all primes ν �= char(k).

We will consider K2 = K(√

ε0) only in the case r is odd. If K2 = Kand P0 /∈ T(F )2, then NK/F (

√ε0) must be −1, it follows that K ′ =

F (√

ε0+ 1√ε0

) = K. Hence#(C−2 ) = 0 in this case. IfK2 �= K, thenK2/F

is not cyclic and K ′ �= K. It follows #(C−2 ) = 1. For a general square free

n, it is also true that #(C−n ) is either 1 or 0. In order to distinguish those n for

#(C−n ) = 1, we introduce, for a given integer e ≥ 0, the set I(e) consisting

of primes dividing rm + 1 for some m satisfying m ≡ 2e (mod 2e+1),and the setN (e) consisting of square-free natural numbers divisible only byprimes in I(e). The following Lemmas give a crucial information about theGalois family Kn.

Lemma 1.6. Suppose that K �∼= F · Fr2 .

(a) For an odd integer n ∈ N , we have dn = 2f(n)ngcd(n,s) where s is the largest

integer with the property that ε0 ∈ (K�)s.(b) For an odd prime ν ∈ I, #(C−

ν ) = 1 always holds.(c) If n is an odd integer in N (e) for some e, then #(C−

n ) = 1.

Proof. (a) Note that [K(µµµn) : F ] = 2f(n), since K �∼= F · Fr2 . For eachodd prime ν ∈ I, since K( ν

√ε0)/K is non-Galois (because µµµν �⊂ K),

we have K(µµµn) ∩ K( ν√

ε0) = K and thus [K(µµµn, ν√

ε0) : K(µµµn)] =[K( ν√

ε0) : K] = ν/ gcd(ν, s). Taking the successive composites of all thefields K(µµµn, ν

√ε0) for ν | n, we obtain [Kn : K(µµµn)] = n/ gcd(n, s) and

[Kn : F ] = [Kn : K(µµµn)][K(µµµn) : F ] =2f(n)n

gcd(n, s).

(b) If Kν = K(µµµν), #(C−ν ) = 1 because f(ν) is always even. If [Kν :

K(µµµν)] = ν, choose σ ∈ Gν such that σ|K �= id, σ|k(µµµν) �= id andσ|2K(µµµν) = id. This is possible because K �∼= F · Fr2 , which implies that

206 Y.-M.J. Chen et al.

K � F (µµµν). After modifying σ by an element of Gal(Kν/K(µµµν)), wemay assume that σ( ν

√ε0) = 1

ν√

ε0. Write elements in C−

ν as ρσ with ρ ∈Gal(Kν/K(µµµν)). Let ζν be a fixed primitive ν-th root of 1. Then ρ( ν

√ε0) =

ζνi ν√

ε0 for certain integer i. It is easy to check that ρσ ∈ C−ν if and only if

ν | i. Therefore #(C−ν ) = 1.

(c) If an odd prime ν ∈ I(e), then f(ν) | 2e+1m for some m. It follows

that 2e+1 ‖ f(ν) because ν � (r2e′m′ − 1) for all m′ | m and e′ ≤ e. We

have therefore 2e+1 ‖ f(n) if n(�= 1) ∈ N (e). Consequently,

f(n)/2 �≡ 0 (mod f(ν))

for all composite n ∈ N (e) with ν | n. This ensures that σ ∈ C−n ⇔ σ|Kν ∈

C−ν for all ν | n. ��

Let d be the product of all distinct primes q dividing r + 1 such thatK( q√

ε0)/K is a constant field extension of degree q. Let d′ = d if 2 | d, andd′ = 2d if 2 � d. Let Id ⊆ I(0) be the subset consisting of all primes dividingrd′m+1 + 1 for some m, and Nd be the set of square-free natural numbersdivisible only by primes from Id. Note we always have gcd(f(n), d′) = 2,for n ∈ Nd.

Lemma 1.7. Suppose that K ∼= F · Fr2 , and n ∈ Nd odd. We have

(a) dn = f(n)ngcd(n,s) .

(b) #(C−n ) = 1.

(c) #(C ′n) =

{1 if gcd(n, d) = 1,

0 otherwise,

where C ′n = {σ ∈ Gal(Kn · Frd′ /F ) : σ |Kn∈ C−

n and σ |Frd′ = τ}.

Proof. (a) Note that, for every odd integer n ∈ Nd, [K(µµµn) : F ] = f(n),since K ∼= F · Fr2 and f(n) is even. If ν � d, then either K( ν

√ε0) =

K or K( ν√

ε0)/K is geometric. It follows that [K(µµµn, ν√

ε0) : K(µµµn)] =ν/ gcd(ν, s). If ν | d, it is clear that [K(µµµn, ν

√ε0) : K(µµµn)] = ν/ gcd(ν, s),

because gcd(ν, f(n)) = 1. Similar to the proof of Lemma 1.6 (a), we have[Kn : K(µµµn)] = n/ gcd(n, s) and

[Kn : F ] = [Kn : K(µµµn)][K(µµµn) : F ] =f(n)n

gcd(n, s).

(b) The proof is similar to that of Lemma 1.6 (b) and (c).(c) If there exists a prime q dividing gcd(n, d), then K( q

√ε0) = K ·Frq

and Frq ⊆ Kn. If σ ∈ C ′n, then σ2 |Frq = id, since σ |Kn∈ C−

n . Then itfollows that σ |Frq = id, because q is odd. This contradicts to the requirementthat σ |F

rd′ = τ . Hence C ′n = ∅. On the other hand, if gcd(n, d) = 1, then

On primitive roots of tori: The case of function fields 207

Kn and F · Frd′/2 are linearly disjoint over F . Since Gal(Kn · Frd′ /F ) isisomorphic to Gal(Kn/F )×Gal(F ·Frd′/2/F ), the pair (σn, τ) correspondsto the unique elemant in C ′

n, where σn is the only element in C−n . Therefore,

#(C ′n) = 1. ��

2. Existence and positivity of the density

Recall that given any set S ⊂ R, the Dirichlet density of S is defined by

den(S) = limz→1+

∑v∈S

1|v|z∑

v∈R1|v|z

.

Our basic analytic tool for obtaining Dirichlet densities is the following

Theorem 2.1. (Bilharz, Clark-Kuwata) Given a countable family{(Kν , Cν)}ν∈I , where Kν is a Galois extension of F and Cν is a union ofconjugacy classes in Gal(Kν/F ). Let kν be the field of constants of Kν .Define cν = #Cν , dν = [Kν : F ] and fν = [kν : k]. Let gν be the genus of

Kν . For any finite subset I of I, we define KI to be the compositum∏ν∈I

Kν

and define the Mobius function by µ(I) = (−1)#I . Also set:

CI = {σ ∈ Gal(KI/F ) : σ|Kν ∈ Cν for all ν ∈ I}.Suppose that the following conditions hold:

(1)∑ν∈I

cν

dν<∞,

(2) There exists a constant A such that gν ≤ Adνfν

for all ν,(3) For each ν, there exists a positive integer aν such that σ|kν = τaν |kν

for every σ ∈ Cν and∑ν∈I

cν

fνr12aν

<∞.

Let S be the set of all prime divisors of F such that their correspondingArtin symbol in Gal(Kν/F ) does not belong to Cν for all ν. Then theDirichlet density den(S) of S exists and is given by the absolutely convergentseries:

den(S) =∑

I

µ(I)#CI

[KI : F ],

where I runs over all finite subsets of I.

Proof. See [4], Theorem 4.1. ��

208 Y.-M.J. Chen et al.

Given a non-torsion point P0 = (x0, y0) ∈ T(F ). Let ε0 be the corre-sponding norm 1 element in K. Let s be the largest integer with the propertythat ε0 ∈ (K�)s. If n is a square-free integer, n/gcd(n, s) will be denotedby n1. The set of all prime numbers not dividing r together with 1 will be de-noted byPr. We will apply Theorem 2.1 to the two families {(Kν , C

−ν )}ν∈I ,

{(Kν , {id})}ν∈Pr introduced in Sect. 1. We first need

Lemma 2.2. (a)∑ν∈Pr

1dν

<∞.

(b) If 0 < θ, then∑ν∈Pr

1fνrθfν

<∞.

(c) gν ≤ Adνfν

for all prime ν, where A is a positive constant.

Proof. (a) It is clear that dν ≥ f(ν)ν1. Hence for almost all ν ∈ Pr,dν ≥ f(ν)ν. Thus it is enough to check the convergence of

∑ν∈Pr

1νf(ν)

.

This follows immediately from a well-known theorem of Romanoff (c.f.[9]).

(b) Since fν ≥ f(ν) for ν ∈ I, hence the conclusion follows from [2],Sect. 3, Lemma 2 on pp. 488.

(c) This is given in [2], [3], Lemma 1 on pp. 488. ��Theorem 2.3. Suppose P0 = (x0, y0) ∈ T(F ) is a non-torsion rationalpoint. Then both den(M−

P0) and den(M+

P0) exist.

Proof. From the definition of C−ν , we have σ ∈ C−

ν =⇒ σ|kν = τ fν/2|kν

for all odd ν ∈ I. Since I ⊂ Pr, Lemma 2.2 (with θ = 1/4) allows us toapply Theorem 2.1 to the family {(Kν , C

−ν )}ν∈I . In view of Corollary 1.3,

this gives the existence of:

den(M−P0

) =∑n∈N

µ(n)#C−n

[Kn : F ].

Next we apply Theorem 2.1 and Lemma 2.2 (with θ = 1/2) to thefamily {(Kν , {id})}ν∈Pr . Using Corollary 1.5, we obtain the existence ofden(M+

P0):

den(M+P0

) =∑

gcd(n,r)=1

µ(n)[Kn : F ]

.

��

On primitive roots of tori: The case of function fields 209

To establish the positivity of the density, we start with the simpler caseof even characteristic

Theorem 2.4. Suppose P0 ∈ T(F ) is a non-torsion rational point, and2 | r. If K �∼= F · Fr2 , then always den(MP0) > 0. On the other hand ifK ∼= F · Fr2 , we have den(MP0) > 0 if and only if

P0 /∈⋃

q|r+1

T(F )q.

Moreover, den(MP0) > 0⇔ den(M−P0

) > 0 for any P0 ∈ T(F ).

Proof. If K �∼= F ·Fr2 , it suffices to show that den(M−P0

) > 0. Let e ≥ 0 besufficiently large such that gcd(s, r2e

+1) = 1. We shall prove that the subsetSe of M−

P0consisting of all prime divisors v with deg v ≡ 2e (mod 2e+1)

already has a positive density. Let I(e) and N (e) be the sets introduced inconnection with Lemma 1.6. For all ν ∈ I(e), we have 2e+1 ‖ f(ν), so thatf(n)/2e+1 = lcmν|n f(ν)/2e+1 holds for n ∈ N (e) and also F

r2e+1 ⊆ Kν .Let σe be the automorphism in Gal(K · F

r2e+1/F ) satisfying σe|K �= id

and σe|Fr2e+1 = τ2e

. Applying Theorem 2.1 to {(K · Fr2e+1 , {σe})} ∪

{(Kν , C−ν )}ν∈I(e) , and using Lemma 1.6, we find the density of the set Se

as given by:

den(Se) =1

2e+2 +∑

n( �=1)∈N (e)

µ(n)#C−n

[Kn : F ]=

12e+2 +

∑n( �=1)∈N (e)

µ(n)2n1f(n)

.

The non-vanishing of this density then follows from an elementary inequalityby Heilbronn, c.f. [6]:

den(Se) =1

2e+2 (1 +∑

n�=1∈N (e)

2e+1µ(n)n1f(n)

)

≥ 12e+2

∏ν∈I(e)

(1− 2e+1

ν1f(ν)) > 0.

The last inequality comes from the fact that gcd(s, r2e+ 1) = 1 implies if

ν ∈ I(e) and ν | s then f(ν) > 2e+1.Suppose now K ∼= F ·Fr2 , we contend that P0 /∈ ⋃

q|r+1 T(F )q implies

den(M−P0

) > 0. Let S′ ⊆M−P0

be th subset consisting of prime divisors hav-ing degree ≡ 1 (mod 2d), where d is the integer introduced in connectionwith Lemma 1.7. It suffices to show that den(S′) > 0. We consider primes

210 Y.-M.J. Chen et al.

in Id, and the set Nd. Then the density of the set S′ is given by (accordingto Lemma 1.7):

den(S′) =12d

+∑

n( �=1)∈Nd

µ(n)#C ′n

[Kn · Frd : F ]=

12d

+∑

n( �=1)∈Ndgcd(n,d)=1

µ(n)dn1f(n)

.

Again by Heilbronn’s inequality, we obtain (note that ν1 = 1 and thatν � r + 1 implies f(ν) > 2):

den(S′) ≥ 12d

∏ν∈Id

(1− 2ν1f(ν)

) > 0.

Finally we observe that if P0 ∈⋃

q|r+1 T(F )q, then clearly den(M−P0

) = 0.On the other hand, because Kq = K if q | r + 1 and P0 ∈ T(F )q, Corollary1.5 says M+

P0= ∅. This completes the proof. ��

The case of odd characteristic, the main result is the following Theorem.Here Theorem 2.1, Lemmas 1.6, 1.7, and Heilbronn’s inequality will be usedrepeatedly in its proof without explicitly mentioning.

Theorem 2.5. Suppose P0 ∈ T(F ) is a non-torsion rational point, and2 � r. We have

den(MP0) > 0⇔{

P0 �∈ T(F )2 if K �∼= F · Fr2 ,

P0 /∈ ⋃q|r+1T(F )q if K ∼= F · Fr2 .

Moreover, for any P0 ∈ T(F ) satisfying K ′ = F (√

ε0 + 1√ε0

) � F · Fr2 ,

den(MP0) > 0⇔ den(M−P0

) > 0.

Proof. It is clear that if P0 ∈ T(F )2 then den(MP0) = 0. We will first showthat

den(M−P0

) > 0⇔{

P0 /∈ T(F )2 if K and K ′ �∼= F · Fr2 ,

P0 /∈ ⋃q|r+1T(F )q if K or K ′ ∼= F · Fr2 .

Suppose K, K ′ �∼= F · Fr2 . We contend that den(M−P0

) > 0 if P0 �∈T(F )2. Let e ≥ 0 be sufficiently large such that gcd(s, r2e

+ 1) = 1. Itsuffices to prove that the subset Se of M−

P0consisting of all prime divisors

v with deg v ≡ 2e (mod 2e+1) already has a positive density.

On primitive roots of tori: The case of function fields 211

If s is even, then K2 = K, K ′ = K and C−2 = ∅. As in the proof of

Theorem 2.4, the density den(Se) is given by:

den(Se) =1

2e+2 +∑

n( �=1)∈N (e)

µ(n)#C−n

[Kn : F ]

=1

2e+2 +∑2�n

n�=1∈N (e)

µ(n)2n1f(n)

≥ 12e+2

∏ν �=2

ν∈I(e)

(1− 2e+1

ν1f(ν)

)> 0.

If s is odd, then K2 �= K, K2/F is not cyclic, and K ′ �= K. If further-more both K, K ′ �= F · Fr2 , then the density in question is:

den(Se) =1

2e+2 −1

2e+3 +∑2�n

n( �=1)∈N (e)

µ(n)2n1f(n)

−∑2�n

n( �=1)∈N (e)

µ(n)4n1f(n)

≥ 12e+3

∏ν �=2

ν∈I(e)

(1− 2e+1

ν1f(ν)

)> 0.

Suppose now K ∼= F · Fr2 , we want to show that P0 /∈ ⋃q|r+1 T(F )q

implies den(M−P0

) > 0. Recall that d is the product of all distinct primesq dividing r + 1 such that K( q

√ε0)/K is the constant field extension of

degree q, and d′ = d if 2 | d, and d′ = 2d if 2 � d. Let S′ ⊆ M−P0

be thesubset consisting of prime divisors having degree≡ 1 (mod d′). It sufficesto prove that den(S′) > 0. We consider primes in Id, and the set Nd. Ifs is even, then K2 = K and K ′ = K hold. Consequently in this case thedensity of S′ is given by:

den(S′) =1d′ +

∑n( �=1)∈Nd

µ(n)#C ′n

[Kn · Frd : F ]

=1d′ +

∑gcd(d′,n)=1

n( �=1)∈Nd

2 µ(n)d′ n1f(n)

≥ 1d′

∏ν�d′ν∈Id

(1− 2

ν1f(ν)

)> 0.

212 Y.-M.J. Chen et al.

If s is odd, then K2 �= K and K ′ �= K, the density of S′ is:

den(S′) =1d′ −

12d′ +

∑gcd(d′,n)=1

n( �=1)∈Nd

2µ(n)d′n1f(n)

−∑

gcd(d′,n)=1n( �=1)∈Nd

2µ(n)2d′ n1f(n)

≥ 12d′

∏ν�d′ν∈Id

(1− 2

ν1f(ν)

)> 0.

To complete the caseK ∼= F ·Fr2 , we observe that ifP0 ∈⋃

q|r+1 T(F )q,

then clearly den(M−P0

) = 0. On the other hand, because M+P0

consists onlyof even degree prime divisors, also den(M+

P0) = 0 if P0 ∈

⋃q|r+1 T(F )q.

Finally suppose K �∼= F · Fr2 but K ′ ∼= F · Fr2 . Then K2 = K · Fr2

and s is odd. Since all even degree prime divisors now splits in K ′/F , weimmediately have den({v ∈ M−

P0: 2 | deg v}) = 0. If it happens that

P0 ∈ T(F )q and q | r + 1, then clearly den({v ∈ M−P0

: 2 � deg v}) = 0.Conversely we contend that if P0 /∈ ⋃

q|r+1T(F )q, then den(M−P0

) > 0.

For this purpose we consider the set of primes in I(0), and the setN (0). Wehave:

den(S0) =14

+∑

n�=1∈N (0)

µ(n)#C−n

[Kn : F ]

=14

+∑2�n

n�=1∈N (0)

µ(n)2n1f(n)

≥ 14

∏ν �=2

ν∈I(0)

(1− 2

ν1f(ν)

)> 0.

Since q | r+1 =⇒ q � r−1provided q is odd, to finish the proof it is enoughto show that in the case K �∼= F ·Fr2 but K ′ ∼= F ·Fr2 , P0 /∈ ⋃

q|r−1 T(F )q

implies den(M+P0

) > 0. Note that den({v ∈ M+P0

: 2 | deg v}) = 0,because K2 = K · Fr2 . Let d′′ be the product of distinct prime numbers qdividing r−1 for which K( q

√ε0)/K is the constant field extension of degree

q. LetI ′′ be the set consisting of prime numbers dividing rmd′′+1−1 for someinteger m ≥ 0, and N ′′ be the set of square-free integers divisible only byprimes in I ′′. Applying Theorem 2.1 to the family {(Kn ·Frd , Cn)}, whereCn = {σ ∈ Gal(Kn · Frd/F ) : σ |Kn= id, σ |F

rd= τ}. Let S′′ ⊆M+

P0be

the subset consisting of prime divisors of degree ≡ 1 (mod d). We obtain

On primitive roots of tori: The case of function fields 213

the density of S′′ as:

den(S′′) =12d

+∑

n( �=1)∈N ′′

µ(n)#C ′′n

[Kn · Frd : F ]

=12d

+∑

gcd(d,n)=1n( �=1)∈N′′

µ(n)2d n1f(n)

≥ 12d

∏ν�d

ν∈I′′

(1− 1

ν1f(ν)

)> 0.

This completes the proof of Theorem 2.5. ��

3. Applications to fundamental units

Let F be a global function field having constant field k = Fr, together witha fixed degree one prime divisor∞. Let K be a given separable quadraticextension of F in which ∞ splits. In analogy with the classical situation,we call such K a real quadratic function field with respect to (F,∞). Thering of functions in F which are regular away from∞ is denoted byO, andthe integral closure ofO in K byOK . One shows easily thatO�

K∼= k�×Z.

A non-constant function ε ∈ K such that εk� generates O�K/k� is called a

fundamental unit of K. In the following we shall fix a fundamental unit εfor K, and let η = N(ε) = σ0(ε)ε ∈ k� denotes its norm, where σ0 is thegenerator of the Galois group of K/F . The order of η ∈ k� is denoted byoη.

Proposition 3.1. Let v be a prime divisor which is inert in K/F , and let ov

be the order of ε modulo v in K(v)�. Then ov divides oη(|v|+ 1).

Proof. Let mv be the maximal ideal of the valuation ring Ov ⊂ Kv. Recallthat v is inert in K/F implies

ε|v| ≡ εσ0 mod mv.

Therefore we have

ε|v|+1 ≡ N(ε) = η mod mv,

and thus

εoη(|v|+1) ≡ 1 mod mv. ��

214 Y.-M.J. Chen et al.

We are particularly interested in the following set of prime divisors of F :

Mε = {v : v is inert in K/F, ε modulo v has order oη(|v|+ 1)}.

To study Mε, we consider the torus T/F which is the quadratic twist of Gm/F

over K. Then let ε0 = εoη corresponds to a non-torsion point P0 ∈ T(F ).Theorems in §2 can apply to P0.

Lemma 3.2. If r is odd, then K ′ = F (√

ε0 + 1√ε0

) �∼= F · Fr2 .

Proof. Note that for any constant field extension,∞ always remains prime.Therefore, if ∞ splits in K/F , then it also splits in KFrd/FFrd for anypositive integer d. (Under constant field extension, a real quadratic functionfield is still real quadratic by definition.) Let the infinite places of K bedenoted by∞+ and∞−. Then the unit ε has divisor of the form g · ∞+ −g · ∞− for some nonzero integer g. The fact that ε is fundamental means|g| equals to the order of the divisor class of∞+ −∞− inside the divisorclass group of K. Since the order of a divisor class does not change underconstant field extensions, ε remains to be a fundamental unit in the realquadratic K ·Fr2 over F ·Fr2 . It follows that the field of constants of K(

√ε0)

cannot be Fr2 . (otherwise, one has Fr2 ⊂ K(√

ε0), K(√

ε0) = KFr2 , andthus ε won’t be fundamental in KFr2 .) Since K ′ ⊂ K(

√ε0), we obtain

K ′ �∼= F · Fr2 . ��

Finally, as an analogue of a recent theorem (Chen-Kitaoka-Yu [3],Roskam [10]), we prove

Theorem 3.3. The set Mε always has a positive density.

Proof. If a given prime divisor v ∈ R is inert in K/F , ov = oη(|v| + 1)implies ε0 modulo v has order |v|+1 in K(v)�. Conversely, if ε0 modulo vhas order |v|+ 1, we contend that ε modulo v must have order oη(|v|+ 1).Suppose this is not the case, there exists prime q > 1, q | oη(|v|+ 1), withεoη(|v|+1)/q modulo v equals 1. It follows q � (|v|+1), therefore q | oη. On theother hand ε(|v|+1)oη/q modulo v equals 1 also implies N(ε)oη/q = 1. Thiscontradicts the assumption that N(ε) has order oη. Thus the set Mε differsfrom the set M−

P0only by a finite set. If 2 | r, Theorem 2.4 immediately

gives den(Mε) > 0. Now suppose 2 � r. Note that ∞ is of degree 1 andsplits in K and hence K �∼= F · Fr2 , also by Lemma 3.2, K ′ �∼= F · Fr2 . Inthis case Theorem 2.5 says that den(Mε) > 0 if and only if P0 /∈ T(F )2. Ifoη is odd, this last condition is satisfied because ε is a fundamental unit. Ifoη is even, P0 /∈ T(F )2 is equivalent to N(εoη/2) �= 1 which follows fromthe definition of oη. Hence we are done. ��

On primitive roots of tori: The case of function fields 215

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