Primes of the form x2 + ny2

Marios Magioladitis

Arbeitsgemeinschaft zur Klassenkorpertheorie

Institute for Experimental Mathematics (IEM)University of Duisburg-Essen

June 2004

2

Arbeitsgemeinschaft zur Klassenkorpertheorie. Organisation: G. Bockle, I. Bouw,C. Diem, G. Frey. Sommersemester 2004.

Lecture 9. Primzahlen der Form x2 + ny2 was given on 29th June 2004 in theInstitute of Experimental Mathematics (IEM).

I would like to thank Prof. David A. Cox (Amherst College) for his immediatereplys to my e-mails.

Marios Magioladitis Homepage: http://www.exp-math.uni-essen.de/˜magiolad E-mail: marios@math.uoc.gr

References

[Cox] D. Cox, Primes of the form x2 + ny2, 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 51.1 Introduction and some historical background . . . . . . . . . . . . . 51.2 Elements of Hilbert Class Field Theory . . . . . . . . . . . . . . . . 61.3 Elements of Quadratic Field Theory . . . . . . . . . . . . . . . . . . 81.4 Some more background on Field Theory . . . . . . . . . . . . . . . . 91.5 Orders in Quadratic Fields . . . . . . . . . . . . . . . . . . . . . . . 10

2 Application of Class Field Theory 172.1 The solution of p = x2 + ny2 for arbitrary n . . . . . . . . . . . . . . 172.2 How our result works in practice . . . . . . . . . . . . . . . . . . . . 21

2.2.1 Primes of the form x2 + 27y2 . . . . . . . . . . . . . . . . . . 222.2.2 Primes of the form x2 + 64y2 . . . . . . . . . . . . . . . . . . 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 inthe form

p = x2 + ny2

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 somecases where Genus Theory fails but doesn’t solve the problem for arbitraryn.

We can write p = x2 +ny2 = (x+√−ny)(x−√−ny). This leads us to restate the

question

Question 1.1.2. Let n be a positive integer. Which primes p can be expressed inthe 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 offractional ideals of OK , and with PK the subgroup of principal fractional ideals i.e.those of the form αOK where α ∈ K∗. We define

Pic(OK) := IK/PK

the (ideal) class group of K. (Pic(OK) is also denoted as ClK in [Neu] and asC(OK) in [Cox]).

Remark 1.2.2. Pic(OK) is a finite group.

Definition 1.2.3. Let L/K be a Galois extension and let p be a prime ideal ofOK . Let P be a prime ideal of OL containing p. The inertial degree of p in Pis 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 ofOK . Let

pOK = Pe1 · · · · ·Pe

k,Pi distinct prime ideals of OLbe 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/Ksuch that: L is the maximum unramified Abelian extension of K. This unique fieldL 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 OKwhich is unramified in L. If P is a prime ideal of OL containing p, then there is aunique 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 element σ of the previous Lemma is called the Artin sym-

bol and is denoted(L/K

P

)since it depends on the prime ideal P of OL.

7

Remark 1.2.8. For every α ∈ OL we have

(L/K

P

)(α) ≡ αN(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 primeideal of OK . Given a prime ideal P of OL containing p, we have:

(i) If σ ∈ Gal (L/K), then

(L/K

σ(P)

)= σ

(L/K

P

)σ−1.

(ii) The order of(L/K

P

)is the inertial degree fP|p.

(iii) p splits completely in L⇔(L/K

P

)= 1.

Proof. Omitted. For (iii) just note that e = 1 since p is unramified in L thus theequivalence 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

as(L/K

p

)since it is independed from the prime ideal P of OL.

Proof. Let P and P′ be two prime ideals of OL containing p. Then P′ = σ(P) forsome σ ∈ Gal (L/K). The last corollary implies that

(L/K

P′

)=

(L/K

σ(P)

)= σ

(L/K

P

)σ−1 =

(L/K

P

).

Definition 1.2.11. Let L/K be an unramified Abelian extension and let a ∈ IK bea fractional ideal with prime factorization

a =

r∏

i=1

piri , ri ∈ Z.

We define the Artin symbol(L/K

a

)to be the product

(L/K

a

)=

r∏

i=1

(L/K

pi

)ri

8

Definition 1.2.12. Let L/K be an unramified Abelian extension. The Artin sym-bol defined above defines a homomorphism of groups, called the Artin (or reci-procity) map, (

L/K

.

): IK → Gal (L/K).

Theorem 1.2.13. (The Artin reciprocity theorem for the Hilbert classfield) 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 isomor-phism

Pic(OK)∼−→ Gal (L/K).

Remark 1.2.14. Obviously: |Pic(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

OK =

{Z[√N ] if N 6≡ 1 mod 4,

Z[√N+12 ] if N ≡ 1 mod 4.

(2) The extensionK/Q is Abelian since Gal (K/Q) ∼= Z/2Z. Thus |Gal (K/Q)| =2.

(3) The discriminant dK of K is defined to be

dK =

{N if N ≡ 1 mod 4,4N otherwise.

Thus, we can write OK = Z[dK+

√dK

2

].

(4) If N < 0 the group of units of OK is

O∗K =

±1,±i if N = −1,±1,±ω,±ω2 if N = −3,±1 otherwise.

9

Proposition 1.3.1. Let K be a quadratic field of discriminant dK and let p be aninteger prime.

(i) If(dKp

)= 0 then pOK = p2, where p is a prime ideal of OK .

(ii) If(dKp

)= 1 then pOK = pp, where p 6= p are prime ideals of OK .

(iii) If(dKp

)= −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 aninteger prime. Then:

(i) p ramifies in K ⇔ p divides dK

(ii) p splits completely in K ⇔(dKp

)= 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 inthe second chapter. We will show that if L/K is an extension then the prime idealsof OK which split completely in L characterize the extension uniquely.

Definition 1.4.1. Let S and T be two sets. We say that S⊂T and and only ifS ⊂ T ∪ Σ for some finite set Σ. Furthermore, S .

= T will mean that S⊂T andT ⊂S.

Definition 1.4.2. Let L/K be a finite extension and let PK be the set of all primeideals of OK . We define

SL/K := {p ∈ PK | p splits completely in L}.

SL/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}.

SL/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

(i) If M is Galois over K, then L ⊂M ⇔ SM/K⊂SL/K .

(ii) If L is Galois over K, then L ⊂M ⇔ SM/K⊂SL/K .(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 freeZ-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 indexf = [OK : O]

is called the conductor of the order.

(b) The discriminant D of O is equal to

D := f2dK ,

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+ fOK .

(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 +fOK 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 frac-tional O-ideal. Then,

a proper⇔ a invertible

Proof. Let a be an invertible ideal of O. Then ab = O for some fractional ideal bof 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 7.4and Lemma 7.5 from [Cox].

Definition 1.5.9. Let O be an order in a quadratic field K. We define the followinggroup

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

Pic(O) := I(O)/P (O)

the (ideal) class group of the order O. (The quotient group is also called thePicard 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 quadraticfield 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 reducedforms of D).

12

Theorem 1.5.12. Let O be an order in an imaginary quadratic field K. Then|Pic(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 forinvertible 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 fractionalideals of O are principal (see Exercise 7.9(b) of [Cox]) but O is not a UFD sincefor example 4 = 2 · 2 = (1 +

√−3)(1−

√−3).

We will define a slightly smaller class of ideals, those prime to the conductor, forwhich unique factorization holds.

Definition 1.5.14. Let O be an order of conductor f in a quadratic field K. Wesay that a nonzero ideal a of O is prime to f if a + fO = O.

Definition 1.5.15. Let a be an ideal of an order O. The norm of a is defined tobe

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, wehave

βO = β(a + fO) = βa + βfO ⊂ a + fOK ⊂ a +O ⊂ Obecause fOK ⊂ 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 principalideals 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).

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(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) thesubgroup of IK(f) generated by principal ideals of the form αOK where α ∈ OKand α ≡ a mod fOK for some a ∈ Z relatively prime to f . Thus,

PK,Z(f) =< αOK | α ∈ OK , exists a ∈ Z s.t. α ≡ a mod fOK , gcd(a, f) = 1 > .

Remark 1.5.21. Let O be an order of conductor f in an imaginary quadratic fieldK then

IK(f) ∼= I(O, f).

Proof. The map a 7→ a ∩ O preserves multiplication and induces an isomorphismIK(f)

∼−→ I(O, f), and the inverse of this map is given by a 7→ aOK . For a detailedproof see Proposition 7.20 from [Cox].

Proposition 1.5.22. Let O be an order of conductor f in an imaginary quadraticfield K. Every ideal of O prime to f has a unique decomposition as a product ofprime ideals of O which are prime to f .

Proof. See Proposition 7.20 and Exercise 7.26 from [Cox].

We can now describe Pic(O) in terms of the maximal order.

Proposition 1.5.23. Let O be an order of conductor f in an imaginary quadraticfield K. Then there are natural isomorphisms

Pic(O) ∼= I(O, f)/P (O, f) ∼= IK(f)/PK,Z(f).

Proof. For the first isomorphism: Note that the map I(O, f) −→ Pic(O) is surjec-tive since every ideal class in Pic(O) contains an ideal of O prime to f . Alsonote that the kernel of this map is I(O, f) ∩ P (O). This obviously containsP (O, f). For the inclusion I(O, f) ∩ P (O) ⊂ P (O, f) consider that an elementin I(O, f)∩P (O) is a fractional ideal αO = ab−1 where α ∈ K∗ and a, b are primeto 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 isomorphismIK(f) ∼= I(O, f) (see Remark 1.5.21). Under this isomorphism P (O, f) maps to asubgroup P ′ of IK(f). For the proof that P ′ = PK,Z(f) see proof of Proposition7.22 from [Cox].

Remark 1.5.24. Let O be an order of conductor f in an imaginary quadratic fieldK then

P (O, f) ∼= PK,Z(f).

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Definition 1.5.25. Let K be a number field. We will call finite primes the primeideals of OK and infinite primes the embeddings of K into C. A real infiniteprime is an embedding σ : K → R and a complex infinite prime is a pair ofcomplex conjugate embeddings σ, σ : K → C, with σ 6= σ.

Definition 1.5.26. Let K be a number field. A modulus m in K is a formalproduct

m =∏

p

pnp

over all primes p (finite or infinite) of K, where the exponents must satisfy thefollowing 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 = m0m∞ wherem0 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-idealsrelatively prime to m (i.e. to m0) and by PK,1(m) the subgroup of IK(m) generatedby the principal ideals of the form αOK where α ∈ OK and α ≡ 1 mod m0 andσ(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(fOK) =< αOK | α ∈ OK , α ≡ 1 mod fOK > .

Remark 1.5.29.

PK,1(fOK) ⊂ PK,Z(f) ⊂ IK(f) = IK(fOK).

Definition 1.5.30. Let m be a modulus of a field K. A congruence subgroupof m is a group H where

PK,1(m) ⊂ H ⊂ IK(m).

Remark 1.5.31. PK,Z(f) is a congruence subgroup for fOK. (See Remark 1.5.29).

Definition 1.5.32. Let m be a modulus of a field K and let H be its congruencesubgroup. 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 modulusdivisible by all ramified primes of the extension. We can extend the Artin symbolby 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 Φm asΦL/K,m to refer explicitly to the extension.

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Theorem 1.5.34. Let m be a modulus of a field K, and let H be a congruencesubgroup for m. Then there is a unique Abelian extension L/K, all of whose ramifiedprimes (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 primesdivide fOK. Then,

IK(f)/PK,Z(f) ∼= Gal (L/K).

Proof. PK,Z(f) is a congruence subgroup for fOK (see Remark 1.5.31) so we canapply Theorem 1.5.34 for the modulus fOK . We have that PK,Z(f) = ker(ΦfOK ).

Definition 1.5.36. Let O be an order of conductor f in an imaginary quadraticfield K. The unique Abelian extension L/K, all of whose ramified primes (finite orinfinite) divide fOK such that

PK,Z(f) = ker(ΦfOK ).

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 exten-sions of K with specified Galois group and restricted ramification.

Theorem 1.5.37. (The Artin Reciprocity Theorem) Let L/K be an Abelianextension and let m to be a modulus divisible by all (finite and infinite) primes ofK that ramify in L. Then:

(i) Φm is surjective.

(ii) If the exponents of the finite primes dividing m are sufficiently large, thenker(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 ofK. 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 restrictionmap. By Theorem 1.5.37 there is a modulus m for which ΦL/K,m,ΦM/K,m are bothcongruence subgroups for m. We can show that r ◦ ΦM/K,m = ΦL/K,m and thenker(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

intermediate field K ⊂ L ⊂ M . Using the first part of the proof it follows thatker(ΦL/K,m) = ker(ΦeL/K,m). The uniqueness part of Theorem 1.5.34 shows that

L = L and the proof is complete. (See also the proof of Corollary 8.7 from [Cox]).

Chapter 2

Application of Class FieldTheory

2.1 The solution of p = x2 + ny2 for arbitrary n

Theorem 2.1.1. Let n > 0 be an integer. There is a monic irreducible polynomialfn(x) ∈ Z[x] of degree h(−4n) such that for every prime p ≥ 3 such that p dividesneither n nor the discriminant of fn(x), then

p = x2 + ny2 ⇔{

(−n/p) = 1 andfn(x) = 0 mod p has an integer solution.

Furthermore, fn(x) may be taken to be the minimal polynomial of a real algebraicinteger α for which L = K(α) is the ring class field of the order Z[

√−n] in the(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 minimalpolynomial 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 knowingthe ring class field of Z[

√−n].

(ii) For historical reasons we mention that(−np

)= 1⇔ p | x2 +ny2, gcd(x, y) =

1.

Lemma 2.1.3. Let L be the ring class field of an order O in an imaginary quadraticfield K. Let τ denote the complex conjugation. Then τ(L) = L and consequentlyL/Q is Galois extension.

17

18

Proof. Let m denote fOK . We note that τ(fOK) = τ(f)OK = fOK . Hence,τ(m) = m. Since

ker(ΦL/K,m) = PK,Z(f)

(see Definition 1.5.36) we have that

ker(Φτ(L)/K,m)(1)= τ(ker(ΦL/K,m)) = τ(PK,Z(f))

(2)= 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 ≡ amod fOK for some integer a i.e. α = a + fOK . Thus, τ(α) = τ(a + fOK) =τ(a) + fOK = a+ fOK = 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 Definition1.5.36). Also, K/Q is Galois and Gal (K/Q) ∼= Z/2Z =< τ >. The fact thatτ(L) = L implies that τGal (L/K)τ−1 = Gal (L/K). Thus, Gal (L/K) is a normalsubgroup 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 Nby a group H and we write G = N oH if

(i) H is a subgroup of G,

(ii) N is a normal subgroup of G,

(iii) N ∩H = {1},(iv) G = NH.

Thus, G = N oH 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 quadraticfield K. Then L/Q is Galois and its Galois group can be written as a semidirectproduct

Gal (L/Q) ∼= Gal (L/K)o (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)o(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 itholds

τ

(L/K

p

)τ−1 =

(L/K

τ(p)

)=

(L/K

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 inGal (L/K) corresponds to the usual action of τ on IK(f). But if a is any ideal inIK(f), then aa = N(a)OK is in PK,Z(f) since N(a) is prime to f . Thus, a givesthe 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 + ny2 ⇔ p splits completely in L.

Proof. Let O = Z[√−n]. The discriminant of O is −4n, and then −4n = f 2dK

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 notdivide f2dk, which implies that p is unramified in K. (See Corollary 1.3.2).

We will prove the following equivalences:

p = x2 + ny2

(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 + ny2 = (x +√−ny)(x −√−ny). If we set

p = (x+√−ny)OK , then pOK = pp, p 6= p is the prime factorization of pOK in OK .

Note that p and p really are prime ideals becauseN(x+√−ny) = N(x−√−ny) = p.

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 writeα = x +

√−ny. Then p = (x +√−ny)OK and p = (x − √−ny)OK . Thus

pOK = (x2 + ny2)OK . It follows easily that p = x2 + ny2.

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(L/K

p

)= 1.

From Corollary 1.2.9(iii) we have that(L/K

p

)= 1 if and only if with p splits

completely in L.

Proof of (4): The first condition says that p splits completely in K and thatsome 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 Exercise5.18 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 extensionof K such that L/Q is Galois. Then there is a real algebraic integer α such thatL = 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

that [L : K(α)] = [L:K][K(α):K] = [L∩R:Q]

[K(α):K] = [Q(α):Q][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 ∩ R : Q(α)] = [L∩R:Q][Q(α):Q] = [L:K]

[Q(α):Q] = [K(α):K][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 proofof 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 minimalmonic polynomial of α over Q and let p be a prime not dividing the discriminantof f(x). Then,

p splits completely in L⇔{ (

dKp

)= 1 and

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 polyno-

mial of α over K. Moreover, note that(dKp

)= 1 ⇔ p splits completely in K

(see Corollary 1.3.2). So we may assume that p splits completely in K. Thisimplies 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 = f1 · · · · · 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 completelyin 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) ≡ 0mod 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) (seeTheorem 1.5.12). Then, if p is an odd prime dividing neither n nor the discriminantof fn(x), by combining Theorem 2.1.6 with Lemma 2.1.8 we have that

p = x2 + ny2 ⇔{ (

−np

)= 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 ofdegree h(−4n) which satisfies the equivalence of Theorem 2.1.1. Let g(x) ∈ K[x] bean irreducible factor of fn(x) over K, and let M = K(α) be the field generated bya root α of g(x). Note that α is an algebraic integer (since is also a root of fn(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 ofα 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

to prove that SM/Q⊂SL/Q (see Proposition 1.4.4).

Using Theorem 2.1.6 we have that SL/Q = {p prime | p = x2 + ny2}. Let Σ bethe set of primes p who divide the discriminant of fn(x) or n. Since fn(x) satisfiesthe equivalence of Theorem 2.1.1 it follows that Σ is a finite set and that SL/Q \Σis the set of primes p which split completely in K and for which fn(x) ≡ 0 mod phas a solution.

To prove that SM/Q⊂SL/Q suppose that p ∈ SM/Q. Then fP|p = 1 for some idealP of L. If we set p = P ∩ OK , then 1 = fP|p = fP|pfp|p. Thus, fp|p = 1,which implies that p splits completely in K (since it’s unramified). Note also thatfn(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 hencefn(x) ≡ 0 mod p has an integer solution. This implies that p ∈ SL/Q. This proves

that SM/Q⊂SL/Q and completes the proof of the theorem.

2.2 How our result works in practice

We now are ready to demonstrate how Theorem 2.1.1 solves the problem we statedat the beginning for specific n. We will determine which primes are represented byx2 + 27y2 or x2 + 64y2.

22

2.2.1 Primes of the form x2 + 27y2

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

L = K( 3√

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.

(ii) L/Q is Galois and Gal (L/Q) ∼= S3. This follows from Lemma 2.1.5 sinceS3∼= (Z/3Z)o (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 6OK . Thus, allprimes 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 OK .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 conse-quently 2 and 3 are the only integer primes that can divide m. Since m is alsopositive and cube-free it must be one of the following eight numbers:

2, 3, 4, 6, 9, 12, 18, 36.

Note that K( 3√

2) ∼= K( 3√

4),K( 3√

3) ∼= K( 3√

9),K( 3√

6) ∼= K( 3√

36),K( 3√

12) ∼=K( 3√

18).Hence, L must be one of the following four fields:

K(3√

2),K(3√

3),K(3√

6),K(3√

12).

All four fields satisfy the conditions given in Remark 2.2.2. We will use numericalcalculations 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 + 27y2 since it holds 31 =22 + 27 · 12. Moreover, 31 divides neither 27 nor the discriminant of f27(x) innone of the cases. By Theorem 2.1.1 the congruence x3 ≡ m mod 31 must havea solution. It is easy to check that this is true only for the case m = 2. It followsthat L = K( 3

√2) as claimed.

We get the following characterization of when p = x2 + 27y2:

Theorem 2.2.5. Let p ≥ 5 be a prime.

p = x2 + 27y2 ⇔{p ≡ 1 mod 3 andx3 ≡ 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(−27p

)= 1 is equivalent to

(−3p

)= 1

i.e. to the congruence p ≡ 1 mod 3 and the proof is complete.

2.2.2 Primes of the form x2 + 64y2

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 sinceS4∼= (Z/4Z)o (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 8OK. Thus, allprime 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 conse-quently 2 is the only integer prime that can divide m. Since m is also positive andfourthpower-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 + 64y2:

Theorem 2.2.9. Let p ≥ 3 be a prime.

p = x2 + 64y2 ⇔{p ≡ 1 mod 4 andx4 ≡ 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.Since the discriminant of f64(x) is −211 the only excluded prime is 2. Finally, we

note that the condition(−64p

)= 1 is equivalent to

(−1p

)= 1 i.e. to the congruence

p ≡ 1 mod 4 and the proof is complete.