Imaginary quadratic fields whose Iwasawa ¸-invariant is equal to 1 by Dongho Byeon (Seoul) ¤y 1 Introduction and statement of results p Let D be the fundamental discriminant of the quadratic field Q( D) and  := ( D ) the usual Kronecker character. Let p be prime, Z the ring of D ¢ p p p-adic integers, andp¸p(Q( D)) the Iwasawa ¸-invariant of the cyclotomic Zp-extension of Q( D). In this paper, we shall prove the following: Theorem 1.1 For any odd prime p, p p X ]f¡X < D < 0 j ¸ (Q( D)) = 1; (p) = 1g À : p D log X Horie [9] proved that forp any odd prime p, there exist infinitely many imagi- nary quadratic fields Q( D) with ¸p(Q( D)) = 0 and the author [1] gave a lower bound for the number of such imaginary quadratic fields. It is knownp that forp any prime p which splits in the imaginary quadratic field Q( D), ¸p(Q( D)) ¸ 1. So it is interesting to see how often the trivial ¸-invariant appears for such a prime. Jochnowitz [10] provedp that for anyp odd prime p, if there exists one imaginary quadratic field Q( D0) with ¸p(Q( D0)) = 1 and ¤2000 Mathematics Subject Classification: 11R11, 11R23. yThis work was supported by grant No. R08-2003-000-10243-0 from the Basic Research Program of the Korea Science and Engineering Foundation. 1 ÂD0 (p) = 1, then there exist an infinite number of such imaginary quadratic fields. p For the case of real quadratic fields, Greenberg [8] conjectured that ¸p(Q( D)) = 0 for all real quadratic fields and all prime numbers p. Ono [11] and Byeon [2] [3] showed thatp for all prime numbersp p, there exist infinitely many real quadratic fields Q( D) with ¸p(Q( D)) = 0 and gave a lower bound for the number of such real quadratic fields. In section 3, we shall prove the following: Proposition 1.2 For any odd primepp, if there is a negative fundamental discriminant D0 < 0 such that ¸p(Q( D0)) = 1 and ÂD0 (p) = 1, then p p X ]f¡X < D < 0 j ¸ (Q( D)) = 1; (p) = 1g À : p D log X In section 4, we shall prove the followings: Proposition 1.3 Let p be an odd prime and Dp0 < 0 be the fundamental discriminant of the imaginary quadratic field Q( 1 ¡ p2). Then  (p) = 1 p D0 p¡1 2 and ¸p(Q( D0)) = 1 if and only if 2 6´ 1 (mod p ), that is, p is not a Wieferich prime. Proposition 1.4 Let p be a Wieferich prime. If p ´ 3 (mod 4), letp D0 < 0 be the fundamental discriminant of the imaginary quadratic field Q( 1 ¡ p) and if p ´ 1 (mod 4), letp D0 < 0 be the fundamental discriminantp of the imaginary quadratic field Q( 4 ¡ p). Then ÂD0 (p) = 1 and ¸p(Q( D0)) = 1. From these three propositions, Theorem 1.1 follows. 2 Preliminaries Let  be a non-trivial even primitive Dirichlet character of conductor f which 2 is not divisible by p . Let Lp(s; Â) be the Kubota-Leopoldt p-adic L-function and O = Zp[Â(1);Â(2); ¢ ¢ ¢]: Then there is a power series F (T;Â) 2 OÂ[[T ]] such that s Lp(s; Â) = F ((1 + pd) ¡ 1;Â); 2 where d = f if p 6 jf and d = f=p if pjf. Let ¼ be a generator for the ideal of O above p. Then we may write F (T;Â) = G(T )U(T ); where U(T ) is a unit of OÂ[[T ]], and G(T ) is a distinguished polynomial: that ¸ is, G(T ) = a0 + a1T + ¢ ¢ ¢ + T with ¼jai for i · ¸ ¡ 1. Define ¸(Lp(s; Â)) be the index of the first coefficient of F (T;Â) not divisible by ¼. Let ! be the the Teichm¨ullercharacter. Lemma 2.1 (Dummit, Ford, Kisilevsky and Sands [5, Proposition 5.1]) Let Dp < 0 be the fundamental discriminant of the imaginary quadratic field Q( D). Then p ¸p(Q( D)) = ¸(Lp(s; ÂD!)): Lemma 2.2 (Washington [13, Lemma1]) Let Dp < 0 be the fundamental discriminant of the imaginary quadratic field Q( D). 2 ¸(Lp(s; ÂD!)) = 1 () Lp(0;ÂD!) 6´ Lp(1;ÂD!) (mod p ): From these lemmas, we can show the following: Proposition 2.3 Let p be an odd prime and Dp < 0 be the fundamental discriminant of the imaginary quadratic field Q( D) such that ÂD(p) = 1. L(1¡p;ÂD) Then p is p-integral and p L(1 ¡ p;  ) ¸ (Q( D)) = 1 () D 6´ 0 (mod p); p p where L(s; ÂD) is the Dirichlet L-function. Proof: By the construction of the p-adic L-function Lp(s; ÂD), ¡1 ¡1 Lp(0;ÂD!) = ¡(1 ¡ ÂD! ¢ ! (p))B1;ÂD!¢! = ¡(1 ¡ ÂD(p))B1;ÂD : where Bn;ÂD is the generalized Bernoulli number. Since ÂD(p) = 1, Lp(0;ÂD!) = 0: 3 Similarly, ¡p p¡1 ¡p Lp(1 ¡ p; ÂD!) = ¡(1 ¡ ÂD! ¢ ! (p)p )Bp;ÂD!¢! =p p¡1 = ¡(1 ¡ ÂD(p)p )Bp;ÂD =p p¡1 = (1 ¡ p )L(1 ¡ p; ÂD) 2 ´ L(1 ¡ p; ÂD) (mod p ): Since ÂD! 6= 1 is not a character of the second kind, Lp(1 ¡ p; ÂD!) and L(1 ¡ p; ÂD) are p-integral (See [14]). By the congruence of Lp(s; ÂD), Lp(1;ÂD!) ´ Lp(0;ÂD!) = 0 (mod p); and 2 Lp(1;ÂD!) ´ Lp(1 ¡ p; ÂD!) (mod p ): L(1¡p;ÂD) Thus p is p-integral and L(1 ¡ p;  ) D 6´ 0 (mod p) () L (1; !) 6´ 0 (mod p2): (1) p p D From the equation (1) and Lemmas 2.1, 2.2, the proposition follows. 2 3 Proof of Proposition 1.2 Let Mk(Γ0(N);Â) denote the space of modular forms of weight k on Γ0(N) with character Â. For a positive integer r ¸ 2, let X N Fr(z) := H(r; N)q 2 M 1 (Γ0(4);Â0) r+ 2 N6=0 be the Cohen modular form [4], where q := e2¼iz. We note that if Dn2 = (¡1)rN, then X r¡1 H(r; N) = L(1 ¡ r; ÂD) ¹(d)ÂD(d)d σ2r¡1(n=d); (2) djn P º where σº(n) := djn d . From Fp(z), we can construct the modular form X H(p; n) n 4 4 Gp(z) := q 2 M 1 (Γ0(4p Q );Â0); p+ 2 ¡n n p ( p )=1;( Q )=¡1 4 where Q is a prime such that Q 6= p. From Proposition 2.3 and the equation (2), if Dp < 0 is the fundamental discriminant of the imaginary quadratic field Q( D) such that ÂD(p) = 1, then H(p; ¡D) L(1 ¡ p;  ) = D p p is p-integral. Using similar methods in Ono [11] and Byeon [2], that is, applying a theorem of Sturm [12] to the following two modular forms X H(p; ln) n 4 4 4l (UljGp)(z) = q 2 M 1 (Γ0(4p Q l); ( )); p+ 2 ¡n n p ¢ ( p )=1;( Q )=¡1 X H(p; n) ln 4 4 4l (VljGp)(z) = q 2 M 1 (Γ0(4p Q l); ( )); p+ 2 ¡n n p ¢ ( p )=1;( Q )=¡1 3 where l 6= p is a suitable prime and comparing the coefficients of q¡D0l of these modular forms, where Dp0 < 0 is a fundamental discriminant of the H(p;¡D0) imaginary quadratic field Q( D0) such that ÂD0 (p) = 1 and p 6´ 0 (mod p), we can obtain the following: Proposition 3.1 Let p be an odd prime. Assume that therep is a fundamental discriminant D0 < 0 of the imaginary quadratic field Q( D0) such that (i) ÂD0 (p) = 1; H(p;¡D0) (ii) p 6´ 0 (mod p). Then there is an arithmetic progression rp (mod ptp) with (rp; ptp) = 1 and ¡rp ( p ) = 1, and a constant ·(p) such that for each prime l ´ rp (mod ptp) there is an integer 1 · dl · ·(p)l for which (i) Dl := ¡dll is a fundamental discriminant, H(p;¡Dl) (ii) p 6´ 0 (mod p). 5 Proof of Proposition 1.2: Let Dl < 0 be the fundamental discriminant in L(1¡p; ) Proposition 3.1. Then  (p) = 1 and H(p;¡Dl) = Dl 6´ 0 (mod p). Dpl p p By Proposition 2.3, ¸ (Q( D )) = 1. By Dirichlet’s theorem on primes in p l p X arithmetic progression, we have that the number of such Dl < X is À log X : 2 4 Proof of Propositions 1.3 and 1.4 To prove Propositions 1.3 and 1.4, we shall use of the following criterion of Gold. Lemma 4.1 (Gold [7]) Let p be an odd prime andpD < 0 be the fundamental discriminant of thep imaginary quadratic field Q( D) such that ÂD(p) = 1. ¯ r Let (p) = PP in Q( D). Supposep that P = (¼) is principal for some integer p¡1 2 r not divisible by p. Then ¸p(Q( D)) = 1 if and only if ¼ 6´ 1 (mod P¯ ). First we shall prove Proposition 1.3. 2 pProof of Proposition 1.3: Wep note that 1¡p is not a square. Let Pp= (p; 1+ 2 ¯ 2 ¯ 2 2 1 ¡ p ) andp P = (p; 1¡ 1 ¡ p ). Then (p) =pPP and P = (1+ 1 ¡ p ), 2 2 P¯ = (1 ¡ 1 ¡ p ). From Lemma 4.1, ¸p(Q( D0)) = 1 if and only if q p (1 + 1 ¡ p2)p¡1 6´ 1 mod ( 1 ¡ 1 ¡ p2 ): This is equivalent to that q q p p (1 + 1 ¡ p2)p ¡ (1 + 1 ¡ p2) 6´ 0 mod ( p2 = (1 ¡ 1 ¡ p2)(1 + 1 ¡ p2)): (3) We see that q q (1 + 1 ¡ p2)p ¡ (1 + 1 ¡ p2) p¡1 à ! p¡1 à ! X2 p X2 p q q ´ + ( ) 1 ¡ p2 ¡ (1 + 1 ¡ p2) n=0 2n n=0 2n + 1 p¡1 à ! p¡1 à ! X2 p X2 p q ´ ( ¡ 1) + ( ¡ 1) 1 ¡ p2 n=0 2n n=0 2n + 1 q ´ (2p¡1 ¡ 1)(1 + 1 ¡ p2) (mod p2); 6 where we have used the fact p¡1 à ! p¡1 à ! X2 p X2 p = = 2p¡1: n=0 2n n=0 2n + 1 Thus the equation (3) is true if and only if 2p¡1 6´ 1 (mod p2), that is, p is not a Wieferich prime and the proposition follows.