1. Generalized Prime Number Theorem.
By Takayoshi MITSUI (ReceivedJanuary 31, 1957)
Let K be an algebraic number field of degree n, K(1),•c, K(ƒÁ1) the real conjugate
fields of K, and K(ƒÁ1+1),•c, K(n) (n=ƒÁ1+2ƒÁ2) the complex conjugate fields of K where
K(p+ƒÁ2)=(p=ƒÁ1+1,•c,ƒÁ1+ƒÁ2), we denote by P the direct product of ƒÁ1 real
lines and ƒÁ2 complex planes, Let ƒÊ be an ideal number of K , and ƒÊ(1),•c,ƒÊ(n) the conjugates of ƒÊ as defined by Hecke [2] , Then V(ƒÊ)=(„ ƒÊ(1)„ ,•c,„ ƒÊ(ƒÁ1)„ ,ƒÊ(ƒÁ1+1),•c,
ƒÊ(ƒÁ1+ƒÁ2) can be regarded as a point in P, and it is a problem to compute the number
of prime ideal numbers ƒÖ such that V(ƒÖ) are in a certain domain in P.
In case n+ƒÁ1=2, Rademacher [7] obtained the following result; let K be a real
quadratic number field, Y1, Y2 positive number, a an ideal of K, ƒÏ a totally positive
integer of K, (ƒÏ, a) t, If we denote by ƒÎ(ƒÏ, a; Y1, Y2) the number of prime numbers
w satisfying the conditions (ƒÖ=ƒÏ(mod a), 0<ƒÖ(1)•…Y1, 0<ƒÖ(2)•…Y2, then we have
(1)
where h is the number of ideal classes, is a totally positive fundamental unit
such that ā>1.
On the other hand, it is known (Siegel [8], walfisz •m10]) that we have
(2)
where the remainder term is uniform with respect to the modulus k, that is , the constants in the remainder term do not depend on k under the assumption k•…(log x) A
for a fixed positive number A.
In the present paper, we shall obtain a result including all above theorems as
special cases. Our Main Theorem will be stated and proved in our final paragraph
•˜ 4, after some preparations in •˜•˜1-3. we begin by generalizing (2) to the case of
arbitrary algebraic number field in •˜ 1. The aim of this paragraph will be attained
by the theorem on ƒÎ(x, _??_), the number of prime ideals p in the class _??_ with Np•…x,
given at the end of •˜1. Now ƒÎ(x, _??_) may be considered as a special case for ƒÉ=1 of a certain sum
(see (3.52)) for Grossencharacter A of K. The estimation of this sum‡”*ƒÉ(ƒÖ) for
the case ƒÉ=1 will be given as Lemma 8 in •˜2, after seven lemmas concerning the
properties of ƒÄ-f unction with Grossencharacter, In •˜2, we shall prove three more lemmas needed to establish the uniformity of the remainder term of our final for 2 Takayoshi MITSUI
mula with respect to the modulus.
In •˜3, we shall generalize the following theorem of Rademacher [6] by means
of which he could establish his above cited result (1); let K be a real quadratic
field, a an ideal, āa a totally positive fundamental unit mod a such that and
define a function W(ƒÊ) for ƒÊ=K as follows;
If we take a totally positive integer o prime to a and denote by ƒÎ(a,ƒÏ; x, v) (0 •…1) the number of totally positive prime numbers ƒÖ of K satisfying the conditions; ƒÖ=p (mod a), NƒÖ•…x, 0•…W(ƒÖ) then we have The theorem, which will be the aim of this paragraph, will be formulated at the beginning. All results of •˜•˜1-2 are utilized in the proof of this theorem, by which our Main Theorem will be then proved in •˜4. The author wishes to express his cordial thanks to Professor S. Iyanaga for his encouragement during the preparation of this paper. 1. Improvement of prime ideal theorem. We consider Hecke's L-function L(s, x) of an algebraic field K with degree n, where x is a character mod a (a means the product of an ideal a and some infinite prime ideals p(i)•‡). We write as usual s=ƒÐ+it, ƒÐ=k(s), t=k(s) and denote by x0 the principal character and by A1, A2,•c positive absolute constants. (For details on the properties of L-functions, cf. Landau [3]). LEMMA 1. If x is a character mod a, then we have in the strip -1/2•…ƒÐ•…4 (1.1) „ L(s, x)(s-1)E(x)„ PROOF. It is obvious that L(s, x)(s-1)E(x) is regular in the strip -1/2•…ƒÐ•…4. First assume that x is a primitive character. On the line ƒÐ=4, we have „ L(s, x)(s-1)E(x)„ •…A3(1+„ t„ ), and on the line ƒÐ=-1/2, we have by the functional equation of L(s, x), „IL(s, x)(s-1)E(x)„ •…A4Na(1+„ t„ )n+1 Finally in the strip -1/2•…ƒÐ<4, we have„ L(s, x)(s-1)E(x)„ •…C1ec2t t, where C1, C2 are positive constants independent of t, ƒÐ. Hence by theorem of Phragmen-Lindelof, we have Generalized Prime Number Theorem. 3 (1.2) „ L(s, x)(s-1)E(x) „ •…A5Na(1+„ t„ )n+1. Next assume x is not primitive and let x0 be the primitive character induced from x. Then we have L(s, x)=L(s, x0) II (1-x0(p)Np-s), in which II(1-x0(p)Np-s)„ •…II(1+Np1/2)•…A6IINp•…A6Na. Therefore (1.1) is obtained. LEMMA 2. In the region „ t„ •†4,ƒÐ•†1-A7/log(„ t„ Na), we have L(s, x) •‚0. PROOF. 1t suffices to prove for t•†4. Suppose that ƒÀ+iƒÁ(ƒÁ•†4) is a zero point of L(s, x). Taking ƒÐ0=1+A/log(ƒÁNa) (A>0 will be determined later) we denote ƒÐ0+iƒÁ by s0. Then In the circle „ s-s0„ •…1/2, we have by Lemma 1 Therefore by lemma ƒÀ of Titchmarsh [9] p. 49, We obtain analogously Taking A sufficiently small, we have Hence, by a well-known inequality we obtain which gives Since we may take A small, the value of the formula in the brackets is surely positive, that is, Thus lemma is proved. LEMMA 3. (Titchmarsh) Suppose that f(s) is regular in the circle „ s-s0„ •… 4 Takayoshi MITSUI and „ f (s)/f(s0)„ 1), „ f•L(s0)/f(s0)„ („ s-s0„ •…r). PROOF. See Titchmarsh [9] p. 50. LEMMA 4. In the region „ t„ •†5, ƒÐ•† 1-A7/8 log(„ t„ Na), we have (1.3) PROOF. Again we may assume t•†5. Let ƒÁ•†5 and put s0=ƒÐ0+iƒÁ. Then L(s, x) is regular in the circle „ s-s0„ •…1/2 and by Lemma 2 L(s, x)•‚0 in the domain defined by the conditions „ s-s0„ •…1/2 and ƒÐ•†ƒÐ0-2ƒÁ•L. Further we have in the circle„ s-s0„ •…1/2 and Therefore, putting A19=max(A17, A18/2), M=A19log(ƒÁNa), ƒÁ=1/2, we have by Lemma 3 („ s-s0„ •…ƒÁ). (1.3) follows at once. LEMMA 5. If x is a complex character mod a, then in the region „ t„ •…10, ƒÐ•†1 -A21/log(Na+1), L(s, x) has no zero point. PROOF. This is similar to the proof of Lemma 2, so we may omit it. LEMMA 6. If x is a real character mod a and x•‚x0, then L(s, x)•‚0 in the region; PROOF. Suppose that lemma is not true, then for any small positive number c there exists a zero point ƒÀ+iƒÁ of L(s, x) such that Now let a be a positive number such that 1•†((a+1)2+4a2)(a+1/2) and put ƒÐ0=1 +c/a log(Na+1), s0=ƒÐ0+iƒÁ. Taking c small, we may assume that two zero points ƒÀ+ iƒÁ, ƒÀ-iƒÁ of L(s, x) are in the circle „ s-s0„ •…1/4. In the circle „ s-s0„ •…1/2, L(s, x) is regular and „ L(s, x)„ <(Na+1)A24, so we have Generalized Prime Number Theorem. 5 („ s-s0„ •…1/2). Therefore by the lemma a of Titchmarsh [9] p. 49 (1.4) where ƒÏ runs through the zero points of L(s, x) in the circle „ s-s0•…1/4, so we have clearly Again taking c sufficiently small, we have so we have from (1.4) Further taking e small, This and the inequality give (1.5) This is a contradiction and proof is completed. LEMMA 7. If x is a real character mod a and x•‚x0, then L(s, x)•‚0 in the region 0<„ t„ •…6, ƒÐ•†1-A28/log(Na+1). PROOF. Let ƒÀ+iƒÁ (0<ƒÁ•…6) be a zero point of L(s, x). Then by Lemma 6, we may assume A22/log(Na+1)•…ƒÁ•…6. Put ƒÐ0=1+A/log(Na+1) (A>0) and consider (1.6) First taking A sufficiently small, we have Next in the circle „ s-s0„ •…1/2 (s0=ƒÐ0+iƒÁ) so we have Finally putting s0=ƒÐ0+2iƒÁ, we obtain 6 Takayoshi MITSUI in the circle •bs0-s•b•…1/2, hence by lemma ƒÀ of Titchmarsh [9] p. 49 Since A may be arbitrarily small, we may assume that A/(A2+A222)<1/4, that is, that the inequality holds. Therefore by (1.6) Further taking A sufficiently small, we obtain, as in the proof of Lemma 2, The proof is thereby completed. From now on we shall denote X=O(Y) for the values X, Y, if there exists a positive absolute constant A such that „ X„ •…AY. LEMMA 8. If x is a non-principal character mod a, then in the region •bt•b•…6, a •†1-A34/log(Na+1), (1,7) L(s, x)-O(log(Na+1.)). PROOF. Let s0=it0+2+A35/log(Na+1) (•bto•b•…5) and consider three concentric circles having center s0 and radii r1=1, r=1+2A25/log(Na+1), and r3=3/2. Fur ther denote by M(rz), M(r2) and M(r3) the maxima of •bL(s, x)•b in these circles renpctively. Applying Hadamard's three circles theorem, we have where M(r3)=O((Na+1)A2) by (1.1), and M(r1)-O(log(Na+1)), log r3/r2/logr3/r1 and so we have M(r3)=O(log(Na+1)). This gives (1..7) at once. LEMMA 9. Let x1, x2 be real primitive characters mod a1 and a2 respective (x1 •‚x2) and define the real characters x1, x2 mod a1a2 (infinite part of a1a2 is the product of the infinite primes contained in a1 or a2) by the conditions; Generalized Prime Number Theorem. 7 Then x1•‚x2. Consequently x1x2 is not principal. PROOF. Suppose that lemma is not true, that is, assume for any ideal 'i prime M to aza;, we have x1(h)=x2(h). Putting b=(a1, a ), we consider two cases ; (1) Suppose that there exist an ideal b and a EEK such that (ab, a1)=(b, a)=1, a-1(modb) and x1(ab)•‚x2(b). Now put a-,b/r (j, r are integers in K), then Q is represented in the form i=r+p1+Ia(i E a f, so Hence there exists ƒÀ* such that Putting a=ƒÀ*/ƒÀ, we have a -1 (mod a1), a* a =1 (mod), therefore a* ab is prime to a1a, so by our assumption at the beginning, which is contrary to the condition x1(ab)•‚x2(b). (2) Suppose that x1(ab) = x2(b) for any b and a such that a =1 (mod b), (ab, a1) =(b, a2)=1. In this case, we may assume that there exists a prime ideal p such that p p_??_bor there is p which is contained in but not in b. (Otherwise, lemma is trivially proved). Now, for any ideal c prime to b there exist a, b such that N a -1 (mod b), (b, a)=1. (cf. Landau [3]). We shall define for such ideal c a function x•L by x•L(c)=x1(b)o This is determined uniquely by c, for suppose that there exist a•L, b•L such that c T a h•L, a•L En 1 (mod b), (b•L, a1)- 1. While by theorem quoted above we obtain ƒÀ, b1 such that c=abz, 9 ==1 (mod b), (ba, a2)=1, therefore by the first assumption so x (c) is uniquely determined. Obviously we see x•L(ac)=x•L(c) for a 1 (mod b), therefore x•L is a character mod b, and in particular x1(ƒÂ) = /1(b) if (b, Oh) = 1. Now if there is p such that p_??_a1 p,l_??_b, then x•L(ƒÀ)n 0, x1(p) = 0, which shows that x1 is not principal, This is a contradiction. If there is contained in a1 but not in b, then we can take a such that Nand for any ideal b prime to ai we see that b ab (mod iii) but /1(b) =x•L(b) W x•L(ab) N 8 Takayoshi MITSUT x1(ali) by the definition of x', which shows that x1 would be a character mod where is obtained from a by removing. This is also a contradiction. Thus the proof is completed. LEMMA 10. If x is a real character mod a and x xa, then for any positive number s we have where B1-B1(s)>0 depends only on s. PROOF. 1t is sufficient to prove lemma under the assumption that x is primitive. Let x1 be real non.-principal primitive character mod az and assume Na>Naz. Fur ther define a function f (s) =Cjc (s)L(s, x)L(s, x1)L(s, xx1). Since xxz is not principal by Lemma 9, f (s) is regular if s - 1, and at s=1 f (s) has a pole with residue p AL(1, x)L(1, xz)L(1, xll) where A is the residue of CK(s) at s=1. For 6> 1 f (s) is represented as Dirichlet series ; then we see a1=1, a10 (n1) and so m=0,1,2,... Hence in the expansion of f (s) ; f (s) = bm(2- s)m (|s-1|<1), we have bo•†1, bm•†0 (m•†0) and holds in the circle l s-2|<2. Now by Lemma 1, we have | L(s, x)1 (1.8) on the circle 1 s-2 1- 3/2. Since f (s) -p/(s -,1) is regular in the circle Is -2 1 c 3/2, this inequality (1.8) holds in this circle and by Cauchy's formula, (1.9) (m•†0). Now for an integer m0 and 9/10 where Generalized Prime Number Theorem. 9 then so we have Now take then obviously Since m0<8A2log N(ƒ¿ƒ¿1)+C, we have (2-ƒ¿)m0=em0log(2-ƒ¿) < 2C4N(ƒ¿ƒ¿1)8A2log(2-ƒ¿)<2C4N(ƒ¿ƒ¿1)8A2(1-ƒ¿). Hence (1.10) Now consider the set of L-functions with real primitive and non-principal characters and take a positive number ƒÃ. Then 1) there exists one having a zero point in the interval (1-ƒÃ/16A2,1)or 2) no such L-function exists. In the case 1) we denote by L (s, x1) that L-function and by ƒ¿1 the zero point. In the case 2) we fix an L-function, L (s, x1) say, and a point ƒ¿1 in (1-ƒÃ/16A2, 1). In both cases we denote by ƒ¿1 the modulus of x1. If we define f(s) by this L (s, X1) and L (s, x), where x is a real primitive character mod ƒ¿ (Nƒ¿>Nƒ¿1), then f(ƒ¿1)•…0, so by (1.10) we have Hence we have for a constant B2=B2(ƒÃ) independent of ƒ¿, x, Further | L(1,xx1)| therefore and the proof is obtained. 10 Takayoshi MITSUI LEMMA 11. If x is a real non-principal character mod ƒ¿, then for any ƒÃ>0, L(s,x) has no zero points in the interval where B3=B3(ƒÃ) depends only on ƒÃ. PROOF. It is sufficient to prove lemma for primitive character x. Now we know that for any character x mod ƒÂ and for ƒÃ>O we have by Lemma 10 (1.11) for B4=B4(ƒÃ). Further by Lemma 8 and Cauchy's formula we have (1.12) |L•L(s, x)| Suppose that lemma is not true, then we can take ƒÃ0>O such that for any b>0 there exist a real primitive character x mod c and s0 for which 1-b/NcƒÃ0 L(s0, x)=O. Now take ƒÃ1 such that ƒÃ0>ƒÃ1>0 and put for B4=B4(ƒÃ4), then we can take a real character x1 mad c1 and s1 such that From (1,12) we have since 1-A34/2 log(Nc1+1)•…1-b1/Nc1ƒÃ0. Hence by (1.11) we have which leads to This is a contradiction and proof is completed. From now on we shall denote by B5, B6,... the positive constants depending only on ƒÃ>O. LEMMA 12. If we take B5 suitably, then for any character mod ƒ¿ L(s, x)(s-1)E(X)_??_O in the region |t|•…6, ƒÐ•†1-B5/Nƒ¿ƒÃ. PROOF. We know that ƒÄK(s)(s-1) has no zero points on the line ƒÐ=1. From this fact and Lemmas 5, 7 and 11 our result follows at once. LEMMA 13. If x is a character mod ƒ¿, then we have Generalized Prime Number Theorem. 11 in the region |t|•…5, ƒÐ•†1-B5/4Nƒ¿ƒÃ. PROOF. Let |t0|•…5, ƒÁ=3B5/4Nƒ¿ƒÃ, ƒÐ0=1+2ƒÁ•L/3 and s0=ƒÐ0+it0, then by Lemma 12 L(s, x)(s-1)E(x)•‚0 in the region |t•…6, ƒÐ•†ƒÐ0-2ƒÁ•L. In the circle s-s0•…1/2 L(s, x)(s-1)E(x) is regular and by Lemma 1, | L(s, x)(s-1)E(x)|<(Nƒ¿+1)B7. Further in this circle we have and Putting B12=max(B9, B11), M=B12Nƒ¿ƒÃ, ƒÁ=1/2, we have by Lemma 3 in the circle |s-s0|•…ƒÁ`. Thus lemma is proved. Now for an ideal a we shall define a segment I and two curves „C2 as follows; t•…5. (B>0) t|•…5. and combine the two pairs of end points of „C1 and „C2 by two segments parallel to real axis. In this path we shall denote by C1, C2 the sum of three segments and two curves respectively. Now from above lemmas it follows that, if we take B, C suitably for ƒÃ>0 and x is a character mod ƒ¿, then in the right-hand side of this path L(s, x) has no zero points and if |t|•…4, (1.13) if |t|•†4. From now on we shall fix a constant A>0 and consider characters of moduli ƒ¿ such that Nƒ¿•…(log x)A. Further we shall use O-symbol only when the constants in it are independent of such characters and of their muduli and denote by c1, c2,... the absolute positive constants. LEMMA 14. Taking P(x, x)=ƒ°(ƒÏ)log NƒÏ log(x/NƒÏ), we have P(x, x)=E(x)x+O(xe-•ãlog x). PROOF. We have (Landau [4]). Put ƒÃ =1/2A and for this ƒÃ take B, C in the definition of path C1, C2 (so B, C depend only on K, A). Then we can replace the path of integral by C1 and C2 and we have P(x, x)=I1+I2+E(x)x+O(•ãx log3 x), | 12 Takayoshi MITSUI where I1, I2 are the integrals on C1 and C2 respectively. Since Nƒ¿ƒÃ•…•ãlog x, we have I1=O(Nƒ¿ƒÃx exp(-c2(log x)1-ƒÃA))=O(x•ãlog x e-c2•ãlog x=O(x e-c3•ãlog x). Dividing this integral at t=e•ãlog x, we have Thus we have I2=O(x loglog x e-c6 •ãlog x)=O(xe-c7 •ãlog x) and finally P(x, x)=E(x)x+O(xe-c3•ãlog x+O(xe-c7•ãlog x)+O(•ãx log3 x). Hence the result is obtained. LEMMA 15. _??_(x, x)=ƒ°x(ƒÏ) log NƒÏ=E(x)x+O(xe-c8•ãlog x). PROOF. This is easily proved, so we shall omit it. LEMMA 16. PROOF. Since we have it suffices to calculate this right-hand side. By Lemma 15 and in the second term This proves the lemma. THEOREM. Let be an ideal class mod ƒ¿ where Nƒ¿•…(log x)A for a fixed positive constant A and ƒÎ(x;_??_) be the number of prime ideals ƒÏ such that NƒÏ•…x and ƒÏ•¸_??_. Then we have where h(ƒ¿) is the number of ideal classes mod ƒ¿, c>0 and the constants in the re Generalized Prime Number Theorem. 13 mainder term depend only on K, A. PROOF. Taking an ideal b belonging to the inverse class of , we have where x runs through all characters mod a. By Lemma 1 6 this left hand side is which gives the proof. 2. Grossencharacter and preliminary lemmas. Let a be an ideal of K, H1, ..., Hr (r = r1 + r.2 1) the totally positive fundamental units mod a, w the number of roots of unity in K, wn the number of the totally positive roots of unity which are congruent to 1 mod a. We shall put _??_=e2ƒÎilw*. Any totally positive unit mod a earl be represented as a product of powers of H1, ..., Hr and . Consider now the matrix and the inverse matrix of a; where ei f (i Ti), = 2 (i r1 + 1). Then Grossencharacter Aƒ¿ of modulus a for ideal number 1 is defined as follows; (2.1) where m1, ..., m1 are rational integers, ar1+1, ..., an are non-negative integers satisfy ing the following conditions; apap+7 2=o, p=r1+1, Sri+r2, (2.2) is a rational integer. (Fox more detail on ideal numbers and Grossencharacter, see Hecke [2]). Now, as in Hecke [2], we write for an ideal a and (a), (ƒÀ) (a)•ß(ƒÀ) (mod ƒ¿) if (a) is prime to a and a/Q is a totally positive unit mod a. (In the notation of •˜1, this a corresponds to the product of a and all infinite prunes.) By this definition, ideals are divided into h(a) classes and these form an abelian group. Let x be a character of this group. if x(ƒÃ)ƒÉ(ƒÃ)=1 for every unit s of K, then we say XƒÉ is a Grossencharacter for ideal. Clearly we see that such character xA has the same 14 Takayoshi MITSUI value for /, v if / is associated with , therefore we can define C-function; (2.3) (d>1), where; means the sum over all ideal numbers not associated with each other. If rA is primitive and non-principal, then (2,4) is an entire function, where (D is the absolute value of discriminant of K.) with (2.5) and a1, ..., a, are 0 or f and determined by x. Furthermore we obtain (2.6) by the functional equation of e(s, 7A). (Hecke [2]), From now on we shall assume that A is non-principal, that is, not all mr, ar1+1, ..., an in the definition of A are zero. Further for later applications we shall define r+ I vectors o,_??_, ..., _??_rin r +1-dimensional euclidean space as follows ; Denoting the length of vector tL by we put (2.7) E=max(•a_??_0•a,•a_??_1•a, ...,•a_??_r•a ) and for any Grossencharacter A (2.S) where lp=ap-ap+r2 (p=r+1, ..., r1+r2) LEMMA 1. If xA is a Grossencharactcr mod a for ideal, then we have in the strip-1/2a3/2 (A1, A2, are absolute positive constants). PROOF. This lemma is similar to Lemma I of •˜1, so we shall have only to sketch the proof. Suppose first xA is primitive, then on the line c 3/2, we have (*) On the line i=-1/2 we have by the functional equation, Generalized Prime Number Theorem. 15 Putting zq=(-1/2+aq+it-2vq)/2 (q=1, ..., r1) and zp(-l+ap+ap+r+itivp)/2 (p= r1+1, ..., r1+r2), we see that the absolute value of F-factor in the right-hand side is By the definition of vz so we have Therefore we have (*) From (*) and (**) follows the desired result as in the proof of Lemma 1 of •˜1. The proofs of next five Lemmas 2, 3, 4, 5, 6, can be obtained in the same way as in Lemma 2, 4, 14, 15, 16 of •˜1, so we omit them. LEMMA 2, If xA is a Grassencharacter mod a for ideal, then in the region a•†1 -A8/log(NaVƒÉ(1+•bt )) , (s, zA) has no zero points. LEMMA 3. In the region a•†1-A9/log(NaVA(1+•bt )), we have From now on we assume that the norms of the moduli of Grossencharacters are less than e and use the O-symbol only when the constants in it are independent of the characters and their moduli. LEMMA 4. Put where ƒÖ runs through prime ideal numbers not associated with each other, then we have 5 Takayoshi MITSUI (We denote by a1, c2, absolute positive constants). LEMMA 5. LEMMA 6. Now we shall consider a character xA which is not always a character for ideal, then we have for the modulus a of A, where, means the sum over prime ideal numbers which are not associated mod a, and ‡” in the right-hand side is the sum over the units not associated mod a. Since xA is considered to be a character mod u of the group of all units, we see that (2.9) where R is the regulator of K, R1z is the absolute value of the determinant Hence by Lemma 6 we obtain LEMMA 7. (2.10) LEMMA s. Let P be an ideal number prime to a, then we have (2.11) PROOF. If y runs through the characters of group mod a of ideals, then we have Hence the result follows immediately from Lemma 7. Now, we can write a Grossencharacter A mod a in the form; (2.12) where Generalized Prime Number Theorem. 17 (2.13) p=r1+1, ..., r1+r2. Obviously these functions Wq, 0p have the properties; Wq(ƒÊƒË)=Wq(ƒÊ)+Wq(ƒË) . 1 if q=i, Wq(_??_)=0, Wq(Hi)= q,2=1, ..., r. 0 if qi, p=r1+1, ..., r1+r2, i=1, ..., r. Since (2.14) we have (2.15) For any ideal number 1 we shall define a vector (2.16) in the r+1-dimensional euclidean space and particularly put q=1, ..., r. (2.17) Then, putting Wo(ƒÊ)=log•bNƒÊ•b, we have from (2.15) (2.18) Since vectorso, are linearly independent, they form a parallelepiped . We shall denote its diameter by da. LEMMA 9. We have (2.19) Ra-1=0(1), Ra=0(Nar). PROOF. Let ƒÅ1, ..., ƒÅr be the totally positive fundamental units and R1 be the absolute value of the determinant and w1 be the number of totally positive roots of unity in K. If we denote by G1 the group consisting of all totally positive units and by Ga the group of all totally positive units such that -1 (mod a), then we have which gives Next let ai be the least positive integers such that 18 Takayoshi MITSUI i=1, ..., r, then we have ai•…c(a)•…Na. Since _??_, ƒÅa11, ..., farr are the independent units generating a subgroup G of Ga, we have therefore Rn•…R1a1...ar=O(Nar). LEMMA 10. By suitable choice of H1, ..., Hr, we have (2.20) da=O(Nar). PROOF. we see da=O(max(•a_??_0•a, ...,•a_??_r•a))= O(max(•a_??_1•a, ..., •a_??_r•a)), since it ofj=Jr+1/n. On the other hand we may assume that (2.21) i=1, ..., r, where 0 au c ace (i > 1). This gives i=1, ..., r, where Therefore we have so da=O(max(•ball , ..., •barr•b)). Since •ba11...arr•b=Ra/R1, we have from Lemma 9 da=O(Ra)=wO(Nar), which completes the proof. Before the estimation of B, we shall prove a lemma. Denote the angle between a vector b and an i-dimensional subspace _??_(1•…i•…r+1) by V i (or _??_" b). LEMMA 11. Let b1, ba be two independent vectors not in an i-dimensional sub space £3 (1•…i•…r), then we have (2.22) (We denote by 5313I the least subspace containing two subspaces 1, £:,). PROOF. If 61 (or 6) is contained in 62 ' (or _??_), (2.22) is trivial, so we may assume that 61 is not contained in 6". Then we can choose i+2 orthonormal vectors c1,..., ci+2 such that After some calculations, we obtain Generalized Prune Namber Theorem. 19 Hence (2.22) follows directly. LEMMA 12. Choosing Hz, ..., Hr as in Lemma 10, we have (2.23) E=O(Na1(r-1)). PROOF. By the definition of E, it suffices to estimate max(•a_??_1•a, ... ,•a_??_r•a). Using the symbol of inner product, we see and Where di=_??_iƒ©+??+i. Put i=0, ..., r, then !J1i is orthogonal to y and is not contained in JJ1, so we see rc/2 di and (2.24) i=1, ..., r. Put now i=2, ..., r,i=2, ..., r, then we have (2.25) i=2, ..., r. Putting _??_i=_??_0_??_..._??_i-1(i=1, ..., r), we see by (2.22) i=2, ..., r, where cz = sin(_??_0ƒ©(_??_1_??_..._??_i)) is independent of a, since Taking az = z/2, c1= sin(_??_0ƒ©_??_i), we have (2.26) sin( gN53q) cq sin aq. q=1, ..., r. If i aerating this process, we have finally Hence by (2.26) sin _??_i•†ci...cr sin i From (2.25), so we have 20 Takayoshi MITSUI Therefore by Lemma 10 we have E = 0(dar-1) = 0(1Var (r-1)), which is the desired result. 3. The numbers of prime ideal numbers satisfying certain conditions. For any ideal number /1 we shall define a vector (3.1) )-(log •bfl(i)•b, ..., log•b,(i)I) in (r+ 1)-dimensional euclidean space, and in particular we shall put Ga q(t) =_??_Ci)(Hq), q1, ..., r, i=1, ...,r+1, for the totally positive fundamental units mod a, H1, ..., Hr, then are linearly independent vectors in r-dimensional subspace _??_(i) and _??_(i)(l) is a linear combination of these vectors ; (3.2) i=1, ..., r+1, where W1(/l), ..., W1(u) are functions defined by (2.13). By these vector , in I we shall define a parallelepiped (al, ... ,ƒ¿r) for real numbers aq such that 0 0•…x•…aq (1•…q•…r) Further let Qr1+1, Qr1+r2 be real numbers such that 0 < 1 1, p=r1+ 1, ..., rl+r2 and p be an ideal number prime to a. For these a, P, aq (1•…q•…r), Qp (r1+1•…p •…r1+r2) we shall now prove the following theorem, which is the aim of this para graph. THEOREM. Let 7r (x 9 a, 3; aq, Qp)=n (x 9 a, P i a1, ..., ar, Nr1+1, ..., 19r1+r2)be the number of prime ideal numbers w which satisfy following conditions; P (mod a), N x, (i)() B(i)(a1, ..., ar), 0 E1) < fJp, p-r1+1, ..., r1+rd and assume that Na•…(log x)A for a fixed positive number A. Then we have (3.3) (i =1, ..., r+1), where w is the number o f the roots o f unity in K and c>0 and the constants in the remainder term depend only on K and A. PROOF. To prove this theorem, we shall first construct a certain function of n variables and its Fourier series. Take now a subset C={i1, ..., i1} of {1, 2, ..., n} and denote {i11, ..., in}={1, 2, ... , n} -C. (0 •…l•…n). First let 1•…l•…n-1. We consider the closed convex set zJ(C) C Rn containing origin as an inner point defined by (3.4) Generalized Prime Number Theorem. 21 where d is a given positive number <1/3. We decompose this set 4(C) into 4l(n l ) sets 4(8'i3, "it), ƒÃ', ƒÃ" =•}1, 1 <_ s <_ 1, 1+1•…t <_ n defined as follows; (3.5) In case 1=0, n, we define 4(C) by x2 -ƒÃxi•…x1•…Exi '=1, ..., n, jz. Now we shall consider the following functions defined in the n-dimensional cube -1/2< x 1<1/2_ (i=1,2, ..., n) 0, if (x1, ..., xn) E 4(C). if (x1, ..., xn) E 4(a'is, 6"it), when l O, n. (3.6) f 0(x19 if (x1, ..., xn) E 4(ai), when 1=0, n. 10 is positive at the inner points of 4(C) and linearly decreasing from 1 to 0 on the half line starting from origin. Further 10 is continuous. In fact, if (x1, ..., xn) is an inner point of 4(E'is, a"it) for example, then the continuity at (x1, ..., xn) is obvious. Suppose that (x1, ..., xn) is contained in 4(Wis, a"it) and 4(i is,, 'f it"), then by (3.5) we have X2sg _ XGSI9 j xZt, x2t and 'I /X t5, > 0, ƒÃ•Œxzs > 0, v"xit, > 0, a"xtit > 0. Therefore we have ƒÅ•Œ xiss = EƒÃ•Œxzs, ƒÅ••Xit, = ƒÃ••lxit, which shows the continuity of 10 at (x1, ..., xn). From f 0 we define now a function f (x1, ..., xn ; C) having whole Rn as the do main of definition by the condition (3.7) f (x1, ..., xn; C) = f 0(x11, ..., xn•Œ; C), where x/ xi (mod 1), ;x0' < 1/2. (i= 1, 2, , n). Then f can be expanded in Fourier series ; (3.8) By the definition (3.6), we have if l 0, n, 22 Takayashi Mrrsuj if 1=0 or ii, where w= --2a1. To obtain I(C; nzi .., rn), therefore, it is enough to consider the integral over 4(6 1,, it) or 4(5i). By (3.4) we see for a continuous function g(x1..., xnn) (3.9) where {yi, ..., = {x11, ..., x } {xz,}, {Y1+2,..., yn} x } {x} and we put x = xis, y = xzt. Therefore we have So finally we have if i 0, n9 (3.10) if 1=0, if 1=n, where To compute further this integral, we shall first assume l 0, n. Observe the n's with subscripts from C ; mil, ..., m1. Some of these m's might be 0. Let I' be the subset of C, such that the rn's with subscripts from I' are different from 0. We put I'=I' i}. On the other hand, F" will denote the subset of {il+r, ..., in} {1, 2, , n} C, such that the m's with subscripts from I" are different from 0. We put again Ijr=I"-{j}. Denote by r', r" the numbers of elements of I', I'' and define r"-1 if j I" r'-1 if a EI', , (3.11) r2 Ti r' if i E I', r•• if j E I", then by (3.9) (3.12) Generalized Prime Number Theorem. 23 Now we put I0•Œ ={i1, ..., il}-I•Œ, I0••{il+1, ..., in}-I••, (3.13) and (3.14) First assume that i•¸I0•Œ, j•¸I0••, then ri=r•Œ, Ii•Œ=I•Œ, rj=r••, Ij••=I••, so we have from (3.12) Therefore (3J5) Next if i•¸I0•Œ, j •¸I••, then ri=r•Œ, Ii•Œ=I•Œ, rj=r••-1, so we have Hence we have (3.16) Similarly (3.17) and finally we have (3.18) From (3.15), (3.16), (3.17) and (3.18) we have (3.19) 24 Takayoshi MITSUI Now if we put in place of respectively in the right hand side of (3.19), then it gives ‡”I(‡™(-i, j)). Since and E(-x, y)=(-1)l-1E(x, y), we have (3.20) Now, we see (3.21) so from (3.19) and (3.20) we have (3.22) I(C; m1, ..., mn)=‡”{I‡™(i, j))+I(‡™(-i, j))} By partial integration, (3.23) which gives our I(C; m1..., mn). By a similar but easier method, we can see that the same result holds for l=0 or n. Thus we have LEMMA 1. In (3.8) we have (3.24) I(C; m1, ..., mn) where M is given by (3.13) and ƒÖ=-2ƒÎi. The right hand-side of (3.24) is equal to (3.25) where Js•Œruns through all subsets of s letters in I•Œ, Jt••runs through all subsets of t letters in I••, and M•Œ, M••are functions defined as follows; For any subsets {j1, Generalized Prime Number Theorem. 25 ..., j k} of {1,2, ...,n}, (3.26) where (3.27) By Lemma 1, we see that (3.23) I(C; ml, ..., m)=I(C; ƒÃ1m1, ...,ƒÃnmn) for any z=•}1 (1•…i•…n) and (3.29) Further we can see easily (3.30) for an absolute positive constant B. This estimation (3.30) will be used later. Now we shall define a function containing real parameters t1, ... , tn as follows (3.31) f(x1, ..., xn; t1, .., to ; C)=f(x1 - t1, ..., xn - tn; C), then denoting its Fourier coefficients by I(C ; m1,..., mn ; t1, ... , tn), we see clearly that (3.32) I( C; 0,..., 0; t1,..., tn)=I(C; 0,..., 0). (3.33) |I(C; m1,..., mn ; t1,..., tn)|=|I(C; m1,..., mn|. Further vie put for C {i 1, ...,I} and define (3.34) where C runs through all subsets of {1, 2,..., n}. This function has the following properties; 0•… F(x1, ..., xn) •… 1 for all points (x1, ..., xn). (3.35) F(x1, ..., xn)=1 if 0•…xi -[xi]•…d. (i=1, ... , n). F(x1, ..., xn)=0 if there is j such that 2d•…xj- [xj]•… 1-d. In fact, the last assertion of (3.35) is obvious by its definition, so we shall prove second assertion, that is, we shall show F(x10, ..., xn0)=1 if 0•…xi0•…d (i=1, ..., n). For the brevity, we put =(x10,... , xn0) and c(C)=(C1,..,Cn) where C1, ..., Cn are 26 Takayoshi MITSUI defined previously for C, then (3.34) can be denoted as follows F(ƒÂ)-‡”cf(ƒÂ -c(C),C). By changing indices, if necessary, we may assume that d•†x10•† ...,•†xn0•†0. Take a set C = {i1, ..., il} and put -c(C)=(y1, ..., yn), then xi0-d if a•¸C. i=i, ..., n. yi={ xi0 if i_?? C. First assume that C = {i, ..., il}_??_ {1, ..., l}, then there exists j such that j =xj0 and yj+1=x0 j+1 -d. Therefore d•…yj+(- yi+i), which shows that for this C we have f(ƒÂ-c(C )) = 0. Hence Let now C = {1, ..., l} and define yi as above, then 0•…- y1•…...•…-yl•…d, 0•…•…yn•…...•…yl+1•…d, yl+1+(-yl)•…d. Hence c(C) is a point in ‡™( -l, L + x) and by the definition of f ( c(C)) we have if l=0. if l_??_, n. if l = n. Therefore Thus the second assertion of (3.35) is proved. Finally we shall show the first assertion of (3.35), i.e. that F(x1, ..., xn,)•…1. Again, we may assume that -d•…x1•…...•…xs•…0•…xs+1•…...•…xt•…d•…xt+i•…...•…xn•…2d. When C ={i1, ..., il} satisfies the conditions C_??_ {t+1, ..., n} or C•¿{1, ..., s}_??_ƒÓ, then, by the definition of f, we have f ( - c(C) a C)=0. Therefore F()=f(-c(C ))9 where C•L runs through all subsets of {i, ..., n} such that C•L•½{t+1, ..., n}, C•L•¿{i, ..., s}=ƒÓ. Now put ƒÂ0 = (y1, ..., yn,), where xi+d if i•…s, yi= xi if s Let {C•L} be the set of subsets C•L such that C•L 1,..., s}, C•L fl {t+ 1, ..., n}= and {C"} be the set of the other subset of {1, ..., n}, then we see Since f is non-negative, this shows that F() c 1. Thus the proof of (3.35) is completed. Now we shall take d so small that is satisfies for non-negative integers k1, ..., kn the conditions 2d(ki+1)•…1., (i =1, ..., n) and define a function (3.36) where then we shall show 0•…F*(x1, ... , xn; k1, ..., kn)•…1 for all (x1, ... , xn). (3.37) F*(x1, ..., xn; k1, ..., kn) = 1 if d•…xi-[xi]•…2d(ki+1) (1•… i•… n)' For the proof of (3.37) we may assume 0•…xi•…1 (1•…i•…n). Put now 1=1,2, ..., n, then by the definition of F we see where y1=xi-2dli (1•…i•…n) and the sum is taken over the systems (ƒÃ1,..,ƒÃn) such that ƒÃi=•}1 (1•…i•…a) and F(y1-ƒÃ1d,..., yn-ƒÃ,d) 0. By the change of the order of y1,..., yn, if necessary, we may assume then since F(y1-ƒÃ1,..., yn-ƒÃnd)= 0 if there exists ƒÃj = a (j > t). We see that z, for which 1•…li•…ki, takes value 1 and -1. (On the other hand, ƒÃi, for which l 0 or k,+1, may take value 1 or -1 alone). Now we shall define a vector = (z1, ..., zn), where and c(ƒÃ1,..., ƒÃt)=(ƒÃ1d,.., ƒÃtd, 0, ..., 0) for a set (ƒÃ1, ..., et). Then we have Since 0•…zi•…d (1•…i•…n), the last sum is equal to 28 Takayoshi MITSUI where C•L runs through all subsets of {1, ..., n} containing all indices i for which ƒÃi =-1 and no indices j for which ƒÃj= 1. (Indices k, for which are not defined, may be contained or may not be contained in C•L). For such C•L, c(ƒÃ1, ..., ƒÃt)+c(C•L) is equal to c(C") defined by a subset C" of {1, ..., n}, therefore where the sum is taken over C belonging to a certain set {C}(ƒÃ1, ..., ƒÃt) determined by (ƒÃ1 ..., ƒÃ) alone. Further we see that if (ƒÃ1, ..., ƒÃt) _??_ e(ƒÅ1, ..., ƒÅt), where ƒÃ, ƒÅi=1 or -1 , then {C} (ƒÃ1, ...,ƒÃt) and {C} (ƒÅ1, ... , ƒÅt, have no common element. Therefore sum {C} of {C}(ƒÃ1, ..., ƒÃt) over all systems (ƒÃ1, ..., ƒÃt) is a family of subsets of {1, ..., n}. Hence we have by (3.35) (3.38) F*(xi, ..., xn; k1, ..., kn)=‡”(ƒÃi) F(ƒÂ- c(ƒÃi, ..., ƒÃt)) =‡”c f(ƒÂ- c(C))•…1. Now assume that d•…xi•…<2d(ki+1) (1•…i•…n), then 0•…1i•…ki and if there is j such that lj=0, J then yj=xj•†d in above notation, so ƒÃj does not appear in (ƒÃ1, ... , ƒÃt). Therefore every ƒÃi takes the value 1 or -1. This shows that the set {C} in the right-hand side of (3,38) is equal to the set {C} of all subsets of {1, 2, ..., n}. Since 0 •…zi•…d (1•…i•…n), we have from (3 .38) and (3.35) Thus the proof of (3 .37) is completed. Now let r1, ..., rn be real numbers such that 0 where d is sufficiently small. Obviously we have 2d(ki+1)•…ri<2d(ki+2) if ri•†2d. (3.39) i=1, .., n. { 2d(ki+1)•…1. For these k1, .., kn, we shall consider the function F * of (3.36). Further we define a function G(x1, ..., xn; r1, ... , rn) as follows; (3.40) By this definition and (3.37) we have G(x1, ..., xn; r1, ..., rn)=F*(x1, .., xn; k1, ..., kn)=1 if d•…xi•…min(2(ki+1)d, ri) (1•…i•…n) and G(x1, ..., xn; r1, ..., rn)=F*(xi, ..., x,; k1, ..., k.n)=0 if there exists at least one xj such that 2d(kj+2) To estimate G-F* for other (x1, ..., xn). we define a function (3.41) Generalized Prime Number Theorem. 29 +F(x1, ..., xn; 2j1, ..., 2js-1, 2ks+2, 2js+1, ..., 2jn) +F(x1, ..., xn; 2j1, ..., 2js-1, 2ks+4, 2js+1, ..., 2jn)}, where s means that the sum over js is removed. It suffices to prove that H(x1, ..., xn; k1, ..., kn)•†1 for 0•…xi•…2d(ki+2) (1•…i•…n) in which there is at least one xj such that 0•…xj•…d or 2d(kj+1)•…xj<2d(kj+2). Now we have by the definition (3.30) of F* and (3.37) (3.42) for 0•…xi•…d(2ki+5) (1•…i•…n). Assume now that 0•…xi<2d(ki+2) (1•…i•…n) and put yi=xi-(2li+1)d, i=1, 2, ..., n, then the left hand side of (3.41) is equal to Since xi-2(ki+2)d•…yi+ƒÃid•…xi (i=1, 2, ..., n), we have so for any (x1, ..., xn) we have (3.43) G(x1, ..., xn; rl, ..., rn)-F* (x1, ..., xn; k1, ..., kn)•…H(x1, ..., xn; k1, ..., kn). Now we shall fix r (0•…r•…n) and denote yp=xp+r (p•†1) for distinction. Let w* be a positive integer and rip ƒÂ=rp+r (p= 1, 2, ..., n-r) be positive numbers such that ƒÂp•…1/w*. Further, taking integers bp, wp (p=1, 2, ..., n-r) such that 0•…bp, wp•…w*-1, we shall define a function from f of (3.31) as follows; (3.44) and similarly define F(xq; yp), F*(xq ; yp ; ki), G(xq ; yp) and H(xq ; yp ; kj) from F, F*, G and H respectively. We have also an inequality (3.45) G-F*•…H for any (x1, ..., xr, y1, ..., yn-r). Let us denote the Fourier coefficients of f (x1, ..., xn; t1, ..., tn; C) by I (C; mi; ti) N for brevity, then f of (3.44) is expanded in the following form In the right hand side we see that 30 Takayoshi MITSUI Hence we have (3.46) where ' means that m1, ..., mr run through all rational integers and l1, ..., ln-r run through rational integers satisfying the condition (3.47) By (3.291. (3.32) and (3.33) we see for C ={i1, ..., il} (3.48) 3.49 I(C; mq, lp ; ti) = I(C ; mq, lp) . From now on let us consider theorem of beginning. For convenience we shall assume K is of n+1 degree and n+1=r1+2r2, r=rl+r2-1. In (3.46) we put xq=Wq(ƒÖ) q=1, ..., r yp=ƒÆp+r1(ƒÖ) p =1, ..., n-r (=r2) and p=1, ..., r2 and sum up the values of (3.46) over prime ideal numbers c which are not associated with each other mod a and satiny following conditions; ƒÖ•ß(mod a), NƒÖ•…x. Denoting this summation by and changing the indices of {l} we have from (3.46) (3.50) In the inner sum of the right hand side, lr1+1, ..., lr1+r2 must satisfy the condition (3.51) Therefore, if we put ap=(|lp|+lp )/2, |lp|-lp)/2, p=r1+1, ..., r1+r2, then lp=ap-ap+r2 and integers ar1+1, ..., an, satisfy the conditions (2.2). Hence, by the definition (2.12) of Grossencharacter, we have (3.52) ‡”*f(Wq(ƒÖ), ƒÆp(ƒÖ ); ti ; C) =ƒÖ*‡”I(C ; mq, lp, ti)‡”*ƒÉ(ƒÖ), where ‡” is regarded as the summation over all Grossencharacters ƒÉ mod a. We shall divide this sum into two parts;ƒÖ *I(C ; 0, 0)‡”*1 and As for the first sum, we have from Theorem of •˜ 1 and (3.48) Generalized Prime Number Theorem. 31 (3.53) (We denote by C1, C2,...positive absolute constants). For the second sum we have from (2.12) and (3.49) Again we shall put lp=mr2+p-1 (p=r1+1,..., r1+r2) and , then we have (3.54) where V=1+Eƒ° m, . Since in the right hand side of (3,54) the value of each term is independent of the signature of m1,..., mn, so we have (3.55) where means the summation over the subsets I•Œ'r•Œ={j1, ..., jr•Œ} of r•Œ letters of C={i1, ..., il} and the subsets I•Œr•Œ•Œ={k1, ..., kr•Œ•Œ} of r" letters of {1, ..., n}-C={il+1, ..., in} and ƒ°" means the summation over m1, ..., mn such that mi=0 if i__??__I•Œr•Œ,,I" r" and mi > 0 if i•¸I•Œr•Œ or i•¸I•Œr". Therefore it is sufficient to estimate ƒ°" for fixed I•Œ=I",.' and i"=i",. Then we have by Lemma 1 (3.56) Therefore from (3.30) we see that it is sufficient to estimate for fixed J = JS•Œ and J•Œ•Œ=Jt•Œ•Œ (3.57) We shall now prove two lemmas. LEMMA 2. Let a, N be positive integers and a•†N, then (3.58) where BN is a positive constant depending only on N. PROOF. We can prove the lemma by induction on N. If N =1, (3.58) is obvious. Assume that (3.58) holds for N=i, then 32 Takayoshi MITSUI This completes the proof. LEMMA3. Let F•†1 and ƒÃ>2b1>0,then (3.59) where , b2>0 and the constants in the right-hand side are independent of F. PROOF. When N=1, we divide the sum into two parts; We have for the first sum Next we define a function Then it is easily seen that f(t) is decreasing for , so we have Thus (3.47) holds for N=1. If N>1, we put U•Œ= 1+F(1+m1+...+mN-1)mN, then U•…U•Œ, so we have which completes the proof. Now we shall consider (3.57). Put J*=I•Œ-J•Œ, J**= I" - J" and then M"(J")-M•Œ(J )=w=u2+v1-(u1+v2). Let us divide the sum T into two parts; When w •† 0, then V=1+E(u1+u2+v1+v2)•…V•Œ=1+2E(u1+v2+1+w), so we have Generalized Prime Number Theorem. 33 Therefore we have by Lemma 2 where V•Œ•Œ=1+2E(w+u1+v2). Since u1v2w(w+u1+v2)•†3(u1v2w)1+1/3, we have from Lemma 3 Similarly we have Since log(1+2E)=O(log Na)=O(log log x) by Lemma 12 of •˜2, we have Therefore from (3.55) and (3.56) Further we have from Lemma 9 of •˜ 2 From this result and (3.53) we have (3.60) and so (3.61) (3.62) If we put 2d(ki+1)=ƒÁi-ƒÁi•Œ, then iƒÁi•Œ|•…2d. Therefore (3.63) From (3.61) we have , so by (3.41) (3.64) 34 Takayoshi MITSUI Hence we have from (3.63), (3.64) and (3.45) Take now then d is sufficiently small for large x, and we obtain (3.65) On the other hand, from (3.2), (3.40) and the definition of G we see that G(Wq(w), Op(w)) =1 if and only if there exists di which is associated with w mod a and satis fies the conditions; , p=r1+l,..., r1+r2. Hence ƒ°*G(Wq(w), Op(w) is equal to the number of prime ideal numbers w such that w•ßp(mod a), Nw•…x,ƒÃ(i)(w)•¸B(i)(ƒÁ1,...,ƒÁr), , p=r1+1, ... , r1+r2. Now we consider ƒ¿1, ..., ƒ¿r, ƒÀr1+r2 in the conditions of our Theorem. We put and define for an non-negative integer k if 0•…k•…tp-1. p=r1+1,...,..., r1+r2. if k=tp. Furthermore we shall denote by ƒÎi(x; a, p; ƒ¿p, ƒÀp; kr1+1,..., kr1+r2), where kp are integers such that 0•…kp•…tp (p=r1+1,..., r1+r2), the number of prime ideal num bers w satisfying the conditions; w•ßp(moda), Nw•…x, ƒÃ(i)(w)•¸B(i)(ƒ¿1,..., ƒ¿r), p=r1+1,..., r1+r2. Then we have from the result just obtained Therefore Generalized Prime Number Theorem. 35 where so the proof of Theorem is completed. Now we have COROLLARY. For ƒ¿q, ƒ¿•Lq such that 0<ƒ¿q-ƒ¿•L•…1 (1•…q•…r) and ƒÀp, such that 0<ƒÀp-ƒÀ•Lp•…1 (r1+1•…p•…r1+r2), the number o f prime ideal numbers w which satisfy the following conditions; w•ßp (mod a), Nw•…x, x(i)(w)•¸B(i)(ƒ¿1,...,ƒ¿r,ƒ¿•L1,...,ƒ¿•Lr), I 9(c) <19p, p=r1+1, .., r1+r2 is equal to In these conditions, B(i)(a1,., ar, a•L,) is an parallelepiped defined by the condition a•Lq xq < aq (1 q r). 4. lain Theorem. THEOREM. Take an ideal number P and an ideal a of algebraic number field K of degree n. (As usual we put n =r1+2r2, r=r1+r2-1). Let Y1,..., Yn be positive numbers such that Yp=Yp ,.2 (p •†r1+1) and Yi S Y/ for a fixed positive number a. Further we assume (p, a)= 1 and Na log 4(Y1... Yn) where A is a fixed positive number. Now let us take positive numbers . Url+r2 such that 0 <_??_ c 1 (p=r1+1, ... , r1+r2) and denote by 1r(a, P ; Yq, fin) = c(a, P i Y,.+i, _??_r1+I, . , r1+r2) the number o f the prime ideal numbers ct satisfying the conditions as follows; c W (moda), ,(q) q=1,..,,r+1 0 c arg w (p) < 2rc_??_, p r1+1, , , r1 +r2, then we have (4.1) where ei =1 (i c r1), = 2 (i •† r1+ 1),h is the number o f ideal cusses, B is the regulator, w is the number o f the roots o f unity in K, c>0 and the constants in the remainder term are dependent only on K,A and a but not on a, P, t5 z, . ., 36 Takayoshi MITSUI PROOF. First we shall consider Pi(Y1,.. Yn), the number of the prime ideal numbers such that w= P (mod a), Yi, (4.2) q=1,..., r+1, q2, p_r1+1,..., rl+r2. Since these conditions are symmetric for i =1, .. r+ 1, it suffices for us to con sider P.r+1(Yi.. Yn), Now, in r-dimensional euclidean space, let D be the set of (y1,...,yr) such that (4.3) then by the definition (3.2) of the vector (r+1)(w), we see that the conditions (4.2) for i = r+ 1 are equivalent to the following ones; (4.4) Taking rational integers i1,...,ir and a positive integer l and using the notation cgcr+1> (1 q <_ r) defined in the beginning of •˜ 3, we define a set (4.5) For a set B=B1(i1,.. 2r) contained in D, we now denote by CB, GB and CB the numbers of prime ideal numbers i which satisfy following conditions (4.6), (4.7) and (4.8) respectively; (4.6) (4.7) where (4.8) where After these definitions we shall show that (4.9) CB CB <_ CB. First we see that Generalized Prime Number Theorem. 37 that is, (4.10) and that the condition ƒÄ(r+1) (ƒÖ)•¸B is equivalent to the following; (4.11) q=1,•c,r. Now let ƒÖ be an ideal number belonging to CB, then from (4.10) and (4.11) follows and by the definitions of Op(ƒÖ) we have p=r1+1,•c, r1+r2, therefore considering (4.11), we see p=r1+1,•c, r1+r2, which show that ƒÖ belongs to CB, that is, CB•…CB. Next take ƒÖ belonging to CB, then from (4.8) and (4.10) that is, and similarly we see that 0•…argƒÖ(p)<2ƒÎvp. Thus (4.9) is proved. Now we represent the point (y1,•c, yr)•¸D as follows (4.12) i =1,•c,r, then we see (4.13) Hence we have Since we have (4.14) log A(Y1•cYn)•…a•LA(log Yr+1)A•…(log Yr+1)A•L where a•L=er+1+(n-er+1)a, A•L=A+Aloga•L. (4.14) show that we can apply Theorem of •˜3 to estimations of CB and CB. Thus we have 38 Takayoshi MITSUI (4.15) (4.16) where (4.17) Now we have (4.18) where The difference of the values of first and last terms of (4.18) is (4.19) where F•••…Y1•cYn and Using the notations •’q(1•…q•…r) and da in •˜2, we have so, putting ƒÂ=da/l, (4.9) is Now we take (4.20) (c>0 will be determined later). Then we see ƒÂ is bounded uniformly for a, since from Lemma 10 of •˜2 da=O(Nar)=O(log Ar(Y1•dYn)). Therefore we obtain from (4.19), (4.15), (4.16) and (4.18) that (4.21) +O(l+rƒÂRaY1•cYn). Next take B=B1(i1,•c, ir) such that B •¿ D•‚ƒÓ, B•¼D and denote by CB the number of prime ideal numbers ƒÖ such that ƒÖ=ƒÏ(mod. a), ƒÄ(r+1) (ƒÖ)•¸B•¿D, „ ƒÖ(r+1)„ Yr+2, 0•…argƒÖ(p)<2ƒÎvp p=r1+1,•c, r1+r2, CB the number of prime ideal numbers ƒÖ such that Generalized Prime Number Theorem. 39 p=r1+1,•c, r1+r2. Then similarly to (4.9) and (4.15), we have CB•†CB and where Theorem of •˜3 can be applied, since it holds and ƒÂis bounded. Furthermore, we obtain (4.22) Put now for B=B1(i1,•d, ir) and let m be the number of B having common point with D, then from (4.2.1) and (4.22) we obtain (4.23) where R1 and R2 have the values at most O(mY1•cYn-c3~ log(l.eYn~)+O(mL rlR(tY1•cYn). Since the volume of the parallelepiped constructed by •’1(r+1),•c,•’r(r+1) is nRa/2r2 and its diameter is so the volume of B is nRa/(2rzlr) and its diameter ‡™ is less than r(1+~•ãr+ 1)ƒÂ. Therefore so R1 and R2 have the values (4.24) 40 Takayoshi MITSUI Now we shall consider the transformation of variables (x1,•c xr) defined by (4.12), that is, i =1,•c, r, then this Jacobian is and we have (4.25) (4.26) where Furthermore by the transformation of which Jacobian is we obtain (4.27) (4.28) We denote by I1, I2 the integrals of the right-hand sides of (4.27) and (4.28) respectively, then I1=I1•L+O(Y1e1•cYrer), I2=I2•L+O(Y1e1•cYrer), where and I2•L is obtained by replacing •}•¢ in I1•L by•}•¢. Further if we put then we have Generalized Prime Number Theorem. 41 I2•L-I=O(Y1•cYn(e2•¢-1))=O(ƒÂY1•cYn), consequently we have (4.29)Ii=I+O(Y1e1•cYrer)+O(ƒÂYr•cYn), i=1, 2. From (4.23), (4.24), (4.27), (4,28) and (4.29) follows (4.30) Since Yi•…Yja, if we put then (4.31) Now we put in (4.20) (4.32) Then we have (4.33) since Ra=O(Nar)=O(log Ar(Y1•cYn)). Thus we obtain by (4.20), (4.30), (4.31), (4.32) and (4.33) (4.34) By similar argument we have for 1•…i•…r+1 (4.35) Hence (4.36) In this left hand side, the number•@ƒÖ such that is contained in Pi(Y1,•c, Yn) and Pj(Y1,•c, Yn). But the number of such ƒÖ is not •@ greater than (4.37) Pj(Y1,•d, Yn)-Pj(Y1,•d, Yi•L,•d, Yn), where we put Yi•L =Yi(1-ƒÃ) for sufficiently small positive number ƒÃ. Therefore (4.37) is at most Taking 42 Takayoshi MITSUI ƒÃ=e-c10•ãlog(Y1•dYn), we see that the number in question is O(Y1•dYne-c11•ãlog(Y1•dYn)). By similar consideration, the number of ƒÖ which are contained at least two Pi(Y1,•d, Yn) and Pj(Y1,•d, Yn) is O(Y1•cYne-c12•ãlog(Y1•dYn)). Hence we obtain (4.38) where ‡”•L means the sum of distinct ƒÖ which are contained in any Pi(Y1,•c, Y n). Now we see from the definition (4.2) while ‡”•L Pi is not less than the number ƒÎ•L of prime ideal numbers ƒÖ satisfying the following conditions; ƒÖ•ßƒÏ (mod. a). 2<ƒÖ(q)•…Yq, q=1,•c, r+1, 0•…arg ƒÖ(p)<2ƒÎp, p=r1+1,•c, r1+r2. Comparing ƒÎ(a, ƒÏ; Yq, vp) and ƒÎ•L, we have Hence we have ƒÎ(a, ƒÏ; Yq, vp)=H(a)I+O(Y1•cYne-e15•ãlog(Y1•dYn)), so the proof is completed. University of Tokyo. Bibliography. [1] T. Estermann, On Dirichlet's L-functions. Jour. London Math. Soc., 23 (1948), 275-279. [2] E. Hecke, Eine neue Art van Zetafunktionen und ihre Beziehungen zur Verteilung der Primzahlen. II. Math. Zejt., 6 (1920), 11-51. [3] E. Landau, Ueber Ideate and Primideale in Idealeklassen. Math. Zeit., 2 (1918), 52-154. [4] E. 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