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Distribution of Units of Real Quadratic Number Fields Y.-M. J. Chen, Y. Kitaoka and J. Yu Nagoya Math. J. Vol. 158 (2000), 167{184 DISTRIBUTION OF UNITS OF REAL QUADRATIC NUMBER FIELDS YEN-MEI J. CHEN, YOSHIYUKI KITAOKA and JING YU1 Dedicated to the seventieth birthday of Professor Tomio Kubota Abstract. Let k be a real quadratic field and k, E the ring of integers and the group of units in k. Denoting by E( ¡ ) the subgroup represented by E of × ¡ ¡ ( k= ¡ ) for a prime ideal , we show that prime ideals for which the order of E( ¡ ) is theoretically maximal have a positive density under the Generalized Riemann Hypothesis. 1. Statement of the result x Let k be a real quadratic number field with discriminant D0 and fun- damental unit (> 1), and let ok and E be the ring of integers in k and the set of units in k, respectively. For a prime ideal p of k we denote by E(p) the subgroup of the unit group (ok=p)× of the residue class group modulo p con- sisting of classes represented by elements of E and set Ip := [(ok=p)× : E(p)], where p is the rational prime lying below p. It is obvious that Ip is inde- pendent of the choice of prime ideals lying above p. Set 1; if p is decomposable or ramified in k, ` := 8p 1; if p remains prime in k and N () = 1, p − k=Q <(p 1)=2; if p remains prime in k and N () = 1, − k=Q − : where Nk=Q stands for the norm from k to the rational number field Q. In [IK] we have shown that `p divides Ip and we observed that in each case the set of prime numbers satisfying Ip = `p has a natural density. K. Masima found that the values in tables there, are connected with the Artin constant 1 A := p 1 p(p 1) = 0:3739558 and showed in [M] that the set of − − · · · decompQosable prime numbers satisfying Ip = `p has a density under the Received October 7, 1999. 1991 Mathematics Subject Classification : 11R45 1Research partially supported by National Science Council, Rep. of China. 167 Downloaded from https://www.cambridge.org/core. IP address: 170.106.34.90, on 26 Sep 2021 at 07:02:42, subject to the Cambridge Core terms of use, available at https://www.cambridge.org/core/terms. https://doi.org/10.1017/S0027763000007364 168 Y.-M. J. CHEN, Y. KITAOKA AND J. YU Generalized Riemann Hypothesis (GRH) following [H]. In this paper, we treat the case where prime numbers remain prime in k following [H], [M]. However, instead of counting prime ideals which are completely decompos- able, we use the Chebotarev Density Theorem by [LO], [S] under the GRH. Our main result is the following. Theorem. Let P(x) be the set of odd prime numbers p x which ≤ remain prime in k and let N(x) be the subset of p P(x) satisfying I = ` . 2 p p Then we have 2 ]N(x) = c0Li(x) + O(x log log x=(log x) ) for a positive constant c0 under the GRH. t Here fields where the GRH is involved are k(ζ2n; p ) for square-free natural numbers n and t = n or 2n, where ζm stands for a primitive m-th x root of unity. The function Li(x) stands for 2 dt= log t as usual. R 2. Algebraic preparation x Throughout this paper, we keep the notation in Section 1. The main results in this section are Theorems 1 and 2. Lemma 1. Let n be a square-free integer ( 1) and suppose that k ≥ 6⊂ Q(ζ ) and suppose 2pn R. Set 2n 2 k(ζ ; 2pn ); if N () = 1; K := 2n k=Q k(ζ ; pn ); if N () = 1; 2n k=Q − and let N := [K : k(ζ2n)] be the extension degree of fields. Then we have N = [K : Q]=2'(2n) n; either if N () = 1 and p k(ζ ), k=Q 2 2n = 8 or if N () = 1 and 2 - n; k=Q − <2n; if N () = 1 and p = k(ζ ); k=Q 2 2n where '(m) is the: Euler function. Proof. Let us recall that for an integer m, and an element a in a field F (ch(F) = 2) 6 which is not contained in F p for every prime divisor p of m, a polynomial xm a is irreducible either if 4 - m, or if 4 m and − j 4a = F 4. − 2 Downloaded from https://www.cambridge.org/core. IP address: 170.106.34.90, on 26 Sep 2021 at 07:02:42, subject to the Cambridge Core terms of use, available at https://www.cambridge.org/core/terms. https://doi.org/10.1017/S0027763000007364 DISTRIBUTION OF UNITS OF NUMBER FIELDS 169 q Let q be an odd prime or 4 and suppose q 2n. We show p = k(ζ2n) first. q q j 2n 2n=q 2 Suppose p k(ζ2n); then Q(p ) = Q(( p ) ) R is a subfield of 2 ⊂ q an abelian field k(ζ2n). Hence any conjugate element of p should be real. q This is a contradiction. Hence p is not in k(ζ2n). Let us show that x2n is irreducible over k(ζ ) if p = k(ζ ). We − 2n 2 2n have only to consider the case of 2 n. Suppose that 2 n, p4 4 k(ζ ) j j − 2 2n and p = k(ζ ); then p 4 k(ζ ) and p 1p k(ζ ) hold. This 2 2n − 2 2n − 2 2n contradicts that p = k(ζ ) since p 1 k(ζ ) holds by 2 n. Thus the 2 2n − 2 2n j above criterion yields that a polynomial x2n is irreducible over k(ζ ) − 2n if p = k(ζ ). Then we have N = [k(ζ ; 2pn ) : k(ζ )] = 2n. If, next 2 2n 2n 2n p k(ζ ), then a polynomial xn p is irreducible over k(ζ ) and then 2 2n − 2n we have N = [k(ζ ; (p )1=n) : k(ζ )] = n. Similarly, xn is irreducible 2n 2n − over k(ζ ) if N () = 1 and 2 - n. 2n k=Q − Remark. In Lemma 1, a rational prime p is unramified in K if p - 2nD0. Proposition 1. Let n, K, N be those in Lemma 1. Let η 1 2 Gal(k(ζ2n)=Q) be an automorphism such that η(ζ2n) = ζ2−n and η induces the non-trivial automorphism of Gal(k=Q). (I) The case of Nk=Q() = 1. There exists an automorphism ρ of order 2 in Gal(K=Q) such that ρ = η on k(ζ2n) if and only if (i) N = 2n, (ii) N = n is odd, or (iii) N = n is even and η(p )p = 1. When ρ exists, it is in the 1 center of Gal(K=Q) and satisfies ρ( 2pn ) = 2pn − and both signs are possible if and only if N is even. (II) The case of N () = 1. If n is odd, then there exists a unique k=Q − automorphism ρ of order 2 in Gal(K=Q) such that ρ = η on k(ζ2n). It is 1 in the center of Gal(K=Q) and ρ(pn ) = pn − . If n is even, then there − is no such automorphism. Proof. Set t = 2n or n according to N () = 1 or 1, respectively. k=Q − If ρ Gal(K=Q) is an extension of η and ρ2 = id:, then setting ρ(pt ) = t 21 2n δp− for some δ K, we have δ = 1 and hence δ is a 2n-th root 2 1 t 2 t t 1 of unity and hence ρ(δ) = η(δ) = δ− and p = ρ (p ) = ρ(δp− ) = 1 t 1 1 2 t δ (δp− ) = δ p. Thus we have δ = 1. − − − Proof of the case (I). Suppose Nk=Q() = 1. Assume that either N = 2n, or N = n is odd, first. We define ξ = 1 by n 1; if N = 2n; ξ := n η(p )p, if N = n; Downloaded from https://www.cambridge.org/core. IP address: 170.106.34.90, on 26 Sep 2021 at 07:02:42, subject to the Cambridge Core terms of use, available at https://www.cambridge.org/core/terms. https://doi.org/10.1017/S0027763000007364 170 Y.-M. J. CHEN, Y. KITAOKA AND J. YU where if N = n, p k(ζ ) holds by virtue of Lemma 1, and η can act 2 2n on p. Let η Gal(K=Q) be an extension of η Gal(k(ζ )=Q); then we 0 2 2 2n have 2n 2n N (η0( p ) p ξn) = η() = 1 2n r 2n 1 and hence η0( p ) = ζN ξn p− for some integer r and a primitive N-th 2n root ζN of unity. Since K = k(ζ2n)( p ) and N = [K : k(ζ2n)], there exists 2n r 2n an automorphism α Gal(K=k(ζ2n)) such that α( p ) = ζN p. Thus an 2 2n 2n 1 automorphism ρ := αη0 is an extension of η and satisfies ρ( p ) = ξn p− and the order of ρ is equal to 2. Secondly, we consider the case where N = n is even. By Lemma 1, we have p k(ζ2n). Take any extension η0 in Gal(K=Q) of η. Since 2n 2n 22n 1=n t 1=n (η0( p ) p ) = η() = 1, we have η0((p ) ) = ζ2n(p )− for some integer t, and tn 1 η(p ) = ζ2np− : If, hence η(p )p = 1, then t is even and since [K : k(ζ2n)] = n and p k(ζ ), we can take α Gal(K=k(ζ )) so that α( 2pn ) = ζt 2pn , 2 2n 2 2n 2n and therefore ρ := αη is what we want.
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