The distribution of the irreducibles in an algebraic number field

J. Aust. Math. Soc. 79 (2005), 369–390

THE DISTRIBUTION OF THE IRREDUCIBLES IN AN ALGEBRAIC NUMBER FIELD ¨ UK, ¨ REBECCA A. ROZARIO and C. SNYDER DAVID M. BRADLEY, ALI E. OZL

(Received 26 January 2004; revised 18 November 2004)

Communicated by W. W. L. Chen

Abstract We study the distribution of principal ideals generated by irreducible elements in an algebraic number field. 2000 Mathematics subject classification: primary 11R27.

1. Introduction In an abstract algebra course, students learn that the concepts of prime and irreducible elements do not coincide in an integral √ domain without unique factorization. Usually, various examples are given in Z[ −5], for instance, showing the existence of irreducibles which are not prime. Of course, as every student knows any prime is irreducible and so generally there are more irreducibles than primes. This difference leads naturally to two questions. First, can one give a characterization of irreducibles in familiar integral domains where unique factorization need not hold, such as the ring of integers in an algebraic number field? Second, how are the irreducibles distributed, again in an algebraic number field? The problem of characterizing irreducibles involves, among many challenges, a good characterization of all the prime ideals in any given ideal class of the ideal class group of the field. This has a particularly nice solution when the Hilbert class field of the number field is an abelian extension of the field of rational numbers Q, for class field theory shows us that the solution involves congruences, modulo certain This manuscript is an extension of the third author’s Masters thesis at University of Maine. c 2005 Australian Mathematical Society 1446-8107/05 $A2:00 + 0:00

369

370

¨ u¨ k, Rebecca A. Rozario and C. Snyder David M. Bradley, Ali E. Ozl

[2]

integers depending on only the field, for the rational primes contained in the prime ideals. (As a minor aside, we give a characterization of the irreducibles and primes in two imaginary quadratic number fields of class number two in the last section of this paper.) In other cases such a satisfactory characterization is not known and probably even nonexistent. In this note, we study instead the distribution of irreducibles. First, we give a little background. Let K be an algebraic number field and denote by M.x/ the number of nonassociate irreducible elements Þ with |N K =Q .Þ/| ≤ x. In the 1960’s, Rémond, [11], showed that M.x/ ∼ C

x .log log x/ D−1 ; log x

as x → ∞;

where C is a positive constant not explicitly given and D is the Davenport constant which is a positive integer depending on only the structure of the ideal class group of K . Now, if we let P.x/ denote the number of nonassociate primes ³ with |N K =Q .³ /| ≤ x, then by a classical density result P.x/ ∼

1 x ; h log x

where h is the class number of the field, that is, the order of the ideal class group. If h > 1 (so D > 1, see Section 2), then there are ‘many more’ irreducibles than primes. If h = 1, however, then the ring of integers is a unique factorization domain and hence the irreducibles and primes coincide. This is consistent with the estimates above once we observe that in this case, C = 1 and D = 1; see the next section for more on these constants. Subsequently, Kaczorowski, [6], gave a major extension of Rémond’s result, which we state here in simplified form: ! D−1 X x x j c M.x/ = m j .log log x/ + O .log log x/ ; log x j=0 log2 x as x → ∞, for some constant c > 0 and complex numbers m j . In particular, m D−1 = C the coefficient in Rémond’s estimate. As in Rémond’s case, the constants depend on K but are not explicitly given. Later, Halter-Koch and Müller in joint work [5] showed, among many results, how to determine the constant C and as a result showed that it depends on only the class group of K . This result prompted us to explore the dependence of some of the other coefficients in Kaczorowski’s estimate on the arithmetic of K . In particular, we consider m D−2 and give an explicit expression for this coefficient. We then apply this to the special case

[3]

Distribution of irreducibles

371

of a number field with cyclic class group in which case we find that m D−2 contains explicit arithmetic information about the field and some of the subfields of its Hilbert class field. (We chose the case of cyclic class group due to the messy combinatorical arguments in the general case. It would still perhaps be of interest to see what happens in general.) Finally, we compute—more precisely, approximate—m D−2 for two imaginary quadratic number fields with class number two. Indeed, this calculation shows that more than just properties of the class group figure into the makeup of m D−2 : 2. A Dirichlet series associated with irreducibles Let K be an algebraic number field, that is, a finite extension of the rational number field, Q, and let K denote its ring of integers. We denote by N .x/ the norm of an element x from K to Q. Also, we denote by N a the norm of an ideal a of K . Furthermore, let Cl = Cl.K / denote the class group of K and h = h K the class number, that is, the order of Cl.K /. In studying the distribution of the irreducibles, we introduce the following function.

O

DEFINITION 1. ¼.s/ = part, ¦ > 1.

O

P .Þ/;Þ irred.

|N .Þ/|−s , where s is a complex number with real

O

The sum runs over the principal ideals generated by irreducible elements of K . We obviously do not wish to count all associates of an irreducible since there are infinitely many when the unit group is infinite, that is, anytime K is not Q or an imaginary quadratic number field. Ultimately, we shall be interested in the ‘summatory’ function given by DEFINITION 2. M.x/ =

P .Þ/;Þ irred.;|N .Þ/|≤x

1, where x is any positive real number.

We shall first determine properties of ¼.s/ and then use a well-known Tauberian theorem to glean information about the distribution of M.x/. To this end, consider the following. Write Cl = {c1 = 1; c2 ; : : : ; ch }. DEFINITION 3. For each positive integer m, let ( ) h Y ki min h k = .k1 ; : : : ; kh / ∈ N0 : c j = 1; k1 + · · · + kh = m ; m =

D

j=1

Q ki min Q Q where ci = 1 means that ciki = 1 and if clii = 1 for some li such that 0 ≤ li ≤ ki for i = 1; : : : ; h, then li = 0 for all i or li = ki for all i. (Here N0 denotes the set of nonnegative integers.)


372

[4]

min

Notice that = guarantees that a product of elements is 1 but no nontrivial subproduct is 1. Hence the product gives a ‘minimal’ representation of 1. Later on it will be more convenient to think of the elements of m as functions in the usual way; namely, ( ) Y X .c/ min : Cl → N0 c = 1; .c/ = m : m =

D

D

c∈Cl

c

DEFINITION 4. The Davenport constant of Cl, denoted by D or D.Cl/, is the largest positive integer m such that m is nonempty.

D

The Davenport constant is defined as above for any finite abelian group. In general, the relation between the Davenport constant and the structure of the group is not known. On the other hand, it is well known (and easy to prove) that the Davenport constant is no larger than the order of the group. We now have the following proposition which gives a connection between irreducibles and prime ideals. First, we denote the set of nonzero prime ideals of K by .

O

P

PROPOSITION 2.1. For any complex s with ¦ > 1 and where whenever ki = 0, ¼.s/ =

D X Y h X

D

m=1 k∈

m

i=1

X ai

P

ai

is defined to be 1

N .ai /−s :

P

∃ pi1 ;:::;piki ∈ ∩ci ai =pi1 ···piki

D

PROOF. For k ∈ m , define ∩ ci ; k = a : a = a1 · · · ah ; ai = pi1 · · · piki ; some pi j ∈ S S where ai = 1, if ki = 0. Now let = k where the union is over all k in m. m By the uniqueness of the factorization of ideals into prime ideals, we see that this union is disjoint. Moreover, by the multiplicativity of the norms, we have

A

P

A

A

D h X XY X

D

m=1 k∈

m

i=1

ai

D

N ai−s =

A

X a∈A

N a−s ;

where ai are as above in the definition of k . Now notice that if a ∈ for some k ∈ m . Thus the ideal class [a] containing a satisfies

D

[a] =

h Y i=1

min

ci ki = 1;

A , then a ∈ A

k

[5]


373

D

by definition of m . Hence a = .Þ/ for some nonzero, nonunit integer Þ in K . But Qh notice that Þ must be irreducible for otherwise [a] = i=1 ci ki = 1, would not be a minimal representation of 1. Conversely, if Þ is irreducible, then .Þ/ ∈ k for some k; namely,

A

.Þ/ =

ki h Y Y

pi j ;

i=1 j=1

for some ki ∈ N0 and pi j ∈

P ∩c: i

Next, we examine the right-hand sum in the proposition above. To this end we define the following family of polynomials. DEFINITION 5. Let k be a positive integer and z1 ; : : : ; zk independent variables. Then X 1 Pk .z/ = Pk .z1 ; : : : ; zk / = z ¹1 · · · zk¹k : ¹1 · · · k ¹k 1 ¹ ! · · · ¹ ! 1 1 k k .¹1P ;:::;¹k /∈N0 j¹ j =k

Moreover, let P0 .z/ = 1. PROPOSITION 2.2. Let k be a positive integer and x 1 ; x2 ; x3 ; : : : be a sequence of P∞ independent variables. Moreover, for j = 1; : : : ; k, let s j = i=1 xij . Then X xn1 · · · xnk = Pk .s1 ; : : : ; sk /: .n 1 ;:::;nk /∈Nk n 1 ≤···≤n k

PROOF. First we introduce some notation. Let x n = xn1 · · · xnk for any n = .n 1 ; : : : ; n k / ∈ Nk : Let T = {n ∈ Nk : n 1 ≤ · · · ≤ n k }. Also, let Sk be the symmetric group on {1; : : : ; k}; for ¦ ∈ Sk , let ¦ n = .n ¦ .n1 / ; : : : ; n ¦ .n k / /. Next, let C = C.¦ / be the conjugacy class of ¦ in Sk , that is, C.¦ / = { ¦ −1 : ∈ S k }. Let ¦ =

k Y

j1 · · · j¹ j

j=1

be a factorization of ¦ into disjoint cycles, where ¹ j ∈ N0 and for each j and i = 1; : : : ; ¹ j , the permutations ji are the distinct j -cycles, say ji = .a ji1 · · · a ji j / with a jil ∈ {1; : : : ; k}; and with the convention that 1-cycles are included so that S j;i {a ji1 ; : : : ; a ji j } = {1; : : : ; k}. Recall that − ∈ C.¦ / if and only if − has the same type of cycle decomposition, that is, if −=

k Y j=1

0j1 · · · 0j¹ 0j

374


[6]

into disjoint cycles with the same conventions as above, then ¹ 0j = ¹ j for j = 1; : : : ; k; (see, for example [2]). Notice then that a conjugacy class in Sk is determined uniquely Pk by a k-tuple, .¹1 ; : : : ; ¹k / ∈ N0k with j=1 j ¹ j = k. Any permutation in the conjugacy class has a cycle decomposition determined by the ¹ j ’s as above. Moreover, recall that k! #C.¦ / = ; ¹1 ! · · · ¹k !1¹1 · · · k ¹k again see [2]. Furthermore, recall that the cardinality of the orbit of n under Sk , Sk n = {n : ∈ Sk }, is equal to |Sk |=|Sk .n/| where Sk .n/ = { ∈ Sk : n = n}, the stabilizer subgroup of n: Moreover, if m ∈ Sk n, then the stabilizer subgroups, Sk .m/ and Sk .n/, are conjugate and thus have the same cardinality. Now for the proof: Notice that 1 X X 1 XX X xm = x m; k! ¦ ∈S k! C ¦ ∈C k k k

m∈N ¦ m=m

m∈N ¦ m=m

P where C is the sum over the conjugacy classes of Sk . Now notice that if we Qk P ¹1 ¹k write ¦ = m∈Nk ;¦ m=m x m = s1 · · · sk , which is j=1 j1 · · · j¹ j as above, then independent of the choice of ¦ ∈ C. Hence X 1 XX X 1 X xm = |C| xm k! C ¦ ∈C k! C k k m∈N ¦ m=m

=

On the other hand,

1 k!

X

m∈N ¦ m=m

.¹1P ;:::;¹k /∈N0k j ¹ j =k

X

xn =

n∈T

k! s ¹1 · · · sk¹k = Pk .s1 ; ·; sk /: ¹1 ! · · · ¹k !1¹1 · · · k ¹k 1

X n∈T

1 X x ; |Sk n| m∈S n m k

since x m = x n for any m ∈ Sk n. Hence X

xn =

X X m∈Nk n∈T m∈Sk n

n∈T

1 1 X X |Sk .n/| x m xm = |Sk n| k! k n∈T m∈N

m∈Sk n

1 X X 1 X X X = |Sk .m/|x m = x k! m∈Nk n∈T k! m∈Nk n∈T ¦ ∈S .m/ m k

m∈Sk n

=

X 1 X X xm 1: k! k ¦ ∈S .m/ n∈T m∈N

k

m∈Sk n

m∈Sk n

[7]


But

P n∈T ; m∈Sk n

375

1 = 1, since only one permutation of m can belong to T . Therefore, X

xn =

1 X X x m = Pk .s1 ; : : : ; sk /; k! k ¦ ∈S m∈N

n∈T

k

¦ m=m

from above, as desired. PROPOSITION 2.3. Let k be a nonnegative integer and c any class in Cl. Then X a

N .a/−s = Pk .z/;

P

∃ p1 ;:::;pk ∈ ∩c a=p1 ···pk

where z j =

P

p∈P ∩c

N p− js , for all Re.s/ = ¦ > 1.

PROOF. For k = 0, both sides are equal to 1, for the left-hand side consists of one term, a = K which has norm equal to 1. Assume k > 0. Write ∩ c = {pn : n ∈ N}. For any n ∈ N, let xn = N pn−s . Then the proposition follows directly from Proposition 2.2, once we observe that N is multiplicative and all series involved converge absolutely, since ¦ > 1.

O

P

We now have the following useful corollary to Proposition 2.3. COROLLARY 2.4. X

¼.s/ =

|N .Þ/|−s =

.Þ/; Þ irred.

where zi j =

P

pi ∈P ∩ci

D X Y h X

D

m=1 k∈

m

Pki .zi1 ; : : : ; ziki /;

i=1

N pi− js .

P −s For the next proposition, write zi1 = = l + gi , where l = pi ∈P ∩ci N pi .1= h/ log.1=.s − 1//, and gi = gi .s/. It is well known that gi .s/ is regular at s = 1. We then have PROPOSITION 2.5. ¼.s/ = if k = .k1 ; : : : ; kh /, then ak;¼ =

k1 X

···

kh h X Y

¼1 =0 ¼h =0 i=1 ¼1 +···+¼h =¼

PD ¼=0

bki ;¼i ;

c¼l ¼ , where c¼ =

with

bki ;¼i =

PD

ki X ¹i1 =¼i

m=max.1;¼/

P

D ak;¼ , where

k∈

m

gi¹i1 −¼i ²k ;¹ ; ¼i ! .¹i1 − ¼i /! i i1


376

[8]

where X

²ki ;¹i1 =

1 k −1

.¹i2 ;:::;¹iki /∈N0i P j¹i j =ki −¹i1

¹i2 ! · · · ¹iki

!2¹i2

¹ik · · · ki i

¹ik

¹i2 zi2 · · · ziki i ;

if ki > 1, and we define ²0;0 = 1, ²1;1 = 1; and ²1;0 = 0. PROOF. First use the definition of the polynomials Pk .z/ to expand ¼.s/ in Proposition 2.3, where the indices of summation are ¹i j for i = 1; : : : ; h and j = 1; : : : ; ki . Hence ¼.s/ =

D X

X

h Y

D

m=1 .k1 ;:::;kh /∈

m

X

1 .l + gi /¹i1 z2i ¹i2 · · · zki i ¹iki ; ¹i1 · · · k!¹iki ¹ ! · · · ¹ !1 i1 ik i /

i=1 .¹i1 ;:::;¹iki P j¹i j =ki

P where .¹i1 ;:::;¹ik / · · · = 1, if ki = 0. Now in the right-hand most sum above, sum over i the ¹i1 first in which case we get X

ki X 1 .l + gi /¹i1 ¹i2 ¹iki ¹i1 .l + g / z · · · z = ²ki ;¹i1 ; i 2i k i i ¹ ¹i1 · · · k! iki ¹ ! · · · ¹ !1 ¹ ! i1 ik i1 i ¹ =0 /

.¹i1 ;:::;¹iki P j¹i j =ki

i1

¹i1 with ² as defined in the statement of the proposition. Next, expand zi1 = .l + gi /¹i1 as ¹i1 X ¹i1 ¼i ¹i1 −¼i l gi : ¼i ¼ =0 i

Then ki ki ¹i1 ki X X X .l + gi /¹i1 1 X ¹i1 ¹i1 −¼i ¼i gi ²ki ;¹i1 l = bki ;¼i l ¼i ; ²ki ;¹i1 = ¹ ! ¹ ! ¼ i1 i ¹ =0 ¹ =0 i1 ¼ =0 ¼ =0 i1

i1

i

i

where the b are defined as above. But then ki h Y X i=1 ¼i =0

bki ;¼i l

¼i

=

k1 X ¼1 =0

···

kh h X Y

bki ;¼i l ¼1 +···+¼h =

¼h =0 i=1

m X

ak;¼l ¼ ;

¼=0

with the a as defined above. P P P P But now k∈Dm m¼=0 ak;¼l ¼ = m¼=0 k∈Dm ak;¼l ¼ . Hence   m D D D X X X X X X  ¼.s/ = ak;¼l ¼ = ak;¼  l ¼ ;

D

m=1 ¼=0 k∈

as desired.

m

¼=0

D

m=max.1;¼/ k∈

m

[9]


377

Now we rewrite the ak;¼ in Proposition 2.5 in a form more convenient for winning an explicit formula for c¼ for ‘large’ ¼. COROLLARY 2.6. ¼.s/ = ak;¼ =

k1 X

···

PD ¼=0

²ki ;ki −½i =

P D−¼

P

¹=max.1;¼/−¼

D ak;¼ , with

k∈

¼+¹

kh h ¹i X Y 1 X ki ! gi¹i −½i ²ki ;ki −½i ; k ! .¹ − ½ /!.k − ¹ /! i i i i i ¹ =0 i=1 ½ =0

¹1 =0 h ¹1 +···+¹h =¹

where (as above)

c¼ l ¼ , where c¼ =

i

X

1 ¹i2 ! · · · ¹iki !

k −1 .¹i2 ;:::;¹ik /∈N0i P i j¹i j =½i

2¹i2

¹ik · · · ki i

¹ik

¹i2 zi2 · · · ziki i :

PROOF. (Sketch) In Propostion 2.5 change variables as follows: let ¹ = m − ¼, let ¹i = ki − ¼i , and let ½i = ki − ¹i1 . From this corollary we extract the following result. PD COROLLARY 2.7. Let ¼.s/ = ¼=0 c¼l ¼ . Then P Qh (i) c D = k∈D D i=1 .1=ki !/. P Qh P Qh P (ii) c D−1 = k∈D D−1 i=1 .1=ki !/ + k∈D D i=1 .1=ki !/ hj=1 k j g j . (iii) If D ≥ 2, then c D−2 =

h h h X Y X Y 1 1 X kjgj + ki ! k∈D i=1 ki ! j=1 k∈D D−2 i=1 D−1 h h X X XY 1 + kj kj gj gj + k j .k j − 1/ ki ! 1≤ j < j ≤h 1 2 1 2 j=1 k∈D i=1 D

1

g 2j

z j2 + 2 2

!! :

2

The proof is a straightforward application of the previous corollary. We further obtain the following expressions for ¼.s/ for some fields with small class number. COROLLARY 2.8. (i) Suppose D = 1 whence h = 1. Then ¼.s/ = l + g1 . (ii) If D = 2 so h = 2, say Cl = {1 = c1 ; c2 }, then 1 1 1 ¼.s/ = l 2 + .1 + g2 /l + g1 + g22 + z22 : 2 2 2 PROOF. In light of the formulas for the c¼ above, it suffices to compute of the groups listed.

D

m

for each


378

[10]

Let Cl = {1 = c1 }. Then we have only one minimal representation of 1, namely 1 = 1, implying that 1 = {1}. Using this with the previous corollary yields (i). Now let Cl = {1 = c1 ; a = c2 }. Then we have two minimal representations min min of 1, namely, 1 = 1, and aa = 1 implying that 1 = {.1; 0/} and 2 = {.0; 2/}; respectively. This yields (ii). min

D

D

D

3. The summatory function M(x) Having established formal properties of the Dirichlet series ¼.s/, we now use wellknown results relating a Dirichlet series to its associated summatory function as in [6]. We present the following weaker form of Kaczorowski’s ‘Main Lemma’ given in [6], which will be sufficiently strong for our purposes. P∞ Let f .s/ = n=1 an n −s be a Dirichlet series where s = ¦ + it with an ; ¦; t real numbers and an ≥ 0. As in [6] we have the following definition.

A

DEFINITION 6. We let be the set of those Dirichlet series f as above satisfying the following three additional properties: (i) For all x; y ∈ R such that 1 ≤ x < y, X an ≤ .y − x/ logc1 y + O.y /; x≤n≤y

for some c1 > 0, < 1 where the constants depend on f only. (ii) There exists a nonnegative integer k and functions g j .s/ for j = 0; : : : ; k, such that k X 1 j f .s/ = g j .s/ log ; s−1 j=0 for ¦ > 1 and such that gk .1/ 6 = 0 and g j .s/ is regular for ¦ > 1 and can be analytically continued to a regular function in the region given by c2 = s = ¦ + it : ¦ > 1 − log.|t| + 2/

R

R

for some c2 > 0. (iii) In the region

R , |g .s/| log j

c3

.|t| + 3/, for some c3 > 0.

PROPOSITION 3.1 (Corollary to Kaczorowski’s Main Lemma). Let f .s/ =

∞ X n=1

an n −s

[11]


379

P

A

be a Dirichlet series in class as defined above. Let S.x/ = n≤x an be the summatory function associated with f .s/. Then for all ž > 0 and all x ≥ ee , ! k−1 X x x j S.x/ = ; e j .log log x/ + O log x j=0 log2−ž x as x → ∞, where the e j are complex numbers given by ej =

k X ¹!

j!

¹= j

g¹ .1/I¹− j ;

with

Z .−1/m 1 Im = e z .log z/m dz; m! 2³i C

C

where is the path of integration consisting of the segment .−∞; −1] on the lower side of the real axis (so that the argument of log z is −³ ), the circumference of the unit circle taken counter-clockwise, and the segment [−1; −∞/ on the upper side of the real axis. The proof may be found in [6] where we take Case I and q = 0 in the Main Lemma. LEMMA 3.2. Let t be any positive real number with t < 1. Then (a) I0 = 0, P∞ m−1 P∞ (b) Im = exp t + n=2 .−1/n−1 .n/t n =n , where = 0:577 : : : is m=1 t Euler’s constant, (c) I1 = 1 and I2 = . R PROOF. Part (a) follows since I0 = C e z dz = 0. With respect to Part (b), consider the formal sum ∞ X

t m Im =

m=0

Z Z 1 1 1 e z e−t log z dz = e z z −t dz = : 2³i C 2³i C 0.t/

But then since I0 = 0, we have ∞ X m=1

X 1 tn Im = .−1/n−1 .n/ = exp t + t0.t/ n n=2 ∞

t

m−1

by [13]. Part (c) follows immediately from (b). COROLLARY 3.3. Let e j be defined as in Proposition 3.1. Then (i) if k ≥ 1, ek−1 = k gk .1/, (ii) if k ≥ 2, ek−2 = .k − 1/gk−1 .1/ + k.k − 1/gk .1/ :

! ;


380

[12]

The proof is immediate from the preceding lemma and proposition. We now apply these results to ¼.s/ to obtain information about M.x/. By [6], using results in [7], ¼.s/ belongs to the class . P We shall state a well-known result about p∈c .1=N ps /, for c ∈ Cl, but first we recall some definitions. Let K be an algebraic number field of degree n over Q with class group Cl.K / = Cl b denote the character group of Cl, that is, the group of homomorof order h. Let Cl phisms from Cl into the multiplicative group C∗ . As usual, we denote the principal character, that is, the constant character 1, by either 0 or simply by 1. Let be an arbitrary character on Cl, then we define the L-series

A

L.s; / =

X .a/ a

N as

.¦ > 1/;

where the sum is over all (nonzero) integral ideals of K . If = 1, the principal character, then L.s; 0 / = K .s/, the Dedekind zeta function of K . As is well known, L.s; / converges absolutely and uniformally on compact subsets in the half plane ¦ > 1. Moreover, since the norm map N is completely multiplicative on the set of ideals of K , we have Y .p/ −1 L.s; / = 1− ; N ps p for all ¦ > 1 and where the product is taken over all (nonzero) prime ideals of K . It is also well known that in the half plane ¦ > 1 − 1=n, the series for L.s; / converges, if 6 = 1, and L.s; / is regular there. On the other hand, K .s/ has a continuation into the same half plane but with a simple pole at s = 1 with (nonzero) residue a K . Furthermore, in the region K given by

R

¦ >1−

cK ; log.|t| + 2/

L.s; / does not vanish, where c K depends on K but not on . Now, since L.s; / is nonzero in the region above, we see that log L.s; / is defined and regular in this region. PROPOSITION 3.4. Let c be an ideal class of Cl. Then ∞ X X 1 X 1 1 1X = log K .s/ + .c/ log L.s; / − ; s Np h h m N pms p∈c m=2 p 6 =1

for ¦ > 1.

pm ∈c

[13]


381

For a proof see, for example [8], (or just about any text on algebraic number theory). P Notice that this proposition allows us to analytically continue p∈c N p−s onto the region K .

R

COROLLARY 3.5. Let gc .s/ =

P

p∈c .1=N p

s

/ − .1= h/ log.1=.s − 1//. Then

∞ X X 1 1 1X gc .s/ = log..s − 1/ K .s// + .c/ log L.s; / − ; h h m N pms m=2 p

hence regular in

R

pm ∈c

6 =1

K.

In particular, ∞ X X 1 1 1X gc .1/ = log a K + .c/ log L.1; / − ; h h m N pm m=2 p pm ∈c

6 =1

where a K is the residue of K .s/ at s = 1. PROOF. Write K .s/ as .1=.s − 1//.s − 1/ K .s/ and then apply log. We now apply this result to M.x/. PROPOSITION 3.6. Let K be an algebraic number field with class number h and associated Davenport number D. Then M.x/ = Dc D h −D

D−2 x X x x ; e j .log log x/ j + O .log log x/ D−1 + log x log x j=0 .log x/3=2

where the e j are given in Proposition 3.1 with g j .s/ = h − j c j .s/. PROOF. The proof is immediate since ¼.s/ =

PD ¼=0

¼ c¼ .s/h −¼ log.1=.s − 1// .

As an immediate corollary we have COROLLARY 3.7. M.x/ ∼ Dc D h −D .x=log x/.log log x/ D−1 . Compare this with [5, Theorem 1]. But we also get the following result. THEOREM 3.8. For D ≥ 2, M.x/ =

x C.log log x/ D−1 + B.log log x/ D−2 + O .E.x// ; log x

382


[14]

where C = Dc D h −D and B = .D − 1/c D−1 .1/h 1−D + D.D − 1/c D h −D , with , Euler’s constant, and where  x  .log log x/ D−3 if D ≥ 3;  log x E.x/ = x  otherwise:  .log x/3=2 4. The special case of number fields with cyclic class group We now investigate the asymptotic behavior of M.x/ when the number field K has cyclic class group Cl of order h > 1. Then we see, by Theorem 3.8, that in order to compute the coefficients C and B, we need to determine c D and c D−1 .s/. First of all, notice that D = h, for we have already observed that D ≤ h for any Cl. But now min since Cl is cyclic generated by c, say, then ch = 1, whence h ≤ D in this case. Now by Corollary 2.7, we need to determine m for m = D = h and m = D − 1 = h − 1. To this end, we cite the following main result of [3].

D

PROPOSITION 4.1. Let S = .a1 ; : : : ; an−k / be a sequence of n − k (not necessarily distinct) elements in Zn = Z=n Z. Suppose 1 ≤ k ≤ n=6 + 1 and that 0 cannot be expressed as a sum over a nonempty subsequence of S; then there exist an integer c coprime to n and a permutation ¦ of the set {1; 2; : : : ; n − k} such that ca¦ .i/ = 1 for Pn−k i = 1; : : : ; n − 2k + 1, and i=n−2k+2 |a¦ .i/ |n ≤ 2k − 2, where |x|n denotes the least positive inverse image of x under the natural homomorphism from the additive group of integers onto Zn . In particular, there are at least n − 2k + 1 terms in S which are relatively prime to n and all congruent to one another modulo n. We use this result to prove the following lemma. LEMMA 4.2. Suppose Cl = hci. Then

D

D

= {k : 1 ≤ k ≤ h; .k; h/ = 1};

where k : Cl → N0 with k .ck / = h and k .cl / = 0 otherwise;

D

D−1

= {½k : 1 ≤ k ≤ h; .k; h/ = 1};

where ½k : Cl → N0 with ½k .ck / = h − 2; ½k .c2k / = 1, and ½k .cl / = 0 otherwise.

D

PROOF. We start by determining the elements of D . Suppose c1 ; : : : ; ch ∈ Cl and Qh min i=1 ci = 1: Then the h sequences S j = .c1 ; : : : ; cˆj ; : : : ; ch / (where c j is omitted)

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383

satisfy the hypotheses of Proposition 4.1 with k = 1. Hence in each S j there are at least h−1 terms which are equal and generating Cl. Hence, we must have c1 = · · · = ch = c and hci = Cl. Thus D is as stated above. Qh−1 min Now consider D−1 . Suppose c1 ; : : : ; ch−1 ∈ Cl and i=1 ci = 1: Then the h − 1 sequences S j = .c1 ; : : : ; cˆj ; : : : ; ch−1 / satisfy the hypotheses above with k = 2 provided h ≥ 6. (For h < 6 the lemma follows by a straightforward calculation.) Hence, assume h ≥ 6 in which case in each S j there are at least h − 3 terms which are equal and generate Cl. But then, without loss of generality, c1 = · · · = ch−2 = c where hci = Cl : Thus ch−2 d = 1 for some d ∈ Cl; whence d = c2 , as desired.

D D

This lemma along with Corollary 2.7 and Theorem 3.8 yields the following proposition. PROPOSITION 4.3. Let K be an algebraic number field with cyclic class group Cl = hci of order h > 1. Then M.x/ =

x C.log log x/h−1 + B.log log x/h−2 + O .E.x// ; log x

where C = '.h/=..h − 1/!h h /, and  B=



'.h/ h − 1  '.h/ 1

+ h−1  a.h/ + .h − 2/!h h h .h − 2/! .h − 1/!

h X

 gck .1/ ;

k=1 .k;h/=1

where a.h/ = 1=2, if h = 3, and a.h/ = 1, otherwise; and where gc is as appears in Corollary 3.5. The proof follows immediately from Corollary 2.7 and Theorem 3.8. (Notice that when h = 3, | 2 | = 1, not '.h/.) Ph We now give an (partially) arithmetic interpretation of k=1 gck .1/. First, we .k;h/=1 introduce some notation. Once again assume K has cyclic class group Cl = hci and let L be the Hilbert class field of K . For each divisor d of h let L d denote the intermediate field in the extension L=K of degree d over K . (Since by class field theory Gal.L=K / ' Cl and Cl is cyclic, L d is uniquely determined.) Notice in particular that L 1 = K and L h = L. Finally, let a L d be the residue of the Dedekind zeta function L d .s/ at s = 1.

D

THEOREM 4.4. Given the assumptions of the previous paragraph, h X k=1 .k;h/=1

gck .1/ =

X ¼.d/ d|h

d

log a L d −

XX m≥2 p h[pm ]i=Cl

1 : m N pm


384

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PROOF. By Corollary 3.5 we have h X k=1 .k;h/=1

h ∞ X X X 1 1 '.h/ gck .s/ = ; log .s − 1/ K .s/ + þ.s/ − h h m N pms p m=2 k=1 .k;h/=1 pm ∈ck

where þ.s/ =

h X X

.ck / log L.s; /:

k=1 6 =1 .k;h/=1

For j = 0; : : : ; h − 1, let j be the character on Cl determined by j .c/ = hj for h a primitive hth root of unity. More generally, let d; j be the character on Cl determined by d; j .c/ = dj , for any positive integer d dividing h. Also let cn . j / =

h X

hjk ;

k=1 .k;h/=1

the usual Ramanujan sum. Then þ.s/ =

h−1 X

ch .− j / log L.s; j /:

j=1

But the Ramanujan sum has the explicit representation (see, for example, [4, page 238]) cn . j / = '.h/

¼.h=.h; j // ; '.h=.h; j //

and thus þ.s/ = '.h/

h−1 X ¼.¹/ X ¹|h

= '.h/

'.¹/

X ¼.¹/ ¹|h

'.¹/

log L.s; j /

j=1 .h; j/=h=¹ h X

log L.s; j / − '.h/ log K .s/:

j=1 p.h; j/=h=¹

Now, by [8, page 230], we have log L d .s/ =

X X

log L.s; ¹; j /:

¹|d j mod v . j;¹/=1

But then by Möbius inversion, X X X log L.s; j / = log L.s; ¹; j / = ¼.¹=d/ log L d .s/: j mod h .h; j/=h=¹

j mod v . j;¹/=1

d|¹

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385

Thus '.h/

X ¼.¹/ ¹|h

'.¹/

= '.h/

h X

log L.s; j /

j=1 .h; j/=h=¹

X X ¼.¹/ ¹ X X ¼.¹/¼.¹=d/ log L d .s/ ¼ log L d .s/ = '.h/ '.¹/ d '.¹/ ¹|h d|¹ d|h ¹|h d|¹

= '.h/

X

log L d .s/¼.d/

¹|h d|¹

d|h

=h

X ¼.d/ d|h

d

X ¼2 .¹/ '.¹/

= '.h/

X d|h

h ¼.d/ log L d .s/ '.h/ d

log L d .s/;

since X ¼2 .¹/ ¹|h d|¹

'.¹/

=

h ¼2 .d/ ; '.¹/ d

see, for example, [1, Lemma 3]. Hence þ.s/ = h

X ¼.d/ d|h

d

log L d .s/ − '.h/ log K .s/:

Now notice that lim+ þ.s/ = h

¦ →1

X ¼.d/ d|h

− h

d

X ¼.d/ d|h

=h

X ¼.d/ d|h

since '.h/ = h

P d|h

log.s − 1/ L d .s/ − '.h/ log.s − 1/ K .s/

d

d

! − '.h/ log.s − 1/

log a L d − '.h/ log a K ;

¼.d/=d. This gives us the result. 5. Examples

The coefficient C of M.x/ depends on the class group of K , more precisely, on the Davenport constant and the order of the class group. On the other hand, the

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coefficient B seems to depend more intrinsically on the arithmetic for the field K . In this section we consider approximating √ B for two imaginary √ quadratic number fields of class number 2, namely, K 1 = Q. −5 / and K 2 = Q. −15 / to see if the B are unequal. But before we carry out the calculations in these special cases, we consider Proposition 4.3 for the case where h = 2. COROLLARY 5.1. Let K be a number field with class number 2. Denote by c the nonprincipal ideal class of Cl. Finally, let L be the Hilbert class field of K . Then 1 x x 1 x M.x/ = ; log log x + .2.1 + gc .1// + / +O 4 log x 4 log x .log x/3=2 where is Euler’s constant and gc .1/ = log a K −

X X 1 1 : log a L − 2 m N pm m≥3 p∈c P

m≡1.2/

P

m m≥3 We need to compute a K , a L , and S = p∈c 1=.m N p /. m≡1.2/ To this end, let F be any algebraic number field. Then the residue of F .s/ at s = 1

is

2r1 .2³ /r2 R F h F ; √ w F |d F | where r1 and r2 are the number of inequivalent real and complex embeddings of F into C, respectively; R F is the regulator of F; h F its class number; w F the number of roots of unity in√ F ; and d F is the discriminant of F. For K 1√= Q. −5 /, r1 = 0, r2 = 1, R K 1 = 1, w K 1 = 2, and d K 1 = −20, and hence a K 1 = ³= 5. √ For K 2 = Q. √−15 /, r1 = 0, r2 = 1, R K 2 = 1, w K 2 = 2, and d K 2 = −15, and hence a K 2 = 2³ = 15: √ √ √ √ The Hilbert class √ fields of Q. −5 / and Q. −15 / are L 1 = Q. −5; 5 / and √ L 2 = Q. −15; 5 /, respectively. To compute a L i in these two cases, we first notice that r1 = 0 and r2 = 2. To compute the other invariants, we shall use the fact that L i are CM-fields, which will allow us to compute the regulators R L , and the fact that Gal.L i =Q/ ' C.2/×C.2/, the Klein four group, which will give us a way to compute the class numbers. √ + + √ To this end, let L = L ∩ R = Q. 5 / in both cases L = L i . Now R L + = log..1 √ 5/=2/ and by [12, Proposition 4.16] (for example) R L = .1=Q/2 log..1 + 5/=2/, where Q = .E L : W L E L + / ∈ {1; 2} with E F and W F the group of units, respectively, roots of unity in F for any number field F. But in our two cases, Q = 1; see [10, Theorem 1]. Thus in both cases √ 1+ 5 R L = 2 log : 2 aF =

O

O

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387

By [8, Proposition 17, page 68] (for example) we see that d L 1 = 202 and d L 2 = 152 . Finally, to compute the class numbers, we use Kuroda’s class number formula: 1 h L = q.L/h 1 h 2 h 3 ; 2 where the h i are the class numbers of the three quadratic subfields of L, and q.L/ = .E L : E 1 E 2 E 3 / with E i the group of units in the quadratic subfields, see, for example, [9]. In our cases, h 1 h 2 h 3 = 2 and since L=K is unramified q.L/ = 1, [10, Theorem 1]. Hence in both cases h L = 1. Therefore, √ ! √ ! 1+ 5 1+ 5 ³2 4³ 2 aL1 = log and a L 2 = log : 10 2 15 2 Next, we need to approximate the two series Si :=

X X m≥3 p∈ci m≡1.2/

1 m N pm

for the fields K i , i = 1; 2 and where Cl.K i / = hci i. Now, since X 1 1 log.z + 1/ 2 = − ; mz m 2 log.z − 1/ z m≥3 m≡1.2/

we see that S=

X X p∈c

m≥3 m≡1.2/

X 1 Np + 1 2 1 : = log − m N pm 2 Np − 1 Np p∈c

We now truncate the series S at N p < x for x > 3 and estimate the truncation error by a little elementary calculus. To this end, we write S = S.x/ + E.x/, where X 1 Np + 1 2 S.x/ := log − Np − 1 Np p∈c 2 N p