Primes of the form x 2 + ny 2. Matthew Bates

Transcription

1 Primes of the form x 2 + ny 2 Matthew Bates

2 Contents 1 The equation Necessity The equation Old knowledge A useful homomorhism Quadratic forms Definitions Proerties of the equivalence classes Class number Class grou Genus theory The need Foundations Proerties Convenient numbers Brief history Counting convenient numbers Class field theory Foreword Preliminaries Quadratic fields The beginning of class field theory Infinitely many n Ring class field theory Relationshi between orders and quadratic forms 41 2

3 Abstract This essay will address the question of which rimes can be reresented as x 2 + ny 2, with x, y Z, for different values of n N. One aim of this essay is to classify all such rimes using simle congruence relations. i.e. to find β 1,..., β r Z and K Z such that, = x 2 + ny 2 β 1,..., β r (mod K) for every n. (I also hoe to show that such a relation exists in the first lace!) Along the way I hoe to introduce the reader to some interesting areas of mathematics. Desite the question s simle aearance, finding the answer is far from simle. To get us into the mood and for logical reasons, we will begin at the beginning (n = 1, 2); with little effort we will see that these cases can be comletely solved using quadratic recirocity. Unfortunately, such a method fails in general (in fact, it will fail for all n 1, 2), hence a new aroach is required. We consider some general theory of quadratic forms in two variables over Z. This will hel us to solve a few more secial cases, most of which were considered and roved by Fermat over 400 years ago; in order to not be beaten by a 400-year-old Frenchman we develo some genus theory of quadratic forms. This will give a comlete solution for many values of n, Euler s so-called convenient numbers (a number is convenient when its rincial genus consists of a single class, whatever that means!). We will then consider how many convenient numbers there are; ultimately coming to the disaointing realisation that there are only 65 (or maybe 66?). Not to be disheartened by the limitations of quadratic form theory in solving our question we ress on. It is clear that to solve the roblem in general we need a new aroach, the new aroach being class field theory. Using Hilbert class field theory we derive a solution for infinitely many n, but not all! Unfortunately, the solution we get is non-constructive, in the sense that we rove that there exists desired congruence relations, but we do not know what they are. To go any further one requires a greater understanding of class field theory and knowledge of modular functions and comlex multilication. 3

4 1 The equation 1.1 Necessity Let us suose that the equation = x 2 +ny 2 with rime, has integer solutions for x, y. Then it is clear that x 2 + ny 2 0 (mod ), and since Z/Z is a field, ( ) 2 x we see n y (mod ) (since y), so n is a quadratic residue modulo. We adot the following standard notation for quadratic residues. Definition 1.1. Legendre symbol Let a Z be odd and an odd rime; then we define the Legendre symbol ( ) 0 a a = 1 a and a is a quadratic residue modulo 1 a and a is not a quadratic residue modulo Using this notation the following set inclusion is easily verbalised. { rime : = x 2 + ny 2 for some x, y Z } { ( ) } n : = 1 (1.1) { ( ) } n The set rime : = 1 is easily determined using quadratic recirocity. Theorem 1.2. (Quadratic Recirocity) [4] Let, q Z be distinct odd rimes, then we have ( ) ( ) = ( 1) ( 1)(q 1) q 4, q Accomanying the law of quadratic recirocity we have the following two sulementary results. Theorem 1.3 (Sulementary Recirocity Laws). First Sulement ( ) 1 = 1 1 (mod 4). Second Sulement ( ) 2 = 1 1, 7 (mod 8). This is an essential tool which we will use throughout this essay. The theorem was conjectured by Euler and Legendre, but they could not rove it. It is notably hard to rove. The first roof (and the following five) were given by Gauss; I don t know these roofs, but I imagine they are not very illuminating. Here is an esecially nice modern roof which makes use of some Galois theory. 4

5 Proof. [5] Let be if 1 (mod 4) and if 3 (mod 4), so that we have = ± 1 (mod 4). Let ζ be a rimitive th root of unity in C and α a rimitive th root of unity in an extension of F q. Such an α exists since F q is a erfect field, hence x 1 has no reeated roots. Let S (Z/Z) be the grou of squares. Hence, S is the unique cyclic subgrou of (Z/Z) of order 1 2 Ṅote q is a square modulo if and only if (q (mod )) mas to the trivial element under the rojection (Z/Z) (Z/Z) S. We will now show that Q( ) is contained in Q[ζ ]. Remark 1.4. This can be seen as a articular instance of the Kronecker Weber theorem, which states that every finite abelian extension of Q is contain in a cyclotomic extension. [6] Proof. (continued) We have Q(ζ ) S = Q( ), where Q(ζ ) S denote the fixed field under S. Let η = k S ζ k. Note for all l S, we have σ l (η) = k S ζ lk = k S ζ k. Hence we see Q(ζ) S Q(η). We now want to show that η is a root of x 2 + x + 1 4, for then we have η = 1± 2, and thus, Q(η) = Q( ). Let T = (Z/Z) S, and η = k T ζk. Now η = σ l (η) for l T, so η is the conjugate of η. Thus η is quadratic. We have that fη Q Q[x] has degree 2 and roots η and η. Thus, f Q η (x) = (x η)(x η) = x 2 (η + η)x + η η Note, η + η = k ζk = 1. Now consider η η; we have two cases corresonding to 1, 3 (mod 4). If 1 (mod 4), then 1 S, so we have a S imlies a S. One sees that η η is a Cauchy roduct and thus has the form 1 k=0 a kζ k. We have a 0 = 0 since there is no s S t T such that s + t 0 (mod ). Thus, 1 η η = a k ζ k k=1 Note, all the a k are equal since η η is fixed under all automorhisms. Furthermore, a k = ( ) Hence, one calculates that all the 1 ak equal 4. Thus, η η = k=1 ζk = 1 4 = 1 4, as required. Use a simler argument for 3 (mod 4). Hence Q(ζ) S Q(η) = Q( ). Finally we see Q(ζ) S = Q(η) since S has index 2 in the Galois grou. We are now ready to finish the roof of quadratic recirocity. Recall, q is a square modulo if and only if (q (mod )) S if and only if σ q fixes. 5

6 Consider the following ring homomorhism, r : Z[ζ] F q [α] ai ζ i (a i (mod q))α i I claim r σ q = Frob r. This is seen by exlicitly calculating the images of ζ, i.e, r σ q (ζ) = r(ζ q ) = α q = Frob(α) = Frob r(ζ), as required. We can see r as a ma from Gal(Q(ζ)/Q) to Gal(F q (α)/f q ). Let τ = 2η+1 = in Z[ζ]. Then q is a square modulo if and only if σ q (τ) = τ (since σ q (τ) = ±τ) if and only if Frob r(τ) = r(τ) if and only if r(τ) F q. Note that r(τ) 2 = r(τ 2 ) = r( ) = (mod q). So r(τ) F q if and only if is a square in F q. This ends the roof of quadratic recirocity. When considering the set { rime : ( n ) = 1}, I claim it is enough to consider (mod 4n). For examle, using quadratic recirocity, one calculates ( ) 1 = 1 1 (mod 4), ( ) 2 = 1 1, 3 (mod 8), ( ) 3 = 1 1, 7 (mod 12), ( ) 5 = 1 1, 3, 7, 9 (mod 20), ( ) 7 = 1 1, 9, 11, 15, 23, 25 (mod 28). I should rove this; I do not know how to. 1.2 The equation Let us consider the equation = x 2 + ny 2 with n Z >0 (1.2) This essay is concerned with which rimes give integer solutions to the above equation. I will follow a similar treatment to that of Cox in his Primes of the form x 2 + ny 2 [1] ; this essay essentially forms a roer subset of the material in that book. Before we begin, let s lay with the equation a bit and see what we can say. To begin with, consider the roduct of two numbers, each of the form x 2 + ny 2. (x 2 + ny 2 )(z 2 + nw 2 ) = (xz ± nyw) 2 + n(xw yz) 2 Hence, we see the roduct of two numbers of the form (1.2) is again of the form (1.2). Is the converse statement true? 6

7 Conjecture 1.5. n N, if ab = x 2 + ny 2 for some x, y Z then a = x 2 + nỹ 2 for some x, ỹ Z? This conjecture turns out to be false for general n. Examle 1.6. Positive examle, n = 2, we have, = 22 = 2 11 = ( )( ). Examle 1.7. Negative examle, n = 3, we have, = 52 = 2 26, but one can not write 2 in the form x 2 + 3y 2. Theorem 1.8. Conjecture 1.1 is true if and only if n = 1 or n = 2. To highlight what goes wrong when n > 2, we consider the factorisation = x 2 + ny 2 = (x ny)(x + ny) in Z[ n]. We now ask when this factorisation is unique, equivalently when Z[ n] is a UFD. We will see that in such cases we can solve (1.2) very easily using quadratic recirocity alone. Theorem 1.9. For n < 0 the ring Z[ n] is a UFD if and only if n = 1 or n = 2. Proof. Assume n < 2, note 2 is irreducible in Z[ n]. If n is even then n = n n = n 2 2 leads to different factorisations; similarly, if n is odd then (1 + n) = (1 n)(1 + n) = 1+n 2 2. To see that these factorisations are actually different observe 2 n and 2 (1 ± n). The if art of this theorem can be seen by showing that Z[ n] is a Euclidean domain when n = 1 or n = 2, (it is actually norm-euclidean), and hence a UFD. These are very standard roofs found in most basic algebra books, see Michael Artin - Algebra [8] for examle. Remark A more interesting and general method of considering when the ring Z[ n] is a UFD is to note that when n 3 (mod 4) Z[ n] is the ring of integers of Q( n). One can show, using Minkowski constants, that the ring of integers of Q[ n] (n > 0), is a UFD if and only if n = 1, 2, 3, 7, 11, 19, 43, 67, 163 [2]. Now we note that of these values, the only ones not congruent to 3 (mod 4) are n = 1, 2; thus Z[ n] is UFD if and only if n = 1, Old knowledge In this section we consider the cases n = 1, 2. The full solutions to such cases were known to Fermat over 400 years ago, although he gave a very different roof to the one we shall give. Fermat s roof relied on a very convoluted recirocity and descent argument [4] ; at the time, Fermat was oblivious to the law of quadratic recirocity and the notion of a UFD, which would have made the roof a lot easier! 7

8 Theorem (Fermat) For an odd rime, for some x, y Z. = x 2 + y 2 if and only if 1 (mod 4) = x 2 + 2y 2 if and only if 1, 7 (mod 8) Proof. I will only give the roof for n = 2, the n = 1 case is analogous. Assume = x 2 + 2y 2. We see x 2 + 2y 2 0 (mod ); since (Z/Z) is a field, ( ) 2 ( ) we see 2 (mod ), i.e. = 1. By quadratic recirocity and the x y 2 first and second sulementary laws, we see( this) imlies 1, 3 (mod 8). 2 Now assume 1, 3 (mod 8), hence = 1, so x = (x + 2)(x 2), but does not divide (x + 2) or (x + 2), hence is not rime. So, since rime is equivalent to irreducible in a UFD, we see = ππ is a non-trivial factorisation. N() = 2, and N(ππ ) = N(π)N(π ) = 2, so, N(π) =. Note, π Z[ 2] so π = a + b 2 for some a, b Z. = N(π) = a 2 + 2b 2 as required. Hence, One sees that the roofs relied critically on the fact that Z[ n] is a UFD, and as we know this is only true for n = 1, 2. The roof fails for all n 1, 2. In order to solve (1.2) for other cases we need a new aroach. We have seen that when n = 1, 2, the two sets in (1.1) are in fact equal. But we will see that in general this is not the case; it is in fact only true in a finite number of cases. 1.4 A useful homomorhism We will now construct a secific homomorhism which will be of great use later. Theorem [1] If D 0, 1 (mod 4) is a non-zero integer, then there exists a unique homomorhism χ : (Z/DZ) {±1} = C 2 ( ) D such that χ([]) = for odd rimes not dividing D. Furthermore, { 1 if D > 0, χ([ 1]) = 1 if D < 0. The roof of this theorem relies heavily on the laws of quadratic recirocity and the Jacobi symbol. So lets define the Jacobi symbol. Definition [9] Let m and M be corime integers with m ositive and odd. Then the Jacobi symbol is defined as follows ( ) M = ( ) M m i i 8

9 where, m = 1... r ; and ( ) M i is the Legendre symbol. Theorem [1] Proerties of the Jacobi symbol, Let m, n, M, N be integers such that (m, M) = (n, N) = 1 with m, n ositive. Then, ( ) ( ) M N 1. = M N (mod m) m m ( ) ( ) ( ) MN M N 2. = m ( ) M 3. = mn ( ) M 4. m m ( M m ) ( M n m ) = ( 1) (M 1)(m 1) 4 ( m M ) 5. If D 0, 1 (mod 4), then m n (mod D) = ( ) D = m ( ) D n The first three statements of this theorem follow easily from the definition of the Jacobi symbol, the fourth statement, which is a version of quadratic recirocity also follows with a bit of work. The result which is most interesting to us is the fifth one. I will rove only this result. Proof. Of fifth statement. Using recirocity we see the two sides are equal to ( m ) D ( 1) (D 1)(m 1) 4 ( 1) (D 1)(n 1) 4 ( n D ) Note, ( m D ) = ( n D ) since m n (mod D). If D 1 (mod 4), then we see (D 1)(m 1)/4 (D 1)(n 1)/4 0 (mod 2) If D 0 (mod 4), then since m n (mod D) we have m = n+kd for some k Z. So, (D 1)(m 1) (D 1)(n+kD 1) (D 1)(n 1)+kD 2 kd (D 1)(n 1) (mod 2) We are now ready to rove that the function χ is a homomorhism. Proof. (Of Theorem 1.6) By Theorem 1.5, art 5, we see that χ is well defined. By Theorem 1.5, art 3 we see χ is multilicative. Hence χ is a homomorhism. The fact that χ is uniquely determined by χ([]), can be seen by the uniqueness of rime factorisation. 9

10 Note, ker(χ) (Z/4nZ), and it has index two. We can now restate the set relation (1.1) using the new notation χ. Theorem Let n Z be non-zero and χ : (Z/4nZ) {±1} be defined as above, if is an odd rime not dividing n, then { rime = x 2 + ny 2 } { rime [] ker(χ) (Z/4nZ) } This may seem like an overcomlication, and one may be confused as to why we use 4n instead of n (makes no difference since 4 is a square), but it will all make more sense when we move on and consider quadratic forms. 2 Quadratic forms 2.1 Definitions Definition 2.1. A quadratic form over a ring R is a homogeneous olynomial of degree two in one or more variables with coefficients in R. We will only concern ourselves with binary quadratic forms over Z, i.e. forms of the tye ax 2 +bxy+cy 2. To make life easier tyograhically, I will occasionally denote such a form by [a, b, c]. Definition 2.2. A form [a, b, c] is said to be rimitive if gcd(a, b, c) = 1. Note, any form is a integer multile of a rimitive form. Definition 2.3. I will define the image of a form f = [a, b, c] to be {f(x, y) : x, y Z}, and I will denote this by Im(f) or Im[a, b, c]. If m Im(f), then we say that m is reresented by f. If m = f(x, y) for some x, y Z with (x, y) = 1, then we say m is roerly reresented by f. Note, if (x, y) = h, then we have that x = hx 0, y = hy 0 and f(x, y) = ah 2 x bh 2 x 0 y 0 + ah 2 y0 2 = h 2 f(x 0, y 0 ). Thus, every number reresented by a form f is a square times a number which is roerly reresented by f. Since we are interested in reresenting rime numbers, we only need concern ourselves with roer reresentations. Note, a subgrou of the matrix grou G M 2 (Z) has ( a ) natural action on the set of forms; that of change of basis, i.e. If A = q G M r s 2 (Z) then f(x, y) A = f(x + qy, rx + sy). We will now define an equivalence relation of the set of forms. Definition 2.4. Two forms f, g are said to be equivalent if they belong to the same orbit under the action of SL 2 (Z). One can verify that this really is an equivalence relation. You might be wondering why I choose SL 2 (Z) as oosed to any other matrix grou. Could I not equally choose GL 2 (Z)? This is what Legendre did, unfortunately this leads to a bit of a mess later on. So, trust me, SL 2 (Z) is the one to ick! 10

11 2.2 Proerties of the equivalence classes Before we go any further, let us examine exactly what a generic form looks like under the action of an element of SL 2 (Z). One is free to calculate that, [a, b, c] r q s = [a 2 +br +cr 2, 2aq +bs+bqr +2crs, aq 2 +bqs+cs 2 ] (2.1) One nice quality of our equivalence is that equivalent forms reresent the same numbers. If f(x 0, y 0 ) = m and f = g A, then g(a(x 0, y 0 ) T ) = m. What s more, any form equivalent to a rimitive form is itself rimitive. I feel that it is about time for some examles. ( ) 1 1 Examle 2.5. We have [2, 2, 3] [2, 2, 3], via the matrix. Note, 0 1 both forms are rimitive. One may wonder if [a, b, c] [a, b, c] in general? No. q r s Examle 2.6. We have [1, 1, 0] [0, 1, 1], using (2.1) we have [1, 1, 0] = [ 2 + r, 2q + s + qr, q 2 + qs] = [0, 1, 1], one sees this has no solutions in SL 2 (Z). (It does in GL 2 (Z), but remember we don t care about that.) Examle 2.7. We know from Examle 2.3 that [2, 2, 3] [2, 2, 3]. Note, 18 is reresented by [2, 2, 3] ( at (x,) y) = (1, 2); now, one calculates 18 is also 1 1 reresented by [2, 2, 3] at (1, 2) 0 1 T = ( 1, 2). That s enough examles for now; let s get back to theory. The next theorem I found very surrising; in fact it still shocks me sometimes. Theorem 2.8. [1] A form f = [a, b, c] roerly reresents a number m if and only if [a, b, c] [m, b, c ] for some b, c Z. Proof. [1] Assume f(, r) = m. Note f(x + qy, rx + sy) = f(, r)x 2 + (2aq + bs + bqr + 2crs)xy + f(q, s)y 2 Since (,( r) = 1, ) there exists q, s Z such that s qr = 1. So, there is a q matrix A = P SL r s 2 (Z) such that [a, b, c] A = [m, b, c ] as required. The converse is trivial. Definition 2.9. The discriminant of [a, b, c] is D = b 2 4ac. We will occasionally use the notation Df or D[a, b, c]. Assume f = g A, then one can calculate that Df = det(a) 2 Dg = Dg. So, equivalent forms have the same discriminant. 11

12 Note, the sign of Df is imortant. If f = [a, b, c], then 4af = [4a 2, 4ab, 4ac] = (2ax + by) 2 (b 2 4ac)y 2 = (2ax + by) 2 Dy 2. One sees that if Df > 0 then f reresents both ositive and negative numbers, in which case we call f indefinite. If Df < 0 then f only reresents ositive numbers if a > 0 and negative numbers if a < 0, in which case we call f ositive (negative) definite. This is not the only information one gets from the discriminant; since D = b 2 4ac we have D b 2 (mod 4), and it follows D 0, 1 (mod 4), (this is actually true in a way more general definition of a discriminant, that of a number field), from this it follows that b is even if and only if D 0 (mod 4). Theorem [1] Let D 0, 1 (mod 4) be an integer, and let m be an odd integer such that (m, D) = 1. Then m is roerly reresented by a rimitive form of discriminant D if and only if D is a quadratic residue modulo m. Proof. If f roerly reresents m, then we may assume f = [m, b, c], thus D = b 2 4mc b 2 (mod m). Suose D b 2 (mod m). Since m is odd we can assume D and b have the same arity (relace b by b + m if necessary). Since D 0, 1 (mod 4) we have D b 2 (mod 4m). Hence D = b 2 4mc for some c Z. Now observe [m, b, c] is a form which roerly reresents m and has discriminant D. 4n I think this is an amazing theorem; I m not sure why. Corollary( ) Let n be an odd integer and an odd rime not dividing n n. Then, = 1 if and only if is reresented by a rimitive form of discriminant 4n. ( ) n Proof. This follows directly from the above theorem and the fact that = ( ). This is a very nice result, and hels us in our aim of solving = x 2 + ny 2. Of course we are still far from any sort of solution; to see where things go wrong we consider a coule of examles. Examle Let = 3. One can calculate that ( ) 5 3 = 1. So, there is a form f of discriminant 20 which reresents 3. Unfortunately, 3 = x 2 + 5y 2 has no solutions! The roblem is that x 2 +5y 2 is not the only form of discriminant 20; we see the form [2, 2, 3] has discriminant 20 and obviously reresents 3; the roblem is [1, 0, 5] [2, 2, 3]. The new roblem becomes if we know is reresented by a form f, of discriminant 4n, then when is f [1, 0, n]?. Before we attemt to answer this, we need a better way of talking about equivalence classes of forms. From now on, we will deal with ositive definite forms only. Definition [1] A ositive definite form [a, b, c] is said to be reduced if b a c and b 0 whenever b = a or a = c. 12

13 One sees that both a and c are always non-negative. We are now ready to state a very imortant theorem. Theorem [1] Every ositive definite form is equivalent to a unique reduced form. Proof. [1] Let f be an arbitrary form. The first ste is to show that f is equivalent to a form [a, b, c] satisfying b a c. Among all forms equivalent to f ick one [a, b, c] such that b is minimal. If a < b, we see g(x, y) = f(x + my, y) = [a, 2ma + b, c ] for some c Z(not imortant). Since a < b, we can choose m such that 2ma + b < b. Contradiction; thus b a. Similarly, we have b c. We can assume a c by the equivalence (x, y) ( y, x). Thus we have b a c. Our next ste is to show such a form is equivalent to a reduced form. By definition the form is already reduced unless b < 0 and a = b or a = c. In these excetional cases, [a, b, c] is reduced, so we need only show that [a, b, c] [a, b, c]. If a = b, then (x, y) (x + y, y) takes [a, a, c] to [a, a, c]. If a = c, then (x, y) ( y, x) takes [a, b, a] to [a, b, a]. Now, we must rove uniqueness, i.e. different reduced forms are not equivalent. If f = [a, b, c] satisfies b a c then it is true (but not obvious as some would have you believe) that f(x, y) (a b +c) min(x 2, y 2 ). Since if x 2 y 2, then ax 2 +bxy+cy 2 ax 2 +bxy+c xy ax 2 b xy +c xy ax 2 b x 2 +cx 2 as required, similarly for x 2 y 2. Thus, f(x, y) (a b +c) whenever xy 0. So, a is the smallest non-zero value of f. We know that equivalent forms reresent the same values, hence if g f, then the outermost coefficients of g match those of f, since Df = Dg be have g ( = [a, ) ±b, c]. It remains to show that g = f. Assume g = f A for some A = q SL r s 2 (Z). Now we see a = g(1, 0) = f(, q) c = g(0, 1) = f(r, s) It follows that (, q) = (±1, 0) and (r, s) = (0, ±1), and since s qr = 1, we have A = ±I 2. In either case g = f. Thus ends this section. 2.3 Class number In this section we will consider how many classes of forms of a given discriminant there are. Suose f = [a, b, c] is reduced and Df = b 2 4ac < 0. Then b 2 a 2 and a c so D = 4ac b 2 4a 2 a 2 = 3a 2, thus D a (2.2) 3 If D is fixed, then b a, hence the above equation imlies that there are only finitely many choices for a, b, c. 13

14 Corollary There is only a finite number of reduced forms of any given discriminant D < 0. Corollary The number of rimitive ositive definite forms of a given discriminant D < 0 is finite. Definition [1] h(d) is defined to be the number of rimitive ositive definite forms of discriminant D, u to equivalence, it is called the class number. Note, since we are rimarily interested in solving = x 2 + ny 2, which has discriminant D = 4n, we will be esecially interested in forms of discriminant D = 4n. Examle One can calculate h(d) for a few small values of D. h( 4) = 1, h( 8) = 1, h( 12) = 1, h( 20) = 2. It is exactly because h( 20) = 2 that we run into a bit of bother when trying to solve x 2 + 5y 2. Note, x 2 + ny 2 is always in reduced form. We can now restate Corollary 2.11 in terms of the class number. Theorem( ) Let n be a ositive integer and an odd rime not dividing n n. Then = 1 if and only if is reresented by one of the h( 4n) reduced forms of discriminant 4n. Theorem Let D 0, 1 (mod 4) be a negative integer, and let χ : (Z/DZ) {±1} = C 2 be defined as above. Then [] ker(χ) if and only if is reresented by one of the h(d) reduced forms of discriminant D. We know that x 2 + ny 2 is always a reduced form of discriminant 4n, so it would be nice if h( 4n) = 1. Unfortunately, this is a rare occurrence. Theorem [1] Let n > 0 be a ositive integer, then h( 4n) = 1 n = 1, 2, 3, 4, 7 Proof. (by Landau) One is free to verify that h( 4n) = 1 for n {1, 2, 3, 4, 7}. Now let us rove there are no other n > 0 with h( 4n) = 1. The basic idea is that for n / {1, 2, 3, 4, 7} we try and find a form not equivalent to x 2 + ny 2 of the same discriminant. We do this by cases. If n is not a rime ower, then n can be written as n = ac where 1 < a < c and (a, c) = 1. Now, the form [a, 0, c] is reduced of discriminant 4ac = 4n. Thus, h( 4n) > 1 when n is not a rime ower. If n = 2 r, for r > 3, then consider [4, 4, 2 r 2 + 1]. It is reduced, since 4 < 2 r with corime coefficients and most imortantly has discriminant (2 r 2 + 1) = 16 2 r 2 = 4n. Thus, h( 4n) > 1 when n = 2 r with r > 3. If n = 8, then [3, 2, 3] is reduced of discriminant 4n, as required. This leaves the cases r = 0, 1, 2, which are known. 14

15 Finally, assume that n = r where is an odd rime. If n + 1 can be written as n + 1 = ac where 1 < a < c, and (a, c) = 1, then [a, 2, c] is reduced and has discriminant 2 2 4ac = 4n. Thus h( 4n) > 1 when n + 1 is not a rime ower. We are left with the case n + 1 = 2 s. Assume s > 5; then [8, 6, 2 s 3 + 1] will do. What about s < 6? Well, these corresond to the cases n = 1, 3, 7, 15 and 31. n = 1, 3, 7 are known. The number n = 15 = 3 5 is a roduct of rimes, hence h( 4 15) > 1. Finally for n = 31, we consider [5, 4, 7]. So, using the above result we immediately see = x 2 + y 2 1 (mod 4) = x 2 + 2y 2 1, 3 (mod 8) = x 2 + 3y 2 1, 7 (mod 12) = x 2 + 4y 2 1 (mod 4) = x 2 + 7y 2 1, 9, 11, 15, 23, 25 (mod 28) This is nice, but we are still a long way from answering the question for all n. To roceed any further, we need a way to characterise = x 2 + ny 2 when h( 4n) > Class grou The aim of this section is to convince one of the surrising fact that the set of forms of a articular discriminant, under the equivalence stated above actually forms a grou under some comlicated comosition law. Let s see what we have. If f(x, y) and g(z, w) are two forms of the same discriminant, D 0, 1 (mod 4), then we say a form F (x, y) of the same discriminant is their comosition rovided f(x, y)g(z, w) = F (B 1 (x, y; z, w), B 2 (x, y; z, w)) where B i (x, y; z, w) = a i xz + b i xw + c i yz + d i yw are bilinear forms. This is a start, but one could question whether this is even well defined, i.e. whether the comosition of two forms is unique. It s not! So we need to restrict the ossibilities for the bilinear forms. This is exactly what Gauss sent a lot of time working out; we will just jum to the answer. Definition [1] Using notation as above, we say a comosition is a direct comosition if a 1 b 2 a 2 b 1 = f(1, 0) and d 1 c 2 d 2 c 1 = g(1, 0) We are now ready to state the main result! Theorem [1] For a fixed discriminant D 0, 1 (mod 4), direct comosition makes the set of classes of forms into a finite abelian grou. 15

16 We call the grou the class grou, and denote it C(D), where D is the discriminant. Unfortunately, actually working out the direct comosition of two forms is insanely hard and messy in general, which makes roving this theorem also hard and messy. Gauss s original roof is so messy and convoluted that it is not worth mentioning. Instead, we will follow an aroach similar to Dirichlet s. Theorem [1] Let f = [a, b, c] and g = [a, b, c ] be quadratic forms with discriminant D and satisfying gcd(a, a, b+b 2 ) = 1. Then there exists a unique B modulo 2aa such that B b (mod 2a) B b (mod 2a ) B 2 D (mod 4aa ) This is essentially roved by the Chinese remainder theorem. Theorem [1] Using notation as above, the direct comosition of f and g is of the form F (x, y) = aa x 2 + Bxy + B2 D 4aa y 2 One can easily calculate that F has discriminant D and is therefore ositive definite. The full roof of this is again quite messy and I am not going to give it (see - Cox [1] ); I only hoe I have made it easier for you to believe. Let us state a few more results about the C(D). Theorem [1] The identity element of C(D) is the class containing x 2 D 4 y2 if D 0 (mod 4), and x 2 + xy + 1 D 4 y2 if D 1 (mod 4). Furthermore, the inverse of [a, b, c] is [a, b, c]. Definition [1] Given D 0, 1 (mod 4) negative, define the rincial form to be x 2 D 4 y2 if D 0 (mod 4) x 2 + xy + 1 D y 2 if D 1 (mod 4) 4 Note, the rincial form is always reduced; also we say [a, b, c] is oosite to [a, b, c]. We will see that the elements of order 2 in the class grou C(D) are imortant. We see such elements are quite easy to find. Theorem [1] A reduced form f = [a, b, c] of discriminant D has order 2 in the class grou C(D) if and only if b = 0, a = b or a = c. Proof. Note, f has order 2 if and only if [a, b, c] [a, b, c]. We have three cases to consider, If b < a < c, then [a, b, c] is reduced and [a, b, c] [a b, c] if and only if b = 0. 16

17 If a = b then (x, y) (x + y, y) gives [a, b, c] [a b, c]. If a = c then (x, y) ( y, x) gives [a, b, c] [a b, c]. We can in fact say more about the elements of order 2. It turns out that we do not actually need to list all the forms in order to determine the number of elements of order 2 in the class grou. We will now define a very imortant number µ. Definition [1] Let D 0, 1 (mod 4) be negative, and let r be the number of odd rimes dividing D. If D 1 (mod 4) then µ := r. If D 0 (mod 4), then D = 4n and we have r if n 3, 7 (mod 8) r + 1 if n 1, 2, 5, 6 (mod 8) µ := r + 2 if n 4 (mod 8) r + 3 if n 0 (mod 8) Theorem [1] The class grou C(D) has exactly 2 µ 1 elements of order 2. Proof. The basic idea of the roof is to count the number of reduced forms with order two. I will briefly show how this is done in the secial case of D = 4n, where n 1 (mod 4). First note that a form of discriminant 4n may be written as [a, 2b, c]. Thus we need to count the number of forms that satisfy 2b = 0, a = 2b or a = c. Since n is odd, we have that r is the number of rime divisors of n. We will only consider the 2b = 0 case. 2b = 0. i.e. the forms [a, 0, c], where ac = n. Since our forms are rimitive we must have gcd(a, c) = 1. Thus there are 2 r choices for a; but we must also have a < c. Hence, we have 2 r 1 forms of this tye. One shows similarly that we have 2 r 1 forms for which a = 2b or a = c. Thus, all together we have 2 r r 1 = 2 r elements of order 2, as required. Remark For a slightly more detailed roof see Cox [1]. 3 Genus theory 3.1 The need Consider the case x 2 + 5y 2 =. One can calculate h( 4 5) = 2. The two forms are given by [1, 0, 5] and [2, 2, 3]. Hence we see, = 1, 3, 7, 9 (mod 20) = x 2 + 5y 2 or = 2x 2 + 2xy + 3y 2 We need a method to decide which are reresented by which equation. Is it a nice slit? Is it an even slit? Lets try out a few small rimes... 17

18 Reresented by x 2 + 5y 2 3 No 7 No 23 No 29 Yes 41 Yes 43 No 47 No 61 Yes 67 No 83 No It aears that = x 2 + 5y 2 1, 9 (mod 20) and = 2x 2 + 2xy + 3y 2 3, 7 (mod 20). Do we always get such a nice artition? We need more definitions. Definition 3.1. We say that two rimitive ositive definite forms of discriminant D are in the same genus if they reresent the same values in (Z/DZ). So, in our examle above (n = 5), we could guess there are two genera, each consisting of a single class. This is great! It means that if what I say is true (I haven t given you much reason to believe me yet!) then we see, = x 2 + 5y 2 1, 9 (mod 20) Before we get carried away, I should say, things are not always so nice... For examle, if n = 14, then one can calculate that h( 4 14) = 4 and that there are two genera, each consisting of two classes. Hence we still have some ambiguity. x y 2, 2x 2 + 7y 2 reresents 1, 9, 15, 23, 25, 29 in (Z/56Z) 3x 2 ± 2xy + 5y 2 reresent 3, 5, 13, 19, 27, 45 in (Z/56Z) We don t let this get us down. Lets study genus theory. 3.2 Foundations Theorem 3.2. [1] Given a negative integer D 0, 1 (mod 4), let ker(χ) (Z/DZ) be defined as before, then: 1. The values in (Z/DZ) reresented by the rincial form of discriminant D form a subgrou H ker(χ). 2. The values in (Z/DZ) reresented by a given form f, form a coset of H in ker(χ). Proof. [1] i) First we show that if a number m is rime to D, and is reresented by a form of discriminant D then [m] ker(χ). This is not too hard to see. 18

19 If m is reresented by a form then we know m = d 2 m 0 where m 0 is roerly reresented. So χ[m] = χ[d 2 m 0 ] = χ[d 2 ]χ[m 0 ] = χ[m 0 ]. Thus we can assume m is roerly reresented, so we have D is a quadratic residue modulo m, i.e. D = b 2 km for some k, b Z. When m is odd, we see ( b 2 ) ( ) km b 2 χ([m]) = = = m m ( ) 2 b = 1 m which roves the claim. So, we see that H (Z/DZ), and when D 0 (mod 4), we see the identity (x 2 + ny 2 )(z 2 + nw 2 ) = (xz ± nyw) 2 + n(xw yz) 2 gives closure. ker(χ). A similar result holds when D 1 (mod 4). Hence H Before we can rove the second art of this theorem we need the following theorem. Theorem 3.3. [1] If f(x, y) is a form and M 0 an integer, then f roerly reresents numbers corime to M. Assuming this we can now return to our original roof. Proof. [1] ii) We will only consider the case when D = 4n since this is the case we are most interested in. We aly the above theorem to M = 4n, we may assume that f = [a, b, c], where gcd(a, 4n) = 1. Note, b is even since f has discriminant 4n, thus we can relace it by 2b, giving af(x, y) = (ax + b y) 2 + ny 2 Since a is corime to 4n, it follows that the values of f(x, y) in (Z/4nZ) lie in the coset [a] 1 H. Now suose that [c] [a] 1 H, then ac z 2 + nw 2 (mod 4n) for some z, w. Using the above identity, we can solve the congruence f(x, y) c (mod 4n). Thus, the coset [a] 1 H consists of exactly the values of (Z/4nZ) reresented by f(x, y). So we see different genera reresent disjoint values in (Z/DZ). Theorem 3.4. [1] The ma Φ : C(D) ker(χ)/h which sends a class to the coset of numbers which it reresents is a grou homomorhism. Proof. That the ma is well defined and surjective follows directly from the above theorem. Now, if f and g are two forms of discriminant D taking values in the cosets H and H resectively, we can assume that F is their direct comosition, so roducts of numbers reresented by f and g are reresented by F. Hence, F reresents values in H H. As required. 19

20 Note that a given fiber Φ 1 (H ), H ker(χ)/h, consists of all classes in a given genus. Thus we can define the genus of H to consist of all forms of discriminant D which reresent the values of H modulo D. We are now ready to state the main theorem of elementary genus theory. Theorem 3.5. [1] Let D 0, 1 (mod 4) be negative, and H < ker(χ). If H is a coset of H in ker(χ) and is an odd rime such that D, then [] H if and only if is reresented by a reduced form of discriminant D in the genus of H. No roof is required for this theorem since it is just a restatement of what we already know. I include it only to summarise our current osition. An interesting and simle corollary is that there are always congruence conditions which characterise when a rime is reresented by some form in a given genus. 3.3 Proerties For our uroses, the most interesting genus is the one containing the rincial form, called the rincial genus, since when D = 4n, the rincial form is x 2 + ny 2. The nicest case is when the rincial genus consists of only a single class, for then we can fully characterise = x 2 + ny 2. This seems to haen quite often for small n, for examle n = 6, 10, 13, 15, 21, 22, 30, giving = x 2 + 6y 2 1, 7 (mod 24) = x y 2 1, 9, 11, 19 (mod 40) = x y 2 1, 9, 17, 25.29, 49 (mod 52) = x y 2 1, 19, 31, 49 (mod 60) = x y 2 1, 25, 37 (mod 84) = x y 2 1, 9, 15, 23, 25, 31, 47, 49, 71, 81 (mod 88) = x y 2 1, 31, 49, 79 (mod 120) We now ask how often the rincial genus consist of a single class? Before we attemt to answer this, we need to know a bit of theory about the genus. Theorem 3.6. [1] Let D 0, 1 (mod 4) be negative, i) All genera of forms of discriminant D consist of the same number of classes. ii) The number of genera of a form of discriminant D is a ower of 2. Proof. [1] Recall Φ : C(D) ker(χ)/h is a grou homomorhism. Hence, the first statement follows since all fibers of a homomorhism have the same size. The second statement is somewhat harder to rove. First note H contains all the squares in (Z/DZ), because if f is the rincial form then f(x, 0) = x 2. 20

21 Thus, every element in ker(χ)/h has order 2. By the structure theorem for finite abelian grous, ker(χ)/h = C m 2 for some m Thus, the image of Φ, being a subgrou of ker(χ)/h, has order 2 k for some k Z. Since Φ(C(D)) tells us the number of genera, we are done. Note that Φ(C(D)) gives us a natural grou structure on the set of genera. Theorem 3.7. [1] Let D 0, 1 (mod 4) be negative, and let µ be defined as above. i) There are 2 µ 1 genera of forms of discriminant D. ii) The rincial genus consists of the classes in C(D) 2 ; the subgrou of squares in the class grou C(D). Before we can rove this theorem, we need a more efficient method of determining when two forms are in the same genus. We introduce the notion of assigned characters. Definition 3.8. [1] Let 1,..., r be the odd rimes dividing D, define the functions χ i, δ, ɛ by: ( ) a χ i (a) = defined when ais corime to i, i δ(a) = ( 1) (a 1)/2 defined when ais odd, ɛ(a) = ( 1) (a2 1)/8 defined when ais odd. These functions will be used in the definition of the assigned character. The functions to be used are determined by the discriminant, D. Definition 3.9. [1] Let D Z be a discriminant. If D 1 (mod 4) then we define χ 1,..., χ r to be the assigned character. If D 0 (mod 4) then we write D = 4n and we define the assigned character deending on n (mod 8) as follows: n Assigned characters Number of assigned characters n 3, 7 (mod 8) χ 1,..., χ r r n 1, 4, 5 (mod 8) χ 1,..., χ r, δ r + 1 n 2 (mod 8) χ 1,..., χ r, δɛ r + 1 n 6 (mod 8) χ 1,..., χ r, ɛ r + 1 n 0 (mod 8) χ 1,..., χ r, δ, ɛ r + 2 I have included the extra column Number of assigned characters to highlight the fact that it is exactly µ as defined above. Since each individual function χ i, δ, ɛ gives a homomorhism (Z/DZ) {±1}, we see that for a fixed D, the assigned characters give a homomorhism, Ψ : (Z/DZ) {±1} µ. 21

22 Theorem [1] The homomorhism Ψ induces the following exact sequence, 1 H (Z/DZ) {±1} µ 1 Where H (Z/DZ) is the rincial genus. Proof. I only give a outline of the roof (for more detail see Cox [1] ). When we only need to worry about χ i, the roof is quite simle. For in such a case we observe that the Legendre symbol induces a surjective homomorhism ( ) {±1} m i with kernel exactly (Z/ m i Z) 2. Thus, by the Chinese remainder theorem, we see Ψ has the desired roerties. This deals with D 1 (mod 4) and n 3, 7 (mod 8) when D = 4n. The remaining cases are dealt with similarly but with added difficulty due to the ossibility of 2 n. We are now ready to rove Theorem 3.7. i) [1] Recall that the number of genera is the order of Φ(C(D)) ker(χ)/h. Note that ker(χ) has index two in (Z/DZ) ; it follows from the revious theorem that ker(χ)/h has order 2 µ 1. Thus it suffices to show that Φ(C(D)) = ker(χ)/h. By definition, Φ mas a form to the coset of values it reresents. Thus we must show that every congruence class in ker(χ) contains a number reresented by a form of discriminant D. Note that every class in ker(χ) contains an odd rime (this can be seen by Dirichlet s theorem [10] ), thus there exists a rime such that [] ker(χ). By definition, this means χ([]) = (D/) = 1; and thus by Theorem 2.10, is reresented by a form of discriminant D. This roves i). ii) [1] For convenience, denote C(D) by C. Recall that Φ : C ker(χ)/h = {±1} µ 1 is a homomorhism; thus we see C 2 ker(φ). This induces the ma C/C 2 {±1} µ 1. (3.1) We know from i) that this ma is surjective; we now consider the order of C/C 2. To do this we note that the squaring ma C C 2 gives us the short exact sequence, 0 C 0 C C 2 0, where C 0 C is the subgrou of elements of order at most two. Hence, the size of C/C 2 is equal to the index [C : C 2 ] which is equal to the order of C 0 ; and we now that the order of C 0 is 2 µ 1 (see Theorem 2.30). Considering equation (3.1) we see we have a surjective homomorhism in which both the domain and the target have the same order, thus we have an isomorhism. Thus, C 2 is exactly the kernel of the ma Φ; and since ker(φ) consists exactly of the classes in the rincial genus, we conclude C 2 = H. We have now roved/seen most of the main theorems of genus theory for rimitive ositive definite quadratic forms. We now consider how often genus theory solves the roblem of reresenting rimes by the form x 2 + ny 2. 22

23 4 Convenient numbers 4.1 Brief history Our treatment of genus theory is somewhat different to that of Gauss when he originally develoed the theory. Gauss defined genera totally in terms of the assigned characters (see Definition 3.9). Given a form f(x, y) of discriminant D, let a be corime to D be reresented by f. Gauss defines the comlete character of f to be the µ-tule resulting from evaluating the assigned characters at a. Two forms of the same discriminant are said to be in the same genus if they have the same comlete character. It can be shown that this definition is equivalent to ours, see Cox [1]. Some may find interest in the knowledge that Gauss s use of the word character is where the modern term grou character comes from. There are in fact many ways of discussing genera. For comleteness I will give some of the most common definitions of genera; but first we need a bit more notation. We say that two forms f, g are equivalent over a ring R if there exists a matrix A GL(2, R), such that f A = g. Theorem 4.1. Let f, g be rimitive forms of discriminant D < 0. The following are equivalent, i) f and g lie in the same genus - i.e. reresent the same values in (Z/DZ). ii) f and g have the same comlete character. iii) f and g reresent the same values in (Z/mZ), for all non-zero m. vi) f and g are equivalent over (Z/mZ), for all non-zero m. v) f and g are equivalent over the -adic integers Z, for all rimes. The roof of these equivalences can be found in Cox [1], they are not imortant for our inquiry. Let us now go back to the task at hand. Genus theory allows us to fully characterise = x 2 +ny 2, but only when the rincial genus consists of a single class. Gauss lists 65 discriminants which satisfy this criteria in his Disquisitiones [11], I resent them here groued according to class number: n with one class in rincial genus h( 4n) 1, 2, 3, 4, 7 1 5, 6, 8, 9, 10, 12, 13, 15, 16, 18, 22, 25, 28, 37, , 24, 30, 33, 40, 42, 45, 48, 57, 60, 70, 72, 78, 85, 88, 4 93, 102, 112, 130, 133, 177, 190, 232, , 120, 165, 168, 210, 240, 273, 280, 312, 8 330, 345, 357, 385, 408, 462, 520, , 1320, 1365, Gauss was not the first to be interested in these numbers; Euler discovered? them earlier in a different context while trying to solve a different roblem. The roblem was to find all such n which satisfy the following criterion: 23

24 Let m be an odd integer corime to n which is roerly reresented by x 2 + ny 2. If the equation m = x 2 + ny 2 has only one solution with x, y 0, then m is a rime number. Euler called such numbers convenient numbers, the name we will also adot. It is an amazing fact that a ositive number n is convenient if and only if forms of discriminant 4n have a rincial genus consisting of a single class. The roof of the equivalence is quite long and requires a good bit of work, I will omit it but the interested reader can consult Cox [1]. Before we move on we feel the need to resent a very imortant classical theorem of quadratic forms, which is essential for the roof of the above equivalence. More than that, it also highlights an area of quadratic form theory which we have comletely ignored, namely the study of the number of reresentations of a number by a form. Theorem 4.2. [12] Let m be a ositive odd integer corime to n > 0. The number of ways that m can be roerly reresented by a form of discriminant 4n is 2 ( ( )) n 1 +. m For the interested reader who desires a roof see - Robert Krzyzanowski [12]. We now consider how many more convenient numbers there are. 4.2 Counting convenient numbers In this section we will consider how many convenient numbers n there are; since this is the same as considering which discriminant 4n has a rincial genus that consists of a single class, we begin by giving a few equivalent definitions for a genus to contain a single class. Theorem 4.3. [12] Let n be a ositive integer.the following are equivalent: i) Every genus of forms of discriminant 4n consists of a single class. ii) Every reduced form of discriminant 4n has order 2. iii) C( 4n) = (Z/2Z) k, for some k > 0. iv) h( 4n) = 2 µ 1. Proof. Denote C = C( 4n). (i ii) Note that the rincial genus is C 2 = {1}, thus every form has order 2. (ii iii) Every form has order 2, i.e. C( 4n) is a grou of exonent 2, hence isomorhic to (Z/2Z) k for some k > 0. (iii iv) We know the number of genera is the index [C : C 2 ], that is 2 µ 1, thus we see h( 4n) = [C : C 2 ] C 2 = 2 µ 1 1 = 2 µ 1. (iv i) Obvious. 24

25 In order to narrow down which numbers could ossible be convenient numbers Euler roved the following theorem. Theorem 4.4. If n is a convenient number, then i) If n is a erfect square then n = 1, 4. ii) If n 3 (mod 4), then 4n is a convenient number. iii) If n 4 (mod 8), then 4n is a convenient number. iv) If n 2 (mod 3), then 9n is a convenient number. v) If n > 1 and n 1 (mod 4), then 4n is not a convenient number. vi) If n 2 (mod 4), then 4n is a convenient number. vii) If n 8 (mod 16), then 4n is not a convenient number. viii) If n 16 (mod 32), then 4n is not a convenient number. Euler attemted to rove all of these results, he made a few errors in some arts, but these were then fixed by others (Grube). These conditions significantly restrict the number of ossible convenient numbers; one also sees that as n gets larger, the restriction gets stronger. It was conjectured that there are only finitely many convenient numbers, but it was not until 1934 when S. Chowla [13] roved that there are finitely many convenient numbers. This theorem has been imroved on since 1934, and it is currently tied u in another big roblem. Theorem 4.5. There are exactly 65 convenient numbers, the largest of which being 1848, if the generalized Riemann hyothesis is true. Else, there is at most one more. The roof of this theorem is non-trivial and makes use of a toics we have not mentioned (L-functions for examle), and for that reason I will omit it. The interested reader can consult Robert Malinowski [12]. An interesting consequence of the finiteness of the set of convenient numbers is the following: Theorem 4.6. There are only finitely many cases where = x 2 + ny 2 can be fully characterised by simle congruence relations modulo 4n (or any other modulus!). Proof. Notice that an equivalent definition of genus imlies that forms in the same genus are equivalent modulo m for all m 0. Hence, there are no congruences a, b, c,... (mod m) which can searate forms in the same genus. There are only finitely many discriminants for which each genus consists of a single class. Our original question of classifying = x 2 + ny 2 seems to have taken a hit; we now know it is not ossible in general to classify all such using simle congruence relations. In the next section we will see that a classification is still ossible using slightly more comlicated congruence conditions. 25

26 5 Class field theory 5.1 Foreword The methods in this section are considerably more modern and abstract than the receding sections, a basic knowledge of algebraic number theory and Galois theory is assumed. Class field theory attemts to describe the finite abelian extensions of a number field by studying roerties intrinsic to the base field. This is a very big task, much too big for our areciation; we will instead concentrate of a secific class of field extensions, finite unramified abelian extensions. This is the so-called Hilbert class field theory. Using Hilbert class field theory we will be able to characterise when a rime is of the form x 2 + ny 2 in terms of the rime s behaviour in the Hilbert class field. This will give us a solution to our roblem for infinitely many n, but not all. Secifically we aim to rove the following theorem. Theorem 5.1. [1] Let n > 0 be square-free with n 3 (mod 4). Then there is a monic irreducible olynomial f n (x) Z[x] of degree h( 4n) such that if is an odd rime with n and f n then: = x 2 + ny 2 {( n ) and f n (x) 0 (mod ) has an integer solution Where f n denotes the discriminant of f n. Furthermore, f n may be taken to be the minimal olynomial of any algebraic integer α such that L = K(α) is the Hilbert class field of K = Q( n). Let us begin. 5.2 Preliminaries A number field is a subfield K of C which has finite degree over Q. It is a fact that such a field can always be written as a rimitive extension of Q, i.e. K = Q(α) for some algebraic α(this is called the rimitive element theorem - see Wikiedia [14]. Given such a field, we denote the integral closure of Z in K by O K, called the ring of integers of K; we call an element of O K an algebraic integer. Theorem 5.2. Let K be a number field, then (i) O K is a subring of C (ii) O K is a free Z-module of rank [K : Q] (iii) O K is a domain with quotient field K (iv) Every non-zero ideal of O K, has finite index (v) O K is a Dedekind domain, i.e. Noetherian, integrally closed in K and every non-zero rime ideal is maximal 26

27 Given an ideal a O K, we defined the norm of a to be N(a) = #(O K /a). When our ideal is rime, the quotient ring (O K /) is a finite field; we call such a field the residue field of. It is clear that in general, O K will not be a UFD, but as a consequence of being a Dedekind domain it does have the following roerty. Theorem 5.3. If a O K is non-zero, then there exists a factorisation of a into rime ideals, i.e. a = n. Such a factorisation is unique u to order. Furthermore, the i s are exactly the rime ideals of O K which contain a. A fractional ideal of O K is a finitely generated O K -submodule of K Theorem 5.4. [1] Let I be a fractional ideal of O K. Then (i) There exists a non-zero d Z such that di O K (ii) I = αa, where α K and a is an ideal of O K (hence the name fractional ideal). Part (i) basically says that we can clear the denominators. Let I K denote the set of fractional ideals of O K and P K the set of rincial fractional ideals, i.e. those of the form αo K for some α K. Theorem 5.5. Let a, b be fractional ideals, then (i) The roduct a.b is also a fraction ideal (ii) There exists a fractional ideal c such that a.c = O K Hence, we see that the set of fractional ideals form a grou under multilication. This grou, I K has a natural subgrou, P K ; since our grou oeration is commutative, we are free to consider the quotient grou I K /P K. I define the ideal class grou of a number field K, C(K), to be I K /P K ; it is a truth universally acknowledged, that C(K) is always finite. Later, we will see that in the case of imaginary quadratic fields, the ideal class grou is closely related to the form class grou defined earlier. We now consider how rime ideals behave in finite extensions. Let L/K be a finite extension, a rime ideal of O K, then O L is a rime ideal of O L, hence has a factorisation, O L = P e1 1...Pen n where the P i are distinct rimes of O L containing. Note, each P i gives rise to a residue field, (O L /P i ), which is a finite extension of the residue field of. Definition 5.6. [1] The integers e i as defined above are called the ramification indices of in P i, also denoted e Pi. The degree of the finite field extension [O L /P i : O K /] is called the inertial degree of in P i, denoted f or f Pi. There is a nice relation between the ramification indices and the inertial degrees. 27