The Ideal Class Group of Quadratic Fields by Ravinder Sra THESIS SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS OF THE DEGREE OF MASTER OF SCIENCE IN MATHEMATICS UNIVERSITY OF NORTHERN BRITISH COLUMBIA April 2023 ©Ravinder Singh Sra, 2023 Dedicated to My mother(Mohinder Kaur) Acknowledgement I would like to thank my supervisor Dr. Edward Dobrowolski who took over the task of helping me finish my thesis. Prince George, BC April 11, 2023 (Ravinder Singh Sra) iii Abstract The ideal class group problem is one of the very interesting problems in algebraic number theory. In this thesis we focused on quadratic fields. We studied the group of units of the rings of algebraic integers and calculated fundamental units in several quadratic fields. We also studied a detailed proof of the analytic Dirichlet class number formula with numerical examples and its relation to binary quadratic forms. In addition, we also presented a detailed proof of Carlitz’s theorem with numerical examples. iv Contents Acknowledgement iii Abstract iv Table of Contents v 1 Introduction 1 1.0.1 Preliminaries 1 1.0.2 Ideals 3 Multiplication of Ideals 4 Domain 14 Unique factorization domain or UFD 15 1.0.4 Integral basis 19 1.0.5 Examples of calculation of integral basis of ideals and their norms . 27 1.0.3 2 Units 30 2.0.1 Units in quadratic number fields 30 2.0.2 Fundamental unit 31 2.0.3 Maple procedure 37 3 Ideal Class Group 41 3.0.1 Legendre symbol 41 Kronecker symbol 43 Ideal class group 45 3.0.2 v 3.0.3 Analytic Dirichlet class number formula 49 Introduction 49 Binary quadratic forms 50 Equivalent forms 50 One-to-one correspondence between classes of equivalent forms and 53 classes of ideals Narrow class group 53 The existence of a unit of norm — 1 in Ok 62 When Ok for does not have a unit of norm —1? 62 3.0.4 Dirichlet’s class number formula 63 3.0.5 Numerical examples of the Dirichlet class number formula 66 4 Carlitz’s theorem 72 4.0.1 Detailed proof of Carlitz’s theorem 72 4.0.2 Examples illustrating Carlitz’s theorem 76 5 Conclusions 80 Bibliography 81 vi Chapter 1 Introduction We assume that reader is familiar with notions of field, ring, and group. For example Q, R, and C are fields. Let E be a field and FQE. Then F is a subfield of E if it is a field with respect to operations from E. For example, Q is a subfield of R, and R is a subfield of C. If E is a subfield of F, E C F, then F is a vector space over E, where elements of F are considered as ’vectors’ and elements of E as ’scalars’. 1.0.1 Preliminaries Definition 1 (Degree of Extension). The dimension of F considered as a vector space over E is denoted by [F : E] and is called the degree ofF over E. Such degree can be finite or infinite. For example [R : Q] = but [C : R] = 2. In this thesis, we will consider only subfields of C that are finite extensions of Q. Definition 2 (Algebraic element). An element a of field K is called algebraic if a is a root of a polynomial with rational 1 Chapter 1 2 coefficients. Definition 3 (Algebraic Extension). An extension K ofQ (finite or infinite) is called algebraic if every element a ofK is algebraic over Q, that is, a is root of a polynomial with rational coefficients. Definition 4 (Adjoining elements to Q). Let a^,a2, be complex numbers. The smallest subfield ofC containing Q and ai,ao, ...a„ is denoted by Q(ai,0C2, ...an). Such field always exists and it is the intersection of all sub¬ fields ofC containing Q and ai,«2,...a„. We say that K is obtained from Q by adjoining ai,a2, . . . , a„. Definition 5 (Algebraic number field). An algebraic number field is a finite degree field extension of the field of rational numbers Q. Note: It is known that 1. Every field extension of Q of finite degree is algebraic, and 2. Every algebraic number field is generated by a single element. For example, it is easy to show that Q(72) = {a + b^Z : a,b G Q}. We have [Q(72) : Q] = 2, the basis of Q(72) over Q is [1, 72], and every element of Q(72) is algebraic. Another example: K = Q(72, x/3) . This field is generated by a single element 0 = 72 + 73. The minimal polynomial of 0 is x4 — 10x2 + 1 , and its conjugates are 9 = 72 + 73, 72 - 73, -72 + 73, and - 72 - 73. Thus, K = Q(72, 73) = Q(0). We have [K : Q] = 4, the basis of K over Q is [1,0,02,03], another basis is [1, 72, 73, 76]. Chapter 1 3 Definition 6 (Cyclotomic field). A cyclotomic field is a number field obtained by adjoining a complex root of unity to Q. Definition 7 (Algebraic integer). An algebraic integer is any complex number that is a root of monic polynomial with coeffi¬ cient in Z. For example, the rational integers are algebraic integers, the numbers a + bs/2 where a, b 6 Z are algebraic integers, but perhaps surprisingly, unlike the rational integers, alge¬ braic integers may have nontrivial denominators. For example, a = 1 is an algebraic integer because it is a root of x2 - x — 1. Definition 8 (Ring of algebraic integers in a held). The set of all algebraic integers in an algebraic field K forms a ring denoted by Ok- Definition 9 (Quadratic fields). A quadratic field is an algebraic number field of degree two over Q. Every quadratic field is of the form Q(\/d), where d is a square free integer other than 0 and 1. Ifd > 0, the corre¬ sponding quadratic field is called a real quadratic held, and for d < 0 an imaginary quadratic held or complex quadratic held. We are interested in the arithmetics in the rings of algebraic integers. 1.0.2 Ideals Note: In this thesis we consider only commutative rings. In what follows we assume that every ring under consideration is commutative. Chapter 1 4 Definition 10 (Ideal). Let iff +, .) be a commutative ring. A subset I of R is called an ideal of R if (L +) is sub¬ group of (R,+) and for every r E R and x E I, rx E I. Notation: I/— 5) in Zfv35]. We shall prove that/2 = (2). We have I2 = (2, 1 + ^/35)(2, 1 + V3^) = (4,2(1 + ^5), 2(1 + V3^), -2(2 - ^5)). Chapter 1 5 Since every generator of I~ is a multiple of 2, we conclude that (2, 1 + \/^5)2 C (2). On the other hand 2 y/— 5 = 4 — 2(2 — y/— 5) G (2, 1 + \/^5)2, and also 2 = 2(1 + V— 5) — 2\/^5. By combining these two equalities, we conclude that 2 G Z2 as a linear combination of generators of Z2 with coefficients from Z[\/^5]. Thus (2) C I~. Since we have showed above that Z2 C (2), we conclude that (2) = Z2. The situation in the ring of rational integers Z is rather trivial. It is easy to show that every ideal of Z is generated by a single element. For example jet Z = (2) = 2Z, and J = (3) = 3Z. Then I J = (2) (3) = (2 x 3) = 6Z. We see that in this case the ideals behave just like num¬ bers. This was an initial idea to use ideals, especially in a number fields. As we will see later, algebraic integers may not have unique factorization property, but ideals do. Propostion 1. The ideals of the ring of integers of an algebraic number field form an Abelian semigroup with unity. It means that the multiplication is commutative and associative. The ring itself is the unity. Definition 11 (Prime ideal). A proper ideal P of an integral domain D* is called prime ideal if for all a,b E D, ab E P implies a EP orb EP. The definition of integral domain and its forms are discussed in the next section. Here we note that Ok is an integral domain. Definition 12 (Maximal ideal). In a ring R, a proper ideal I is called maximal ideal, if there exist no other proper ideal J of Chapter 1 6 R such that I C J G R. That is if I GJ G R, then either 1 = J or J = R. Propostion 2. If p is a prime ideal in Ok then p Q Q = p^. Hence p contains a unique prime number p. Proof It suffices to notice that p Q Q is a prime ideal in Z which is easy to prove. Propostion 3. Every prime ideal of Ok is a maximal ideal of OkProof Let p be a prime ideal in Ok- We know then that the quotient ring A = 0^/p is an integral domain, that is, has no zero divisors. The ideal p is maximal if and only if A is a held. Hence, it remains to show that every nonzero element of A is invertible, that is, if a G Ok \ p then there exists P G Ok, such that (a + p)(p + p) = 1+p. Let f(x) = d 1- cm be the smallest degree monic polynomial in Z[x] , such that /(a) G p. Such polynomials certainly exist, since a is an algebraic integer so, g(a) = 0 G p for some monic polynomial g{x) in Z[x]. We have a'" + cia'"-1 d a(a'" 1 d F cm G p. Thus +p = — cm + p. Let p be the unique prime number in p. Then p cannot divide cm, since otherwise we would have a'"-1 d F cm-i G p because p is a prime ideal not containing a, but this contradicts the fact that m was minimal. Hence, there are integers a and b < such that acm mod p. Let p = d — bp=\ Fcm-i). Then (a + p)(p + p) = (ap + p) = l+p which finishes the proof. Note: Every maximal ideal in any domain is a prime ideal but the converse is in general false. However as we have proved above, the converse is true in the ring of algebraic integers Ok- Chapter 1 7 Propostion 4. Tfp is a prime ideal of Ok, and a b are ideals in Ok, then ab=>p Dawp j b. Further, if ip D pi . . ,pr, where all ideals p, are prime then p = pi for some i. Proof The first part follows directly from the definition of prime ideal. The second uses the fact that prime ideals are maximal. Theorem 1. Let K be an algebraic number field. Then every proper ideal of Ok can be expressed uniquely up to order as a product of prime ideals. Note: This s a very important theorem. In general,as we mentioned above, algebraic integers do not factor uniquely into a product of irreducible (or prime) elements, but the ideal do. This is the reason we are using ideals in studying algebraic number fields. The proof of this theorem can be found in most textbooks on algebraic number theory, for example in ( [1], Theorem 8.3.1). Definition 13 (Fractional ideal). Let D be an integral domain, and K be quotient field ofD. A nonempty subset A of K with following three properties: 1. a,P,G A => a + P G A. 2. 0tEA,rED^raEA. 3. There exists a nonzero y G D such that yA C D, is called fractional ideal ofD. Note: For clarity, we will refer to ordinary ideals of D as integral ideals, so every integral ideal of D is also a fractional ideal but not vice versa. If A is a fractional ideal of D then Chapter 1 8 A = |z, where y is nonzero and y G D and 1 is integral ideal of D. For example, let A={^nez) so A is a nonempty subset of Q. Clearly A has properties (1) and (2). Also, 25A = Z so that (3) holds. Hence A is a fractional ideal of Z. One more example, ^Z is a fractional ideal of Z The multiplication of fractional ideals is defined in exactly the same way as multiplica¬ tion of integral ideals. However, while integral ideals form a semigroup, the fractional ideals form a group. We have Theorem 2. The set of fractional ideals of an algebraic number field K form an Abelian group under multiplication. Proof The identity element is Ok as in the case of semigroup of integral ideals. It is easily seen that the product of fractional ideals is a fractional ideal, their multiplication is commu¬ tative and associative. However, the existence of an inverse ideal requires proof. We need to prove that if I is a fractional ideal of K then there exists a fractional ideal J of K, such that IJ=OK. We will somewhat modify an expository paper by Keith Conrad ( [5]). Suppose that Z is a fractional ideal of K. Then for some y G K, yZ is a proper ideal of OkConsequently, yZ = pi . . . p,„, where p/, i = 1 . . . m are prime ideals. The next step is to show that each prime ideal of Ok is invertible. Let p be prime ideal of Ok- Define p by P = {ye^:YpC OK}. Chapter 1 9 Then this definition directly implies that p is a fractional ideal. Indeed, the closure under addition or subtraction is clear, as is the closure under multiplication by elements form OkFurther, for any 8 from p, 8p C Ok- We show first that p D OkThe inclusion p D Ok follows directly from definition of p . It remains to show the existence of x G p \ Ok- For this, let x be any nonzero element of p. Then W c p. Suppose that the ideal (x) factors as (x) = qi...qr into a product of prime ideals. If r = 1 we get P (a) = qi- But p and qi are maximal as prime ideals. Hence p = (x). Consequently, j 6 p, and ± Suppose then that r 2. Then (x) = qi .. . qr C q2 • • • qr and by unique factorization of ideals the inclusion is sharp. Therefore there exists y G qi • • • qr \ (x). Now, we will use the fact that p D Ok to show that p is the inverse of p . Pick x G p that is not in Ok- Then (x)p C Ok, so P Cp + Wp C OKClearly, p + (x)p is an ideal. Since p is maximal, there are two possibilities: 1. p + (x)p = Ok, and 2. p = p + (x)p. OK- Chapter 1 10 The first possibility gives p + (x)p = p(C^ + (x)) = OKThis means that p = Ok + (x) is the inverse of p . It remains to rule out the second possibility. It implies that xp Cp. We will see later that every ideal has integral basis, that is there are elements cq , . . . , a„ of p, such that every element of p can be uniquely expressed as a linear combination of these elements with coefficients in Z. This gives ai oc;( This implies that x is an eigenvalue of the characteristic polynomial of the matrix . This polynomial is monic and so x is an algebraic integer in Ok, which is not the case. We conclude that prime ideals are invertible fractional ideals. Returning now to any nonzero fractional ideal I we know that yl is an integral ideal for some nonzero yE Ok- Hence yl = pi . . . p, is a product of prime ideals which are invertible. Therefore is a fractional ideal that is the inverse of I. Definition 14 (Norm of ideal). Let I be a nonzero (proper) ideal, then norm of I is defined by N(1) = \Ok/1\ = [Ok-I]. Chapter 1 11 Thus, the norm of a non zero ideal I of a ring Ok is the size of finite quotient ring Ok / I. The norm of the zero ideal is taken to be zero. Propostion 5. If J is an ideal in Ok then N(J) = \Ok/J\ is finite. Proof. In Section 1.0.4 we prove [Theorem 4] that Ok has an integral basis. Let [K : Q] — n and {cq ,a2, ... ,a„} be such a basis. It is easy to show that JO Z is an integral ideal in Z. Since Z is PID, there exists a positive integer a, such that JO Z = ff and obviously (a) C J. There exist natural homomorphism ([) : Ok/ (a) —> Ok/ J defined by (|)(x + (a)) = x+J. Clearly, 0 is onto. It is suffices to show that Ok /{a} is finite. \-mnan\mi 6 Z, ; = 1, . . . ,n}. Ok = {mi C is an embedding then <5 is an isomorphism of K onto Each such isomor¬ phism is completely defined by the value of o(0) which must be one of the conjugates 0(. There are exactly n such embedding, so we have n embedding Oi , <52, • • , &n with <5, = 0, for all i = 1,2,... ,n. Definition 15 (Norm of an element). Let K be an algebraic number field and let Ci,C2, , be a complete set of embeddings of K into C. IfdEK then Nk/qCL is defined by n A^/Q(a) = n^(< Z=1 Note : If [K : Q] = n, a 6 K and deg a = m, then [Q(a) : Q] = m, and m\n because [£:Q] = [£:Q(a)][Q(a):Q]. If oci = a, «2 • • , Om are conjugate of a then because, it can be shown that each conjugate a, will occur the same number of times among ai(a),O2(a),...,o„(a), that is times. Chapter 1 13 Corollary 1. In the case when a e K, [K : Q] = deg a = n we get N(a) = aia2...a„, where cq = a, 0C2, . . . , are the conjugates of a. The situation is easy when K is a quadratic field. Corollary 2. Let a be element of a quadratic number field K = Qjx/df Then there are ex¬ actly two embeddings <51 , and G2 ofK into C, given by Oi ( y/d) = \/d, and d) = — s/d. If a = a + b\/d then the norm of a equals N(cl) = a1 (a)o2(a) = aa, where we denoted <52 (ot) by a. If CL Q then CL is its algebraic conjugate, if CL E Q then CL = CL. Thus we have N(a + bs/d^ = a2 — db2, N(2) = 4 because a = 2.b = d = 0, 7V(y/10) = — 10 because a = 0,b = l,d = 10 Definition 16 (Principal ideal). A principal ideal is an ideal I in ring R that is generated by single element a ofR.we write I = (a). We have I = {ra : r E R} For example, for positive integer 5 the set 5Z = {0,±5,±10,...} is an ideal of Z and 5Z is principal ideal generated by 5 (or— 5) so that Chapter 1 14 5Z = (5) = (—5) Propostion 7. Principal fractional ideals form a group under multiplication. It is a subgroup of the group of fractional ideals. Proof. Let K be a field, I and J be principal ideals. Principal ideals are a subset of the fractional ideals, it is sufficient to show that I and J are closed under multiplication and inverse. Since I and J are principal ideals, al = (bfcJ = (d) for a, b,c,d 6 Ok where a,c I is multiple of and J is multiple of 0. and their product I J is multiple of which is again a fractional ideal thus closed under multiplication. Z-1 is multiple of|which is again a fractional ideal thus closed under inverse. 1.0.3 Domain Definition 17 (Integral domain). An integral domain is a nonzero commutative ring in which the product of any two nonzero elements is nonzero. Definition 18 (Unit in an integral domain). An element u of an integral domain D is said to be a unit if there exist some element u [ such that uu~l = 1. Definition 19 (Irreducible element in an integral domain). An irreducible element of an integral domain is a nonzero element that is not invertible and is not the product of two non invertible elements. Definition 20 (Prime element in an integral domain). An element p is said to be prime element of an integral domain D if p and if p\ab then either p\a or p\bfora,b 6 D. 0, p is a nonunit Chapter 1 15 Unique factorization domain or UFD Definition 21 (Associate element). Two elements a and b in an integral domain D are called associated if b = an, where u is unit in D, then a = bu~x . Example. In Z, we have 30 = (2)(3)(5) = (—2)(—3)(5) = (2)(-3)(-5) = (-2)(3)(-5). The elements 2 and —2, 3 and —3, 5 and —5 are associated elements. Definition 22 (Unique factorization domain). An integral domain D is called unique factorization domain, or UFD, if every nonzero, nonunit element a of D can be expressed as product of irreducible elements and a every two such factorization the number of factors is the same and corresponding elements are associated. For example, Z is UFD. This is known as the fundamental theorem of arithmetic. Note: If a]O2 .. .an = b\b2-- -bn and a, is associated with bi for each i = 1,2, ... ,n then bi = aim ... bn = anun with units «i, u?, ... ,un then bib-i . . ,bn = ft[U2 . . . u„)a\a2 . . .an = vai«2 • an, so we say that D is UFD, if every nonzero, nonunit element of D can be written in a unique form c = vai«2 where v is unit in D and ai,«2, • ,an are irreducible elements of D. Definition 23 (Principal ideal domain). A principal ideal domain, or PID, is an integral domain in which every ideal is principal i.e. can be generated by a single element. Propostion 8. Suppose that a principal ideal domain R is not a field. Then an ideal I = (p) is maximal if and only if p is an irreducible element. Proof. (=>) Let I = (p) be a maximal ideal. If p = 0 then (0) would be the only proper ideal in R. This implies that for any we Chapter 1 16 have (/?) = R which implies that for some 5 G R, sp = 1, so every nonzero p is invertible and 7? is a field, against our assumption. Similarly, p cannot be a unit, because then p-1 G R. and 1 = p-1p G I, so I = R, which is not the case. Now, suppose that p = ab. Then a\p. Hence (p) = pRQaR = (a). However 7 is a maximal ideal, hence either 1 = aR or aR = R Suppose I = aR then a E P, because 1 G R. Hence a = pq with some qER. We getp = ab = pqb. Since the cancellation law holds in a domain, we get 1 = qb. Hence b is a unit. Now suppose that aR = R. Since 1 G R, we get 1 = as, with some s E R. Hence, a is unit. So in both cases, p = ab either a is unit or b is unit, Thus p = ab implies that a or b is a unit. Therefore p is irreducible. (<=) Let p be an irreducible element of R. Then (p) R, since (p) = R would imply that p is a unit. Suppose that (p) C J C R. Now, J is an ideal in R, but R is a PID, so J = (#) for some qER. This implies that q\p. However p is irreducible, so p = qs, where s is a unit of R. Then (p) = (qs) = Rsq = (Rs)q = Rq= (qf which proves that (p) is maximal. We are going to prove now that every PID is a UFD ring. The next proposition is the first step in this direction. Propostion 9. In every PID the ascending chain of ideals («1) C (a2) C (a3)... stabilizes, that is (an) = (am) for all n m, staring at some m. This is called the ascending chain condition on principal ideals. Proof If (a\ ) C (af) C . . . is an ascending chain of ideals in R, let A = |J/Li (o/)- We claim that A is an ideal. Suppose a,b G A, then a G (af and b G (af), for some j,k 1. Either Chapter 1 j k or k 17 j, suppose k we know that a j. Then (af) C (ak), so that a,b G (ak). Since (u/J is an ideal — b G (ak) C A and ra G (ak) C A for some r G R. Therefore, A is an ideal and R is PID, A = (c) for some c G R. Since A = U(o,), we know that c G (a,;) for some n. consequently, (c) C (an) and for each i n =A = (c) C (a„) C (a,) C Therefore, («/) = (a„) for each i n. Corollary 3. Let R be a PID ring. Then every nonzero nonunit element a is divisible by an irreducible element. Proof. Let a\ be nonzero nonunit element of R, and let 1 = («i ) . If 7 is a maximal ideal then ai is irreducible, so aj |ai, and we are done. If I is not maximal then there exist an ideal L G R such that f C h C R. If /2 ia an ideal of R then f = (02) with some ai G R. If I2 is maximal then «2 is irreducible (ai) C (o2) => If I2 is not maximal then there exist an ideal f ER such that h £ L C R, and | ap By the previous proposition this process must stop at some maximal ideal {amf Then am | a\ , and am is irreducible. Corollary 4. An element in PID is prime if and only if it is irreducible. Proof. Let 7? be a PID, and p a prime element in 7?. Suppose that p = ab with a,b G R. So p\ab. Since p is prime element, then p\a or p\b. If p\a then a = pm, for mER. So p = pmb. Hence p — pmb = 0 and p( 1 — mb) = 0, but p is nonzero, so 1 — mb = 0, mb = 1 and b is a unit. Similarly, we can show that a is unit if we start with p\b. Thus p is irreducible. Conversely, suppose p is irreducible and p\ab. Let {p) be the ideal R generated by p. By Proposition 7, (p) is maximal ideal because p is irreducible. Every maximal ideal is a prime ideal. Now, p\ab implies that ab E (p). Since Chapter 1 18 (p) is prime, we conclude that a E (p) or b E (p). Thus a\p or p\b. Therefore p is a prime element. Corollary 5. Every nonzero, nonunit element in PID is product of irreducible elements. Proof. Suppose there exists a nonzero nonunit element a E D, which is not a product of irreducible elements. Then a = b[C[ where b\ , ci are nonunits. Now either bi or ci cannot be written as a product of irreducibles, say bi = d[ is one which cannot be. Then d\ is reducible and d\ = bici where neither bi nor C2 is a unit. Now either bi or C2 cannot be written as a product of irreducibles, let us say that bi = ^2 is this element. Note that (di) C (^2)Now the element J2 is reducible di = &3C3, where neither /23 nor C3 is a unit and either /23 or C3 (assume bf) cannot be written as product of irreducibles. Set d^b?,. Now (df) C (df) C (df), continuing in this way, we can construct a strictly ascending chain of ideals, this is not possible by Proposition 9 Theorem 3. Every PID is a UFD. Proof. Let R be PID. Let a be a nonzero, nonunit element of R. By Corollary 5, a is a product of irreducible elements. Suppose that a = p\P2- Pm = Q}Q2 - q^ where each pi and cp are irreducible elements. Now we show that p, is associate of some qi and m = n. Suppose that n > m. Now pi |o, so pi divides one of the elements q\ , q2, . . . , qn By Corollary 4, pi is a prime element, since it is irreducible. Without loss of generality, Chapter 1 19 suppose that pi |^i . Hence q\ = pi%i , with some xi G R But gi is irreducible, so xi must be unit. Now a = p\p2-- .pm = q\q2--- qn implies that PlP2---Pm = PlXiq2---qm- So P2P3 • • Pm = x\q2 - qn- By applying same argument to p2 and q2 we get P3P4---Pm = X]X2q3---qn,--- By proceeding like this, at the end we get 1= . .Xmqm+\qm+2 1 = {x\Xt -Xmqm+}. • qn- qn—l^qni which implies qn is a unit, a contradiction. Thus n m. Similarly we show that m f n, so m = n. Further pi and qi are associated. Note : The converse is not true, i.e. every UFD is not a PID. For example, consider the ring Z[x] . The ideal (2,x) is not principal: suppose (2,x) = (a) for some a. Since this ideal contains the even integers, a must be some integer and in fact it must be 2. But (2) does not contain polynomials with odd coefficients, so (2,x) / (2). However, Z [x] is UFD ring, which can be shown by Euclidean algorithm. 1.0.4 Integral basis Definition 24 (An integral basis of Ok)A set of algebraic integers ai,«2...a5 G K is called integral basis of Ok if every algebraic integer can be written uniquely in the form ^bsas, Y=&iai +b2OoA where b] ^2 ,---bs G Z. We also refer to an integral basis of Ok as an integral basis of K. In a theorem below we prove that every algebraic number field has at least one integral basis. Chapter 1 20 Let K be an algebraic number field and [K : Q] = n. Then K = Q(0) for some 9 6 K. Let Ci , o2 , • • , be the embedding of K into C, and o; = 0, for / = 1, 2 . . . , n, 0 = 0j , 02, . . . , 0„ are the conjugates of 0. Definition 25 (Discriminant of an element). be elements ofK. Their discriminant is defined by , a2. . . . , Let 2 ci(ai) ci(a2) . ai(a„) D(ai,a2,...,a„) = c„(ai) a„(a2) . It can be shown that D(ai , a2, ... , a„) 0 if and only if ai , a2 . . . , a„ are linearly inde¬ pendent over Q. Theorem 4. Every number field has an integral basis. Proof. We will follow the exposition from ( [10] , Section 2.4). Let [£ : Q] = n and suppose that cq , a2 . . . , a„ be a basis of K over Q. By multiplying these elements by appropriate non zero integers, we can obtain another basis whose elements are algebraic integers. Hence in what follows we assume that ai ... , a„ are algebraic integers. We shall prove that a basis with algebraic integers which has the smallest discriminant in fact an integral basis. Suppose that cq , . . . , a„ has the smallest discriminant and for a contradiction, suppose it is not an integral basis of K. It means there exists an algebraic integer a G K that is not linear combination of CQ . . . , a„ with integral coefficients, since ai . . . , a7i is also basis over Q, we have a = t'l «i H 1- c„a„ with ci , . . . , cn G Q and at least one of c, is not an integer. Without loss of generality, suppose that cy 0 Z. Then c = a\ + r with some ay G C and 0 < r < 1( r is the fractional part of ci) If we replace ai by a — aicti = rai 4 of K over Q. we obtain another basis {a — oci , a2, . . . ,a„} Chapter 1 21 2 roi(ai) Gi(a2) 0 «2 tfi(M = ro„(ai) o„(a2) ••• o„(a„) The last matrix is obtained by subtracting Ci multiple of the z-th column for i = 2 . . . n from first column. We obtain a contradiction because = r2D(ai,...,a„) < D(ai, . . . ,a„). Therefore, {ai, . . . , a„} is an integral basis of K. Theorem 5. Let K be a quadratic field Q(Vd) . If d = 1 mod 4 then an integral basis ofK is {1, 1+2^ }, otherwise it is {1, \/d}. Proof. Let d be a squarefree integer. Let a G Q(v^). Then a can be written as a = r+^, where r,s,t are integers with gcd(r,s,f) = 1 and t>\, and if a is an algebraic integer then a satisfies G Z with So t\2r and t2^ — s~d also =L We shall show that (t, r) = 1. Suppose (t, f) = m. m2\r2 — s2d, hence m2\s2d. Further, m\r so m2\r2. Together with m\r2 — s2d this gives m2\s2d. Then m\r and But (m,s) = 1 because (r,s,t) = L It follows that m2|rZ, however d is a squarefree integer. Hence m = 1. Now t\2r implies 112, thus either t = 1 or 2. Chapter 1 22 If / = 1 this will yield the element of Z(\/^). The case t = 2 can only occur if 4| r2 — ds2 Then d must be a quadratic residue modulo 4, and . since d is squarefree we must have d = 1 mod 4, and also r = s mod 2. Thus a = r~f^ with r = s mod 2. This yield an integral basis If d 1 mod 4 we must have t = 1, so a = r + sVd. Hence {1, \/d} is an integral basis in this case. Definition 26 (Free abelian group). An abelian group G is called free abelian group of rank n if there are n elements ai , a2, .. . , a„ in G such that G = Zai + Zao + • • • + Za„ and every element p of G can be expressed by unique linear combination of the form p = + m2^2 + + mnan, mi G Z for i = "• l,2,...,n. Theorem 6. Let [K : Q] = n and let J be a nonzero ideal of Ok- Then J has an integral basis of n elements. Proof We have shown that Ok has integral basis. Let {ai , a2, . . . , a„} be such a basis. Then Ok = Zai + Za2 H 1- Za„ = Z". We say that {Ok,+} is free abelian group of rank n. It is known that every subgroup H of a free abelian group G is also free a abelian group. Further, if H has finite index in G than rank)//) = rank(G). We have proved that [Ok : K] is finite, therefore, J is free abelian group of n elements. So J has integral basis with n elements. Lemma 1. Let J be a nonzero ideal of Ok- Suppose that {ai , a2 , . . . , a„ } is an integral basis of Ok- Then for every i, 1 0, k^ G Z, i = 1, 2, ... , n}. By Lemma [1], Si / (|), because /nia, G Sp Choose Pi from Si to be any element whose coefficient ki of «i is the smallest. Let Pi = mi«i +C12CC2 3 Fci„a„. Then mi divides all coefficients kj of elements in Sp This Chapter 1 24 is because J is closed to linear combinations with integer coefficients. Euclidean algorithm to find the greatest common divisor mi . Then consider the set of element in J of the form kj G Z. i — 2, 3 , . . . , n { . Si = {^2^2 H” ^3^3 T ’ T knCLfl | V > ' ' Again Si 0 because G Si. Let again m2 be the smallest coefficient ki for all elements in Si. We continue this process, until we determine all elements Pi, P2 • • • , Pn- Claim : {Pi, P2, . , P;i} is integral basis of J. • • Proof. Let P = Z>iOCi + biVo 3 Vbnan G J with b\,b2, • • • ,bn G Z. Then b\ is multiple of mi, say b\ = z\m. Then P — ci pi G Si, P — ciPi = bf^i 3 Now, bi' is multiple of m2, say bf = Zim2- Then P — zi P 1 — Z2P2 \-bn'an. Si, etc. At the end we get, P - Cl Pl -C2P2 c„P„ = 0. Moreover, the set {Pi , P2, . . . , P,J is linearly independent over Z. This is because {ai . «2- • is linearly independent. If ciPi +C2P23 Pc„P„ = 0 with ci,c2,...,c„ G Z. Then Q = 0 because «i occurs only in Pi , then ci = 0 because di occurs only in P2, etc. Lemma 3. We have \Ok/J\ — mpni . . .m„, where m\,mi,. .. ,mn are defined in the previous lemma. Proof. It suffices to show that the set of cosets of the form ncti 3- ridi 3- where 0 ri m, for each i = 1, 2, . . . , n, is a complete set or distinct cosets of J in Ok- • 3~ rnan WJ, (1-1) a,J Chapter 1 25 Completeness : Let zi«i + "2^2 5 FZnOt« +J be any coset of J in Ok Elements, Pi , P2 . . . , are in J. We have Zi = k/mi + r, with 0 r, < m. Then ZiOli + Z2Ot2 + + J = Z1(X1 + Z2Ot2 + fl (Xi + Zl'oti + + + Zn + Zn(Xn — A^iPi + J +J By successively interacting an appropriate multiple of P,, i = 1,2, .. .,n, we get Z1«1 +Z2«2 4 ^rnan + J with 0 FzA +J = nai +r2«2 4 r, < mi, i=l,2,...,n These show the completeness of the cosets in (1.1). Now we show that all cosets in (1.1) are distinct. Suppose that noci + r2«2 H + J where all 0 \-rnan+J = siOti +\2«2 H r, < , mi, 0 < Si < mt for i = 1 , 2, . . . n. Then (n - si)«i + (r2 - s2)a2 5 This implies that n F (rn - sn)a„ G J. — is multiple Sj of mi , but |n — si | /7,2} is also such basis, because 1 + /7 1 + /7 2 -1 (1 + /7,2) = S»WM = 2/7 1 + /7 2 1-/7 2 \/W = and det = 16x7, = -1 =4x7 2- Example 2 Let K = Q(/iT) and J = (3 + 7/11, 5 + 8/11). Nk (3 + 7/11) = -530, Nk (5 + 8/11) = -679 (679,530) = 1. Hence 1 G J, so J = OkThe integral basis is {1 , /11} and N( J) = 1. Example 3 Let K = Q(/ll) and J = (3 + /11,6 + 9/11) Nk (3 + 7/11) = -530 and Nk (6 + 9/11) = -855. J = {(a + /,/TT) (3 + 7/11) + (c + d/Tl ) (6 + 9/11) |a, b,c, d e Z} = { (3a + 22b + 6c + 99<7 ) + (2a 3b 9c 6t7) /11|a, b,c,d G Z}. First element: gcd(3,77,6,99) = 1. Take b = 2, a = — 51, c — d — 0, we get l + (7(-51) + 3(2))/ll = 1-351/11. Second element: 3a + 22b + 6c + 99d = 0 => a = ^y-b let b = 3k,a= —22k — 2c — 33d. — 2c — 33d, Chapter 1 29 Now 7a + 3b + 9c + 6d = 7(— 77b — 2c — 33d) + 9k + 9c + 6d = 53Ok-5c-225d, gcd(53O,5,225) = 5. Hence the second element is 5a/1T. The integral basis is (1 — 351 /1T, 5 /11) or simpler = 100 X 11. /7/=4xlLW) = /BP=5. Example 4 Let K = Q(/17), J = (3 - 2/17, 3 + 9 /17). The integral basis of Ok is {1, We have 3 -2/17 = 5 -4(±±^) and 3 + 9/17 = -6+ 18(^4^). then / = = (1-14/66/. } because 17=1 mod 4. ^ + 4 and J= (5 -4£, -6+ 18/ = (5 -4/ -1 + 14/ So J= {(a + b/(l-14/ + (c + J/(66/|a,b,c,JGZ} = {(a— 14b + 264c/) + (—14a— 13b + 66c + +66d)£,\a,b,c,d G Z}, then a = 14b - 264d and -14a - 13b + 66c + 66d = — 14(14b 264^7) 13b + 66c + 66d - - = 209b + 66c + 3762d. Since gcd(209, 66, 3762) = 11, we get 1 / as the second element, we get basis of J as {1 — 14^, 11/. Another basis is {1 — 3/ 11/ and N(J) = 11. Chapter 2 Units Definition 27 (Unit). If a is an element of the ring of integers Op of an algebraic number field F, a is called a unit if there exist a nonzero element b in Op such that ab = 1. Op may contain an infinite n umber of units. Theorem 8. (Dirichlet 's unit theorem ) Let K be an algebraic number field of degree n. Let r be the number of real embedding ofK into C and 2s the number of complex embedding ofK. Then Op contains r + s — 1 units Ei,E2---U+5-1 such that each unit of Op can be expressed uniquely in the form ps"1 . . .E^Zp where p is a root unity in Op and ni... .nr+s-i are inte¬ gers. 2.0.1 Units in quadratic number fields Dirichlet’s unit theorem allow us to describe all units in quadratic number Helds Q(y/d). All unis have the form £ = where - is a primitive root of unity in Ofs/df When d > 0 i.e. K = Q(s/d) is real quadratic field, r = 2,s = 0,t = r+s 30 — 1 = 1. We get Chapter 2 £ 31 = ^"r|* = ±r]^, because ^= — 1 is only real primitive root of unity Ok- Thus, every real quadratic field has infinitely many units. The unit r| is called the fundamental unit in Ok • For example; If <7 = 3, =>r = 2,5 = 0,t = r + 5— 1 = 1, then £ = 2 + If J = -3, 3)”, where n = 0, 1,2. => r = 0,5 = l,t = r + s— 1=0, then £ = ±( 2.0.2 Fundamental unit Theorem 9. Let d > 1 be a squarefree integer, and K = Q(v^)1. Then in Ok exists the smallest unit T| that is greater than 1, r| > 1. 2. Every unit of Ok is the form u = ±r|'! with n 6 Z. Proof 1. If d 1 mod 4 then Ok = {a + b\/d\a,b E Z}. If d=\ mod 4 then Ok = { :a = We will restrict the proof to the case d m°d 2, a, b E Z}. 1 mod 4. The proof for d = 1 mod 4 is almost the same and we will skip it. Let U be the group of units of Ok, and define 5 = {u E U : u > 1}, 5 1 = {u E S : u — a + bs/d,ab > 0,a,b E Z}, and S- = {u E S : u = a + b^d,ab <0,a,b E Z}. Then S = 5 U S . We shall show that there are no units r| > 1 in S- . For this suppose thatT] > 1 is in S_. Thenr| = a + b^d and ab < 0. Let T| =a — bs/d. Then r)T) = a2 — b2d = ±1, because it is a unit in Z. Hence T]-1 = ±r|. Now, a(— b) > Chapter 2 32 0, so r|-1 = |a| + \b\y/d. However T]-1 < 1. This leaves the only possibility that r| = |n|, a contradiction. We conclude that all units T] > 1 are in 5+. Clearly S+ has the smallest unit T| > 1 because a and b are positive integers. Therefore, the smallest unit T| > 1 exists in Ok- 2. Now we prove that unit in Ok is of the form u = ±r|" with n G Z. Consider first case u > 1. By the choice of T|, we get u r| > 1. There exists a positive integer n, such that T|"+1 > u T|" Hence, r| > 1. By the choice of T|, we conclude that = 1, so u = T|”. Now consider the case 0 < u < 1. Then m-1 > 1, and we get u integer n, so u = T]-”, ' = T|” with a positive — n G Z. Finally, the case u < 0, can be reduced to previous case by considering the unit —u > 0. Clearly, — u = T|” with same n G Z. Therefore, every unit in Ok is of the form ±T|", n G Z. (Trivially, it is also true for m = ±1) Definition 28 (Fundamental unit). Let d > 1 be a squrefree integer and K = Q^d\ Then the unit T| > 1 described in the Theorem 9 is called the fundamental unit ofK. The fundamental unit of K = Q(a/^) can be found by expansion of \/ d into a continued fraction. s/d = a0 + «1 4 Chapter 2 33 ai,i= 1, 2, 3 . . . are defined by ao = [>/<7J , *0 = «0 — L^J ={^} = fractional part of s/d. If an and xn are already defined, then we define an+\ = [j-] and x„+i = {^-}. The continued fraction for any irrational number is infinite. For brevity we write s/d = [<70,01,02, «3, ], rather then writing compound fraction. In 1770, Lagrange proved ••• that continued fractions for quadratic irrationalities are always periodic. In case of \/d we get Vd = [«0;, «1 , «2, ,O/,O1,<12,O3, . . .] = [«©;, «1 , «2, ,O/] which means that cq , <72 , . . . , 0/ is the period of the expansion and / is its length. The finite part [ 1 be squarefree integer, K = ^y/d}, and suppose that [<7o; «i«2 is the continued fraction expansion ofs/d. Denote the fundamental unit of K by T|. Then 1. Ifd^l mod 4 or d=l mod 8 then T| = hiy + ki-i s/d. 2. If d = 5 mod 8 and there are positive odd divisors A ofhi-i and B ofki-\, such that A < 2h^_v and B < 2(^)3 and A3 + 3AB2d = 8/?/_i and 3A2B + B3d = 81/_| then T| = A+^d otherwise = T| In either case 2V(t|) = (— l)b + ki-\sfd. Chapter 2 34 Determining the fundamental unit q of Oq, for positive squarefree integer d. =0, Step 1: h-x = Po = O,0o = l,oo = [Vd],ho = = 1- Step 2: Calculate Pn. Qn,an,hn,kn for n = 1,23.... Pn = Pn— 1 4“ ^n—lQn—1 7? = 1,2,3..., P„ n — ~q 7 , 0 — 1,2,3..., eW [5^],n= 1,2,3.., o„ = kn — 4“ 2, P — 1,2,3..., kn = ctnkn—\ + kn-2, n = 1,2,3..., Stop at Nth step when Pn = P[,Qn = QiStep 3: Put I = N —1 Step 4: If d = 2 mod 4, d = 3 mod 4, or d = 1 mod 8 then r| = h^+ki^Vd.N^ = (-l)z 1 /3 Step 5: If d = 5 mod 8 find all positive odd divisor A of h/=i less than 2^0^ and all positive odd divisor B of kj-\ less than Uki-i/m)^3 . If for some pair {A^B) we have A3 + 3AB2d = Sh,_^3A2B + B3d = 8^_i then q= A+^ ,N(r\) = (-l)z; otherwise we have X] = hi-l+kt_l\/d,N(T[) = (-l)z. For example, Suppose d = 13 so 13 = 5(mod&) and V13 = 3.6055 oto = \/l 3. _ _ J, _ J, 71j+3 _ 3+713 1 n1 ao-[ao] 4 713-3 713-3 713+3 _4 1 713+1 1+713 = L = = 2= „_ 3 „ 4 ai-[ai] 1 _ 3++13 _t 1 a2-[a2] 1+713-1 i i 2+yi3_1 a3-[a3] 3 713-1713+1 713+2 2+713 3 713-2 713+2 _2 713+1 1+713 713-1 713+1 = 4 Chapter 2 35 1 n5 = Cl4-[a4] = 1 6 — a5-[a5] ,/z 1 4 ^13+3 3+^13 = 713— 3 713+3 = 1 l+^-l _ _ 1 713+3 _ 3+77 _ 1 1 4 3+77-6 77-3 77+3 <70 = [a0] = 3, <7i = [ai] = 1,02 = [a?] = 1,03 = [«3] = 1, <74 = [«4] = 1, <75 = [«5] = 6, <76 = [ag] = 1. /1—1 = 1 , /10 = 3, h„ = anhn—\ 7 hn—2 /11 = <7i/?o + /1-1 = 4 /1? = «2/11 + /io = 7 /13 = <73/12 7 /11 = 11 /14 = <74/13 + /12 = 18 /15 = «5/147/13 = 119 /16 = «6^5 + /44 =137 /1—1 = 1, /10 = 1 /1« = ^nkn—i 7 kn—i 1 k\ = a\kn + k-i = 1 ^2 = «2^1 7 /10 = 2 k^ = a^k^ 7 /11 =3 k^ = a^kj 7 /12 = 5 /15 = <75/14 7 £3 = 33 k^ = a^ks 7 /14 = 38 N = 6J = N-i = 5Ju-i = 18,^-! =5, A is odd, A|/i/_i, 1 A 2/1^^ => A|18, 1 A 5.3, => A = 1 or 3. Bis odd,B|^_i,l fimdamentahmit == proc( d, steps ) P. Q, a, n, I, T. A. B, SerA, SerB, m, H, K := 0: /:[-l] := 1 P[0] == O:0[O] := 1 :a[0] floor(V7) : 74 ®1 floor(V^) :A|0] 1 T>— 0. prinif^ n Pn Qn an hn kn n"); local /i — for ^from 1 tojfepjdo -P[n— 1] -I- a(n P[ n] Q[»] floor L Jrr -/ I CM a[n] 1- floor —— — 1] Q\n— 1] : ’ : —— ) //[ n] >" a[n| /;| n 1 1 4- n 2 ] . 4 n] a[ 11 ] A| n 1 ] 4- 4 11 2 ] : printfl'' “oSd %5d %8d %8d °o8d °o8d n”tn,P|n], pjn], of n], /r[n]. ip’!) : ifiz> 1 and P| n] = P| 1 ] and @| u] = 1 ] then break end if end do / 1— n 1 : ifr/mod4 2 or", 1 8 hn kn 1 3 unit - a +b sqrt(d), 1 2 3 14 17 3 > Jundameniahmit^ 8 2, 20 ) Pn 1 2 unit an Qn 9 9 = a +b sqrt (d) , 1 1 18 18 163 2943 where a and b are 9 "N(unit)=", -1 18 325 1 Chapter 2 40 > fioidainenrolimit( 9 7, 20 ) Pn 1 2 3 4 5 6 7 8 9 10 11 12 unit an Qn 9 7 8 3 5 4 5 3 8 7 9 9 = a +b sqrt (d) t hn 16 3 11 8 9 9 8 11 3 16 1 16 kn 1 5 1 1 1 1 1 1 5 1 18 1 10 59 69 128 197 325 522 847 4757 5604 105629 111233 where a and b are "N(unit>", 1 5604 — 1 6 7 13 20 33 53 86 483 569 10725 11294 569 Chapter 3 Ideal Class Group 3.0.1 Legendre symbol Legendre symbol is a multiplicative function with values —1,0, 1. Definition 29 (Legendre symbol). Let p be an odd prime. An integer a is a quadratic residue modulo p if it is congruent to a perfect square modulo p and a quadratic nonresidue modulo p otherwise. The Legendre symbol is a function of a and p defined as if a is a square modulo p and a 0 mod p if a is a nonquadratic modulo p if a = 0 mod p Properties of the Legendre symbol; Suppose p and q are odd primes, and a and b are integers not divisible by p, the following properties for Legendre symbol hold; 1. If a = b mod p, then ") = 0) 41 Chapter 3 42 if p = 1 mod 4 if p = — 1 mod 4 7 (?)(?)=(-»w The last property is known as the famous quadratic reciprocity law. For example, consider (^) = = (— now consider n) = (^) (^) (||) (y) = 0) = -l= (— consider = m=m = mm=K-D=-i consider =(-D™(^)=m = ® = (S) = 1- Thus, (^) = (£)(£) = (-l)(-l)l = l. Take one more example, consider ® = (^) (109) (iw) (109)’ , x 2^1 *09-1 /109\ 2 now consider / 3 J\ = (—1) 2 •j Chapter 3 43 consider (^) = (-1)^' ' (^) =m=«)= (?)=iconsider (^) = ThuS, (122) - (^5) (j^g) Kronecker symbol Let n be a nonzero integer with prime factorization ek n^u.p^£1 ...pk\ where u is unit, and p, are primes. Let a be an integer. The Kronecker symbol is defined by For odd p,, the number is simply the usual Legendre symbol. The case when p, = 2, we define (") by if a is even if a = ±1 mod 8 ifa = ±3 mod 8. The quantity ($) = 1 when u = 1. When u— —1, we define it by a ^1 — 1, if a < 0 1, if a 0 Chapter 3 44 Finally, when w = 0, /nx 0 1, if a = ±1 0, otherwise. = Basic properties of the Kronecker symbol: 1. (^) = ±1, if gcd(a,n) = 1, otherwise (J) = 0. 2. (^) ~ (») («) unless w = — 1, one of a,b is zero and other one is negative. 3. ( w) = («) ( v) unless n = — L one of u, v is zero and other one has odd part congruent to 3 mod 4 4. For u > 0, we have whenever a = b mod u If a, b have same sign, the same also holds for u < 0. 5. For a ^3 mod 4, a / 0, we have u = v mod < = (") whenever 4|a| a=2 mod 4 otherwise. Examples of Kronecker symbol Consider (£) = (3^) = (|) (2) = (-1)1 = -1 One more example, Chapter 3 45 consider (£) = (^) = (|) (J) = 1.1 = 1 3.0.2 Ideal class group Definition 30 (Discriminant of algebraic number field). Let K be an algebraic number field of degree n. Let {r|i,T|2, • • ,1],,} be an integral basis for K. Then D(r|1 , T|2 , ... ,T|„) is called the discriminant ofK and denoted by d(K\ Theorem 10. Let K be a quadratic number field. Let d be the unique squarefree integer such that K = Q(a/J). Then discriminant d(Kj ofK is given by d(K) = < 4d, ifdfii mod 4 if d = 1 mod 4 d, Proof If d 1 mod 4, an integral basis for K is {1, d(K} = 1 Vd If d = 1 mod 4, an integral basis for K is {1, l+Jd d^K) = 2 1-Vd so that = (-2Vd)2 = 4d. ^-4} so that = ( —Vd)2 = d. 2 Definition 31 (Ideal class group). In number theory, the ideal class group of algebraic number field K is the quotient group where Jk is the group of fractional ideals of the ring of integers K, and Pk is the subgroup of principal ideals in Jk, is denoted by H(K). Chapter 3 46 If all ideals of the ring of integers of an algebraic number field are principal, then the ring is a principal ideal domain. In such case the ideal class group is trivial. For example, Z, Z[i], Z[co], where i is forth root of unity and (0 is cube root of unity, are all principal domains so the corresponding fields have ideal class number one, they have trivial class group. We are going to prove that the ideal class group for any number field K is finite. For this we will need following theorems from geometry of numbers. Theorem 11 ( [1], Theorem 12.5.1). Let K = Q(0) be an algebraic number field of degree n = r+2s, where 0 has r real conjugates and s pairs of nonreal complex conjugates. Let A be an integral or fractional ideal of Ok- Then there exists an element a(^ 0) G A, such that N(Ay\d[K)\ For the proof we need following lemmas. Lemma 4. Let 5(R") be a centrally symmetric convex body of volume V(5) contains a lattice point 2". Then S 0. The following theorem about linear forms is formulated in the format that is suitable for Theorem 11. Lemma 5. (Minkowski’s linear forms theorem) Let A = [ajk]nxn he a complex matrix, such that a^ G Rfor j = 1 , 2, ... , r and k = 1, 2, . . . , n and aj+sk = ajk for j = r + 1, . .. ,r + s;k = 1,2,. . . ,n. Chapter 3 47 Suppose that positive real numbers 81 , ... , 5„ satisfy’ the following conditions |det(a#-)| §1 • • 8„ and 8j = 8j+s,j = r+l,...,r+s. Then the system of linear in equations n Y^jkSk 87, j = 1,2,. ..,n k=l has a solution in integers yi , . . . ,yn, not all zero. Theorem 12. Let K = Q(0) be an algebraic number field of degree n = r + 2s, where 0 has r real conjugates and s pairs ofnonreal complex conjugates. Let A be an integral or fractional ideal of Ok- Then there exists an element a(^ 0) G A such that IW)! Proof. Let 0] , 02, • • ; 0« be the conjugates of 0. We reorder 0i , 02, . .. , 0„ in such a way that 01 , 02, . . . , 0r G R and 0,+i , 0,+2, ...,0„GC\ R. The complex conjugate of 0 is also an alge¬ braic conjugate of 0,we can further order 0/+i , 0r+2, • • • , 0» so that 0,+j+i = 0,+i , . . . , 0„ = 9r+2s = 0r+5 where r + 2s = n. Let Gj , . . . be n monomorphism Gj : K C is chosen so that o,(0) = 0,. Hence ar+i+i = 6r+t(t = 1, ... ,5). Let {ai,...,an} be a basis for A. We define linear forms Lj(x)(j = 1,2,..., nf where x = (xi,...,x„), by n T/(x) = F r-=i Chapter 3 48 (j,k= 1,2, ... ,n). These forms satisfy Minkowski’s linear forms theorem with Moreover, | det(u#-)| = | det(G7(a^)| = y/|D(A)| = N(A) a/KO 0 Let 5; = ( | det(K)| V2”, j = 1, 2, . . . , n Then s /2 V MWCO1/2 = \ 71/ Sr |det(o^)| so, by Minkowski’s linear forms theorem, there exist integers yi,j2, ,yn, not all zero, such that IW N(A)l/n\d(K)\l/2n,j = 1,2, ...,n Choose m G 1 , 2, . . . , n such that 6m = I, where I denotes the identity monomorphism from K to K . Set n Ct = ^m(y) = n ^mW'fyk = k=l so that a G A and a / 0. The conjugates of a are ^kyk k=l Chapter 3 49 n 0 which differ greatly in nature. Case d<0 Then det(A) = -^d > 0. If a > 0 then we must also have c > 0 and the symmetric matrix A is called positive definite. This is because in such as > 0 for any nonzero x y GR Chapter 3 52 It follows that in such case f(x,y) > 0 whenever (x,y) f (0,0). If d < 0 and a < 0 then c < 0 and f(x,y) < 0 for any (x, y) matrix is negative definite. However, in this case (0, 0) then we say corresponding — f(x,y) is positive definite. Hence there is no need to study such forms separately, so we assume that a > 0 ( and consequently c > 0) in the case d < 0. As we have realized, the set of integers that can be represented by a given form f is exactly the same for any form in the equivalence class containing f. In order to count the classes of equivalent forms it is convenient to find in such class a special representative, called canonical form, or reduced form. Definition 34 (Reduced positive form). A positive definite form is called reduced if \b\ a c. Proposition.(This is Theorem 2.4 in [3]) Every positive definite form with discriminant d < 0 is equivalent to a unique reduced form. Note: The uniqueness requires a convention that the coefficient of xy is nonnegative in reduced forms of two exceptions ax2 + bxy + ay2 and ax2 + axy + cy2 which otherwise would have two reduced forms. With fixed discriminant d < 0, it follows easily from this theorem that number of classes for d < 0 is finite. Case d>0 In this case f(xyy) can represent positive and negative integers, and we call such form indef¬ inite. The definition of reduced form must be modified. Definition 35 (Reduced indefinite form). Suppose that f(x,y) = ax2 + bxy + cy2 has discriminant d > 0. Then we say f is reduced if 0 0. One-to-one correspondence between classes of equivalent forms and classes of ideals Recall that the class group of ideals of an algebraic number field K is given by ‘K where Ik is the multiplicative group of fractional ideals in Ok and Pk is its subgroup of principal ideals. Also recall two principal ideals (o) and (&) are identical if and only if b = ua, where u is unit in Ok- We shall now focus on quadratic number fields K = where d is squarefree integer, d fL Narrow class group Definition 36 (Positive principal ideals). Let K be a quadratic field, and I be a principal (fractional) ideal of Ok- If I = (a) and N(a) > 0, we say that I is positive. Propostion 12. The positive principal (fractional) ideals of Ok form a group, denoted by Pf Proof Since (a)(b) = (ab), the theorem follows by multiplication of norm, N(ab) = N(a)N(b). Note: 1. If d < 0 then P^ = P. This is because all norms of nonzero algebraic integers are positive N(a + bs/d) = a2 + \d\b2. Chapter 3 54 — 2. Suppose that there is a unit u in Ok with negative norm, N(u) = 1. Then the ideals (a) and (ua) are the same. If N(a) is negative then Notin') = N(u)N(a) is positive. Hence we can choose a generator of any principal ideal with a positive norm. Hence again Pf = PK. 3. If d > 0 and there is no unit u in Ok with N(u) = — 1 then P^ is a proper subgroup of Pk of index 2. That is [Pk : P£] = 2. Obviously in this case there exists an element b E Ok with N(b) < 0, and the ideal (b) has no generator with positive norm. It is easy to see that in this case there are exactly two cosets of P^ in Pk, namely P^ and bP^, , as shown in the following Lemma. Definition 37 (Narrow ideal class group). The narrow ideal class group of a quadratic field K is defined by rK Lemma 6. P^ is a subgroup of Pk, of index 2, that is, [Pk : P^] = 2. Proof. Let n E Ok be element of negative norm. Obviously such element exists, for example bl+^f Then N(n) = a2 — b2d or (N(n) = a2 + ab — b2 ^-). So,if b is sufficiently large then N(n) < 0. Clearly P^ is subgroup of Pk, and it has two cosets: P^ and (n)P^. This is because if an ideal J E Pk\ Pk^ then J = (a) and N(a) < 0, but n = a + bs/d, (or n = a + J = (n) (^) , and (^) E P^, so J E (n^P^. So, there are exactly two cosets, Pf and (n)P^. In [3] on page 101 Theorem 6.19 we find the following statement. Propostion 13. If d > 1 is a squarefree integer, K = ^\/d) and Ok has no unit with norm -1 then the ideals class group H(K) is isomorphic to the subgroup of squares of the narrow class group H+(Kf Chapter 3 55 This statement is actually false. In an article in MathOverflow [9] the example of K = Q(a/210) was considered. In fact = ZS®Zn where H(K)+ = Z2® ^2® ^2- However, the squares of H(K)+ form a trivial group, so h(K) would be equal to 1, while in fact, h^K) = 4. We will calculate these values in Chapter 4. Propostion 14. h(K)~ = 2h(K} for any quadratic field K = Q(\/tZ) if d > 1 when there is no unit in Ok with norm —1. Proof. First proof We have P(K} and P+(K) are normal subgroup of I{K), P+(K) C P^K). So by the third isomorphism theorem for groups we get l(K) p+W P+(K) I(K) P(K) Hence, PiKyP+m '' Since [PK : P^] = 2 as shown in Lemma 6, we get h+(K) = 2h(K^ Second direct proof Let S = {JiP+(Kfi. . . , JmP+(Kf} be the set of all distinct cosets of P+(K) in H+(Kf Now, fi)JiP+(K) must be equal to one of the cosets in S. and it is not equal to JiP+(K). With¬ out loss of generality suppose (n)J]P+(K) = JzP+(K\ We can continue this process, now starting with J^P+(K), etc...until we obtain a new set of m cosets of the form 5 = { fP+ (K) fn)fP+(Kfi..., JkP+ (K) , in) JkP+ (K) } . We conclude that m = 2k. By Lemma 6, JrP(K) = JP+(K) \JffJP+(K) for r = 1,2, .. . ,k. Chapter 3 Hence H(K) 56 — has k cosets of P(K) where H+(K) has m — 2k cosets of and the theorem follows. The following theorem relates the number of equivalence classes of binary quadratic forms and the ideal class number of an algebraic number field. Theorem 13. The narrow ideal class number h+(K) of an algebraic number field K is equal to the number of equivalence classes of binary quadratic forms ax2 +bxy + cy2 of discrimi¬ nant d(K). Thus h(K), the class number of K is equal to the number of equivalence classes of binary quadratic forms if Ok has a unit with norm — 1 and h^K) equals to the half of the number of equivalence classes of binary quadratic forms otherwise. Note : In fact, much more than Theoreml3 is known: The equivalence classes of binary quadratic forms form a group under "composition" of forms, which however we are not studying in this thesis. This group is isomorphic to H (K). In what follows we are exploring D.B. Zagier exposition [11]. The isomorphic is in the following way: Let J be an ideal (integral or fractional) in Ok and let (a, 0) be its integral basis then we associate with the following quadratic form (xa + y0)(xa/+y0/) NO) see [11] page(93). Here a' and 0' are algebraic associate of a and 0 respectively. The quadratic form at the right hand side has integral coefficient, because since a G J we have and similarly W)I(P). Example : In Example 1 of Section 1.0.5 we have J = (3 + 5\/7, 5 — 9\/7) we have found integral basis ( 1 + y7, 2) and N(J) = 2. Hence the corresponding form is (x(l+V7)+v2Xx(l—/7)+y2) _ _3%2 _|_ 2^ _|_ 2y2 and its discriminant D = 22 — 4 x — 3x2 = 28. Chapter 3 57 which is same as the field discriminant of K = Q(^/7) i-e. d(K) =4x7 = 28. On the other hand, given a binary quadratic form ax2 4- bxy + cy2 with discriminant D < 0, we map it into a fractional ideal in the way 9 9 ax~ + bxy + cy- —> J = Z + Z b 4- \/~D 2a (3.1) or, alternatively, we can map the form into an integral ideal ax2 + bxy + cy2 —> f = Z(2a) + Z(i>+ \/D) (3.2) Clearly, J' = (2a)J, so both ideals are equivalent modulo principal ideals in 3.1 and 3.2. It is easy to check that the sets on right hand sides are ideals indeed and their integral basis are (1, b+^ ) and (2a,b 4- y/D) respectively. Propostion 15. J = Z(2a) + ^(b + x/D), where D = b2 — 4ac, and a, ft, c G Z is in fact an ideal of Ok, K = Qfv^). Proof Case-1 D = 0 mod 4. In this case Ok = {x + y^D : x,y G Z}. It is clear that J is a group under addition. It re¬ mains to show that it is closed under multiplication by elements from Ok- Let n G J, then n = zi (2a) + Z2(b + VD) with integers zi and Z2- The product (x+y^D)n can be rearranged as (x + yy/tyn = zix(2a) + z\2ay(b + y/D) - ~iby(2a) + z2x(b+x/D~) + -jybfb + VO) - Z22yc(2af where each component belongs to J. This complete the first part of case-1. Case-2 D = 1 mod 4. Now Ok = { : x,y G Z , x = y mod 4}. If x and y are even we can reduce to case- 1 , so suppose that x and y are odd. Notice also that D = 1 mod 4 implies that also b is odd. Now we can rearrange as popows x + y^D 2 x+yy/b +z2^ + >fD^ = Chapter 3 58 z\a(x-by) + ziay(b +Vd) +z2 . This is an element of J because x — by is even and (b + Vd) + z2yc(2a) G Z, since x, y and b are odd Note: The construction of J provides also its integral basis, which is {2a, b + Vd}. This allows to calculate the norm of J. We use this fact in calculation of the norms of ideals in examples for Carlitz’s theorem. In case of d > 0 we modify maps 3.1 and 3.2 by multiplying the integral basis of the ideals by any element with negative norm. For this purpose X = \/d will do, as NiX) = —Vd x y/d = —d. So, we have the map ax2 + bxy + cy2 —> ZX + Z^^X. L. Dirichlet derived the formulas for ideal class number of quadratic number fields by count¬ ing the number of equivalence classes of quadratic forms. However, in the next section, the calculations of using these formulas are presented. Here, we just show two easy algorithms for counting the equivalence classes for binary quadratic form directly. In the case if d < 0 the form ax2 + bxy + cy2 is reduced if \b\ hence \b\ a c, see [3],Proposition [2.1] This restriction together with D = b2 — ^ac, D = d{K), K = ambiguous forms leads to the following procedure and the elimination of Chapter 3 59 > ReducedPositiveForms :=proc(i) ^description "Numbers and lists unequivalent reduced positive forms with discriminant Delta" locals, b, c. n : n := 0 : for a from max |6|, I do to floor if(#" + d) mod 4=0 and _ (*2 + a) . ‘ modo = 0 then 4-n if not( (b < 0 and c = a I or ( b < 0 and h =-a) ) then « t= a + 1 : print(n, "form", a, b,c) end if: end if: end do: end do: end proc: In this case the number of classes of quadratic forms equal h(K\ In the case of d > 0 we have > IndefiniteReducedForms ;=proc(A local S, a, b, c : for b from 1 to lloor(S) do fS—b\ ,, ( S + b , « -— do for to floor a from ceil I 2 2 J \ J —-— -— 7 I 4«— r if floor(c) c then print (a, b, c) : print ( -a, b,-c) end if: end do end do: end proc: This procedure just lists the reduced forms of discriminant D = d(K). Then we still need to group the reduced forms in chains as shown in the proof of Proposition 3.6 in [3]. We form pairs of equivalent reduced forms according to the rule (a,b,c) ~ (c,b' , c') in which Chapter 3 60 c < 0 and a and c' are positive. For d = 210, D = 4 x 210. We can use cycles to determine the structure of the group H^K). In order to do this we take any form each cycle and convert it into an ideal of Ok, for K = ^Vd), according to the formula (a,b,c^Z(2a) + Z{b + y/D). Our procedure gives the following results: Finally, d=2 1 0, so d(k)= 840 This is the example showing that a proposotion is false > It rcAerfi 5 (840) 5.20, -22 -5. 20, 22 10,20,-11 -10, 20, 11 11,20, -10 -11,20, 10 22. 20. -5 -22,20, 5 3.24. -22 -3. 24, 22 6. 24. I ] -6.24, 11 11,24, —6 - -11,24.6 22. 24, -3 -22,24,3 1.28, -14 -1,28, 14 2,28. -7 -2,28, 7 7, 28. -2 -7,28,2 14, 28, -I -14,28, 1 Chapter 3 61 Cycles ~ 1. (5,20,-22) (-22,24,3)- (3,24,-22)~(-22,20,5) 2.(-5,20,22)~(22,224,-3)~(-3,24,22)-(22,20,-5) 3. (-6,24, ll)~(ll,20=-10)~(-10.20,ll)~ (11.24,-6) 4. (6,24,-11)~(-11.20,10)~(10,20.-11)~(-11.24,6) 5. (1,28,-14)— (-14,28,1) 6. (-1,28.14), (14.(28.-1) 7. (2,28,-7)~(-7,28,2) a8 (-2,28, 7)~(7,28,-2)| Fundamental unit q=29 -2^ 210 ,.V(r|) 1 .Hence (X) 8, /»(/0=4. = = The first two cycles give the same ideal = (10,20 + X4 x 210), the cycles [3] and [4] give /2 = (6,24 + X4 x 210), [5] and [6] give 73 = (1,28 + X4 x 210) and [7] and [8] give 74 = (2,28 + X4x 210). Since these ideals are representatives of classes modulo principal ideals, we can find equiv¬ alent ideals by dividing both generators by an integer. Thus 71 = (10,20 + 2X210) = (5,2X210) = (2)(5,X210) = (4) (25, 5X210,210) = (4) (5, 5 X210) because gcd(25, 210) = 5 further 72 ~ (1,2X210) = 0^=1. 72 = (12,24 + 2^210) = (12, 2^210) ~ (2)(6,X210), so 72 = (4) (36, 6X2 1 0. 210) = (24) ( 1 , v'2 1 0) so 7^ = 1. 73 = (2,28 + 2X210) = (2,2X210) = (2)(1, \/210) = (2), so73 = 1. 74 = (4,28 + 2X210) = (4,2X210) (8)(l,X210) = (8) so742 = 1. - (2)(2,X2W), 7^ = (4) (4, 2X210, 210) = (4) (2, 2X210) = We used the fact that if an ideal contains a unit is equal to Therefore contains four elements; neutral and three of order 2. Hence H(K) = Z2®Z2, which is the four Klein group. As for H+(K), it contains eight elements, so it can be isomorphic to one of Zg, Z4®Z2, or Z2®Z2® Z2. The squares of these groups form Z4, Z2, or one element group respectively, so non of them is Z2 ® Z2, showing that Proposition 13 is false. Chapter 3 62 The existence of a unit of norm — 1 in Ok For df\ mod 4, we have N(a + bVd) = a2 — db2 Hence there is no unit norm — 1 if and only if the negative Pell equation x2 — dy2 = — 1 has no solution. It is known that the negative Pell equation has solution if and only if period of continuous fraction expression of is odd. For d = 1 mod 4, the corresponding equation is x2 — dy2 = —4. By considering parity of x and y. we conclude that this equation is equivalent to x2 — dy2 = — 1 for d 1 mod 4. so it is more general. — When Ok for Q(^f) does not have a unit of norm 1? This is important question because for such real d > 0, h(K) h+{K) In general solution to this question is unknown, however there are partial results. For exam¬ ple in [1] we find the following facts; Theorem 14. Let d be a prime with d = 1 mod 4. Then the fundamental unit ofO^^ has norm — 1. Theorem 15. Let d be a prime with d = 5 mod 8. Then the fundamental unit of has norm — 1. Theorem 16. Let p and q be distinct primes such that — p = q= 1 mod 4, ( ) = \qj Then the fundamental unit of O^y^-^ has norm — 1. —1 O^^^ Chapter 3 63 3.0.4 Dirichlet’s class number formula Dirichlet’s class number formula was conjectured by Jacobi in 1832 and was proved by Dirichlet in 1839. There are many expositions of this proof with variety of shortcuts and simplifications. Here, we follow the detailed proof from [11], The proof is quite sophisti¬ cated and we describe only major steps. The main idea is to relate the class number h(D) of classes of quadratic forms of discriminant D with the number of solution of the equation 7 7 t ax + bxy + cy = n, (3.3) where n is positive integer. We consider only primitive forms (a, b. c) that is with gcd(a, b. c) = 1 and for which D is a fundamental discriminant that is either D = 0 mod 4 or D = 1 mod 4. This corresponds to field discriminants K = ^s/d} where d is squarefree integer and field discriminant d(K) = D = 4d if d 1 mod 4 or D = d if d = 1 mod 4. For a form f(x,y) = ax2 + bxy + cy2 denoted also by (n, b. c), let R[n) denote the number of solutions in integers (x, y) of the equation 3.3 counted in a way described below. For a given form f(x,y) x — ox2 + bxy + cy2 with matrix A — — ax' + Py', y = yr' + Sy', where M = 6 there are substitution and MT AM = A. So after this substitution we have f(x,y) = f(x',yf where We say that such substitution is an automorphism of f, and consider the solutions (x, y) and (x',y') of (3.3) as equivalent. The number R(n) denotes the number of non equivalent solu¬ tions. The set of such matrices M forms a group. Uf = {M G SL2^ : MTAM ~ A}. There exist a bijection between the set of solutions of the Pell’s equation t2 Uf, given by — u2D = 4 and Chapter 3 64 t- bu 2 —cu an t+bu 2 EUf. The equation t2 — u2D = 4 gives all the units with norm 1 of Ok for K = Q(v^) = Q(\/D). The units are of the form for both cases D = 0 mod 4 and D = 1 mod 4. Thus for D < 0 the number of automorphisms w = \Uf\ is finite,and Uf is cyclic. We have 6 if D = -3 w(D) = < 4 if D = -4 2 if D < —4 which is exactly the number of units in Ok for the corresponding field K. For D > 0, Uf is infinite and Uf = In either case /?(/,«) is finite. Further, let MD) R(n) = E R(fi,n) (3.4) i=i where /i,/2, ••• ,//;(£>) denote any representative of non equivalent binary primitive integer with discriminant D. The next major step is showing that Rff = E„,|„Zd(zi)’ where Xd = (^) is the Kronecker symbol. While it is difficult to obtain a closed form for R(n), it is possible to do so for the average of R(n) when n changes 1 to infinity. We have lim 1 N E v n—l M = where L(s,Xd) denotes, the so called L— series with character Xd, = L7=i In 3.5 we just take 5=1. Further, for any primitive form of with discriminant D and for n > 0 we have 2q_ , »V|D| logep , fD for D<0 (3.6) for D>0 Chapter 3 65 where £o = is the smallest positive solution of 3.5 with to > 0 and uq > 0. All solutions of 3.5 are of the form ±£q where n G Z. Clearly £o = T|, the fundamental unit. By connecting (3.5) and (3.6) we get D<0 for logen^l’^) (3.7) D>0 for It remains to calculate L(l, %p). We are following [11] and we will only outline the main steps. By using Gauss sum G= 77=1 we get 1 N ^t^e~2^N. XW = u n=l This leads to k=l K k=l n=l It foils to .. M1,X) = In our case \ 1 V’ - / \ 7t TC/? L ~ 7t77 \ “ Vv ' is a real character (Kronecker symbol) and this formula simplifies to 71 1^1 |PH L «XW f°r D<°’ 71=1 and 1 Tin x(»)logsin— VD £ for D > 0. 77=1 For D > 0 in the book ( [2]) a simplification is given. Namely for D > 0, we have Xd = Xd(D — n), and also sin(7t — ^) = sin^). By combining this remark and equations Chapter 3 66 (3.4), (3.5), and (3.7) , the formula for h(D) in case D > 0 becomes o/ - «<§4 nJT If the fundamental unit of Ok has norm 1, then h(K) = where h(D) is the number of classes of quadratic forms and h(K) is the class number of K^\/DY We get 1 = “hln n'J 5 JlTT Xo(«)logsin(— ) (3.8) withr] = Eq. If Ok has fundamental unitr| with 7V(r|) = —1 then £o = T|2 so logEo = 21ogr|, again we get same formula 3.8. The case D < 0 is somewhat easier and equations (3.4), (3.5), and (3.7) give — w(D~) = h(K) = 2|^| E X /7=1 "U This formula is the same for the number of classes of quadratic forms and classes of ideals. 3.0.5 Numerical examples of the Dirichlet class number formula Dirichlet’s class number formula for d < 0 Let K be quadratic field of discriminant d. Then = W(D) y|t/| 1 2| discriminant D = 3 and w(D) = 6 Chapter 3 67 Put these values in the formula, (4 h^ = = -l{l(-3)+2(-3)} = — 1{1(1) + 2(— 1)} = -!{-!} = 1 Calculation of by counting the number of classes of quadratic forms by procedure "positive" d— 3, sod(K)=-3 > ReducedPositi veForms( 3 ) L 1. 1 So h( K)= 1 Example 2 — W = 2.|-40| &O Suppose d = 10 => discriminant D = —40 and w(D) = 2. ^{im+2m+-+39(^)} = {i m +3 m cn +9 m + n +7 + 13 (4°) + 17 (4°) + 19 (4°) +21 (4°) +23 (4°) + 27(^)+29(<)+31(^)+33(^^ = ^{Kl) + 3(-l)+7(l) + 9(l) + 11(1) + 13(1) + 17(-1) + 19(1) +21(1) + 23(1) + 27(-l) + 29(-l) + 31(-l) + 33(-l) + 37(1) + 37(1) + 39(-l)} = ^{120-200} =2 Calculation of "positive" by counting the number of classes of quadratic forms by procedure Chapter 3 68 d=-10, d(K>-40 > ReducedPositiveForms( 40 ) l,"form", 1,0, 10 2, "form’', 2, 0, 5 So h(K)=2 Example 3 Suppose d = — 14 =+ discriminant D = —56 and w(D) = 2. + 19(1) +23(1) +25(1) +27(1) + 29(—l) + 31(—1) + 33(—l) + 37(—l) + 39(1) + 41(-l) + 43(-l)+45(l)+47(-l) + 51(-l) + 53(-l) + 55(-l)}. = {224 -448} =4 Calculation of h(K) by counting the number of classes of quadratic forms by procedure "positive" d=-14, d(K)=-56 > ReducedPositiveForms ( 56 ) So h(K)=4 I, "form". 3. —2, 5 2. "form", 1,0, 14 3. "form". 2, 0, 7 4. "form". 3, 2, 5 Chapter 3 69 Drichlet’s class formula for d > 0 logsin®. where T| is fundamental unit. Example 1 Suppose d = 2, we can calculate T] = 1 + >/2 and discriminant D = 8 logsin (?) + (1) logsin = + (I) logsin (t) } = i^loCp sin2 ai =1 liogr = ^(1+72)2 cl sin2 23 J _ -1 1 . 1— COS J , r log(l+V2) 2 1— cos -* =2 I loot = log(lW2)21O^TT^ - logHWV^M- J -1 _ 1 loaf^l ^±11f = ~ log(lW2)2 °W2+1 ^2+1 = Ml^)2log(1 + ^) 2 = Calculation of by counting the number of classes of quadratic forms by procedure "in¬ definite" d-2, d(K>-8. N(q>--I > lndefiniteReducedFonns(8 ) 1.2. -1 -1.2. I These two forms make one cycle ( U.- 1 ) ~ - 1 . 2. 1 ). Hence h( K>= I Example 2 Suppose d = 3, we can calculate q = 2 + >/3 and discriminant D=12 Chapter 3 70 + (t) ,Og sin + (f) log Sin 12 = log(2~W3) { (t) log + (y)logsin^ + (y) log sin = M^ lndefiniteReducedForms( 1 2 ) 1.2. -2 - 1.2.2 2, 2. - 1 -2.2. 1 Cycles (1.2.-2) ~( 2. 2. 1 ) and (2^,-1) ( I. 2, 2). -— — We have two cyles and there is no unit with negatve norm, hence h(K)~2/2=l. Example 3 Suppose d = 6, we can calculate T] = 5 + 2 \/6 and discriminant D = 24 < = log(5+2y6) ( T ) 1Qg sin 24 + ( T) 1Qg sin >+ ( ) 1+ (T) 1Qg sin 24 T 1Qg sin + ( ) log sin g + ( ) log sin § + ( ) log sin § + ( ) log sin ^ ^ ^ + (^)1ogsin|^ + (^)logsin^ + (^)logsin^} ^ = Iog(5+2V6) {log sin 54 + log sin S - log sin 24 - 1Qg sin -1 1O„( sin 24 sin g 1 = -* sin sin cos cos 1 r 6— 4 1 3 log(5+276) l°gl cos g— log(5+276) _ “ —1 cos^ A 1 = log(5+276) 10g 1 z v2 } Chapter 3 71 =1 1og J /Hl = Iog(5+276) ° '-76+2' 2_ zd_ log! | = log(5+2/6) 61 10+476 _ ' “I Ingf 1 _1 = .log(5+276) 5+276' = 1. Calculation of h(K) by counting the number of classes of quadratic forms by procedure "in¬ definite" d=6, d( K 1=24. Noir 1 = > hidefiniteReducedFornvi\ 24 | 1,4. -2 -1,4,2 2, 4, - 1 -2,4. I "cycles ( 1 .4.-2) - ( - 2. 4. 1 } and (2,4,- 1 ) - ( - 1 . 4. 2 ). 2 V ( q ) - I . so A ( A' ) = — = I . Chapter 4 Carlitz’s theorem 4.0.1 Detailed proof of Carlitz’s theorem Definition 38 (Order of a group). The order of a group is the number of element present in that group. We denote order of a group by |G|. Theorem 17 (Carltiz’s theorem). Let K be an algebraic number field. Then h(K) = 1 or 2 if and only if whenever a nonzero nonunit a £ Ok can be written as a = ua\a2-..as = uib[b2-..bt, where u and wi are units and a\,a2, ...bt are irreducible elements of Ok then s = t. Proof. Case I, h[K) = 1 If h(K) = 1 then every ideal in Ok is principal ideal, we already know that every P.I.D ring is a U.F.D. ring. Hence s = t. Case II, h(K) = 2. Now suppose that h{K) = 2. 72 Chapter 4 73 Let a be anon unit in Ok, and suppose that a = wen 02 • • where u is a unit, andai,a2, • • are irreducible elements in Ok- Then (a) = (a1)(a2)...(a„). Some of the ideals (aj, («2),. • • , (a„) in (a) may be prime ideals. Suppose that (ai ) . . . , (a^) are prime ideals, but ((Xa+1 ),..., (a,, ) are not. We know that the each integral ideal is product of prime ideals, so (o5) = pSl . . ,ps for 5 = k + 1 , . . . , n, and all ms 2 where all ideals pSi are prime. Hence (a) = where m is the sum of mSj and all ideals are prime. We shall show that for any choice of i and j, pipj is an integral principal ideal. The ideal class group of K has exactly two classes, H(K)=I(K)/P(K) = {1,A} where we denoted by 1 the neutral element P(K), and the class of ideals A is the coset of all non principal ideals Since (aj = pSl . . .ps , and as is irreducible, none of the ideals. pSl ps can not be principal, because if pSj = (4>) with not a unit then would divide as, but as is irreducible, so we would have (as) = pSj = but this is not possible, because mSj 2. Claim: Any product of two non principal ideals pi and pj is integral principal ideal. Proof. Denote by [p;] the class of ideals represented by p^. Then [p,] = A and [pj] = A, Chapter 4 74 because pi and pj are not principal ideals so [pi ] + 1 and [pj] 1. Hence, [p,][p7] = [aP;] = A2 = 1, because h(K) = 2. So pipj is a principal ideal. This ideal is integral because pi and pj are integral ideals. By matching the ideals p\ , . . . , pm in pairs in whatever manner we conclude that m must be even, so (a) = (ai)(a2) (^/2), where ay , . . . , a^, Tti , . . . , ^„/2 are irreducible elements of Ok- Note that if m were odd, we would have (tt) = (fli)...(^)(7ti)...(7tr)72,', but (a) is a principal ideal, and p, is not principal. This leads to a contradiction = (aKf1...^1), Pi = because pt is not principal. We conclude that (a) = (ai)...(a^(7ti)...(7tm/2), so (a) = (ai...ajtKi...Km/2). Hence a = ua] . . .«^1. . where u is a unit. Therefore a factors into k + m/2 irreducible factors. Since the factorization of ideal (a) into prime ideals is unique, every such factorization must have exactly k prime principal ideals and m pairwise non principal ideals factors. Hence a always factors into k + m/2 irreducible elements. Chapter 4 75 Case III h(K) > 2. Part-1 Suppose that H(K) has a class of order m > 2. Let A be such class and the order of A, |A| = m > 2. By Hecke’s Lemma [8] there is a prime ideal p\ in A, so A = [pi] . Since A"! = /, then A"1 = [pi]m = Ip'"] = 1, so p"1 = (Ki) is a principal ideal. We claim that 7t is an irreducible element of Ok- For the contradiction suppose that K = ab with nonunits a.b G OkBy the theorem on unique factorization of ideals, we conclude that (a) = pk, (b) = p\ with integers k, I with k + 1 = m,k \pn> 1. Then we have Ak = [pk] = |(a) | = 1, but the order of A is m > k. and we get a contradiction. In any group, the order of an inverse element is equal to the order of the element, so we have |A~ 1 1 = |A| = m. Similarly to A, A-1 also contains a prime ideal, say p2- Hence p™ = (712) , and by the same argument as above 7^ is irreducible. Now AA-1 = 1 implies that [pi][ps] = 1, so [P1P2] = 1, and we conclude that pipz = (7:3). Again, we claim that 713 is irreducible. For a contradiction, suppose 713 = ab with nonunits a,b G Ok- Then P1P2 = {ab) = (a)(b) Since p\ and P2 are not principal, (a) p\ or p2, {b) (4.1) p\ or p2- Hence {a) factors into at least two prime ideal factors and so does (b). Then L.H.S. of (4.1) has two prime ideal factors, while the R.H.S. has at least four ideal factors, a contradiction. So we have (pip2)m = P™Pi = (^1)^2) = (^1^2). and {p\P2)m = O3)"1 = (71'3 )• Chapter 4 76 So (jt'") = (KiTIo). Hence it"' = M7li7l2, with some unit u. We have m > 2 irreducible elements on the L.H.S. and two on the R.H.S. Part-2 Suppose that every element of H(K) has order 2, except of the class 1. We conclude that H(K') = ®"=i ^2- Since H(K) > 2, then n 2, Therefore H(K) contains a subgroup isomorphic to Z2 ® Z2. Hence there are classes Ai ,A2 in H(K) such that |Ai | = |A2| = 2, and AjA2 = A3, |A3| = 2. Similarly as Part-1, let pi be a prime ideal in Ai for i = 1,2,3, Then pj = (tt,) for i = 1,2,3. with irreducible elements 7ti , 7l2, 7l3. NowAiA2 = A3, and AiA2A3 =a3a3 = A2 = 1. Hence [pi][p2][p3] = 1, and p\P2P3 = (ft), with some 7t G Ok- By the same argument as in Part-1, we conclude that n is irreducible, P1P2P3 = (ft), and P1P2H = (ft2)- Thus (711 ft2ft3) = ft2. Hence 7t2 = u7li7l27l3, and we have different number of irreducible elements at both sides. 4.0.2 Examples illustrating Carlitz’s theorem Example 1 K = Q(\/— 5). This is most popular example in most textbooks on introductory number theory. Chapter 4 77 It is known that h(K) = 2, which can be checked by the method shown in Chapter 3. The only units in Ok are ±1. We have (2) (3) = (1 + a/— 5)(1 — >/— 5). Clearly all factor on both sides are irreducible, for example if — 2 = (o + by/^5) (c + t/>/^5) then 2 = (a — b\^5)(c d >/— 5) , so 4 = (a2 + 5Zr)(c2 + 5d2). Hence b = d = 0, a = ±2 and c = ±1 or a = ±1 and c = ±2. So that a + b\/^5 or c + d >/—5 is unit. So 2 is irreducible. In a similar way we can show that 3 and 1 + a/— 5 and 1 — >/— 5. Example 2 K = Q(>/— 13) here again h(K} = 2. The group units is {±1}. We have (2)(7) = (1 + ^13)(1-a/^I3). We conclude again that these elements are not associated. Further, if 7 = (a — /x/-13)(c + Ja/—13), then 7 = (a - a/^13) (c - dx/^13) and 49 = (a2 + 13/>2)(c2 + 13r/2). The right hand side can not factor as 7 x 7 and we conclude that one of that b = d = 0 and one of factors is a unit. The case for factor 2 is completely analogous. Also 1 ± >/—13 = (a + 7»/— 13) (c + d >/—13) leads to 14 = (a2 + 13Z?2)(c2 + 13d2) and we rule out factorization as 2 x 7 on right hand side. Hence one of the factors there is a unit. Example 3^ = Q(\/15). — — >/15) x (1 + \/15) = (2) x (7). The elements 2 or 7 are not associated with >/15 — 1 or a/15 + 1, because these elements have different norms; Here also h(K) = 2. We have ( 1) x (1 AM2) = 22 = 4, AM7) = 72 = 49, but Nk ( 1 ± a/15 ) = 14. Q Q Q It remains to show that these elements are irreducible. Suppose that 2 = (a+by/XS^c+dy/ 15). Since (^(x+y/^) = x— y/15 is an automorphism of K and identity on Q we get 2 = (a — b^XS^c — d>/15). Hence 4 = (a2 - X5b2)(c2 — 15d2). We cannot have a2 — 157>2 = ±2 because ±2 is not a square modulo 15 so one of the factors is a unit and 2 is irreducible in Ok- Chapter 4 78 Similarly, 7 is irreducible, because a2 — 15b~ = ±7 has no solution in integers, since ±7 is not a square modulo 15. Finally, if 1 ± /15 = (a + by/lS^c + c/x/15) then — 14 = (+2 — 15&2)(c2 — 15<72). For the reasons stated above, right hand side cannot factor as ±2 x ±7, so adjoin one of the factor there must be a unit. Hence also elements 1 ± /15 are irreducible. In the first three examples, we showed factorization of same length (equal 2), but with not associated factors. Now we are going to follow Part 2 of the proof Carlitz’s theorem to show a factorizations of distinct lengths for a field with h[K) = 4. Example 4 Let^ = Q(/210), so D = 4x210. Let Ji = 10Z + (20 + 2/210)Z, H is the ideal from example on page 60 of chapter 3. Let Ai = [Ji], We need to find a prime ideal in this class. We have [Ji] = [10Z + (20 + 2/210)Z] = [10Z + 2/2WZ] = [5Z + /2W] = [(5, /216)]. Let Pi = (5, /210). We claim that Pi is prime ideal. For this, notice that the construction of Pi implies that {5, >/210} is its integral basis,and Pp = (5) so we have N(Pi) = \J = 5. Since 5 is a prime number, we conclude that Pi is prime. Let 72 be the ideal corresponding to the form (10,24,-11) so 72 = (12, 24 + 2/210) and we getA2 = [72]. Hence A2 = [(6, 12 + /216)] = [(12 - /216) (6, 12+ /216)] = [(6(12- /216), 66)] = [12-/216,11] = [1 - /216,11], Let P2 = (1 - x/21b, 11). We can check that {1 — -/210, 11} is an integral basis of P2, and N(P2) = \J = U- Hence P2 is a prime ideal. As 73 we take the ideal corresponding to the form (2,28,— 7), so 73 = (4,28 + 2/210) = (4,2/210) = (2)(2,/2W). Let A 3 = [73], Hence A3 = [(2, /210)]. Let P3 = (2, /210). We check that = 2, and Pj = (2) so P3 is prime as well. Following the Part 2 of the proof of Carlitz theorem, we conclude P[P2P3 = (5,/2W)(l -/210,ll)(2,/2T0) = (10, /210) (1 - /210, 11) = (10- 10/216, 110, /216-210, 11/216) Chapter 4 79 = (10 + 7210, 110, 7210 — 210,117210) (by adding forth generator to the first) = (10 + 7210, 110, 220, 11 7210) (by subtracting first generator from the three) = (10, +7710, HO, 117710) = (10+7710,117710) (because 110 = -(10+ 7710) (10- 7710) = (10 + 7710). For the last equality we have (10+ 7710, 117210) = (10- 77To)(io+77w, ii^io)(i5-^) = (-110,110(7210-21))(I5Z^) = (110)(-l,^W-21)(^-^) = (H°)(IUZ^), because (-1,7710-21) = OK. = (io + TTio). We claim that P? = (31 + 27710). For this we have = ( 1 - 7710, 11)2 = (211 - 27710, 11-117710, 121). Now 211 - 27710 = (61 - 47710) (31 + 2 7710) , 11 -117^0 = (41 -372W)(31 +27210), 121 = (31 -27710) (31+27710). Hence P2 = (31 +27710)/, where /= (61 -47710,41 - 37710,31 -27710). The norms of the generators of /are 192, —10 x 11 and 121 respectively. Since gcd(192, 19 x 11, ll2) = 1, we conclude that J = Ok- Hence P2 = (31 +27210). Finally, P^Pl = (10+ 72W)2. Therefore, (2) (5) (31 + 27710) = u(10+ 72W)2 where u is a unit in Ok . On the left hand side we have three irreducible factors, while on the right hand side two. Moreover, we check by direct calculation that u = 1. The irreducibility of the factors is guarantied by the proof of Carlitz’s theorem, but can also be checked directly. Chapter 5 Conclusions The application of binary quadratic forms appears to be a promising approach to study the class number problem for quadratic Helds. I will try to study this approach more in depth in the future, also the problem of finding conditions when a quadratic field has a fundamental unit with norm -1 seems very interesting. 80 Bibliography [1] §.Alaca and K.S.Wlliams, Introdctory Agebraic Number Theory,Cambridge University Press 2004. [2] Z.I.Borevich and I.R.Shafrevich, Number Theory, Academic Press, New York and Lon¬ don, 1966 [3] D.A. Buell, Binary Quadratic Forms: Classical Theory and Modern Computations, United states of America. [4] L. Carlitz, A characterization of algebraic number fields with class number two, Proc. Amer. Math. Soc. 11 (1960), pp.391-392. [5] K. Conrad, Ideal Factorization, public domain: https://kconrad.math.uconn.edu/blurbs/gradnumthy/idealfactor.pdf [6] H. Davenport, Multiplicative Number Theory, Charendon Press Oxford 1986. [7] B. Green, Algebrac Number Theory, {http : course-archve.maths . x . ac . uk/view_mat erial/48320} [8] E.Hecke, Uber die L-funktionen und den Dirichletschen Primzahlsatzfur einen beliebigen Zahlkorper, Nachr.Akad. Wiss. Gottingen. Math.-Phys. KI. Ila (1917), pp. 299-318. [9] MathOverflow (public domain) {https://mathoerfolw . net /quest ions/172432/ class-number-for-binary\\ -quadrat i c-forms-discriminant-delt a-t o-class-number} [10] I. Stewart and D.Tal, Algebraic Number Theory and Fermat's Last Teorem, 3-d edition, A K Peters, Ltd., 2001 [11] D. B. Zagier, Zetafunktionen und qadratische Korper, Springer-Verlag 1981. 81