Journal of Number Theory 130 (2010) 2147–2156

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Journal of Number Theory

www.elsevier.com/locate/jnt

Note on divisor function for quaternion algebras

Guangshi Lü 1, Honggang Xia ∗,2

Dept. of Math., Shandong University, Jinan Shandong, 250100, China

a r t i c l e i n f o a b s t r a c t

Article history:Received 5 February 2010Available online 2 June 2010Communicated by Wenzhi Luo

MSC:11N3711F70

Keywords:Divisor functionQuaternion algebraAsymptotic formula

Let a be an integral ideal in a quaternion algebra U over ra-tional numbers Q which ramifies precisely at p and ∞, andd(a) be its divisor function. Recently, Kim and Zhang proved

a quaternion analogue of the classical formula∑∞

n=1d2(n)

ns =ζ 4(s)ζ(2s) , and then established an asymptotic formula for the sum∑

N(a)�x,a �=0, Ol(a)=O d2(a). In this note we improved the upperbound for the error term in the asymptotic formula, and then con-sidered the quaternion analogue of another well-known formula∑∞

n=1d(n2)

ns = ζ 3(s)ζ(2s) .

© 2010 Elsevier Inc. All rights reserved.

1. Introduction and main results

Let p be any fixed prime and U = U(p) be the quaternion algebra over the rational field Q whichramifies precisely at p and ∞. Suppose that a is an ideal in U, and then we can canonically associateit with two orders,

Ol = Ol(a) = {u ∈ U, ua ⊂ a},Or = Or(a) = {u ∈ U, au ⊂ a}.

They are called the left order and the right order respectively. An ideal a is integral if Ol(a) andOr(a) are maximal and contain a. A product ab is called proper if Or(a) = Ol(b). We assume that

* Corresponding author.E-mail addresses: gslv@sdu.edu.cn (G. Lü), xia@sdu.edu.cn (H. Xia).

1 Supported by the National Natural Science Foundation of China (Grant No. 10971119), Shandong Province Natural ScienceFoundation (Grant No. ZR2009AM007), and IIFSDU.

2 Supported by SRF for ROCS, SEM (Grant No. [2009]8) and Young Scholars Foundation by the School of Mathematics,Shandong University.

0022-314X/$ – see front matter © 2010 Elsevier Inc. All rights reserved.doi:10.1016/j.jnt.2010.03.008

2148 G. Lü, H. Xia / Journal of Number Theory 130 (2010) 2147–2156

all products in this note are proper and all ideals are integral. We say that b|la, that is, b divides a

from the left, if there exists c such that bc is proper and equal to a. Similarly we define b|ra. Givenan integral a, Kim and Zhang [3] introduced the divisor function

d(a) = �{(b, c): a = bc

}.

They further showed that d(a) is multiplicative, i.e. if a = bc with (N(b), N(c)) = 1, then d(a) =d(b)d(c), where N(b) and N(c) are the norms of the ideals b and c.

Let O be a maximal order in U. Define the zeta-function of the maximal order O as

ζO(s) =∑a�=0

Ol(a)=O

N(a)−s,

where N(a) is the norm of the ideal a, and the sum is over non-zero integral ideals which satisfyOl(a) = O. For the zeta-function of a maximal order, Kim and Zhang [3] proved an interesting quater-nion analogue of the well-known formula for the classical divisor function

∞∑n=1

d2(n)

ns= ζ 4(s)

ζ(2s), (1.1)

where d(n) is the divisor function and ζ(s) is the Riemann zeta-function (see e.g. Ivic [2, Chapter 1,Section 7]). More precisely, they proved the following beautiful formula.

Theorem A.

∑a�=0

Ol(a)=O

d2(a)N(a)−s = ζ 4O(s)

ζO(2s).

As a corollary, they proved an interesting asymptotic formula for the summatory function of d2(a).

Theorem B. For any ε > 0, we have

∑N(a)�x

a�=0, Ol(a)=O

d2(a) = R(x) + O(x

53 +ε

),

where R(x) = x2(A log3 x + B log2 x + C log x + D) for some constants A, B, C and D.

In this short note we first focus on the estimation of the error term in the asymptotic formula inTheorem B. By connecting this question with the classical four-dimensional divisor problem, i.e. thesummatory function of the arithmetical function d4(n) (see e.g. Ivic [2, Chapter 13]), we are able toprove a better result.

Theorem 1.1. For any ε > 0, we have

∑N(a)�x

a�=0, Ol(a)=O

d2(a) = R(x) + O(x

32 +ε

),

where R(x) = x2(A log3 x + B log2 x + C log x + D) for some constants A, B, C and D.

G. Lü, H. Xia / Journal of Number Theory 130 (2010) 2147–2156 2149

In addition, there is another well-known identity involving the divisor function d(n), namely

∞∑n=1

d(n2)

ns= ζ 3(s)

ζ(2s). (1.2)

In the classical case, formulae (1.1) and (1.2), as well as the summatory functions of their coeffi-cients, are usually studied simultaneously (see e.g. Ivic, Chapter 14, Section 2). Therefore it seemsinteresting to investigate the quaternion analogue of formula (1.2). To this end, we have the followingresults.

Theorem 1.2.

∑a�=0

Ol(a)=O

d(a2

)N(a)−s = ζ 3

O(s)

ζO(2s)L(s),

where L(s) is an infinite product over primes which converges absolutely for Re s > 4/3.

Theorem 1.3. For any ε > 0, we have

∑N(a)�x

a�=0, Ol(a)=O

d(a2) = R1(x) + O

(x

13996 +ε

),

where R1(x) = x2(A1 log2 x + B1 log x + C1) for some constants A1 , B1 , and C1 .

In Section 2, we shall follow the arguments of Kim and Zhang [3] to establish Theorem 1.2. Forall the basic facts and conclusions on quaternion algebras used in Section 2, please refer to Kim andZhang [3] and Reiner [5]. In Section 3, by connecting the asymptotic formulae in Theorems 1.1 and 1.3with the classical divisor problems, we complete the proof of Theorems 1.1 and 1.3. In fact, our resultscome from the best known results on three-dimensional and four-dimensional divisor problems (seee.g. formula (13.14) and Theorem 13.2 in Ivic [2]).

2. Proof of Theorem 1.2

In order to avoid repetition, we shall follow most of the notations used in Kim and Zhang [3].Recall that U = U(p) be a quaternion algebra over the rational field Q which ramifies precisely at pand ∞. Let q denote a rational prime. For the case q �= p, let a be an ideal with norm q2n+m and(qn)|maxa, i.e., (qn)|a with n maximal. We say that a is primitive if n = 0.

Define S(c) = #{a primitive | a = p1p2 · · ·pc, p1 = pc, c � 2}, D(c) = #{a primitive | a = p1p2 · · ·pc,

p1 �= pc, c � 2}, here the pi′s are prime ideals with norm q.

Lemma 2.1.

S(c) ={

qc−1 + 1 if c is even;qc−1 + q if c is odd.

(2.1)

Proof. By Proposition 13 in [3], we have

S(c) + D(c) = (q + 1)qc−1. (2.2)

2150 G. Lü, H. Xia / Journal of Number Theory 130 (2010) 2147–2156

On the other hand, we get the recursive relation as follows

S(c) = S(c − 2)q + D(c − 2)(q − 1). (2.3)

From (2.2) and (2.3), we complete the proof of this lemma. �Define further

p(m, c) = #{a

∣∣ a primitive, N(a) = qm,(qc)∣∣

maxa2}.

Lemma 2.2.

p(m, c) =

⎧⎪⎪⎨⎪⎪⎩

qm + 1 if c = 0, m odd;qm + q if c = 0, m even;(q − 1)(qm−c−1 + qc−1) if m > 1, m odd, 0 < c � (m − 1)/2;(q − 1)(qm−c−1 + qc) if m > 1, m even, 0 < c � (m − 1)/2.

Proof. First we remark several special cases for p(m, c).If m = 1,

p(m, c) = p(1,0) = q + 1.

If m > 1, m odd, c = (m − 1)/2, then we have

p(m, c) = (q + 1)(q − 1)qc−1.

On the other hand if m > 1, c = 0, we can conclude that

p(m,0) = qS(m − 1) + (q − 1)D(m − 1) = (q − 1)(q + 1)qm−2 + S(m − 1).

Now let’s suppose m > 1, 0 < c < (m − 1)/2, and a be the ideal with norm qm and (qc)|maxa2. So

we can write a = a1a2 · · ·acac+1 · · ·am−c−1am−ca−1c a

−1c−1 · · ·a−1

1 , and

a2 = a1 · · ·acac+1 · · ·am−c−1am−ca−1c · · ·a−1

1 a1 · · ·acac+1 · · ·am−c−1am−ca−1c · · ·a−1

1

= (qc)a1 · · ·acac+1 · · ·am−c−1am−cac+1 · · ·am−c−1am−ca

−1c a

−1c−1 · · ·a−1

1 .

Here ai, i = 1, . . . ,m, are maximal ideals with norm q. From the expressions of a and a2 abovewe can see that ac �= a

−1c+1,ac �= am−c,am−c �= a

−1c+1. In the expression of a2 there is the piece

· · ·ac+1 · · ·am−c−1am−cac+1 · · · , so there are S(m−2c +1) possibilities for this piece and the conditionam−c �= a

−1c+1 is satisfied. Now for ac there are q + 1 − 2 = q − 1 choices and for ai, i = 1, . . . , c − 1,

q choices each. Finally we have p(m, c) = qc−1(q − 1)S(m − 2c + 1). This completes the proof ofLemma 2.2. �

Now we begin to prove Theorem 1.2. Since d(a) is multiplicative, we have

∑a�=0

O (a)=O

d(a2)N(a)−s =

( ∞∑m=0

∑N(a)=pm

O (a)=O

d(a2)N(a)−s

) ∏q �=p

( ∞∑m=0

∑N(a)=qm

O (a)=O

d(a2)N(a)−s

). (2.4)

l l l

G. Lü, H. Xia / Journal of Number Theory 130 (2010) 2147–2156 2151

Therefore it suffices to consider

∞∑m=0

∑N(a)=qm

Ol(a)=O

d(a2)q−ms, q prime.

First we consider the case q �= p. We divide the sum into two parts according to whether m is oddor even

∞∑m=0

∑N(a)=qm

Ol(a)=O

d(a2)q−ms =

∞∑m=0

[m/2]∑n=0

∑N(a)=qm

(qn)|maxa

d(a2)q−ms

=∞∑

k=2

k∑n=0

k−n−1∑c=1

p(2k − 2n)d(a2)q−2ks

+∞∑

k=1

k∑n=0

k−n∑c=1

p(2k − 2n)d(a2)q−(2k+1)s

+ (terms with m = 0,1,2).

In the above sum a has the property that (qn)|maxa, N(a) = qm , (q2n+c)|maxa2. Now by Proposition 12

in Kim and Zhang [3], we have

∞∑m=0

∑N(a)=qm

Ol(a)=O

d(a2)q−ms

=∞∑

k=2

k−2∑n=0

k−n−1∑c=1

(q − 1)(q2k−2n−c−1 + qc)[q2n+c+1 − 1

q − 1(4k − 4n − 2c + 1) − 2(2n + c + 1)

q − 1

+ 2(q2n+c+1 − 1)

(q − 1)2

]q−2ks

+∞∑

k=2

(q2 + q

)(q2k−1 − 1

q − 15 − 2(2k − 1)

q − 1+ 2(q2k−1 − 1)

(q − 1)2

)q−2ks

+∞∑

k=2

(q2k+1 − 1

q − 1− 2(2k + 1)

q − 1+ 2(q2k+1 − 1)

(q − 1)2

)q−2ks

+∞∑

k=2

k−2∑n=0

(q2k−2n + q

)[q2n+1 − 1

q − 1(4k − 4n + 1) − 2(2n + 1)

q − 1+ 2(q2n+1 − 1)

(q − 1)2

]q−2ks

+∞∑

k=1

k−1∑n=0

k−n∑c=1

(q − 1)(q2k−2n−c + qc−1)[q2n+c+1 − 1

q − 1(4k − 4n − 2c + 3) − 2(2n + c + 1)

q − 1

+ 2(q2n+c+1 − 1)

(q − 1)2

]q−(2k+1)s

2152 G. Lü, H. Xia / Journal of Number Theory 130 (2010) 2147–2156

+∞∑

k=1

(1 + q)

(q2k+1 − 1

q − 13 − 2(2k + 1)

q − 1+ 2(q2k+1 − 1)

(q − 1)2

)q−(2k+1)s

+∞∑

k=1

k−1∑n=0

(q2k−2n+1 + 1

)[q2n+1 − 1

q − 1(4k − 4n + 3) − 2(2n + 1)

q − 1+ 2(q2n+1 − 1)

(q − 1)2

]q−(2k+1)s

+ 1 + 3(q + 1)q−s + (6q2 + 8q + 5

)q−2s

= · · ·= q2s(q5 + q6s + 3q7s + 3q8s + q9s + 2q4+s + 3q5+s + 3q3+2s + 5q4+2s − 2q3+3s

+ q4+3s + q5+3s − 6q2+4s − 6q3+4s − q4+4s − q5+4s − q1+5s − 8q2+5s − 6q3+5s

+ q4+5s − 2q1+6s − 2q2+6s + 3q3+6s + 5q1+7s + 2q1+8s)/((q − qs)4(q + qs)3(

q2s − 1)2)

.

Note: After dividing the sum on m into two parts we split each sum into three parts since the formulaof p(m, c) for c = 0 is special. The last three terms are corresponding to the cases that m = 0,1,2respectively. The work omitted in the dots could be done by hand but it is rather tedious and ele-mentary. We get the result with the help of Mathematica.

Hence by some elementary arguments, we find that for the case q �= p

∞∑m=0

∑N(a)=qm

Ol(a)=O

d(a2)N(a)−s = (1 + q−s)(1 − q1−2s)

(1 − q1−s)3(1 − q−s)2

(1 + O

(1

p3σ−3

)). (2.5)

In addition, for the case q = p, by Lemma 5 and Proposition 7 in Kim and Zhang [3], we have

∞∑m=0

∑N(a)=pm

Ol(a)=O

d(a2)N(a)−s =

∞∑m=0

(2m + 1)p−ms = ps(1 + ps)

(ps − 1)2. (2.6)

Recall that Lemma 16 in [3] gives

ζO(s) = (1 − p1−s)ζ(s)ζ(s − 1). (2.7)

Therefore from (2.4)–(2.7), we find that

∑a�=0

Ol(a)=O

d(a2)N(a)−s = ζ 3

O(s)

ζO(2s)L(s),

where L(s) is an infinite product over primes which converges absolutely for Re s > 4/3.

3. Proof of Theorems 1.1 and 1.3

In this section, we complete the proof of Theorems 1.1 and 1.3. Recall that the classical divisorfunction is defined by the identity

ζ k(s) =∞∑ dk(n)

ns,

n=1

G. Lü, H. Xia / Journal of Number Theory 130 (2010) 2147–2156 2153

for Re s > 1, the k-dimensional divisor problems investigate the estimation of �k(x), the error termin the asymptotic formula for

∑n�x dk(n). Our results are based on the best known results on three-

dimensional and four-dimensional divisor problems, namely

∑n�x

d3(n) = x(a1 log2 x + b1 log x + c1

) + �3(x), (3.1)

∑n�x

d4(n) = x(a log3 x + b log2 x + c log x + d

) + �4(x), (3.2)

where a, b, c, d, a1, b1, c1 are some constants, and �3(x) and �4(x) satisfy

�3(x) � x4396 +ε, �4(x) � x

12 +ε. (3.3)

One can find these results in Chapter 13 of Ivic [2], or can refer to the original papers Kolesnik [4]and Heath-Brown [1].

Proof of Theorem 1.1. From Theorem A, we have

ζ 4O(s)

ζO(2s)=

∑a�=0

Ol(a)=O

d2(a)N(a)−s :=∞∑

n=1

an

ns. (3.4)

By (2.7), we find that

∞∑n=1

an

ns= (1 − p1−s)4ζ 4(s)ζ 4(s − 1)

(1 − p1−2s)ζ(2s)ζ(2s − 1). (3.5)

Note that for Re s > 2,

ζ 4(s)ζ 4(s − 1) =∞∑

n=1

d4(n)

ns

∞∑m=1

d4(m)

ms−1

=∞∑

n=1

( ∑n=kl

d4(k)d4(l)l

)n−s :=

∞∑n=1

bn

ns. (3.6)

Therefore for Re s > 2, we can rewrite (3.5) as

∞∑n=1

an

ns=

∞∑n=1

bn

ns

∞∑n=1

cn

ns, (3.7)

where for Re s > 1,

∞∑ cn

ns= (1 − p1−s)4

(1 − p1−2s)ζ(2s)ζ(2s − 1)

n=1

2154 G. Lü, H. Xia / Journal of Number Theory 130 (2010) 2147–2156

= (1 − p1−s)4

1 − p1−2s

∞∑n=1

μ(n)

n2s

∞∑n=1

μ(n)

n2s−1

= (1 − p1−s)4(

1 − p−2s) ∏q �=p

(1 − q−2s)(1 − q1−2s),

and for σ > 1

∞∑n=1

|cn|nσ

< +∞. (3.8)

By (3.2) and partial summation, we have

∑n�x

nd4(n) =x∫

1−0

t d

(∑n�t

d4(n)

)

= x2(α log3 x + β log2 x + γ log x + δ) + O

(x

32 +ε

). (3.9)

From (3.6), we obtain

∑n�x

bn =∑n�x

( ∑n=kl

d4(k)d4(l)l

)=

∑k�x

d4(k)∑

l�x/k

d4(l)l. (3.10)

By (3.9) and (3.10), we have

∑n�x

bn =∑k�x

d4(k){(x/k)2(α log3(x/k) + β log2(x/k) + γ log(x/k) + δ

) + O((x/k)

32 +ε

)}

= x2(α1 log3 x + β1 log2 x + γ1 log x + δ1) + O

(x

32 +ε

). (3.11)

From (3.4) and (3.7), we have

∑N(a)�x

a�=0, Ol(a)=O

d2(a) =∑n�x

an =∑n�x

∑n=kl

bkcl

=∑l�x

cl

∑k�x/l

bk.

Then (3.11) gives

∑N(a)�x

a�=0, Ol(a)=O

d2(a) =∑l�x

cl{(x/l)2(α1 log3(x/l) + β1 log2(x/l) + γ1 log(x/l) + δ1

) + O((x/l)

32 +ε

)}

= x2(A log3 x + B log2 +C log x + D) + O

(x

32 +ε

),

where we have used (3.8). This completes the proof of Theorem 1.1. �

G. Lü, H. Xia / Journal of Number Theory 130 (2010) 2147–2156 2155

The proof of Theorem 1.3 is similar to that of Theorem 1.1. So we may be brief.

Proof of Theorem 1.3. By Theorem 1.2 and (2.7), we have

∑a�=0

Ol(a)=O

d(a2)N(a)−s = ζ 3

O(s)

ζO(2s)L(s) = (1 − p1−s)3ζ 3(s)ζ 3(s − 1)

(1 − p1−2s)ζ(2s)ζ(2s − 1)L(s), (3.12)

where L(s) is an infinite product over primes which converges absolutely for Re s > 4/3. Then wehave

∑N(a)�x

a�=0, Ol(a)=O

d(a2) =

∑n�x

( ∑n=kl

uk vl

), (3.13)

where

un =∑n=kl

d3(k)d3(l)l, (3.14)

and for Re s > 4/3, the Dirichlet series

∞∑n=1

vn

ns= (1 − p1−s)3

(1 − p1−2s)ζ(2s)ζ(2s − 1)L(s) (3.15)

absolutely converges.By (3.1) and (3.14), we apply partial summation to obtain

∑n�x

un =∑n�x

∑n=kl

d3(k)d3(l)l =∑k�x

d3(k)∑

l�x/k

d3(l)l

=∑k�x

d3(k){(x/k)2(α2 log2(x/k) + β2 log(x/k) + γ2

) + O((x/k)1+ 43

96 +ε)}

= x2(α3 log2 x + β3 log x + γ3) + O

(x

13996 +ε

). (3.16)

On collecting (3.13), (3.15) and (3.16), we have

∑N(a)�x

a�=0, Ol(a)=O

d(a2) =

∑l�x

vl{(x/l)2(α3 log2(x/l) + β3 log(x/l) + γ3

) + O((x/l)

13996 +ε

)}

= x2(A1 log2 x + B1 log x + C1) + O

(x

13996 +ε

).

This completes the proof of Theorem 1.3. �

2156 G. Lü, H. Xia / Journal of Number Theory 130 (2010) 2147–2156

References

[1] D.R. Heath-Brown, Mean values of the zeta-function and divisor problems, in: Recent Progress in Analytic Number Theory,Symposium Durham, vol. 1, Academic, London, 1979, pp. 115–119.

[2] A. Ivic, The Riemann Zeta-Function, John Wiley & Sons, New York, 1985.[3] H. Kim, Y.C. Zhang, Divisor function for quaternion algebras and application to fourth moments of L-functions, J. Number

Theory 129 (2009) 3000–3019.[4] G. Kolesnik, On the estimation of multiple exponential sums, in: Recent Progress in Analytic Number Theory, Symposium

Durham, vol. 1, Academic, London, 1979, pp. 231–246.[5] I. Reiner, Maximal Orders, Academic Press, London, 1975.