ORDERS IN QUADRATIC IMAGINARY FIELDS OF SMALL CLASS NUMBER

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1 ORDERS IN QUADRATIC IMAGINARY FIELDS OF SMALL CLASS NUMBER JANIS KLAISE Abstract. In 2003 M. Watkins [Wat03] solved the Gauss Class Number Problem for class numbers u to 100: given a class number h, what is the comlete list of quadratic imaginary fields K with this class number? In this roject we extend the roblem by including non-maximal orders - subrings O of the ring of integers O K that are also Z-modules of rank 2 admiting a Q-basis of K. These orders are more comlicated than O K since they need not be Dedekind domains. The main theoretical art is the class number formula linking the class numbers of O and O K. We give a comlete roof of this and use it together with the results of Watkins to find the orders of class number h = 1, 2, 3 by solving the class number formula in a Diohantine fashion. We then describe a different algorithm that was suggested and imlemented during the writing of this roject and succesfully finds all orders of class number u to 100. Finally, we motivate the roblem by its alication to j-invariants in comlex multilication. Contents 1. Introduction 1 2. The Class Number Formula 4 3. Examle Calculations An Algorithm An Alication to Comlex Multilication 20 References Introduction Gauss in his Disquisitiones Arithmeticae [Gau86, ] came u with several conjectures about imaginary quadratic fields in the language of discriminants of integral binary quadratic forms. In articular, he gave lists of discriminants corresonding to imaginary quadratic fields with given small class number and believed them to be comlete without roof. This became known as the Gauss Class Number Problem - for each h 1 rovide a comlete list of imaginary quadratic fields with class number h. The case h = 1 was solved indeendently by Heegner [Hee52], Baker [Bak66] and Stark [Sta67]. Subsequently, h = 2 was solved again by Baker [Bak71] and Stark [Sta75]. The cases h = 3 and h = 4 were solved by Oesterlé [Oes85] and Arno [Arn92] resectively. Wagner [Wag96] solved the cases Date: March 27,

2 ORDERS IN QUADRATIC IMAGINARY FIELDS OF SMALL CLASS NUMBER 2 h = 5, 6, 7 and Arno et al. [ARW98] solved the cases for 5 h 23 odd. Finally, Watkins [Wat03] solved the roblem for all h 100. In this roject we will extend these calculations to general orders of imaginary quadratic fields that corresond to non-fundamental discriminants defined below. Every quadratic field K has a discriminant d K which is defined in terms of the integral basis of the ring of integers O K. The discriminants d K are what we call fundamental discriminants: Definition. An integer D is called a fundamental discriminant if either D 1 mod 4 and is squarefree or D = 4d with d 2, 3 mod 4 with d squarefree. Thus the fundamental discriminants are exactly the discriminants of quadratic fields. d It is well known that if Q is a quadratic field with d a squarefree integer and O K = [1, w K ] is an integral basis, then if d 1 mod 4, the discriminant is d K = 4d and we can take w K = d and if d 1 mod 4, then the discriminant is d K = d and we can take w K = 1+ d 2 See for examle [ST01, ]or [Coh96,. 223]. More succintly, we always have an integral basis O K = [1, w K ] with w K = d K+ d K 2. Now Gauss originally included in his lists what we sometimes call non-fundamental discriminants. Together with the fundamental ones, they give a very easy descrition of what we mean by a discriminant in general: Definition. An integer D is called a discriminant if D 0, 1 mod 4 and is non-square. The question is, are these non-fundamental discriminants connected with quadratic fields in a similar way as the fundamental discriminants are as discriminants of quadratic fields? The answer is yes - they corresond to discriminants of orders in quadratic fields which will be the main objects of study in this roject. There are many definitions of an order, some of which exclude certain roerties and rove them later. Here we give a comrehensive definition that includes the roerties we are interested in: Definition. An order O of a quadratic field K is a subring O O K that is also a free Z-module of rank 2 = [K : Q] so that it contains an integral basis of K. In articular, we call O K the maximal order since it contains every other order. Moreover, since both O and O K are free Z-modules of the same rank, we have that the index f = [O K : O] is finite. We call f the conductor of O. Then if O K = [1, w K ] as above, one can show that O = [1, fw K ] See for examle [Cox89,. 133]. Since an order O admits an integral basis of K, we can define the discriminant of O in exactly the same way as for O K - the square of a 2 2 determinant with rows being the integral basis under the 2 automorhisms of the field K. In the same way as for O K, we can show that the discriminant of O is indeendent of the basis used and calculating it using the basis [1, fw K ] gives the discriminant D = f 2 d K

3 ORDERS IN QUADRATIC IMAGINARY FIELDS OF SMALL CLASS NUMBER 3 where d K is the fundamental discriminant of O K. So in fact every such discriminant D satisfies D 0, 1 mod 4 and so is consistent with our revious definition of a discriminant. But more is true. Any discriminant D is a discriminant of some order in a quadratic imaginary field; we summarize this in the following roosition: Proosition. [Coh96,. 224] If K is a quadratic field of discriminant d K then every order O of K has discriminant D = f 2 d K where f 1 is an integer. Conversely, if D 0, 1 mod 4 and is squarefree, then D = f 2 d K uniquely for some fundamental discriminant d K and there exists a unique order O Q d K with discriminant D. In Section 2 we will develo a theory of ideals of O and arrive at a notion of the class grou C O of O analogously to the usual class grou C O K of the field K. The main result we derive is the following class number formula linking the class numbers of O and O K : Theorem. [Cox89,. 146] Let O be the order of conductor f in an imaginary quadratic field K. Then ho = h O K f [ 1 O K : O ] 1 f 1.1 Furthermore, h O is always an integer multile of h O K. Here is the Legendre-Kronecker symbol: Definition. For an odd rime, let be the Legendre symbol and for = 2 define the Kronecker symbol: dk = 2 0 if 2 d K 1 if d K 1 mod 8 1 if d K 5 mod Our aim is then to caclulate recisely which orders or equivalently discriminants have a given class number h = h O u to h 100. In Section 3 we use the class number formula 1.1 and the Watkins table of fundamental discriminants [Wat03,. 936] to exlicitly solve the formula as a Diohantine equation for f, d K given h O = 1, 2, 3. In Section 4 we describe a different algorithm that was suggested and imlemented during the course of writing this roject. It succesfully finds all airs f, d K given the desired class number h O = 1, In the last section we motivate the roblem of finding all discriminants of given class number by its alication to j-invariants in the theory of comlex multilication.

4 ORDERS IN QUADRATIC IMAGINARY FIELDS OF SMALL CLASS NUMBER 4 2. The Class Number Formula In order to rove the class number formula given above, we need to understand more about the theory of ideals in a non-maximal order O. In articular, the difficulties arise because O is not a Dedekind domain, hence unique factorization of ideals into rime ideals fails and not every non-zero fractional ideal is invertible so they don t form a grou under multilication. To retrieve these familiar roerties characteristic of the maximal order O K we need to restrict ourselves to smaller subsets of ideals a O with some secial roerties. Throughout this section we closely follow the book [Cox89, ]. We reroduce the comlete roof of the class number formula filling in the more imortant details left out in the book and cutting out the less imortant where ossible. Definition 1. An ideal a O is called roer if O = {β K : βa a}. Note that it is always true that O {β K : βa a} since a is an ideal of O, but equality need not occur. We will use Definition 1 in the context of fractional ideals. Recall that a fractional ideal of O is a an O-submodule a K such that there exists r O \ {0} with ra O. All of the fractional ideals take the form αa where α K and a is an ideal of O [Cox89,. 135]. Recall also, that for the maximal order O K, the quotient O K /a for any non-zero ideal a is finite, so we can define the norm of and ideal to be N a = O K /a. The roof of the finiteness of the quotient adats in the case of a non-maximal order O, so our definition of a norm extends to all orders. We know that all non-zero fractional ideals a of O K are invertible in the sense that there exists another fractional ideal b of O K such that ab = O K. We now show that the same is true in any order O if we restrict to roer fractional ideals. In fact, the notions of roer and invertible are equivalent. Proosition 2. Let O be an order of a quadratic field K and let a be a fractional O-ideal. Then a is roer a is invertible. Proof. = Suose a is invertible. Then there exists fractional O-ideal b such that ab = O. Now let β K satisfy βa a, then we have so β O and hence a is roer. βo = β ab = βa b ab = O For the other imlication we need a technical lemma. Lemma 3. Let K = Q τ be a quadratic field and let ax 2 + bx + c bet the minimal olynomial of τ with gcd a, b, c = 1. Then [1, τ] is a roer fractional ideal of the order [1, aτ] K.

5 ORDERS IN QUADRATIC IMAGINARY FIELDS OF SMALL CLASS NUMBER 5 Proof. Note that [1, aτ] is indeed an order since aτ is an algebraic integer. Given β K, if β [1, τ] [1, τ], then this means β 1 [1, τ] and β τ [1, τ]. In other words we have β = m + nτ for somem, n Z βτ = mτ + nτ 2 = cn bn a + a + m τ Since gcd a, b, c = 1, we have βτ [1, τ] a n and so {β K : β [1, τ] [1, τ]} = [1, aτ]. Using this we can rove the other imlication. = Suose a is roer. Note that a is a Z-module of rank 2 so we can write a = [α, β] for some α, β K. Now define τ = β α so that a = α [1, τ]. Let ax 2 + bx + c bet the minimal olynomial of τ with corime coefficients. Then by Lemma 3, we must have O = [1, aτ]. Now consider the non-trivial automorhism of K taking τ to its conjugate τ. Since τ is a root of the minimum olynomial, alying the lemma once more gives that a = α [1, τ ] is a fractional ideal of [1, aτ] = [1, aτ ] = O. We will show that aa = Nα a O which shows that a is invertible. Note that aaa = aαα [1, τ] [1, τ ] = N α [a, aτ, aτ, aττ ] But we also have τ + τ = b/a and ττ = c/a, so the above exression becomes since gcd a, b, c = 1. aaa = N α [a, aτ, b, c] = N α [1, aτ] = N α O In articular, this roosition also shows that in the maximal order O K every non-zero fractional ideal is roer. Now that we know that roer fractional ideals are invertible, we can define the ideal class grou C O of an order O. Definition 4. Let O be an order. Denote by I O the set of all roer fractional. Then I O is a grou under multilication and contains the rincial O-ideals denoted P O I O as a subgrou. We define the ideal class grou to be the quotient C O = I O /P O Note that if O = O K is the maximal order, C O K is the usual ideal class grou of the number field K. From now on we will use the notation I K and P K for I O K and P O K resectively. Next we state a major theorem relating integral binary quadratic forms to ideals. This is a classical result which allows to set u a 1 1 corresondence between ideal classes in C O and classes of forms as defined below. We are only interested in one alication of this theorem, so we do not dwell on the details for long. The basic definitions and results are summarized below.

6 ORDERS IN QUADRATIC IMAGINARY FIELDS OF SMALL CLASS NUMBER 6 Definition 5. Terminology of quadratic forms A binary integral quadratic form is an exression of the form f x, y = ax 2 + bxy + cx with a, b, c Z. A form is said to be rimitive if gcd a, b, c = 1 An integer m is reresented by a form f x, y if the equation f x, y = m has an integer solution in x and y. In addition, if gcd x, y = 1, then it is said that f reresents m roerly. Two forms f x, y and g x, y are said to be roerly equivalent if there exist integers, q, r, s with f x, y = g x + qy, rx + sy and s qr = 1 and this gives an equivalence relation on the set of forms. A form f is said to be ositive definite if f x, y > 0 for all x, y. The discriminant of a form is defined to be D = b 2 4ac. A rimitive ositive definite form ax 2 + bx + c is said to be reduced if b a c and b 0 if either b = a or a = c Fact 6. Proerties of quadratic forms Equivalent forms have the same discriminant and the notion of ositive definiteness is invariant under equivalence [Cox89,. 25]. So define C D to be the set of roer equivalence classes of rimitive ositive definite forms of discriminant D. Every rimitive ositive definite form is roerly equivalent to a unique reduced form [Cox89,. 27]. There is a finite number of reduced forms of fixed discriminant D [Cox89,. 29]. Thus C D is finite and we write h D for the number if classes of rimitive ositive definite forms of discriminant D. If D 0, 1 mod 4 is negative, then C D can be made into an abelian grou under certain comosition rules for forms, called the form class grou [Cox89,. 50]. Theorem 7. [Cox89,. 137] Let O be the order of discriminant D in an imaginary quadratic field K. Then 1 If f x, y = ax 2 + bxy + [ cy 2 is a rimitive ositive definite quadratic form of discriminant D, then a, b + ] D /2 is a roer ideal of O. [ 2 The ma f x, y a, b + ] D /2 induces an isomorhism between the form class grouo C D and the ideal class grou C O. 3 A ositive integer m is reresented by a form f x, y iff m is the norm N a of some ideal a in the corresonding ideal class in C O. It is not hard to come u with the idea of this corresondence between quadratic forms and ideals, the hard art of the theorem is checking all the details that come with this corresondence, for examle, that equivalent forms are maed to ideals in the same class, so the roof involves a lot of exlicit calculations. One immediate consequence of this theorem is that calculating the class number of an order O is equivalent to counting the number reduced forms of discriminant D. I have imlemented this myself to see that it gives the correct results, however the actual algorithm is quite inefficient when comared to analytic methods.

7 ORDERS IN QUADRATIC IMAGINARY FIELDS OF SMALL CLASS NUMBER 7 For us, we are interested in alying art 3. of Theorem 7 to deduce the following fact about ideal classes in C O which will lay a crucial role in establishing the formula. Proosition 8. Let O be an order of an imaginary quadratic field. Given an integer M 0, every ideal class in C O contains a roer O-ideal with norm relatively rime to M. Proof. We first show the following fact about quadratic forms. Given a rimitve form f x, y = ax 2 + bxy + cy 2 with gcd a, b, c = 1 and an integer M, f x, y roerly reresents numbers corime to M. In other words, we can always find x, y corime with f x, y corime to M. To see this, let be a rime dividing M. Note that sincef x, y is rimitive, does not divide at least one of a, b, c. If a and c, then b so taking x, y corime to will make f x, y corime to. If on the other hand a res. c, then take x, y with x and y res. x and y so that f x, y is again corime to M. Now, by the Chinese Remainder Theorem, we can choose x, y in this way for every rime M. At this oint, if x, y are not corime, divide through by any common factors. Now we use art 3. of Theorem 7: since f x, y reresents numbers corime to M, the corresonding ideal class to f x, y contains ideals with norm corime to M. In order to rove the formula, we need to translate roer O-ideals in terms of O K -ideals. We do this through the concet of ideals rime to the conductor. Definition 9. Let O be an order of conductor f and let a O be a nonzero ideal. We say that a is rime to f if a + fo = O. The following lemma summarizes give two imortant roerties of such ideals. Lemma 10. Let O be an order of conductor f. Then Proof. 1 An O-ideal a is rime to f its norm N a is corime to f. 2 Every O-ideal rime to f is roer. 1 Define a ma m f : O/a O/a to be multilication by f. Then we have a + fo = O m f is surjective m f is an isomorhism But by the classification of finite Abelian grous, m f is an isomorhism f is corime to the order N a of O/a. 2 Let β K satisfy βa a. Then we have β O K and βo = β a + fo = βa + βfo a + fo K But we also have fo K O so that βo O giving β O roving a is roer. It follows from this lemma that O-ideals rime to f form subset of I O that is closed under multilication. Thus we can the define the subgrou of fractional ideals they generate and denote it by I O, f I O and as usual we also have the subgrou generated by the rincial ideals deonted P O, f I O, f. It turns out we can describe the class grou C O in terms of these two subgrous.

8 ORDERS IN QUADRATIC IMAGINARY FIELDS OF SMALL CLASS NUMBER 8 Proosition 11. The inclusion I O, f I O induces an isomorhism I O, f /P O, f = I O /P O = C O Proof. The ma I O, f C O taking a fractional ideal rime to f to its ideal class is surjective by Proosition 8 and the kernel of this ma is I O, f P O. This clearly contains P O, f so we need to show the other inclusion, then the result follows by the First Isomorhism Theorem. An element of the kernel is a rincial fractional ideal αo = ab 1 with α K and a, b O-ideals rime to f. Define m = N b.then we have mo = N b O = b b so that mb 1 = b. This gives us mαo = amb 1 = a b O which shows that mαo P O, f. But then αo = mαo mo 1 is in P O, f as well. We now extend the definition of ideals rime to conductor to ideals of the maximal order O K. This will enable us to derive a relation between ideals of O rime to the conductor f and ideals of O K. Definition 12. Let m > 0 be an integer. An ideal a O K is rime to m if a + mo K = O K. As in the Lemma 10, this is the same as saying gcd N a, m = 1. We can thus similarly define a subgrou of fractional ideals I K m I K generated by O K -ideals rime to m. The next roosition relates such ideals of O and O K. Proosition 13. Let O be an order of conductor f. Then Proof. 1 If a is an O K -ideal rime to f, then a O is an O-ideal rime to f of the same norm. 2 If a is an O-ideal rime to f, then ao K is an O K -ideal rime to f of the same norm. 3 The ma a a O induces an isomorhism I K f = I O, f with an inverse ma a ao K. 1 Let a O K be rime to f. Consider the comosite ma O O K O K /a This has kernel a O, so the ma O/a O O K /a is injective. So since N a is rime to f, so is N a O showing that a O is rime to f. To see that the norms are equal, we show that the injective ma is also surjective. Since a is rime to f, multilication by f gives an automorhism of O K /a as in Lemma 10. But we also have fo K O, so the ma is surjective. 2 Let a O be rime to f. We have ao K + fo K = a + fo O K = OO K = O K so that ao K is rime to f. For norms see the argument in art 3.

9 ORDERS IN QUADRATIC IMAGINARY FIELDS OF SMALL CLASS NUMBER 9 3 We rove 2 claims: ao K O = a when a is an O-ideal rime to f 2.1 a O O K = a when a is an O K -ideal rime to f 2.2 We start with claim 2.1; if a is an O-ideal rime to f, then: ao K O = ao K O O = ao K O a + fo a + f ao K O a + afo K a + ao = a The other inclusion is obvious so equality follows. For claim 2.2, if a is an O K -ideal rime to f, then: a = ao = a a O + fo a O O K + fa a O O K + a O O K since fa fo K O and fa a O K also. The other inclusion is again obvious, so equality follows. In articular, these two claims and art 1. give the norm statement in art 2. This also gives a bijection between O K -ideals and O-ideals rime to f. This extends to an isomorhism I K f = I O, f since the mas a a O and a ao K reserve multilication. Using this we can describe the grou C O urely in terms of the maximal order. To do this, we need to make one more definition - this time of a subgrou of I K f. Definition 14. Denote by P K,Z f the subgrou of I K f generated by rincial ideals αo K with α O K satisfying α a mod fo K for some integer a corime to f. Proosition 15. Let O be an order of conductor f. Then there are isomorhisms C O = I O, f /P O, f = I K f /P K,Z f Proof. The first isomorhism was roved in Proosition 11. By roosition 13, the ma a ao K induces an isomorhism I O, f = I K f and under this isomorhism, P O, f I O, f mas to some subgrou P I K f. So we want to show that P = P K,Z f. We first rove the following claim for α O K : α a mod fo K, a Z, gcd a, f = 1 α O, gcd N α, f = = If α a mod fo K, then it follows that N α a 2 mod f, so then gcd N α, f = gcd a 2, f = 1. Also, since fo K O, we have α O. = Let α O = [1, fw K ] have norm corime to f. This means α a mod fo K for some a Z. Since gcd N α, f = 1 and N α a 2 mod f, we must have gcd a, f = 1.

10 ORDERS IN QUADRATIC IMAGINARY FIELDS OF SMALL CLASS NUMBER 10 Now, P O, f is generated by ideals αo where α O and N α is corime to f, so P is generated by corresonding ideals αo K and by the claim this means P = P K,Z f. We need one last fact to rove the main theorem. This gives an exression for the number of units in a quotient of O K. Proosition 16. Let f be a ositive integer. Then Φ f = f f O K /fo K = Φ f Φ K f is the Euler Phi-function and Φ K f = f f 1 1. Proof. First we rove the following claim. Let O K be a rime ideal. Then O K / n = N n N n Recall that the ideals of O K / n are in 1 1 corresondence with ideals I of O K such that n I. That is I = k for 0 k n which gives n + 1 ideals of O K / n : O K / n, / n,..., n 1 / n, 0 It is clear from this list that there is a unique maximal ideal, namely / n. Since every non-unit of a non-zero ring lies in some maximal ideal, we have [x] O K / n is a unit x /. This gives O K / n = O K / n / n The first summand is simly N n. To comute the second one, we use the Third Isomorhism Theorem: which roves the claim. / n = O K/ n O K / = N n N = N n 1 Now let a O K be any ideal and consider its rime factorization a = r By the Chinese Remainder Theorem we have O K /a r = O K / ni i i=1 i=1 ni i. Using this together with claim 2.5, we calculate r r O K /a = O K / ni i = N i ni 1 1 = N a 1 1 N i=1 i=1 i N a 2.6 Now consider ideals of the form a = O K for a rational rime. We want to rewrite formula 2.6 with these ideals. There are three cases: 1 slits so that O K = 1 2 with 1 2 and N 1 = N 2 = so that N O K = 2 and 1 1 = = 1 N O K

11 ORDERS IN QUADRATIC IMAGINARY FIELDS OF SMALL CLASS NUMBER 11 2 is inert so that O K = with N = 2 and 1 1 = 1 1 N 2 = O K 3 is ramified so that O K = 2 with N = so that again N O K = 2 and 1 1 = 1 1 N O K We can summarize the three cases in one single exression: 1 1 = N O K where is the Legendre symbol Kronecker symbol for = 2 so that it is equal to +1, 1, 0 in the cases resectively [Cox89,. 104]. Putting this all together, for a ositive integer f, we have shown that O K /fo K = f = Φ f Φ K f Finally, we come to the main theorem: Theorem 17. [Cox89,. 146] Let O be the order of conductor f in an imaginary quadratic field K. Then ho = h O K f [ 1 O K : O ] 1 f Furthermore, h O is always an integer multile of h O K. 2.7 Proof. We have roved so far that h O = C O = I K f /P K,Z f h O K = C O K = I K /P K Also, I K f I K and P K,Z I K f P K which induces a homomorhism I K f /P K,Z f I K /P K with kernel I K f P K /P K,Z f. Thus, the following sequence is exact: 0 I K f P K /P K,Z f I K f /P K,Z f I K /P K 2.8 Also, I K f /P K,Z f = C O and I K /P K = C OK. By Proosition 8, every class in C O K containts an O K -ideal with norm corime to f. This makes the ma C O C O K surjective which shows that h O K h O. In articular, using the sequence 2.8, we get h O h O K = I K f P K /P K,Z f We will calculate the order on the RHS relating this quotient to O K /fo K. Let [α] O K /fo K, then we have αo K + fo K = O K, so αo K is rime to f, so lies in I K f P K. Also, if α β mod fo K, we can find u O so

12 ORDERS IN QUADRATIC IMAGINARY FIELDS OF SMALL CLASS NUMBER 12 that uα uβ mod fo K. So both uαo K and uβo K lie in P K,Z f and since αo K uβo K = βo K uαo K, the ideals αo K and βo K lie in the same class in I K f P K /P K,Z f. This means that the ma is a well-defined homomorhism. φ: O K /fo K I K f P K /P K,Z f [α] [αo K ] φ is surjective. Let αo K I K f P K. Then αo K = ab 1 for a, b O K -ideals rime to f. Let m = N b and in Proosition 11 we saw that b = mb 1 so that mαo K = ab giving mα O K and so mαo K is also an O K -ideal rime to f. Note that mo K P K,Z f, thus [αo K ] = [mαo K ] = φ [mα] roving surjectivity. Next we find the kernel of φ. We define the two mas d: {±1} Z/fZ O K 2.9 ±1 [±1], ±1 ψ : Z/fZ O K O K/fO K 2.10 [n], ɛ [ nɛ 1] We will use these to show that the following sequence is exact: 1 {±1} d Z/fZ O K O K /fo K ψ φ I K f P K /P K,Z f Now if n is an integer corime to f and ɛ O K, then ψ [n], ɛ = nɛ 1 mod fo K so [n], ɛ ker ψ nɛ 1 1 mod fo K Thus ker ψ = [±1], ±1 = im d. ɛ n mod fo K ɛ = ±1 n = ±1 mod f By the definition of P K,Z, we immediately get im ψ ker φ. To show the other inclusion, let [α] ker φ. Thus αo K P K,Z which means αo K = βo K γ 1 O K for some β, γ satisfying β b mod fo K and γ c mod fo K with b, c corime to f. This means [b], [c] Z/fZ. Now, α = βγ 1 ɛ for some ɛ O K. Then ψ [b], 1 = b mod fo K = β mod fo K ψ [c], ɛ = cɛ 1 mod fo K = γɛ 1 mod fo K ψ [b] [c] 1, ɛ = βγ 1 ɛ mod fo K = [α] We have shown that [b] [c] 1, ɛ 1 Z/fZ O K mas to [α] O K/fO K roving the other inclusion. Thus, by the exactness of 2.11, we have I K f P K /P K,Z f = O K/fO K {±1} Z/fZ O K

13 ORDERS IN QUADRATIC IMAGINARY FIELDS OF SMALL CLASS NUMBER 13 We calculated in Proosition 16 that O K /fo K = Φ f Φ K f and we also know that Z/fZ = Φ f. Thus ho h O K = I K f P K /P K,Z f = Φ K f [ O K : O ] roving the theorem. In the following section we use this formula to exlicitly calculate all discriminants with class numbers h = 1, 2, 3. It is worth mentioning that there are analytic roofs of this theorem, using the analytic class number formula See [Coh96,. 233] for an outline. In addition, one can derive essentially the same formula for the number of elements in C O for general number fields not just imaginary quadratic using canonical exact sequences, but calculating the orders of O K /fo K and O/fO is harder See [Neu99, 12]. 3. Examle Calculations. We will make use the following well-known result about units in quadratic imaginary fields: Proosition 18. [ST01,. 77] Let K be an imaginary quadratic field with discriminant d K and the ring of integers O K. Then the grou of units O K is as follows: O K = { ±1, ±ω, ±ω 2} for d K = 3 and ω = e 2πi 3 O K = {±1, ±i} for d K = 4 O K = {±1} for all other d K < 0 Some observations: If [ O K = 2, then O K : O ] = 1 since ±1 are always units. If [ O K = 4, then O K : O ] = 2 since ±i / [1, fi] for f > 1 If O K = 6, then [ [ ] O K : O ] = 3 since ±e 2πi 3, ±e 4πi 3 / 1, fe 2πi 3 for f > 1 Recall the integer valued function we first introduced in Proosition 16: Definition 19. Define Φ K f = f f So the roblem of finding all orders of class number h reduces to solving Φ K f = h O h O K [ O K : O ] 3.2 for f, d K which uniquely determine the order O by its discriminant D = f 2 d K. In rincile, if we wanted to find all orders of class number h 100, we could solve Φ K f = m for m 300. But note that in the secial cases where [ O K : O ] 1, we already know what d K is and the job of finding f becomes easier. We give a few useful roerties of the function Φ K.

14 ORDERS IN QUADRATIC IMAGINARY FIELDS OF SMALL CLASS NUMBER 14 Proosition 20. Φ K is multilicative. Proof. Let f = e en n be a rime decomosition of the conductor. Then Φ K f = f 1 1 f = e en n = n i=1 ei i = Φ K ei i n 1 1 i=1 In articular, since Φ K e = 1 e dk following roerties: i 1 i 1 i 1 i = e 1 dk we have the If e f, then { e 1 Φ f always e Φ K f if d 3.3 If f, then Φ K f 3.4 Examle 21. We find all orders of class number h O = 1. Since h O K h O we immediately have h O K = 1. There are 9 maximal orders of class number 1 [Sta67] whose discriminants are: 3, 4, 7, 8, 11, 19, 43, 67, If [ O K : O ] = 1, need to solve Φ K f = 1. We consider a rime f. Case 1: = 1 = 1 1 = = 2 = d K 1 mod 8 = d K = 7 Case 2: = 1 = is imossible Case 3: = 0 = 1 is imossible So we have found one non-maximal order with f, d K = 2, 7 and discriminant D = 28. If [ O K : O ] = 2, need to solve Φ K f = 2 with d K = 4. Case 1: = 1 = 1 2 = = 2 or = 3 If = 2, then 4 2 = 0 is false. If = 3, then 4 3 = 1 is false. Case 2: = 1 = is imossible Case 3: = 0 = 2 = = 2 = d K = 0 mod 2 is true So we have an order with f, d K = 2, 4 and discriminant D = 16.

15 ORDERS IN QUADRATIC IMAGINARY FIELDS OF SMALL CLASS NUMBER 15 Lastly, if [ O K : O ] = 3, need to solve Φ K f = 3 with d K = 3. Case 1: = 1 = 1 3 = = 2 = d K 1 mod 8 is false Case 2: = 1 = = = 2 = d K 5 mod 8 is true Case 3: = 0 = 3 = = 3 = d K 0 mod 3 is true So we have found another two orders with f, d K = 2, 3, 3, 3 and discriminants D = 12, 27 resectively. This means that there are 13 orders of class number 1 of which 4 are non-maximal. Examle 22. To find the orders of class number h O = 2 we must have h O K = 1 or h O K = 2. The 9 maximal orders of class number 1 are listed in 3.5. There are 18 maximal orders of class number 2 [Bak71] with discriminants as follows: 15, 20, 24, 35, 40, 51, 52, 88, 91, 115, 123, 148, 187, 232, 235, 267, 403, So if h O K = 1, then Φ K f = 2 4, 6 deending on the size of the unit grou. If h O K = 2, then Φ K f = 1 as the unit grou has always size 2. We start with the latter case. If h O K = 2, need to solve Φ K f = 1. Consider a rime f. We saw in Examle 21 that the only solution is f = 2 with d K 1 mod 8 which gives d K = 15 and so an order with discriminant D = 60. Now suose h O K = 1 and the unit grou has size 2. Need to solve Φ K f = 2. Case 1: = 1 = 1 2 = = 2 or = 3 If = 2, then d K2 = 1 = 1 mod 8 = d K = 7 f = 2, however, is imossible as Φ K 2 = 1 2. f = 4 is also ossible since = 2 Φ K f by Proerty 3.3. Then indeed Φ K 4 = 2 and we have found an order with discriminant D = = 112. If = 3, then d K3 = 1 = 1 mod 3 = d K = 8, 11. Indeed Φ K 3 = 2 which gives two orders with discriminants D = = 72 and D = = 99. Note that f cannot contain a higher ower of 3 since = 3 Φ K f. We also have the ossibility f = 2 3. Then d K 1 mod 8, 1 mod 3 = d K 1 mod 24 which is not satisfied by any of the 9 fundamental discriminants. Finally, f = is imossible as Φ K 12 = 4 2. Case 2: Case 3: = 1 = is imossible = 0 = 2 = = 2 = d K 0 mod 2 = d K = 8. Indeed Φ K 2 = 2 which gives an order with discriminant D = = 32. By roerty 3.3, f cannot containt any higher ower of 2.

16 ORDERS IN QUADRATIC IMAGINARY FIELDS OF SMALL CLASS NUMBER 16 Now suose the unit grou has size 4, i.e d K = 4 and we want to solve Φ K f = 4. 4 Case 1: = 1 = 1 4 = = 2, 5 but note that 4 2 = 0 which is a contradiction. So f = 5 is the only ossibility. 4 Case 2: = 1 = = = 3 and no higher ower of 3 is ossible. Case 3: = 0 = 4 = = 2 and so f = 2, 2 2 are ossible. 4 So Φ K 2 = 2 4. Φ K 3 = 4 which gives an order with discriminant D = = 36. Φ K 4 = 4 which gives an order with discriminant D = = 64. Finally, Φ K 5 = 4 which gives an order with discriminant D = = 100. Note that we do not need to consider any other roducts f of the rimes 2, 3, 5 since Φ K is multilicative so any such roducts will ensure Φ K f > 4. Finally, suosse the unit grou has size 6, i.e. d K = 3 and we want to solve Φ K f = 6. 3 Case 1: = 1 = 1 6 = = 2, 3, 7 but note that 3 2 = 1 and 3 3 = 0 which are contradictions. So f = 7 is the only ossibility. 3 Case 2: = 1 = = = 2, 5 and f = 2 2 is also ossible. 3 Case 3: = 0 = 6 = = 2, 3 but again 3 2 = 1 and so f = 3 is the only ossibility. We have Φ K 7 = 6 which gives an order of discriminant D = = 147. Φ K 2 = 3 6. Φ K 4 = 6 which gives an order of discriminant D = = 48. Φ K 5 = 6 which gives and order of discriminant D = = 75. Φ K 3 = 3 6. Since Φ K is multilicative, we are done since any other roduct f of the rimes 2, 3, 5 will have Φ K f > 6. To summarize, we have found 11 non-maximal ideals of class number 2 and there are no others. As a list f, d K : 2, 8, 2, 15, 3, 4, 3, 8, 3, 11, 4, 3, 4, 4, 4, 7, 5, 3, 5, 4, 7, 3 In articular, this calculation corrects an error in [?,. 408] which gives an incomlete list of orders of class number 2, ommiting the order 3, 8. Notation: Write e n to mean e n but e+1 n. The equation Φ K f = m for m odd is articularly easy to solve since there are not a lot of otions for the rime divisors of f. If = 1, then 1 m and so = 2 f if d K 1 mod 8. If If = 1, then + 1 m and so = 2 f if 3 m and d K 5 mod 8. = 0, then m and so if e m and d K then d f for all d e.

17 ORDERS IN QUADRATIC IMAGINARY FIELDS OF SMALL CLASS NUMBER 17 Examle 23. We use this to find all orders of class number h = 3. There are 16 maximal orders of class number 3 [Oes85] with discriminants: 23, 31, 59, 83, 107, 139, 211, 283, 307, 331, 379, 499, 547, 643, 883, If h O K = 3, need to solve Φ K f = 1 and we know from Examles 21 and 22 that this gives f = 2 with d K 1 mod 8. This corresonds to two orders with discriminants D = = 92 and D = = 124. Next, if h O K = 1 and the unit grou has size 2, this corresonds to Φ K f = 3. From the discussion above, 2 f iff d K = 1 mod 8 or d K = 1 mod 5 and 3 f iff 3 d K. From 3.5, there are no d K with unit grou of size 2 and 3 d K. Also, if d K 1 mod 8, then Φ K 2 = 1 3. On the other hand, if d K 5 mod 8, then Φ K 2 = 3 as required. There are 5 such maximal orders which give rise to 5 non-maximal orders of conductor 2: 2, 11, 2, 19, 2, 43, 2, 67, 2, 163 Now suose the unit grou has size 4, i.e. d K = 4 so we want to solve Φ K f = 6.\ 4 4 Case 1: = 1 = 1 6 = = 2, 3, 7 but none of these have = Case 2: = 1 = = = 2, 5 but none of these have = 1 4 Case 3: = 0 = 6 = = 2, 3 but 4 3 = 1 so only f = 2, 2 2 are ossible. However Φ K 2 = 2 and Φ K 4 = 4 so we do not get any new orders. Finaly suose the unit grou has size 6, i.e. d K = 3 so we want to solve Φ K f = 9. From the discussion above for m odd, f has ossible factors 2, 3, 9 but no higher owers of either 2, 3. Φ K 2 = 3 9 and Φ K 3 = 3 9 but together they give Φ K 6 = 9, an order 6, 3. Also Φ K 9 = 9 which gives an order 9, 3. We have found 9 non-maximal orders of class number 3 and there are no others: 2, 11, 2, 19, 2, 23, 2, 31, 2, 43, 2, 67, 2, 163, 6, 3, 9, 3 These examles demonstrate that in rincile it is ossible to devise an algorithm that solves these Diohantine equations. It would be easy in the case Φ K f = m where m is odd as already noted, but in the case m is even, more ossibilities for the rime divisors of f need to be considered. Then several results for different m need to be combined in calculating the discriminants of any one order which makes this less straightforward. In the following section we describe a different algorithm

18 ORDERS IN QUADRATIC IMAGINARY FIELDS OF SMALL CLASS NUMBER 18 based on the idea of comaring Φ K f with the Euler Phi function Φ f together with a lower bound to turn this into an exaustive search roblem. 4. An Algorithm In this section we look at the roblem of finding all orders with a given class number from a different ersective which uses the class number formula derived in Section 2 as an uer bound for an exhaustive search. This algorithm was first suggested by Andrew V. Sutherland at MIT and later imlemented in Sage by William Stein at the University of Washington. We give a descrition of the algorithm and the outut in a condensed format listing the number of orders with given class number h and the maximum discriminant in absolute value for each h u to 100. Recall the function Φ K f = f f 1 1 we defined earlier. This bears 1 1. In fact, remarkable similarity to the Euler Phi function Φ f = f f Proosition 16 and Theorem 17 show that these two functions are intimitely linked in roving the class number formula. Now let f, d K reresent the order O with conductor f and discriminant D = f 2 d K where d K is the fundamental discriminant of the maximal order O K in which O lies. Also, let w = [ O K : O ]. Changing notation so that h O K = h d K and h O = h f 2 d K we have h f 2 h d K d K = w Φ K f h d K w Φ f > h d K l f 4.1 w where l f is a lower bound for the Euler Phi function [Cla,. 6] given by f for f loglogf+ l f = loglogf 0 for f = 1 Note that l f is an increasing function of f In addition to this we use the Watkins s table of fundamental discriminants [Wat03,. 936] which gives a list with 100 entries h, d K max, n where h is the class number h = 1,..., 100, d K max is the maximum fundamental discriminant in absolute value and n is the number of the fundamental discriminants of class number h. The strategy is now clear. Given the maximum desired class number hmax in our case hmax = 100 we will use a double loo: the outer loo will run over the fundamental discriminants d K while the inner loo will run over conductors f as long as hd K w l f hmax is satisfied. We will create a dictionary with keys h the class numbers and entries a list of f, d K reresenting orders of that class number. Here is the full descrition: 1 Find the biggest fundamental discriminant in absolute value d K max in Watkins s table for h in the range 1, hmax. 2 Outer loo over fundamental discriminants d K in the range d K max, 2, for each calculate h d K. 3 Inner loo over conductors f starting from f = 1 until hd K w l f hmax is satisfied.

19 ORDERS IN QUADRATIC IMAGINARY FIELDS OF SMALL CLASS NUMBER 19 4 For each f, d K in the inner loo calculate h = h f 2 d K. 5 Test if h hmax; if so, aend f, d K to our dictionary with the key h. The correctness and the termination of the algorithm follow from ste 3. We know the bound on the fundamental discriminant d K max from the Watkins s table but the conductor f at first sight can be arbitrarily big. However, by the inequalities in 4.1 we only need to check f while hd K w l f hmax since otherwise h = h f 2 d K > hmax is out ouf our desired bound hmax. Since l f is an increasing function of f, this means that the algorihtm will terminate in a finite number of stes. The algorithm was run by John Cremona at the University of Warwick taking 19.5 hours to return a comlete dictionary of orders with class number u to 100. There are a number of imrovements to make the algorithm more efficient, however it is imortant to note that this was essentially a one-off calculation the results of which have been saved. Here we resent a Watkins-style table of the total number of orders maximal and non-maximal and the biggest discriminant in absolute value for each class number h 100: h # large h # large h # large h # large Using this table anyone who wishes to find all orders of a given class number h only needs to check all discriminants u to the largest one given in the table and the result can be double checked against the number of orders also given. If only fundamental discriminants are of concern, refer again to [Wat03,. 936] for the

20 ORDERS IN QUADRATIC IMAGINARY FIELDS OF SMALL CLASS NUMBER 20 number of these. There is a striking similarity between the third column of this table and the corresonding one in Watkins s table listing the maximum fundamental discriminant in absolute value. In fact, these columns are identical excet for 10 cases which are highlighted in the table 9 on which are curiously multiles of 163. On closer scrutiny, it turns out that these 10 discriminants ar the only ones that are bigger than the fundamental bound given by Watkins. This means if Watkins had included non-fundamental discriminants in his calculations, he would have missed only 10! discriminants which is a miniscule error taking into account the number of new discriminants we get when extending to non-fundamental ones. It is temting to conjecture that for any clas number, there is at most one non-fundamental discriminant bigger in absolute value than the biggest fundamental one, but more data should be gathered. At resent it seems out of reach unless eole look into Watkins s method and otimize it to solve the roblem for h > 100 in reasonable timeframes. This concludes the main body of the roject. We have found an answer to the roblem we set out to solve with some collaboration. The next and final section is a bonus section which sets out to motivate the roblem by its connection with comlex multilication of ellitic curves. 5. An Alication to Comlex Multilication In this section we will give a very brief survey of the Klein j-invariants which are imortant in the theory of comlex multilication for ellitic curves but also have connections with class field theory and have some very nice roerties in connection with discriminants of given class number that we found in earlier sections. This should be considered as an informal introduction to the toic. For the initiated reader, a good reference is [Cox89, Ch. 3]. The j-invariant is defined as an invarinat of a lattice in C : Definition. [Cox89,. 200] A lattice is an additive subgrou L C generated by two comlex numbers ω 1, ω 2 which ar linearly indeendent over R. We write L = [ω 1, ω 2 ]. We will also need ellitic functions defined for these lattices: Definition. [Cox89,. 200] An ellitic functions for a latice L is a function f z defined on C excet for isolated singularities such that 1 f z is meromorhic on C. 2 f z + ω = f z for all ω L. An imortant ellitic function that is used in defining j-invarinats is the Weierstrass -function defined for a lattice L and z L: z; L = 1 z 2 + ω L\{0} 1 z w 2 1 w 2 5.1

21 ORDERS IN QUADRATIC IMAGINARY FIELDS OF SMALL CLASS NUMBER 21 We will need two constants related this function to define j-invariants, so let g 2 L = 60 g 3 L = 140 ω L\{0} ω L\{0} In general, ellitic functions deend on the choice of lattice, but there exists a conveniet equivalence relation on lattices which means that equivalent lattices have basically the same ellitic functions: Definition. [Cox89,. 205] Two lattices L, L are called homothetic if there exists λ C \ {0} so that L = λl. The j-invariants first aear when classifying lattices u to homothety. Setting L = g 2 L 3 27g 3 L 2 we can check this is non-zero for any lattice, so that we define: Definition. [Cox89,. 206] The j-invariant of a lattice L is the comlex number j L = 1728 g 2 L 3 L 4 1 ω The following theorem makes it clear why these j-invariants matter: Theorem. [Cox89,. 206] If L, L are lattices in C, then j L = j L L and L are homothetic. What has this got to do with orders in imaginary quadratic fields? We know from our earlier results that if a O is a roer fractional ideal of an order O K, then a = [α, β] for some α, β K. Since K is a subset of C and it is an imaginary quadratic field, α and β are linearly indeendent over R. So the ideal a = [α, β] gives a natural lattice in C so that j a is defined. We will see some remarkable roerties of these numbers soon. 1 ω 6 Time doesn t ermit to discuss the theory of comlex multilication in full detail, but the idea is to ask the question for what comlex numbers α C we have that αz is a rational function in z? The answer [Cox89,. 209] is surrising indeed: first of all, any ellitic function f z is rational in z and z, and then if f αz is rational in f z for some α C \ R, it follows that this is true for elements of an entire order O in some imaginary quadratic field. This henomenon is called comlex multilication. Comlex multilication is an intrinsic roerty of the lattice and one can check that homothethy doesn t change this, so we can tlak about the homothety classes of lattices. We have the following beautiful result: Corollary. [Cox89,. 212] Let O be an order in an imaginary quadratic field. There is a 1 1 corresondence between the ideal class grou C O and the homothety classes of lattices with O as their full ring of comlex multilication.

22 ORDERS IN QUADRATIC IMAGINARY FIELDS OF SMALL CLASS NUMBER 22 In articular, the class number h O gives us the number of homothety classes of lattices with O as their full ring of comlex multilication. An examle is in order. Consider, for examle, lattices which have comlex multilication by 3. This means that we have an order O containing 3 in the field K = Q 3, so either O = Z [ 3 ] or O = Z [ e 2πi/3]. Both of these have class number 1, so u to homothety, the only lattices of this kind are [ 1, 3 ] and [ 1, e 2πi/3]. Ultimately, we want to calculate j-invariants of lattices. This involves homogeinity roerties of the constants g 2, g 3 defined earlier, so we will not cover this in detail. But as an examle, consider comlex multilication by i = 1. U to homothety, the only lattice admitting this is L = [1, i]. To comute j L = j i, we note il = L, so it follows by the Theorem above that g 3 L = g 3 il = i 6 g 3 L = g 3 L so that g 3 L = 0 and consequently j i = These j-invariants, as already mentioned, have some remarkable roerties. First of all, if we let τ H be a comlex number in the uer-half lane, we can define the j-function in the obvious way by j τ = j [1, τ]. Then for q = e 2πiτ, we have a q-exansion of j with integer coefficients [Cox89,. 297]: j τ = 1 q q q q q q This is one reason why the factor 1728 aears in the definition of j-invariants - it is the smallest factor ensuring that all coefficients in the exanison are corime integers. The fundamental facts about j-invariants which give an even deeer insight into the connection with imaginary quadratic fields are as follow: 1 If τ is a quadratic number i.e τ is a root of az 2 +bz+c = 0, gcd a, b, c = 1, D = b 2 4ac < 0, then j τ is an algebraic integer. 2 The degree of j τ is h = h D, the class number of the discriminant. 3 There is a secial case: j τ Q h D = 1 Thus, if we wanted to comute the j-invariants of all 13 class number h = 1 orders, we could use the q-exansion in 5.2, summming u enough terms and taking the nearest integer value. As an examle, for τ = i, q = e 2πi2 = e 2π so that we have q q q q q q so that indeed j i = 1728 rovided we know something about the convergence of the q-exansion. Here we give a full table of j-invariants for class number h = 1 which is originally due to a scheme by Weber discussed in [Cox89, ] but

23 ORDERS IN QUADRATIC IMAGINARY FIELDS OF SMALL CLASS NUMBER 23 one can easily get these by the series aroximation we just saw in action. The 13 j-invariants are as follows: D τ j τ = j O i = i = It is no coincidence that most of these numbers are erfect cubes, but we will not discuss this henomenon here. But these j-invariants exlain the aarently bizzare henomenon of almost integer numbers. Consider e π 163 = There is no reason why a transcendental number made u of seemingly arbitrary constants should be this close to a whole number. But the miracle turns out to be a simle consequence of comlex multilication. Let z = and consider the j-invariant j z = as already calculated. We also have 1 q = e 2πiz = e π 163 and q 1, so in fact it is no surrise that e π 163 is very close to What is more interesting is that this examle shows that the q-exansion converges very raidly. There are many more roerties of j-invariants we have not discussed, in articular the connection to class field theory. In articular, if τ is a quadratic integer so that Q τ is an imaginary quadratic field, it turns out that the field extension Q τ, j τ is the Hilbert Class Field of Q τ i.e. the maximum abelian unramified extension of Q τ - here unramified is rather technical, so we do not define it. In articular, since we know that for orders with class number h = 1 the j-invariants are integers, an immediate consequence is that if the ring of integers O Qτ is a unique factorization domain, the field Q τ is its own Hilbert Class Field. References [Arn92] S. Arno, The Imaginary Quadratic Fields of Class Number 4, Acta Arith. 40, , [ARW98] S. Arno, M. L. Robinson, F. S. Wheeler, Imaginary Quadratic Fields with Small Odd Class Number, Acta. Arith. 83, , [Bak66] A. Baker, Linear Forms in the Logarithms of Algebraic Numbers I, Mathematika 13, , [Bak71] A. Baker, Imaginary Quadratic Fields with Class Number 2, Ann. Math. 94, , [Cla] P. L. Clark, Arithmetical Functions III: Orders of Magnitude, htt://math.uga.edu/~ete/4400arithmeticorders.df