The Splitting of Prime Ideals in the Maximal. Order of a Quadratic Field

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1 International Mathematical Forum, Vol. 13, 2018, no. 11, HIKARI Ltd, The Splitting of Prime Ideals in the Maximal Order of a Quadratic Field Lionel Bapoungué Université de Yaoundé 1 École normale supérieure BP 47 Yaoundé - Cameroun Copyright 2018 Lionel Bapoungué. This article is distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Abstract Let K be a quadratic field, O K its maximal order and O f an order of index f in K. Write M f for the set of associated complete module M with O f and let β be a mapping from M f to M 1. If fo K is an ideal of O K such that fo K M O K and if q O K O K /fo K is the canonical mapping, it is shown that (Z/fZ) (O K /fo K ) = (Z/fZ) and q induces an isomorphism ker (β) (O K /fo K ) / (Z/fZ) where A denotes the group of units of a ring A. Next, considering the prime number p N and denoting by J f and J K the respective subgroups of M f and M 1, we use the mapping δ J f J K to describe the prime ideals which are divisors of fo K. Mathematics Subject Classifications: 11J68, 11J86 Keywords: Quadratic fields, orders of quadratic fields, discriminant, ideals of quadratic fields, splitting of prime ideals 1 Introduction Let K = Q( d), the quadratic field associated with the square-free integer d Z ; write O K = Z + Zω for its maximal order with 1 + d, if d 1(mod 4) ω = { 2 d, if d 2 or 3 (mod 4)

2 494 Lionel Bapoungué of discriminant d, if d 1 (mod 4) Disc(O K ) = { 4d, if d 2 or 3 (mod 4). Let O f = Z + Zfω be an order of index f N in K of discriminant Disc(O f ) = f 2 Disc(O K ). Write M f for the set of associated complete modules M with O f. [2, Chapter II] shows that : M f is an abelian group for the product of modules, the identity element is O f and the inverse of M is M 1 1 = N(M) Mσ, where N(M) and M σ are respectively the norm and the conjugate of M ; the norm mapping M N(M) is a homomorphism of the group M f to the multiplicative group Q +. It is also wellknown in [2, Chapter II] that if: f = 1, M 1 is the set of associated complete modules with O K : the complete modules associated with O K are subgroups of K of rank 2 ; on the over hand, they are modules over O K in usual sense. f > 1, the group M f is more complicated because if M M f, it is a module over the ring O f with the property that O f is the biggest subring O of O K such that M is an O-module. M M f is called integral module if M O f. In the opposite case, M is called fractional module. By analogy with the ideals, a complete module M associated with O and integral is called an O-ideal. In particular, M is an ideal of O ; but there are ideals of O which are not O-ideals, for instance, the ideal fo which is associated with O K. An O-ideal I is said to be O-irreducible if I O and if I = JL, then J = O or L = O where J and L are O-ideals. Let I and J be two O-ideals ; we shall say that I divides J I J, if there exists an O-ideal L such that J = IL. Further, I J if and only if J I. An O-ideal I is said to be Oprime if I = I 1 I 2, then I 1 I or I 2 I where I 1 and I 2 are O-ideals. Two O-ideals A and B are called relatively prime if A + B = O. Throughout this paper, all ideals will be non-zero. We shall employ the following result proved in [5, Theorem 5.6] (see also [6]). Proposition 1. The (non-zero) ideals of the order O are the Z-modules of the form where c = D b2 4a Z. b + D I = d (Z + Z 2a )

3 Splitting of prime ideals in the maximal order of a quadratic field 495 In paper [1], we have studied the factorization in a non-maximal order of a quadratic field in which miscellaneous discussions was clearly devoted to the definition and the theorem below. Definition 1. Let O f be an order of index f in K. Write A f for the set of elements of O f whose norm is relatively prime to f. An element v A f is said to be: (i) irreducible if x, y A f and if v = xy, then x O f or y O f where O f is the group of units of O f ; (ii) k-prime if x, y A f and if v xy, then v x k or v y k where k is the exponent of O f. Theorem 1. The k-th powers of the elements of A f have a unique representation as a product of elements which are irreducible and k-prime. This paper is a logic continuation of [1]. More precisely, the goal of this work is to study the splitting of prime ideals in the maximal order of a quadratic field. In 2, we prove that if q O K O K /fo K is the canonical mapping, then and q induces the mapping (Z/fZ) (O K /fo K ) = (Z/fZ) θ ker(β) (O K /fo K ) (Z / fz) where A denotes the group of units of a ring A (Theorems 4 and 5) ; for the proofs we shall use the following result proved in [3, Chapter VI]: Theorem 2. Let G be a free commutative group of finite type, {e 1,, e n } a base of G and H a non-zero subgroup of G. Then, there exists a base {f 1,, f n } of G and a i Z, 1 i r (1 r n) such that: G = n Zf i and H = i = 1 r Za i f i with a 1 a 2 a r. i = 1 In the remainder of 2, we give the fundamental property of the mapping θ (Theorem 6). Writing for J f = {I + fo = O I is an O ideal} and J K = {I + fo K = O K I is an ideal of O K }, the respective subgroups of M f and M 1, we prove in 3 some lemmas concerning O-ideals. The paper is ended in 4 with a description of the prime ideals which are divisors of fo K (Theorem 7), the discussions involving the following results proved respectively in [2, p. 159] and [4, p. 84]:

4 496 Lionel Bapoungué Proposition 2. Let K be a quadratic field, O f an order of index f in K with Disc(O f ) = D. Then the modules A O f associated with O f of norm m N are exactly furnished by (s, u, v, w) N 2 Z 2 such that: v + D A = us (Z + Z ), D = v 2 4uv, u v < u, gcd(u, v, w) = 1, us 2 2u = m, the quadruple (s, u, v, w) corresponding to A. Theorem 3. A denotes a Dedekind ring of characteristic zero, K its field of fractions, L a finite extension of K of degree n and B the integral closure of A in L. Let m be a non-zero prime ideal of A. Then, Bm is an ideal of B and it has an expression of the form q e Bm = P i i, where the P i s are precisely distinct prime ideals D of B such that D A = m, the e i s are positive integers and the product sign denotes multiplication of ideals. Moreover, q i=1 e i f i = [B/Bm A/m] = n i=1 where f i called the residual degree of P i over A is written for the dimension of B/P i over A/m. 2 The mapping θ For M M f we define the mapping β M f M 1 by β(m) = MO K. Then, clearly, if M σ is the conjugate of M, then β(m σ ) = (β(m)) σ and if I is an Oideal (i.e. I O), then β(i) is an ideal of O K. In the ensuing of this work, we shall assume that O = O f is a fixed order. Proposition 3. Let K be a quadratic field, O K its maximal order, O an order of K and M M f. Then: (i) N(M) = N(MO K ). (ii) If β M f M 1 is a mapping defined by β(m) = MO K, then β(m) M 1 and β is a homomorphism of groups. Proof. (i) Appealing to section 1, N(M) is characterized by M 1 = 1 N(M) Mσ M σ = N(M)O

5 Splitting of prime ideals in the maximal order of a quadratic field 497 whence, MO K M σ O K = N(M)OO K so that MO K (MO K ) σ = N(M)O K showing that N(M) = N(MO K ). (ii) β(m) M 1 because O K MO K = MO K. Next, we have β(o) = OO K = O K and if M 1, M 2 M f, then β(m 1 M 2 ) = M 1 M 2 O K = M 1 O K M 2 O K = β(m 1 )β(m 2 ) which shows that β is a homomorphism of groups. From now on we shall leave the details of arguments of ω defined in 1. Lemma 1. Add to the notations and hypotheses of proposition 3 the assumption that M verifies MO K = O K. Then: (i) fo M O K ; (ii) αo K stabilizes M, where α is a divisor of the index f. Proof. (i) If M M f such that MO K = O K, we may write : 1 = n i=1 m i x i with m i M, x i O K, whence n f = fm i x i M i=1 because fx i fo K O K and as fωx i O K, we have n fω = fωm i x i M i=1 so that fo K M. But fo K M because from fo K = M we have fo K O K = MO K. As we have assumed that MO K = O K, we deduce that fo K = O K hence f = 1 and M MO K = O K. It follows that fo M O K. (ii) Now we show that M is stabilized by αo K. Taking n = 2, G = O K and H = M in theorem 2, we see that there exist e 1, e 2 O K and a 1, a 2 N such that : fo K = Zfe 1 + Zfe 2 M = Za 1 e 1 + Za 2 e 2 O K = Ze 1 + Ze 2. Therefore, there exist f 1, f 2 N such that f = a 1 f 1, f = a 2 f 2. Set d = gcd (f 1, f 2 ) and f = αd. Then, for i, j = 1, 2, we have e i e j Ze 1 + Ze 2. Therefore, it suffices to see that αa j e i M and this is true since a 1 α and a 2 α hence αa j e i Za i e i, so that αe 1 M M and αe 2 M M ; in other words αo K stabilizes M. Theorem 4. With the notations and hypotheses of proposition 3, if q O K O K /fo K is the canonical mapping, then (Z/ fz) (O K /fo K ) = (Z/fZ).

6 498 Lionel Bapoungué Proof. First, it is clear that q(m) is a non-zero additive subgroup of O K /fo K. Next, from lemma 1 and the notations of its proof, as M is stabilized by αo K, it follows that αo K stabilizes also O α (the order of index α) and since α f this is only possible if α = f, therefore d = 1. Then, we define q(m) by Since gcd(f 1, f 2 ) = 1, we have q(m) = Za 1e 1 + Za 2 e 2 Zfe 1 + Zfe 2. q(m) Z/f 1 Z Z/f 2 Z Z/f 1 f 2 Z, in concrete terms, q(m) = Zα 1 + Zα 2 where α 1 = q(a 1 e 1 ) and α 2 = q(a 2 e 2 ). Next, by Bezout s identity, there exist integers g and h such that 1 = gf 1 + hf 2. Set α 1 = gf 1 α 1 + hf 2 α 1. Then, modulo f 1 Z we have α 1 hf 2 α 1 ; similarly, α 2 gf 1 α 1 (mod f 2 Z). Take α 1 = f 2 γ and α 2 = f 1 γ such that γ = hα 1 + gα 2 ; then, γ is a generator of q(m). Further, as MO K = O K, we have γ (O K /fo K ) ; this implies that the ideal of O K /fo K generated by q(m) is the whole integer O K /fo K. But fx = 0 for any x O K /fo K ; hence the ring Z/fZ injects itself into O K /fo K ; therefore, taking units of both sides of (Z/fZ) (O K /fo K ), we obtain (Z/fZ) (O K /fo K ). Moreover, if n Z/fZ is a unit in O K /fo K, there exists γ O K such that nγ = 1. Write γ for γ = u + vω ; then, = 1 n(u + vω) 1 = 0 ; since fx = 0, we have n(u + vω) 1 = 0 fx = 0 and writing x for x = r + sω, it follows that nγ = 1 n(u + vω) 1 = f(r + sω) nu 1 + nvω = fr + fsω. Equating coefficients of ω on both sides of the last equality, we obtain nu fr = 1. This is an instance of Bezout s identity, which shows that gcd(n, f) = 1 so that n (Z/fZ). It follows that (Z/fZ) (O K /fo K ) = (Z/fZ). We now go on to show how the mapping θ ker(β) (O K /fo K ) (Z/fZ) can be constructed. Clearly, M is a complete module because M O K. Theorem 5. With the notations and hypotheses of theorem 4, q induces the mapping θ ker(β) (O K /fo K ) /(Z/fZ). Proof. Let γ 1 and γ 2 be two generators of q(m). Set γ 1 = m 1 γ 2, γ 2 = m 2 γ 1 ; then, from the multiplication of these two last expressions, we deduce that 1 = m 1 m 2 in O K /fo K and m 1 (Z/fZ) (O K /fo K ) = (Z/fZ). This shows that

7 Splitting of prime ideals in the maximal order of a quadratic field 499 q(m) is characterized by an element of (O K /fo K ) / (Z/fZ). Thus, we have constructed the mapping θ. Definition 2. Let γ be a generator of q(m) and γ M (Z/fZ) of γ. Then the mapping θ is defined by θ(m) = γ M. Lemma 1. With the same notations as above, let γ O K be, γ (O K /fo K ) and O M the ring of stabilizing of M. Assume that M = fo K + Zγ. Then MO K = O K and O = O M where O denotes an arbitrary order. Proof. First of all, MO K = fo K + γo K is an ideal of O K and as γ is a unit (mod fo K ), we have 1 γa (mod fo K ) i.e. there exists b O K such that 1 = γa + fb MO K so that MO K = O K. We prove now the last equality. We have fωγ fo K M hence O M = fo K + γo M so that O O M. Conversely, let α O M, α = r + sω ; then sω O M therefore sωγ M ; write sωγ for sωγ = fk + nγ ; then (sω n)γ = fk fo K γo K = fγo K that is (sω n)γ = fγt for t O K and sω n = ft fo K O hence sω O and α O. This proves that O M O. Therefore O = O M. Theorem 6. The mapping θ is an isomorphism. Proof. Let M 1, M 2 M f be. If q(m 1 ) = Zγ 1 and q(m 2 ) = Zγ 2, then q(m 1 M 2 ) = q(m 1 )q(m 2 ) = Zγ 1 γ 2 which shows that θ is a homomorphism. Next, let M = fo K + Zγ where γ Z is a unit modulo f; then, as fω M and since 1 M because 1 = γh + fl Zγ + fo K, we see that O M. But it is clear that, since fo K O, we have M O. Therefore M = O. So θ is injective. It remains to show that θ is surjective. Clearly M is a complete module : M O K, next as q(m) = Zq(γ), we see that θ(m) = γ and finally the result follows from lemma 1. Corollary 1. The kernel of β is given by the set ker(β) = {M = fo K + Zγ γ O K representing γ (O K /fo K ) / (Z/fZ) }. Definition 3. The function f Φ(f) = Card(O K /fo K ) is often called Euler s Φ-quadratic function. Remark 1. We have already used the function f Φ(f) in [1]. 3 Some lemmas concerning O-ideals This section is devoted to the study of some lemmas concerning O-ideals. First of all, we prove the following lemma:

8 500 Lionel Bapoungué Lemma 2. Let K be a quadratic field, O f = Z + Zfω an order of index f in K and I an ideal of O f such that I + fo f = O f. Then, I is an O f -ideal. Proof. Let O g be an another order of index g in K. If I is stabilized by O g, then we have gωi I. Since I + fo f = O f, there exist a I and b O f such that 1 = a + fb ; whence, gω = gωa + gωfb gωi + O f O f. Now, write gω for gω = m + nfω with m, n Z ; then, m + nfω = gωa + gωfb m + nfω = gω(a + bf). Therefore, m = 0 and nf = g(a + bf). But, as a + bf = 1, we have nf = g hence f g so that O g O f. Next, appealing to the sets J f and J K defined in 1, we go on to prove the following lemma. Lemma 3. Let K be a quadratic field, O K its maximal order and O an order of index f in K. Let J f and J K be. If I J K then: (i) I O + fo = O ; (ii) I O J f ; (iii) (I O)O K = I. Proof. (i) If I J K then according to the definition of J K we have O = O K O = (I + fo K ) O I O + fo O which shows that I O + fo = O. (ii) Since I O is an ideal of O, from (i) and lemma 2, we see that I O is an Oideal so that I O J f. (iii) Clearly, (I O)O K I. Conversely, we have I = IO = I((I O) + fo) I(I O) + fio K (I O)O K + fio K and as fi I and fi fo K O, it follows that I (I O)O K + (I O)O K = (I O)O K. Therefore (I O)O K = I. In the ensuing discussion we shall consider the mapping δ from J f to J K, which is the restriction of β.

9 Splitting of prime ideals in the maximal order of a quadratic field 501 Proposition 4. Let δ be a mapping from J f to J K. Then δ(i O) = (I O)O K and δ 1 (I) = I O. Proof. This results from the fact that clearly δ(j f ) J K and from lemma 3 so that (I O)O K = I δ(i O) = I. Remark 2. As δ(nm) = nδ(m) for any n N, it suffices to hit on the ideals of O K, therefore to hit on the prime ideals of O K. But, if I + fo = O then, according to proposition 4 and (iii) of lemma 3, we have δ 1 (I) = I O. It remains to hit on the prime ideals which are divisors of fo K. This is the aim of the next section. 4 The prime divisors ideals of fo K In this, we give a description of the prime divisors ideals of fo K. Theorem 7. Let K = Q( d), the quadratic field associated with the square-free d Z, O K its maximal order with D K = Disc(O K ), O f an order of index f in K with D = Disc(O f ), fo f an ideal associated with O K, M M f and δ a mapping from J f to J K. Let I O K be a prime such that I + fo K = O K and p N a prime satisfying I Z = pz. Then, the prime divisors ideals of fo K are: I = O K p when p remains prime in O K ; when p ramifies in O K ; when p splits in O K. I 2e+1 = O K p e I I = p 2e+1 (Z + Z pe r + D 2p 2(e+1) ) Proof. Taking n = 2, O K = B, I = D, A = Z and m = pz in theorem 3, we see that there exist precisely distinct prime ideals I of O K such that I Z = pz. Moreover, the formula q e i f i = 2 i=1 of the same theorem entails q 2 and the following three cases: p remains prime, p ramifies and p splits.

10 502 Lionel Bapoungué Case 1: p remains prime. In this case, we have q = 1, e 1 = 1, f 1 = 2 and this entails clearly I = O Kp. Now, we intend to study case 2 and case 3 by making use of the fact that f may be expressed as f = p e g with p g. Case 2: p ramifies. In this case, we have q = 1, e 1 = 2, f 1 = 1. Then, taking A = M, u = p 2e+1, s = 1 and v = rp e+1 in proposition 2, we see that M = p 2e+1 (Z + Z pe+1 r + D 2p 2e+1 ) O with m = N(M) = p 2e+1 and proposition 1 shows that M is an ideal I of O with w = v2 D. As p is prime, let us first consider the case in which p is odd. In this 4u case, we can choose r such that r = { 0, if D K 0 (mod 2) 1, if D K 1 (mod 2). Then, with the above relations, we have v 2 D = r 2 p 2e+2 p 2e g 2 D K with p D K and p 2 D K ; therefore according to the choice of r, u and 4 divide v 2 D = p 2e+1 (r 2 p g2 D K p ), because D K 0 (mod 2) D K 0 (mod 4). This shows that w Z. Moreover, as p g2 D K, this imposes gcd(u, v, w) = 1 and since β J p f J K is a mapping, δ(i) is an ideal of O K of norm N(I) = p 2e+1 : it s only I 2e+1 = O K p e I. Now, when p = 2, as above, we have A = M, u = 2 2e+1, s = 1, v = 2 2e+1 r and M = 2 2e+1 (Z + Z 2e+1 r + D 2.2 2e+1 ) O K with r = { 0, if D K 0 (mod 8) 1, if D K 1 (mod 4)

11 Splitting of prime ideals in the maximal order of a quadratic field 503 and proposition 1 shows that M is an O-ideal I of norm m = N(I) = 2 2e+1 with w = v2 D ; thus, we may write D 4u K = 2 ε α with α 1 (mod 2), and this imposes ε = 2 and ε = 3. If : ε = 2, then D K = 4d and α = d 3 (mod 4) ; whence 1 αg 2 0 (mod 2), hence w Z and 1 αg 2 1 α 2 (mod 4), therefore w 1 (mod 2) and gcd(u, v, w) = 1 ; ε = 3, then 2αg 2 0 (mod 2), 2αg 2 0 (mod 4), therefore w = αg 2 1 (mod 2) and gcd(u, v, w) = 1. For these two values of ε, we deduce that 4u = 2 2e+3 divides v 2 D = 2 2e+2 (r 2 2 ε 2 αg 2 ). So, we conclude as in the case in which p 2 that I 2e+1 = O K 2 e I. Case 3: p splits. In this remaining case we have : q = 2, e 1 = e 2 = 1, f 1 = f 2 = 1. Then, taking A = M, u = p 2e+1, s = 1 and v = rp e in proposition 2, we also obtain M = p 2e+1 (Z + Z pe r + D 2p 2e+1 ) M f of norm m = N(M) = p 2e+1, and proposition 1 shows that M is an O-ideal I of norm m = N(I) = p 2e+1 with w = v2 D. As above, we study the following two cases : p = 2 and p > 2. 4u When p > 2 i.e. p is odd, clearly, as D K is a square modulo p, g 2 D K is also a square modulo p ; then we can choose r such that 0 < r < p, r 2 g 2 D K (mod p), where r and D K are of the same parity. As r 2 g 2 D K (mod p 2 ), then replacing r by s = r + 2p, we see that s verifies 0 < s < p 2, s 2 g 2 D K + 4pr (mod p 2 ), where s and D K are of the same parity. Therefore, r 2 g 2 D K 0 (mod p 2 )(otherwise p 4r therefore p r). Consider now the number r satisfying 0 < r < p 2, r D K (mod 2), r 2 g 2 D K (mod p), r 2 g 2 D K 0 (mod p 2 ). Since r 2 g 2 D K (mod p), we have u (v 2 D) = p 2e (r 2 g 2 D K ) and according to the parity of r, 4 (v 2 D). Hence w Z and p w because r 2 g 2 D K 0 (mod p 2 ) so that gcd(u, v, w) = 1. Moreover, v = rp e < p e+2 p 2e+1 = u ; it follows that p e I M f. Further, as and D K are of the same parity, we have r+g D K p e I = p 2e+1 (Z + Z pe r + D 2p 2e ) O K. 2 O K because r

12 504 Lionel Bapoungué Now, when p = 2, we have D K 1 (mod 8), therefore g 2 D K 1(mod 8). As in the case when p is odd, we have A = M, u = 2 2e+1, s = 1, v = 2 e r and w = v 2 D 4u. Then, I = 2 2e+1 (Z + Z 2e r + D 2.2 2e+1 ) M f and I O f of norm m = N(I) = 2 2e+1. Next, we can choose the number r such that r = { 3, if g2 D K 1 (mod 16) 1, if g 2 D K 1 (mod 16). Then 4u = 2 2e+3 divides v 2 D = 2 2e (r 2 g 2 D K ) because r 2 g 2 D K (mod 8), hence w Z and as g 2 D K 1 (mod 16), we have g 2 D K 1 (mod 16) ; it follows that w 1 (mod 2). Moreover, v = 2 e r < 2 e+2 2 2e+1 = u. Then 2 e I M f and 2 e I O K. This ends the proof of the theorem. Remark 3. I/O f p e O K, but I/O f p e O f, therefore I/O f p e is not integral. Corollary 2. With the notations and hypotheses of theorem 7, δ is surjective. Proof. Since δ J f J K is a mapping, from theorem 7, we deduce that, when : - p remains prime in O K, the prime divisors ideals I of fo K are I = O K p so that I = O K p = δ ( O f p) ; - p ramifies in O K, the prime divisors ideals I of fo K are I 2e so that I/O f p e M f verifies I 2e = δ (I O f p e ) = δ (I)δ (O f p e ) 1 = I ; - p splits in O K, the prime divisors ideals I of fo K are I = p 2e+1 (Z + Z pe r + D 2p 2(e+1) ) hence, δ (p e I) O K and as the norm of δ (p e I) is p, we have So β is surjective. I = β(p e I) or I σ = β(p e I) σ. Theorem 8. With the notations and hypotheses of theorem 7, δ is an isomorphism.

13 Splitting of prime ideals in the maximal order of a quadratic field 505 Proof. This results from proposition 4 and corollary 2. References [1] L. Bapoungué, Factorisation dans un ordre non maximal d un corps quadratique, Expo. Math., 20 (2002), [2] Z. I. Borevitch and I. Chafarevitch, Théorie des Nombres, Gauthier-Villars, Paris, [3] P. Dubreil and M. L. Dubreil-Jacotin, Leçons D algèbre Moderne, Dunod, Paris, [4] P. Samuel, Théorie Algébrique des Nombres, Édition revue et corrigée, Hermann, Paris, [5] T. Takagi, Elementary Number Theory, Kyoritsu, Tokyo, (in Japanese) [6] H. C. Williams and M. C. Wunderlich, On the parallel generation of the residues for the continued fraction factoring algorithm, Math. Comp., 48 (1987), Received: August 11, 2018; Published: November 4, 2018

On a special case of the Diophantine equation ax 2 + bx + c = dy n

Sciencia Acta Xaveriana Vol. 2 No. 1 An International Science Journal pp. 59 71 ISSN. 0976-1152 March 2011 On a special case of the Diophantine equation ax 2 + bx + c = dy n Lionel Bapoungué Université