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(1)

LXX.2 (1995)

On an additive function on the set of ideals of an arbitrary number field

by

Tianxin Cai (Hangzhou)

1. Introduction. In [1] K. Alladi and P. Erd˝os showed that if βα(n) = P

p|npα then

X

n≤x

β1(n) ∼ π2 12 · x2

log x.

Many papers have been written concerning the additive function β1(n), see for example [4], [5], [6]. The best result is due to the author [2] and A. Mercier [9]; they have given the asymptotic formula

(1) X

n≤x

β1(n) = x2 nXm

i=1

di/ logix + O(1/ logm+1x) o

with arbitrary fixed m ≥ 1 and

(2) di=

i−1 X

v=0

(−2)v

v! ζ(v)(2) (i − 1)!

2i , 1 ≤ i ≤ m.

J. M. De Koninck and A. Ivi´c [5] also got (1), but they did not give the expression of (2).

In 1989, P. Zarzycki [10] studied the distribution of values of an additive function Bα(a) on the Gaussian integers given by

Bα(a) =X p|a

N (p)α

with fixed α > 0; the asterisk means that the summation is over the non- associate prime divisors p of a Gaussian integer a and N (a) = N (x + iy) = x2+ y2 is the norm of a. This function is a generalization of the function βα(n). Zarzycki obtained the asymptotic formula for the summatory func- tion P

a∈DBα(a) by using the complex integration technique where D is a certain set of Gaussian integers. The main result he obtained is the following

[97]

(2)

Theorem 1. For x → ∞, X

N (a)≤x

Bα(a) = ζ(1 + α)L(1 + α, χ4)

1 + α ·x1+α log x

 1 + O

 1 log x



where χ4 denotes the non-principal Dirichlet character modulo 4.

In [3], I announced that by using an elementary technique one could prove the following

Theorem 2. For x → ∞, X

N (a)≤x

Bα(a) =

N

X

n=1

n−1

X

v=0

(−1)v(1 + α)v

v! ξ(v)(1 + α)



× (n − 1)!x1+α (1 + α)nlognx + O

 x1+α logN +1x



with any fixed positive integer N , where ξ(s) = ζ(s)L(s, χ4) and ξ(v)(s) is the v-th derivative of ξ(s).

In the present paper, we consider the function Bα(a) defined on the integral ideals of an arbitrary number field K by the formula

Bα(a) =X

p|a

N (p)α,

where the asterisk means that p runs over prime ideals of K. We obtain the following

Theorem 3. For every number field K, X

N (a)≤x

Bα(a) =

N

X

n=1 n−1

X

v=0

(−1)v(1 + α)v

v! ζK(v)(1 + α) (3)

× (n − 1)!x1+α (1 + α)nlognx + O

 x1+α logN +1x



with any fixed positive integer N , where ζK is the Dedekind zeta function of K and ζK(v) its v-th derivative.

Clearly Theorem 3 contains Theorem 2 since every ideal of Q(

−1) is principal.

2. Auxiliary lemmas

Lemma 1. Let K be a number field. Then π(x, K) := X

N (p)≤x

1 = Li x + O(xe−c

log x)

(3)

with some c > 0. Moreover , for arbitrary fixed N ≥ 1, π(x, K) =

N

X

n=1

(n − 1)!x lognx + O

 x

logN +1x

 . For the proof, see [8], Theorem 191, and [7], §4, formula (5).

Lemma 2. For arbitrary fixed N ≥ 1 and α > 0, X

N (p)≤x

N (p)α=

N

X

n=1

(n − 1)!

(1 + α)n · x1+α lognx + O

 x1+α logN +1x

 .

P r o o f. We have X

N (p)≤x

N (p)α=

x

R

2−ε

tαdπ(t, K)

= xαπ(x, K) −

x

R

2−ε

αtα−1π(t, K) dt

=

N

X

n=1

(n − 1)!x1+α lognx

x

R

2−ε N

X

n=1

(n − 1)!αtα

lognt dt + O

 x1+α logN +1x



= 1

1 + α

N

X

n=1

(n − 1)!x1+α lognx − α

N

X

n=2

 1

(1 + α)2 + 1 (1 + α)3 + . . . + 1

(1 + α)n

 (n − 1)!x1+α lognx + O

 x1+α logN +1x



=

N

X

n=1

(n − 1)!

(1 + α)n · x1+α lognx+ O

 x1+α logN +1x

 .

Lemma 3. Let 2 ≤ Q < logL+1x, for arbitrary fixed L ≥ 1. Then

N

X

n=1

(n − 1)!

(1 + α)n · 1 lognx/Q

=

N

X

n=1

n−1

X

v=0

(1 + α)vlogvQ v!

 (n − 1)!

(1 + α)nlognx + O

 logNQ logN +1x

 . P r o o f. By Newton’s binomial formula,

N

X

n=1

(n − 1)!

(1 + α)nlognx/Q =

N

X

n=1

(n − 1)!

(1 + α)nlognx



1 −log Q log x

−n

(4)

=

N

X

n=1

(n − 1)!

(1 + α)nlognx

N −1

X

v=0

−n v



(−1)v log Q log x

v

+ O

 logNQ logN +1x



=

N

X

k=1

1 logkx

X

n+v=k n≥1, v≥0

n + v − 1 v

 (n − 1)!

(1 + α)n logvQ + O

 logN x logN +1x



=

N

X

k=1

(k − 1)!

(1 + α)klogkx

k−1

X

v=0

(1 + α)vlogvQ

v! + O

 logNQ logN +1x

 .

3. Proof of Theorem 3. Let P (a) denote a prime ideal factor of a in K of the largest possible norm. We should obtain the corresponding summation (3) for N (P (a))α instead of Bα(a). In fact, let

(4) X

N (a)≤x

=X

1+X

2. Here and below P

1 means N (a) ≤ x and P (a) is unique,P

2 means N (a)

≤ x and P (a) is not unique. By Theorem 202 of [8], one has

(5) X

N (a)≤x

1 = O(x), hence

(6) X

2(Bα(a) − N (P (a))α) = O(xα/2log x) X

N (a)≤x

1 = O(x1+α/2log x).

On the other hand,

(7) X

1(Bα(a) − N (P (a))α)

= X

1

n X

pkka, p=P (a) k≥2

N (p)α+ X

p|a, p6=P (a)

N (p)αo

X

N (pk)≤x k≥2

N (p)αX

1 pkka

1 + X N (p)≤x1/2

N (p)α X

1 p|a, p6=P (a)

1

X

N (pk)≤x k≥2

N (p)α X

N (a)≤x/N (pk)

1 + X N (p)≤x1/2

N (p)α X

N (a)≤x/N (p)

1

= On

x X N (pk)≤x

k≥2

N (p)α−1o + On

x X

N (p)≤x1/2

N (p)α−1o

= O x1+α/2 log x

 ,

where in the last equality we use (5) and Lemma 2.

(5)

Now we make two steps:

(i) We first sum over all N (a) ≤ x with N (P (a)) ≤ x/ logM +1x, M = [(N + 1)/α], including those with P (a)Q(a) | a, N (Q(a)) = N (P (a)). Then

X

N (a)≤x

N (P (a))α X

N (a)≤x

xα

logα(M +1)x = O

 x1+α logN +1x

 .

(ii) Next we sum over all N (a) ≤ x such that N (P (a)) > x/ logM +1x.

Assume a = pkpαk−1k−1. . . pα11, N (pk) > N (pk−1) ≥ . . . ≥ N (p1). Let q = pαk−1k−1. . . pα11. Noting that N (pk−1) ≤ N (q) < N (pk), if x is large enough, by Lemmas 2 and 3, one has

X

N (pkq)≤x

N (pk)α= X

N (q)<logM +1x

X

x/ logM +1x<N (p)≤x/N (q)

N (p)α

= X

N (q)<logM +1x

 N

X

n=1

(n − 1)!

(1 + α)n · y1+α logny + O

 y1+α logN +1y



x/N (q)

x/ logM +1x

= X

N (q)<logM +1x

1 N (q)1+α

N

X

n=1

(n − 1)!

(1 + α)n · x1+α

lognx/N (q)+ O

 x1+α logN +1x



= X

N (q)<logM +1x

1 N (q)1+α

N

X

n=1

n−1 X

v=0

(1 + α)vlogvN (q) v!

 (n − 1)!

(1 + α)n · x1+α lognx + O x1+α(log log x)N

logN +1x



=

N

X

n=1

n−1 X

v=0

(1 + α)v v!

X

N (q)<logM +1x

log N (q) N (q)1+α

 (n − 1)!

(1 + α)n · x1+α lognx + O x1+α(log log x)N

logN +1x



=

N

X

n=1 n−1

X

v=0

(−1)v(1 + α)v

v! ζK(v)(1 + α)(n − 1)!

(1 + α)n · x1+α lognx + O x1+α(log log x)N

logN +1x

 .

In the last equality we use the formulas (cf. [8], Theorem 141) X

a

1

N (a)s = ζK(s), Re s > 1,

(6)

and

X

a

logvN (a)

N (a)s = (−1)vζK(v)(s), Re s > 1.

Noting that N can be any positive integer, by (4), (6) and (7), we complete the proof of Theorem 3.

References

[1] K. A l l a d i and P. E r d ˝o s, On an additive arithmetic function, Pacific J. Math. 71 (1977), 275–294.

[2] T. X. C a i, On the mean value estimate of a certain arithmetic function, Chinese Sci. Bull. 29 (1984), 1481–1484 (in Chinese).

[3] —, On an additive function on the Gaussian integers, Adv. in Math. (Beijing) 20 (2) (1991), 250.

[4] J. M. D e K o n i n c k and A. I v i ´c, Topics in Arithmetical Functions, Notas Mat., Amsterdam, 1980.

[5] —, —, The distributions of the average prime divisors of an integer , Arch. Math.

(Basel) 43 (1984), 37–43.

[6] P. E r d ˝o s and A. I v i ´c, Estimates for sums involving the largest prime factor of an integer and certain related additive functions, Studia Sci. Math. Hungar. 15 (1980), 183–199.

[7] E. L a n d a u, Handbuch der Lehre von der Verteilung der Primzahlen, Leipzig, 1909.

[8] —, Einf¨uhrung in die elementare und analytische Theorie der algebraischen Zahlen und der Ideale, Leipzig, 1927.

[9] A. M e r c i e r, Comportement asymptotique deP

p≤xpa{x/p}k, Canad. Math. Bull.

30 (1987), 309–317.

[10] P. Z a r z y c k i, On a certain additive function on the Gaussian integers, Acta Arith.

52 (1989), 75–90.

DEPARTMENT OF MATHEMATICS HANGZHOU UNIVERSITY

HANGZHOU, P.R. CHINA

Received on 14.5.1992

and in revised form on 31.1.1995 (2363)

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