Primes of the form x2 + ny2

Primes of the form x2 + ny 2 Marios Magioladitis Arbeitsgemeinschaft zur Klassenkörpertheorie Institute for Experimental Mathematics (IEM) University of Duisburg-Essen June 2004 2 Arbeitsgemeinschaft zur Klassenkörpertheorie. Organisation: G. Böckle, I. Bouw, C. Diem, G. Frey. Sommersemester 2004. Lecture 9. Primzahlen der Form x2 + ny2 was given on 29th June 2004 in the Institute of Experimental Mathematics (IEM). I would like to thank Prof. David A. Cox (Amherst College) for his immediate replys to my e-mails. Marios Magioladitis Homepage: http://www.exp-math.uni-essen.de/˜magiolad Email: [email protected] References [Cox] D. Cox, Primes of the form x2 + ny 2 , Pure and Appl. Math., Wiley, 1989. [Neu] J. Neukirch, Algebraic Number Theory, Springer Verlag, 1999. [Mil] J.S. Milne, Class Field Theory, Lecture Notes v3.1, 1997. Available at: http://www.jmilne.org/math/CourseNotes/math776.html Contents 1 Prelimaries 1.1 Introduction and some historical background 1.2 Elements of Hilbert Class Field Theory . . . 1.3 Elements of Quadratic Field Theory . . . . . 1.4 Some more background on Field Theory . . . 1.5 Orders in Quadratic Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 5 6 8 9 10 2 Application of Class Field Theory 2.1 The solution of p = x2 + ny 2 for arbitrary n 2.2 How our result works in practice . . . . . . 2.2.1 Primes of the form x2 + 27y 2 . . . . 2.2.2 Primes of the form x2 + 64y 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 17 21 22 23 3 . . . . 4 Chapter 1 Prelimaries 1.1 Introduction and some historical background We will answer the following question: Question 1.1.1. Let n be a positive integer. Which primes p can be expressed in the form p = x2 + ny 2 where x and y are integers? (a) Classic Number Theory solves the problem for only few cases. (b) Genus Theory solves the problem for a large but finite number of n’s. (c) Cubic and Biquadratic reciprocity enables us to solve the problem on some cases where Genus Theory fails but doesn’t solve the problem for arbitrary n. We can write p = x2 + ny 2 = (x + question √ √ −ny)(x − −ny). This leads us to restate the Question 1.1.2. Let n be a positive integer. Which primes p can be expressed in the form √ √ p = (x + −ny)(x − −ny) where x and y are integers? 5 6 1.2 Elements of Hilbert Class Field Theory Definition 1.2.1. Let K be a number field. We will note with IK the group of fractional ideals of OK , and with PK the subgroup of principal fractional ideals i.e. those of the form αOK where α ∈ K ∗ . We define P ic(OK ) := IK /PK the (ideal) class group of K. (P ic(OK ) is also denoted as ClK in [Neu] and as C(OK ) in [Cox]). Remark 1.2.2. P ic(OK ) is a finite group. Definition 1.2.3. Let L/K be a Galois extension and let p be a prime ideal of OK . Let P be a prime ideal of OL containing p. The inertial degree of p in P is defined to be f = fP|p := [OK /p : OL /P]. Definition 1.2.4. Let L/K be a Galois extension and let p be a prime ideal of OK . Let pOK = Pe1 · · · · · Pek , Pi distinct prime ideals of OL be the prime factorization of pOK in OL . We say that: p ramifies in L if e > 1. p is unramified in L if e = 1. p splits completely in L if e = f = 1. Theorem 1.2.5. Let K be a number field. There is a finite Galois extension L/K such that: L is the maximum unramified Abelian extension of K. This unique field L is called the Hilbert class field of K. Lemma 1.2.6. Let L/K be a Galois extension and let p be a prime ideal of O K which is unramified in L. If P is a prime ideal of OL containing p, then there is a unique element σ ∈ Gal (L/K) such that for all α ∈ OL σ(α) ≡ αN (p) mod P, where N (p) = |OK /p| is the norm of p. Proof. Omitted. See Lemma 5.19 of [Cox]. Definition 1.2.7. The σ of the previous Lemma is called the Artin sym  element since it depends on the prime ideal P of OL . bol and is denoted L/K P 7 Remark 1.2.8. For every α ∈ OL we have   L/K (α) ≡ αN (p) P mod P, where p = P ∩ OK . The Artin symbol has the following useful properties: Corollary 1.2.9. Let L/K be a Galois extension, and let p be an unramified prime ideal of OK . Given a prime ideal P of OL containing p, we have: (i) If σ ∈ Gal (L/K), then  L/K P  L/K σ(P)  =σ  L/K P  σ −1 .  is the inertial degree fP|p .   = 1. (iii) p splits completely in L ⇔ L/K P (ii) The order of Proof. Omitted. For (iii) just note that e = 1 since p is unramified in L thus the equivalence follows immediately from (ii). See Corollary 5.21 from [Cox]. Remark 1.2.10. If the extension L/K is Abelian the Artin symbol can be written  L/K since it is independed from the prime ideal P of OL . p as Proof. Let P and P′ be two prime ideals of OL containing p. Then P′ = σ(P) for some σ ∈ Gal (L/K). The last corollary implies that         L/K L/K L/K L/K −1 = σ σ = . = P′ σ(P) P P Definition 1.2.11. Let L/K be an unramified Abelian extension and let a ∈ IK be a fractional ideal with prime factorization a= r Y i=1 We define the Artin symbol  L/K a   L/K a pi ri , ri ∈ Z. to be the product  = r r  Y L/K i i=1 pi 8 Definition 1.2.12. Let L/K be an unramified Abelian extension. The Artin symbol defined above defines a homomorphism of groups, called the Artin (or reciprocity) map,   L/K : IK → Gal (L/K). . Theorem 1.2.13. (The Artin reciprocity theorem for the Hilbert class field) Let K be a number field and L its Hilbert class field. The Artin map   L/K : IK → Gal (L/K). . is surjective, and its kernel is exactly PK . Thus the Artin map induces an isomorphism ∼ P ic(OK ) −→ Gal (L/K). Remark 1.2.14. Obviously: |P ic(OK )| = |Gal (L/K)| = [L : K]. 1.3 Elements of Quadratic Field Theory We will state briefly some elements of the Quadratic Field Theory √ Let K = Q( N ), where N 6= 0, 1 is a square-free integer. (1) The ring OK of algebraic integers of K is ( √ if N ≡ 6 1 mod 4, Z[ √ N ] OK = if N ≡ 1 mod 4. Z[ N2+1 ] (2) The extension K/Q is Abelian since Gal (K/Q) ∼ = Z/2Z. Thus |Gal (K/Q)| = 2. (3) The discriminant dK of K is defined to be  N if N ≡ 1 mod 4, dK = 4N otherwise. h i √ Thus, we can write OK = Z dK +2 dK . (4) If N < 0 the group of units of OK is   ±1, ±i ∗ ±1, ±ω, ±ω 2 OK =  ±1 if N = −1, if N = −3, otherwise. 9 Proposition 1.3.1. Let K be a quadratic field of discriminant dK and let p be an integer prime.   (i) If dpK = 0 then pOK = p2 , where p is a prime ideal of OK .   (ii) If dpK = 1 then pOK = pp, where p 6= p are prime ideals of OK .   (iii) If dpK = −1 then pOK is a prime ideal of OK . Furthermore, the primes in (i)-(iii) give all nonzero primes of OK . Proof. See Proposition 5.16 from [Cox]. Corollary 1.3.2. Let K be a quadratic field of discriminant dK , and let p be an integer prime. Then: (i) p ramifies in K ⇔ p divides dK   (ii) p splits completely in K ⇔ dpK = 1 Proof. Follows immediately from Proposition 1.3.1. 1.4 Some more background on Field Theory We will state briefly some elements of Field Theory that they will be necessary in the second chapter. We will show that if L/K is an extension then the prime ideals of OK which split completely in L characterize the extension uniquely. ˙ and and only if Definition 1.4.1. Let S and T be two sets. We say that S ⊂T . ˙ and S ⊂ T ∪ Σ for some finite set Σ. Furthermore, S = T will mean that S ⊂T ˙ T ⊂S. Definition 1.4.2. Let L/K be a finite extension and let PK be the set of all prime ideals of OK . We define SL/K := {p ∈ PK | p splits completely in L}. SeL/K := {p ∈ PK | p is unramified in L, fP|p = 1 for some prime ideal P of OL }. Remark 1.4.3. Note that SL/Q = {p prime | p splits completely in L}. SeL/Q = {p prime | p is unramified in L, fP|p = 1 for some prime ideal P of OL }. Proposition 1.4.4. Let L and M be finite extensions of a number field K. Then 10 ˙ L/K . (i) If M is Galois over K, then L ⊂ M ⇔ SM/K ⊂S ˙ L/K . (ii) If L is Galois over K, then L ⊂ M ⇔ SeM/K ⊂S . (iii) If L and M are Galois over K, then L = M ⇔ SM/K = SL/K . Proof. Notice that (iii) is an immediate consequence of (i). For the proof (i) and (ii) see Proposition 8.20 from [Cox]. 1.5 Orders in Quadratic Fields Definition 1.5.1. An order O in a quadratic field K is a subset of K such that (i) O is a subring of K containing 1, (ii) O is a finitely generated Z-module, (iii) O contains a Q-basis of K. Remark 1.5.2. (i) Since O is clearly torsion-free we can replace (ii) and (iii) by “O is a free Z-module of rank 2”. (ii) OK is always an order of K. Furthermore, OK contains any order O of K, so we can say that OK is the maximal order of K. Definition 1.5.3. Let O be an order in a quadratic field K. (a) The index f = [OK : O] is called the conductor of the order. (b) The discriminant D of O is equal to D := f 2 dK , where dK is the discriminant of K. √ √ √ Remark 1.5.4. The discriminant of Z[ −n] = Z + −nZ is (−2 −n)2 = −4n. Remark 1.5.5. Let O be an order in a quadratic field K. (a) O = Z + f OK . (b) O is always a Noetherian ring. It is a Dedekind domain if and only if f = 1. (Thus, unique factorization doesn’t hold for orders). Proof. Omitted. 11 √ √ Remark 1.5.6. Let K = Q( −n) where n is square-free. Then Z[ −n] = Z + f OK where f = 1 or 2. Proof. Follows immediately from the previous remark. Remark 1.5.7. In general, given any ideal a of O we have O ⊂ {β ∈ K : βa ⊂ a} . If equality holds i.e. O = {β ∈ K : βa ⊂ a} we will call a proper. This definition extends for fractional ideals. Proposition 1.5.8. Let O be an order in a quadratic field K and let a be a fractional O-ideal. Then, a proper ⇔ a invertible Proof. Let a be an invertible ideal of O. Then ab = O for some fractional ideal b of O. Let β ∈ K such that βa ⊂ a. Then, we have βO = β(ab) = (βa)b ⊂ ab = O. So β ∈ O. This yields that a is proper. For the other direction see Proposition 7.4 and Lemma 7.5 from [Cox]. Definition 1.5.9. Let O be an order in a quadratic field K. We define the following group I(O) := {a | a invertible fractional ideal of O}. Also we define P (O) := {αO | α ∈ K ∗ } the subgroup of principal fractional ideals of O. Finally, we define P ic(O) := I(O)/P (O) the (ideal) class group of the order O. (The quotient group is also called the Picard group). Remark 1.5.10. I(OK ) = IK and P (OK ) = PK Definition 1.5.11. Let O be an order of discriminant D in an imaginary quadratic field K. We define the class number of O to be h(O) := h(D), where h(D) is the class number of the discriminant D (i.e. the number of reduced forms of D). 12 Theorem 1.5.12. Let O be an order in an imaginary quadratic field K. Then |P ic(O)| = h(O). Proof. Omitted. See Theorem 7.7 from [Cox]. Remark 1.5.13. Unique factorization doesn’t hold for proper ideals √ (not even for invertible ideals in quadratic fields). For example consider K = Q( −3) and O = √ Z[ −3] which is an order of conductor 2. Unique factorization for proper (i.e. invertible) ideals of O would imply that O is a UFD since all invertible fractional ideals of O are principal (see√Exercise √ 7.9(b) of [Cox]) but O is not a UFD since for example 4 = 2 · 2 = (1 + −3)(1 − −3). We will define a slightly smaller class of ideals, those prime to the conductor, for which unique factorization holds. Definition 1.5.14. Let O be an order of conductor f in a quadratic field K. We say that a nonzero ideal a of O is prime to f if a + f O = O. Definition 1.5.15. Let a be an ideal of an order O. The norm of a is defined to be N (a) = |O/a|. Lemma 1.5.16. Let O be an order of conductor f and let a be an ideal of O. (i) a is prime to f ⇔ gcd(N (a), f ) = 1. (ii) a is prime to f ⇒ a is proper. Proof. Proof of (i) Omitted. See Lemma 7.18 from [Cox]. Proof of (ii) Let β ∈ K such that βa ⊂ a. This implies that β ∈ OK . Thus, we have βO = β(a + f O) = βa + βf O ⊂ a + f OK ⊂ a + O ⊂ O because f OK ⊂ O and a is a O-ideal. So β ∈ O. This proves that a is proper. Definition 1.5.17. Let O be an order of conductor f . We will denote by I(O, f ) the subgroup of fractional ideals generated by the ideals of O which are prime to f. Also we will denote by P (O, f ) the subgroup of I(O, f ) generated by the principal ideals αO where α ∈ O has norm N (α) prime to f. Thus, I(O, f ) :=< a ✁ O | a prime to f >, P (O, f ) :=< αO | α ∈ O, gcd(N (α), f ) = 1 > . Remark 1.5.18. (i) P (O, f ) is a subgroup of I(O, f ). 13 (ii) I(O, f ) is a subgroup of I(O). (See Lemma 1.5.16(ii)). Definition 1.5.19. Let f be a positive number. We define the group IK (f ) :=< a ✁ OK | a prime to f > . Definition 1.5.20. Let f be a positive number. We will denote by PK,Z (f ) the subgroup of IK (f ) generated by principal ideals of the form αOK where α ∈ OK and α ≡ a mod f OK for some a ∈ Z relatively prime to f . Thus, PK,Z (f ) =< αOK | α ∈ OK , exists a ∈ Z s.t. α ≡ a mod f OK , gcd(a, f ) = 1 > . Remark 1.5.21. Let O be an order of conductor f in an imaginary quadratic field K then IK (f ) ∼ = I(O, f ). Proof. The map a 7→ a ∩ O preserves multiplication and induces an isomorphism ∼ IK (f ) −→ I(O, f ), and the inverse of this map is given by a 7→ aOK . For a detailed proof see Proposition 7.20 from [Cox]. Proposition 1.5.22. Let O be an order of conductor f in an imaginary quadratic field K. Every ideal of O prime to f has a unique decomposition as a product of prime ideals of O which are prime to f . Proof. See Proposition 7.20 and Exercise 7.26 from [Cox]. We can now describe P ic(O) in terms of the maximal order. Proposition 1.5.23. Let O be an order of conductor f in an imaginary quadratic field K. Then there are natural isomorphisms P ic(O) ∼ = I(O, f )/P (O, f ) ∼ = IK (f )/PK,Z (f ). Proof. For the first isomorphism: Note that the map I(O, f ) −→ P ic(O) is surjective since every ideal class in P ic(O) contains an ideal of O prime to f . Also note that the kernel of this map is I(O, f ) ∩ P (O). This obviously contains P (O, f ). For the inclusion I(O, f ) ∩ P (O) ⊂ P (O, f ) consider that an element in I(O, f ) ∩ P (O) is a fractional ideal αO = ab−1 where α ∈ K ∗ and a, b are prime to f . Note that N (b)αO = aN (b)b ⊂ O. Thus, N (b)αO ∈ P (O, f ). Finally, αO = N (b)αO · (N (b)O)−1 ∈ P (O, f ). This completes the proof. For the second isomorphism: Note that the map a 7→ a ∩ O induces an isomorphism IK (f ) ∼ = I(O, f ) (see Remark 1.5.21). Under this isomorphism P (O, f ) maps to a subgroup P ′ of IK (f ). For the proof that P ′ = PK,Z (f ) see proof of Proposition 7.22 from [Cox]. Remark 1.5.24. Let O be an order of conductor f in an imaginary quadratic field K then P (O, f ) ∼ = PK,Z (f ). 14 Definition 1.5.25. Let K be a number field. We will call finite primes the prime ideals of OK and infinite primes the embeddings of K into C. A real infinite prime is an embedding σ : K → R and a complex infinite prime is a pair of complex conjugate embeddings σ, σ̄ : K → C, with σ 6= σ̄. Definition 1.5.26. Let K be a number field. A modulus m in K is a formal product Y m= pnp p over all primes p (finite or infinite) of K, where the exponents must satisfy the following conditions: (i) np ≥ 0 and at most finitely many are nonzero. (ii) np = 0 wherever p is a complex infinite prime. (iii) np ≤ 1 wherever p is a real infinite prime. If np = 0 for all the exponents we set m = 1. We can also write m = m 0 m∞ where m0 is an ideal of OK and m∞ is a product of distinct real infinite primes. Definition 1.5.27. We will denote by IK (m) the group of fractional OK -ideals relatively prime to m (i.e. to m0 ) and by PK,1 (m) the subgroup of IK (m) generated by the principal ideals of the form αOK where α ∈ OK and α ≡ 1 mod m0 and σ(α) > 0 for every real infinite prime σ dividing m∞ . Thus, IK (m) = {a ∈ IK | a prime to m0 }. Remark 1.5.28. Let K be an imaginary quadratic field. Then, PK,1 (f OK ) =< αOK | α ∈ OK , α ≡ 1 mod f OK > . Remark 1.5.29. PK,1 (f OK ) ⊂ PK,Z (f ) ⊂ IK (f ) = IK (f OK ). Definition 1.5.30. Let m be a modulus of a field K. A congruence subgroup of m is a group H where PK,1 (m) ⊂ H ⊂ IK (m). Remark 1.5.31. PK,Z (f ) is a congruence subgroup for f OK . (See Remark 1.5.29). Definition 1.5.32. Let m be a modulus of a field K and let H be its congruence subgroup. A generalized ideal class group for m is the quotient IK (m)/H. Definition 1.5.33. Let L/K be an Abelian extension and let m be a modulus divisible by all ramified primes of the extension. We can extend the Artin symbol by multiplicativity to give an homomorphism of groups Φm : IK (m) −→ Gal (L/K) which is called the Artin map for L/K and m. Sometimes we will denote Φ m as ΦL/K,m to refer explicitly to the extension. 15 Theorem 1.5.34. Let m be a modulus of a field K, and let H be a congruence subgroup for m. Then there is a unique Abelian extension L/K, all of whose ramified primes (finite or infinite) divide m such that H = ker(ΦL/K,m ). Proof. Omitted. Remark 1.5.35. Let L/K be the Abelian extension all of whose ramified primes divide f OK . Then, IK (f )/PK,Z (f ) ∼ = Gal (L/K). Proof. PK,Z (f ) is a congruence subgroup for f OK (see Remark 1.5.31) so we can apply Theorem 1.5.34 for the modulus f OK . We have that PK,Z (f ) = ker(Φf OK ). Definition 1.5.36. Let O be an order of conductor f in an imaginary quadratic field K. The unique Abelian extension L/K, all of whose ramified primes (finite or infinite) divide f OK such that PK,Z (f ) = ker(Φf OK ). is called the ring class field of the order O. The importance of Theorem 1.5.34 is that it allows us to construct Abelian extensions of K with specified Galois group and restricted ramification. Theorem 1.5.37. (The Artin Reciprocity Theorem) Let L/K be an Abelian extension and let m to be a modulus divisible by all (finite and infinite) primes of K that ramify in L. Then: (i) Φm is surjective. (ii) If the exponents of the finite primes dividing m are sufficiently large, then ker(Φm ) is a congruence subgroup for m and consequently the isomorphism ∼ IK (m)/ ker(Φm ) −→ Gal (L/K) shows that Gal (L/K) is a generalized ideal class group for the modulus m. Proof. See Theorem 8.2 from [Cox]. Corollary 1.5.38. Let K be a number field and L, M be Abelian extensions of K. Then, L ⊂ M if and only if there is a modulus m, divisible by all primes of K, ramified in either L or M , such that PK,1 (m) ⊂ ker(ΦM/K,m ) ⊂ ker(ΦL/K,m ). 16 Proof. We assume that L ⊂ M . Let r : Gal (M/K) → Gal (L/K) be the restriction map. By Theorem 1.5.37 there is a modulus m for which ΦL/K,m , ΦM/K,m are both congruence subgroups for m. We can show that r ◦ ΦM/K,m = ΦL/K,m and then ker(ΦM/K,m ) ⊂ ker(ΦL/K,m ) follows immediately. For the other direction assume that PK,1 (m) ⊂ ker(ΦM/K,m ) ⊂ ker(ΦL/K,m ). Then, under the map ΦM/K,m : IK (m) → Gal (M/K) the subgroup ker(ΦM/K,m ) ⊂ IK (m) maps to a subgroup H ⊂ Gal (M/K). By Galois theory H corresponds to an e ⊂ M . Using the first part of the proof it follows that intermediate field K ⊂ L ker(ΦL/K,m ) = ker(ΦL/K,m ). The uniqueness part of Theorem 1.5.34 shows that e e and the proof is complete. (See also the proof of Corollary 8.7 from [Cox]). L=L Chapter 2 Application of Class Field Theory 2.1 The solution of p = x2 + ny 2 for arbitrary n Theorem 2.1.1. Let n > 0 be an integer. There is a monic irreducible polynomial fn (x) ∈ Z[x] of degree h(−4n) such that for every prime p ≥ 3 such that p divides neither n nor the discriminant of fn (x), then  (−n/p) = 1 and 2 2 p = x + ny ⇔ fn (x) = 0 mod p has an integer solution. Furthermore, fn (x) may be taken to be the minimal polynomial of a real √ algebraic −n] in the integer α for which L = K(α) is the ring class field of the order Z[ √ (imaginary quadratic) field K = Q( −n). Finally, if fn (x) is any monic integer polynomial of degree h(−4n) for which the above equivalence holds, then fn (x) is irreducible over Z and is the minimal polynomial of a primitive element of the ring class field L described above. Remark 2.1.2. (i) The last part of the theorem √ shows that knowing fn (x) is equivalent to knowing the ring class field of Z[ −n].   = 1 ⇔ p | x2 + ny 2 , gcd(x, y) = (ii) For historical reasons we mention that −n p 1. Lemma 2.1.3. Let L be the ring class field of an order O in an imaginary quadratic field K. Let τ denote the complex conjugation. Then τ (L) = L and consequently L/Q is Galois extension. 17 18 Proof. Let m denote f OK . We note that τ (f OK ) = τ (f )OK = f OK . Hence, τ (m) = m. Since ker(ΦL/K,m ) = PK,Z (f ) (see Definition 1.5.36) we have that (1) (2) ker(Φτ (L)/K,m ) = τ (ker(ΦL/K,m )) = τ (PK,Z (f )) = PK,Z (f ) = ker(ΦL/K,m ). Equality (1) holds due to the definition of the Artin map. Equality (2): Let αOK ∈ PK,Z (f ). It holds that τ (αOK ) = τ (α)OK . Furthermore, we have that α ≡ a mod f OK for some integer a i.e. α = a + f OK . Thus, τ (α) = τ (a + f OK ) = τ (a) + f OK = a + f OK = α. So τ (αOK ) = αOK and this proves equality (2). Due to corollary 1.5.38 this implies τ (L) = L. From the definition of the ring class field we know that L/K is Galois (see Definition 1.5.36). Also, K/Q is Galois and Gal (K/Q) ∼ = Z/2Z =< τ >. The fact that τ (L) = L implies that τ Gal (L/K)τ −1 = Gal (L/K). Thus, Gal (L/K) is a normal subgroup of AutQ (L). This is equivalent with the fact that L/Q is Galois. Definition 2.1.4. We say that a group G is a semidirect product of a group N by a group H and we write G = N ⋊ H if (i) H is a subgroup of G, (ii) N is a normal subgroup of G, (iii) N ∩ H = {1}, (iv) G = N H. Thus, G = N ⋊ H if and only if we have an exact sequence 1 −→ N −→ G −→ H −→ 1. Lemma 2.1.5. Let L be the ring class field of an order O in an imaginary quadratic field K. Then L/Q is Galois and its Galois group can be written as a semidirect product Gal (L/Q) ∼ = Gal (L/K) ⋊ (Z/2Z) where the nontrivial element of Z/2Z acts on Gal (L/K) by sending σ to σ −1 . Proof. Lemma 2.1.3 implies that L/Q is Galois. So we have an exact sequence: 1 −→ Gal (L/K) −→ Gal (L/Q) −→ Gal (K/Q)(∼ = Z/2Z) −→ 1. Since τ ∈ Gal (L/Q) we have that Gal (L/Q) is the semidirect product Gal (L/K)⋊ (Z/2Z) where the nontrivial element of Z/2Z acts by conjugation by τ . Lemma 1.2.6 (the uniqueness part) implies that for every prime ideal p of OK it holds       L/K L/K L/K τ τ −1 = = . p τ (p) p 19 (For the first equality see Corollary 1.2.9(i)). Thus, under the isomorphism IK (f )/PK,Z (f ) ∼ = Gal (L/K), conjugation by τ in Gal (L/K) corresponds to the usual action of τ on IK (f ). But if a is any ideal in IK (f ), then aa = N (a)OK is in PK,Z (f ) since N (a) is prime to f . Thus, a gives the inverse of a in the quotient IK (f )/PK,Z (f ) and the lemma is proved. √ Theorem 2.1.6. Let n > 0 integer and L √ the ring class field of the order Z[ −n] in the (imaginary quadratic) field K = Q( −n). For every prime p ≥ 3 not dividing n we have, p = x2 + ny 2 ⇔ p splits completely in L. √ Proof. Let O = Z[ −n]. The discriminant of O is −4n, and then −4n = f 2 dK where f is the conductor of O. (See Definition 1.5.3(b) and Remark 1.5.4). Let p be an odd prime not dividing n. Then p does not divide −4n i.e. p does not divide f 2 dk , which implies that p is unramified in K. (See Corollary 1.3.2). We will prove the following equivalences: p = x2 + ny 2 (1) ⇔ pOK = pp, p 6= p and p = αOK , α ∈ O. (2) ⇔ pOK = pp, p 6= p and p ∈ PK,Z (f ) (3) ⇔ pOK = pp, p 6= p and p splits completely in L (4) ⇔ p splits completely in L √ √ Proof√of (1): Suppose that p = x2 + ny 2 = (x + −ny)(x − −ny). If we set p = (x+ −ny)OK , then pOK = pp, p 6= p is the prime√ factorization of pO √ K in OK . p really are prime ideals because N (x+ −ny) = N (x− −ny) = p. Note that p and √ Also note x + −ny ∈ O, and that p 6= p since p is unramified in K. (See Corollary 1.3.2). Conversely,√ suppose that pOK = pp, √ where p = αOK , α ∈ O. √ We can write α = x + −ny. Then p = (x + −ny)OK and p = (x − −ny)OK . Thus pOK = (x2 + ny 2 )OK . It follows easily that p = x2 + ny 2 . Proof of (2): Since p doesn’t divide f , we have p = αOK , α ∈ O ⇔ p ∈ P (O, f ) ⇔ p ∈ PK,Z (f ). The last equivalence follows from Remark 1.5.24. Proof of (3): The isomorphism IK (f )/PK,Z (f) ∼ = Gal (L/K) given by the Artin map shows that p ∈ PK,Z (f ) if and only if From Corollary 1.2.9(iii) we have that completely in L. L/K p  L/K p = 1.  = 1 if and only if with p splits Proof of (4): The first condition says that p splits completely in K and that some prime of K containing p splits completely in L. Because the extension L/Q is 20 Galois (see Lemma 2.1.3) this is equivalent with the second condition. (See Exercise 5.18 from [Cox]). This completes the proof of the theorem. Lemma 2.1.7. Let K be an imaginary quadratic field and let L be a finite extension of K such that L/Q is Galois. Then there is a real algebraic integer α such that L = K(α). Proof. Because L/Q is Galois we first note that [L : Q] = [L : L ∩ R][L ∩ R : Q] = 2[L ∩ R : Q]. [L : Q] = [L : K][K : Q] = 2[L : K]. This implies that, [L ∩ R : Q] = [L : K]. We will prove that for α ∈ L ∩ R it holds L ∩ R = Q(α) ⇔ L = K(α) First we note that K(α) ⊂ L. If we assume that L ∩ R = Q(α) then we have [L:K] [L∩R:Q] [Q(α):Q] that [L : K(α)] = [K(α):K] = [K(α):K] = [K(α):K] = 1. For the other direction we note that Q(α) ⊂ L ∩ R. If we assume that L = K(α) then we have that [L:K] [L∩R:Q] = [Q(α):Q] = [K(α):K] [L ∩ R : Q(α)] = [Q(α):Q] [Q(α):Q] = 1. Now, if α ∈ OL ∩ R satisfies L ∩ R = Q(α) then α is a real integral primitive element of L over K and the proof of the Lemma is complete. Lemma 2.1.8. Let K, L, α be as in Lemma 2.1.7. Let f (x) ∈ Z[x] be the minimal monic polynomial of α over Q and let p be a prime not dividing the discriminant of f (x). Then, (   dK = 1 and p p splits completely in L ⇔ f (x) ≡ 0 mod p has an integer solution. Proof. Since [Q(α) : Q] = [K(α) : K], f (x)  ∈  Z[x] is also the minimal polynodK mial of α over K. Moreover, note that = 1 ⇔ p splits completely in K p (see Corollary 1.3.2). So we may assume that p splits completely in K. This implies that pOK = pp, p 6= p, since K is an imaginary quadratic field. Thus, Z/pZ ∼ = OK /p. Note now, that f (x) ≡ 0 mod p has an integer solution ⇔ f (x) ≡ 0 mod p has a solution in OK . Hence, we have to prove that p splits completely in L ⇔ f (x) ≡ 0 mod p has a solution in OK . Let f = f 1 · · · · · f t be the factorization of f mod p over OK /p. If we assume that f (x) ≡ 0 mod p has a solution in OK then one f i is of degree 1. But then, because L/Q is Galois, all f i are of degree 1. Thus, p splits completely in L i.e. also p splits completely in L. Hence, p splits completely in L. If we assume that p splits completely in L then all f i are of degree 1. Thus, f (x) ≡ 0 mod p has a solution in OK and the proof is complete. 21 Proof. (of Theorem 2.1.1) By Lemma 2.1.3 the extension L/Q is finite and Galois. So there is a real algebraic integer α such that L = K(α) (see Lemma 2.1.7). Let fn (x) ∈ Z[x] be the minimal polynomial of α over K. Since O has discriminant −4n (see Remark 1.5.4) the degree of fn (x) is [L : K] = h(O) = h(−4n) (see Theorem 1.5.12). Then, if p is an odd prime dividing neither n nor the discriminant of fn (x), by combining Theorem 2.1.6 with Lemma 2.1.8 we have that 2 2 p = x + ny ⇔ (  −n p  = 1 and fn (x) ≡ 0 mod p has an integer solution. To prove the last part of the theorem, let fn (x) ∈ Z[x] be a monic polynomial of degree h(−4n) which satisfies the equivalence of Theorem 2.1.1. Let g(x) ∈ K[x] be an irreducible factor of fn (x) over K, and let M = K(α) be the field generated by a root α of g(x). Note that α is an algebraic integer (since is also a root of f n (x)). If we can show that L ⊂ M , then h(−4n) = [L : K] ≤ [M : K] = deg(g(x)) ≤ deg(fn (x)) = h(−4n), which will prove that L = M = K(α) and that fn (x) is the minimal polynomial of α over K and hence over Q. It remains to prove that L ⊂ M . Since L/Q is Galois (see Lemma 2.1.3) it suffices ˙ L/Q (see Proposition 1.4.4). to prove that SeM/Q ⊂S Using Theorem 2.1.6 we have that SL/Q = {p prime | p = x2 + ny 2 }. Let Σ be the set of primes p who divide the discriminant of fn (x) or n. Since fn (x) satisfies the equivalence of Theorem 2.1.1 it follows that Σ is a finite set and that SL/Q \ Σ is the set of primes p which split completely in K and for which fn (x) ≡ 0 mod p has a solution. ˙ L/Q suppose that p ∈ SeM/Q . Then fP|p = 1 for some ideal To prove that SeM/Q ⊂S P of L. If we set p = P ∩ OK , then 1 = fP|p = fP|p fp|p . Thus, fp|p = 1, which implies that p splits completely in K (since it’s unramified). Note also that fn (x) ≡ 0 mod P has a solution in OM since α ∈ OM and g(α) = fn (α) = 0. But fp|p = 1 implies that [OM /P : Z/pZ] = 1 i.e. OM /P ∼ = Z/pZ, and hence fn (x) ≡ 0 mod p has an integer solution. This implies that p ∈ SL/Q . This proves ˙ L/Q and completes the proof of the theorem. that SeM/Q ⊂S 2.2 How our result works in practice We now are ready to demonstrate how Theorem 2.1.1 solves the problem we stated at the beginning for specific n. We will determine which primes are represented by x2 + 27y 2 or x2 + 64y 2 . 22 2.2.1 Primes of the form x2 + 27y 2 The first step is to determine which ring class field is involved: √ √ Proposition 2.2.1. The ring class field of the order Z[ −27] in K = Q( −3) is √ 3 L = K( 2). To prove the proposition we will use the following facts: √ Remark 2.2.2. Let L be the ring class field of K = Q( −3). (i) L/K is a cubic Galois extension since [L : K] = h(−4 · 27) = 3. Thus, Gal (L/K) ∼ = Z/3Z. ∼ S3 . This follows from Lemma 2.1.5 since (ii) L/Q is Galois and Gal (L/Q) = S3 ∼ = (Z/3Z) ⋊ (Z/2Z) with Z/2Z acting non-trivially. √ √ (iii) Z[ −27] = Z[3 −3] is an order of conductor 6 since −3 ≡ 1 mod 4, so L corresponds to a generalized ideal class group of the modulus 6O K . Thus, all primes ideals of OK that ramify in L must divide 6OK . √ Lemma 2.2.3. If M is a cubic extension of K = Q( −3) with Gal (M/Q) ∼ = S3 , √ then M = K( 3 m) for some cube-free positive integer m. Proof. Kummer Theory. See Lemma 9.6 from [Cox]. √ Lemma 2.2.4. Let K be an imaginary quadratic field and L = K( 3 m) be a cubic extension of K, where m is cube-free positive integer. Let p be a prime ideal of O K . Then p ramifies in L. Proof. See Exercise 9.5 from [Cox]. √ So the ring class field of K is in the form L = K( 3 m) for some cube-free positive integer m. Now, we will use the ramification of L/K to restrict m. From Remark 2.2.2(iii) we know that all ramified primes divide 6OK and consequently 2 and 3 are the only integer primes that can divide m. Since m is also positive and cube-free it must be one of the following eight numbers: 2, 3, 4, 6, 9, 12, 18, 36. √ √ √ √ √ √ √ 3 3 3 3 3 3 Note that K( 3 2) ∼ = K( 4), K( 3) ∼ = K( 9), K( 6) ∼ = K( 36), K( 12) ∼ = √ 3 K( 18). Hence, L must be one of the following four fields: √ √ √ √ 3 3 3 3 K( 2), K( 3), K( 6), K( 12). All four fields satisfy the conditions given in Remark 2.2.2. We will use numerical calculations concerning the polynomial f27 (x) = x3 − m to determine L. We have 23 the following cases: m Discriminant of f27 (x) 2 −22 · 33 3 −35 6 −22 · 35 12 −24 · 35 We note that the prime number 31 is of the form x2 + 27y 2 since it holds 31 = 22 + 27 · 12 . Moreover, 31 divides neither 27 nor the discriminant of f27 (x) in none of the cases. By Theorem 2.1.1 the congruence x3 ≡ m mod 31 must have a solution. It√is easy to check that this is true only for the case m = 2. It follows that L = K( 3 2) as claimed. We get the following characterization of when p = x2 + 27y 2 : Theorem 2.2.5. Let p ≥ 5 be a prime.  p ≡ 1 mod 3 and p = x2 + 27y 2 ⇔ x3 ≡ 2 mod p has an integer solution. √ √ Proof. By Proposition 2.2.1 the ring class field of Z[ −27] is L = K( 3 2). Since √ 3 2 is a real algebraic integer, the polynomial f27 (x) may be taken to be x3 − 2. Since the discriminant of f27 (x) is −22 · 33 (see above)  the only excluded primes   are 2 and 3. Finally, we note that the condition −27 = 1 is equivalent to p i.e. to the congruence p ≡ 1 mod 3 and the proof is complete. 2.2.2 −3 p =1 Primes of the form x2 + 64y 2 The first step is to determine which ring class field is involved: √ Proposition 2.2.6. The ring class field of the order Z[ −64] in K = Q(i) is √ L = K( 4 2). To prove the proposition we will use the following facts: Remark 2.2.7. Let L be the ring class field of K = Q(i). (i) L/K is Galois and [L : K] = h(−4 · 64) = 4. (ii) L/Q is Galois and Gal (L/Q) ∼ = S4 . This follows from Lemma 2.1.5 since S4 ∼ = (Z/4Z) ⋊ (Z/2Z) with Z/2Z acting non-trivially. √ (iii) Z[ −64] = Z[8i] is an order of conductor 8 since −1 ≡ 3 mod 4, so L corresponds to a generalized ideal class group of the modulus 8O K . Thus, all prime ideals of OK that ramify in L must divide 8OK . √ Lemma 2.2.8. If M/K is a cyclic extension of degree 4 then M = K( 4 m) for some fourthpower-free integer m. 24 Proof. Kummer theory. See Proposition 10.1 of [Mil]. √ So the ring class field of K is in the form L = K( 4 m) for some fourthpower-free positive integer m. Now, we will use the ramification of L/K to restrict m. From Remark 2.2.7(iii) we know that all ramified primes divide 8OK and consequently 2 is the only integer prime that can divide m. Since m is also positive and fourthpower-free it must be one of the following three numbers: 2, 4, 8. √ It follows that L = K( 4 2) as claimed since in all cases we get the same field. We get the following characterization of when p = x2 + 64y 2 : Theorem 2.2.9. Let p ≥ 3 be a prime.  p ≡ 1 mod 4 and p = x2 + 64y 2 ⇔ x4 ≡ 2 mod p has an integer solution. √ √ Proof. By Proposition 2.2.6 the ring class field of Z[ −64] is L = K( 4 2). Since √ 4 2 is a real algebraic integer, the polynomial f64 (x) may be taken to be x4 − 2. 11 Since the discriminant of prime is 2. Finally, we    f64(x) is −2 the only excluded −64 = 1 is equivalent to −1 = 1 i.e. to the congruence note that the condition p p p ≡ 1 mod 4 and the proof is complete.