Composition and Bhargava's Cubes

Transcription

1 Composition and Bhargava's Cubs Florian Bouyr Contnts 1 Introduction 1 A classical viw on Gauss composition.1 Som dnitions Rduction of positiv dnit forms Dirichlt composition Th group of positiv dnit binary quadratic forms Gnral cas Bhargava's cubs A nw look at Gauss composition Cubs of intgrs Anothr proof of Thorm Binary cubic forms Pairs of binary quadratic forms Pairs of quatrnary altrnating -forms Diagrams Conclusion and Furthr Work A Appndix A.1 Calculations to th proof of A.1.1 Solving th 18 quations A.1. Showing I 1, I, I 3 ar fractional idals A. Calculation to prov in Thorm A.3 Proof of th bijction btwn C(Z Sym Z ; D) and C(Col 3 1,1; O) Introduction In 1801, whil studying binary quadratic forms, Gauss constructd a composition law in his Disquisitons Arithmtica [Gauss(1986)]. This composition law gav a group structur to th st of quivalnc classs of primitiv binary quadratic forms of a givn discriminant, which was rmarkabl as this was don bfor th notion of a group xistd. A usful consqunc of this group structur was discovrd by Dirichlt around 1838, whn h showd a bijction btwn this st and th st of idal classs of quadratic ordrs. This had a twofold impact: in on dirction it gav a simplr mthod of showing that composition did turn th aformntiond st into a group; whil in th othr dirction it gav a tool to comput and undrstand th idal classs, a tool which is still usd today. It sms natural to ask if w can nd othr sts of quivalnt forms and quip thm with a gnralisd composition law, which will not only turn th st into a group, but also giv us a tool to xplor dirnt numbr lds or rings. In 00 Bhargava startd to answr this qustion in a sris of four articls, whr h nds fourtn composition laws which can b usd to nd information on numbr rings and thir class groups. H nds ths composition laws by considring dirnt siz cubs of intgrs and crating dirnt forms from thm. This ida was rvolutionary as with this h managd to shd som 1

2 light on quartic and quintic numbr lds. Du to th impact thy had th cubs ar now rfrrd to as Bhargava's cubs. This papr is dividd in two main parts, in sction w will look at binary quadratic forms. W will spnd som tim on th rduction thory of positiv dnitiv forms, look at Dirichlt's composition, bfor proving th corrspondnc btwn quivalnc classs of primitiv binary quadratic forms and idal classs of orintd ordrs of quadratic lds. Th aim of this sction is to both show th dpth of information w hav from binary quadratic forms, w will only covr a minut fraction of it, and to introduc th main ida which links binary quadratic forms and idal classs. Th scond part, that is Sction 3, will follow Bhargava's rst, out of th four, paprs. W will look at cubs of intgrs, and construct dirnt objcts which will b in corrspondnc with dirnt moduls of a quadratic ring. Sinc w will b xploring quadratic rings, among our constructions w will rcovr Gauss's composition law, but w shall look approach it from a dirnt angl than in Sction. Th v objcts w will b looking at ar: binary quadratic forms; cubs of intgrs; binary cubic forms; pairs of binary quadratic forms and quatrnary altrnating -forms. Notation. In this papr if D < 0 thn D will b th notation for i D. A classical viw on Gauss composition.1 Som dnitions To start this sction w ar going to follow [Cox(1989)] and [Lmmrmyr(010)] to introduc th concpt of quadratic forms and how this ld Gauss to crat a composition law. W start by dning our objct of intrst, namly: Dnition.1. A binary quadratic form is a polynomial in two variabls of th form f(x, y) = ax + bxy + cy with a, b, c Z. Furthrmor w say a binary quadratic form is primitiv if gcd(a, b, c) = 1. Strictly spaking th abov dnition should b calld intgral binary quadratic forms but sinc in this papr w will only dal with intgral forms, that is whr th cocints ar ovr Z, w will drop th adjctiv intgral. In this papr, to rprsnt th binary quadratic form ax + bxy + cy, w will altrnat btwn th notation (a, b, c) (as usd originally by Gauss [Gauss(1986), Sction 153]) whn w ar not worrid about th valu of x and y, and th notation f(x, y) (as usd by Cox [Cox(1989)]). W not that any binary quadratic form is an intgr multipl of a primitiv binary quadratic form. An important numbr rlatd to a binary quadratic form is th discriminant: Dnition.. Th discriminant, D, of a binary quadratic form (a, b, c) is th intgr D = b ac. If w considr th group SL (Z), that is th st of all two by two matrics with dtrminant 1 and intgr ntris quippd with th multiplication opration, w can dn an action on th st of binary quadratic forms as follow. Lt f(x, y) b a binary quadratic form and ( ) r s S = SL t u (Z). Thn w can dn a nw binary quadratic form f S (x, y) = f(rx + sy, tx + uy). Dnition.3. W say two binary quadratic forms f(x, y) and f (x, y) ar quivalnt if thr xists S SL (Z) such that f S (x, y) = f (x, y). A numbr m is rprsntd by a binary quadratic form f(x, y) if thr xists (x 1, y 1 ) Z \ {(0, 0)} such that f(x 1, y 1 ) = m. Nxt w not a fw invariants of th action of SL (Z). Proposition.. Lt f(x, y) and f (x, y) b two quivalnt binary quadratic forms thn: 1. f(x, y) and f (x, y) hav th sam discriminant,. f(x, y) and f (x, y) rprsnt th sam numbrs,

3 3. lt g and g b th gcd of th cocints of f(x, y) and f (x, y) rspctivly. Thn g = g. Proof. First w not that to vry binary quadratic form corrsponds a two-by-two matrix. Namly if f(x, y) = ax + bxy + cy, thn th matrix ( ) a b A = b c is such that f(x, y) = x T Ax, whr x = (x, y) is a column vctor. W can s that dt A = ac b = D, whr D is th discriminant of f. By dnition f = f S = (Sx) T A(Sx) = x T (S T AS)x and sinc dt S = 1 w hav that dt(s T AS) = dt A, i.., th discriminant of f is th sam as th discriminant of f. Th binary quadratic form f can b considrd as a map from Z \ {(0, 0)} Z mapping (x, y) f(x, y). Ltting f = f S w gt th following commutativ diagram: Z \ {(0, 0)} S = Z \ {(0, 0)} f f Z Sinc SL (Z) is a group, vry lmnt S SL (Z) has an invrs, hnc S rally givs a bijction as showd in th diagram. So th imag of f is th sam as th imag of f, that is, thy rprsnt th sam numbrs. Lt g th gcd of th cocints of f (x, y). Thn notic that th numbrs that f (x, y) rprsnts ar all divisibl by g. Also not that f(1, 0) = a, f(0, 1) = c and f(1, 1) = a + b + c, so using th scond statmnt of this thorm w know that g a, g c and g (a + b + c), hnc g b. Hnc g g, and a symmtrical argumnt givs g g, so w hav g = g Th last part of th thorm givs in particular that if f(x, y) is primitiv, thn so is any binary quadratic form quivalnt to f(x, y). W not that af(x, y) = (ax + by) Dy, using this idntity w can s that if D is positiv thn f(x, y) can rprsnt both positiv and ngativ intgrs. On th othr hand if D is ngativ thn (ax + by) Dy is always positiv so f(x, y) can only rprsnt positiv intgrs if a is positiv or only ngativ intgrs if a is ngativ. This lads to th following dnition: Dnition.5. A binary quadratic form (a, b, c) is calld: positiv dnit if D < 0 and a > 0, ngativ dnit if D < 0 and a < 0, indnit if D > 0.. Rduction of positiv dnit forms In this subsction w will only study dnit binary quadratic forms. Sinc th study of ngativ dnit binary quadratic form can b dducd from th study of positiv dnit quadratic forms, w will not worry about thm. Furthrmor w assum all binary quadratic forms ar primitiv. W will look at primitiv indnit binary quadratic forms in a latr subsction. Th nxt qustion to aris is how to dtrmin whn two binary quadratic forms ar quivalnt? Lagrang cam up with a rduction algorithm [Cox(1989), p35] which turns vry binary quadratic form into a uniqu quivalnt form with minimal cocints. Thorm.6 (Dnition). Evry (primitiv positiv dnit) binary quadratic form is quivalnt to a uniqu binary quadratic form Q = (a, b, c) whr Q has th proprty that { b a c in all cass, b 0 if b = a or a = c. Th form Q is said to b (Lagrang) rducd. 3

4 Rmark. Whil thr is a notion of rducd form in th indnit cas, th rducd form is not uniqu. That is, an indnit binary quadratic form can b quivalnt to svral rducd forms. For this rason th thory of indnit binary quadratic form is quit intrsting, but it would tak us too far ald to what w want to do. Proof. For th proof of xistnc of such a quadratic form w will follow th constructiv proof in [Lmmrmyr(010), p9] whil for th uniqunss part w will hav a clos look at [Cox(1989), p7]. Existnc: Lt ( ) ( ) T n 1 n 0 1 =, U = SL (Z) and lt our binary quadratic form b (a, b, c). Not that T n turns (a, b, c) into (a, b an, an bn+c) and U turns (a, b, c) into (c, b, a). W ar going to apply th following algorithm: 1. If b > a, thn apply T n, with an appropriat n, to (a, b, c) to rduc b mod a so that b a.. If a > c, thn apply U to (a, b, c). 3. If b a c, thn stop; ls go to 1. This algorithm trminats whn b a and a c, i.., whn th rst two rquird conditions ar mt. This algorithm has to trminat, bcaus at vry loop of th algorithm w ar making b smallr and sinc b N w can not kp rducing b indnitly. W now nd to considr th two cass: if b = a, or a = c with b < 0. In th rst cas th transformation T 1 taks (a, a, c) to (a, a, c) which is of th rquird form, whil in th scond cas w us th matrix U to chang (a, b, a) into (a, b, a). This provs th xistnc of rducd binary quadratic forms. Uniqunss: Bfor w procd w show that if f(x, y) = (a, b, c) is rducd, thn th smallst numbr it rprsnt is a. Lt f(x, y) b rducd. Notic that f(±1, 0) = a and f(0, ±1) = c. Suppos that x y. Thn x y so f(x, y) ax +bxy+cx ax bxy +cy x (a b +c). Similarly if y x thn f(x, y) y (a b +c), so putting ths togthr w gt f(x, y) min{x, y }(a b +c). Hnc if xy 0 thn f(x, y) a b + c, and sinc f(x, y) is rducd a b + c c > a. So a is th smallst numbr rprsntd by f(x, y). If furthrmor w say a numbr is proprly rprsntd if gcd(x, y) = 1, thn, whn a c, w s that c is th nxt smallst proprly rprsntd numbr of f(x, y). Lt f(x, y) = (a, b, c) b rducd with a < c < a b + c, w dal with th cas b = a or a = c in th nxt paragraph. Suppos that g(x, y) = (a, b, c ) is an othr rducd binary quadratic form which is quivalnt to f(x, y), thn sinc th rprsnt th sam numbrs (Proposition.) thy must hav th sam rst cocint, i.., a = a. W know a c, (as g(x, y) is rducd) so suppos a = c thn g(±1, 0) = g(0, ±1) = a. In that cas, sinc g(x, y) is quivalnt to f(x, y), f(x, y) must also hav four dirnt solutions to f(x, y) = a. This is a contradiction as w know: if xy 0 thn f(x, y) a b + c > a; and f(0, y) = cy > a; so (±1, 0) ar th only two solutions to f(x, y) = a. Hnc a < c which, sinc c is th nxt smallst numbr proprly rprsntd by g(x, y), mans that c = c. Sinc f(x, y) and g(x, y) ar quivalnt thy hav th sam discriminant hnc, sinc a = a, c = c, w hav b = ±b. By assumption g(x, y) = f(rx + sy, tx + uy) with ru st = 1, so a = g(1, 0) = f(r, t) and c = g(0, 1) = f(s, u). This implis that (r, t) = (±1, 0) and (s, u) = (0, ±1). So f = g id or f = g id whr id is th idntity matrix and id th diagonal matrix with 1 as ntris. In ithr cas, this implis that b = b, i.., f(x, y) = g(x, y). If b = a or a = c thn w no longr hav th inquality a < c < a b + c. In both cass w still hav that a is th smallst numbr rprsntd by f(x, y), hnc a = a. W show that in both cass c = c. If a = c and a < c thn g(x, y) = a mans (x, y) = (±1, 0) but f(x, y) = a has at last four dirnt solutions which is a contradiction so w must hav that a = c implis a = c, i.., c = c. If b = a and a < c thn sinc w still hav th inquality a < a b + c, w can us th sam argumnt as bfor to show that a < c and conclud that c = c, as c is still th nxt smallst numbr proprly rprsntd by g(x, y). So in all cass w hav a = a, c = c which mans b = ±b as f(x, y), g(x, y) hav th sam discriminant. Sinc ithr b = a or a = c w hav b 0 and ithr b = b = a or c = c = a = a. Hnc b 0 and b = b, i.., f(x, y) = g(x, y). This thorm is usful in th study of quivalnc classs of binary quadratic forms as it givs a natural rprsntativ of such class. If w x a discriminant D < 0 and dnot by [f(x, y)] (or [(a, b, c)])

5 th quivalnc class of f(x, y), which w can rprsnt by its rducd binary quadratic form, w can look at th st of all quivalnc classs of binary quadratic forms with discriminant D. W will lt h(d) dnot th numbr of quivalnt classs. Thorm.7. Lt D < 0 b xd, thn h(d) is qual to th numbr of rducd forms of discriminant D. Furthrmor h(d) is nit. Proof. Th rst part follows from th prvious thorm. For th scond part x D < 0 and lt f(x, y) = (a, b, c) b a rducd binary quadratic form of discriminant D. Thn D = ac b a a = 3a, hnc a D/3. Sinc b a and a, b Z thr ar only a nit numbr of possibilitis for a and b. Furthrmor sinc c = D b a w hav that c is dtrmind onc a and b ar known. So for a xd D < 0 thr ar only a nit numbr of rducd binary quadratic forms with discriminant D..3 Dirichlt composition Stpping back to includ indnit binary quadratic forms, w us th sam notation [f(x, y)] to dnot th quivalnc class of f(x, y). W hav a st of quivalnc classs of primitiv binary quadratic forms, so on might ask is can w nd a binary opration to turn this st into a group? Gauss answrd this qustion by dning how to compos two binary quadratic forms of a givn discriminant D [Gauss(1986), sction 35]. Unfortunatly it is rgardd by many mathmaticians to b dicult, somtims vn vry dicult to undrstand and work with [Shanks(1989)]. It is oftn asir to follow Dirichlt's basic ida (rst apparing in [Dirichlt(1871), 10th Supplmnt]), which originally was somwhat mor rstrictiv as it imposs som conditions to th two forms to b composd, but dos hav th advantag of having an xplicit formula. W can slightly modify Dirichlt composition to ovrcom th rstriction imposd. Dirichlt thn found a group isomorphism btwn th st of quivalnc classs of primitiv binary quadratic forms of a xd discriminant and quivalnc classs of propr idals of xd ordrs of a quadratic ld, hnc asily stablishing that th quivalnc class of binary quadratic forms form an ablian group. Intuitivly, sinc binary quadratic forms rprsnt an innit sts of numbrs, givn two binary quadratic forms w want to nd a third on that rprsnts th pairwis products of th numbrs th two binary quadratic forms rprsnt. For a historical point of viw w ar going to giv Gauss' dnition of composition rst. Not. For this subsction w assum that D 0 and that our binary quadratic forms ar primitiv. Dnition.8. Lt f(x, y) and g(x, y) b two binary quadratic form. Dn a dirct composition to b a form F (x, y) = ax + bxy + cy such that f(x, y)g(z, w) = F (B 1 (x, y; z, w), B (x, y; z, w)), a 1 b a b 1 = f(1, 0), a 1 c a c 1 = g(1, 0), whr B i (x, y; z, w) = a i xz + b i xw + c i yz + d i yw, with a i, b i, c i, d i Z, ar bilinar forms. As w can s, this dnition is not vry asy to work with as it dos not giv a way of automatically crating a dirct composition. For this nd, Dirichlt's composition is much asir to us and is oftn rgardd as th way to do composition nowadays. To dn Dirichlt's composition w nd th following lmma. Lmma.9. Lt f(x, y) = (a, b, c) and g(x, y) = (a, b, c ) b two binary quadratic form of discriminant D and lt gcd(a, a, b+b ) =. ( Not that b and b must hav th sam parity sinc f(x, y), g(x, y) hav th sam discriminant). Thn thr is a uniqu intgr B modulo aa B b mod a, B b mod a, B D mod aa. such that: 5

6 Proof. This proof loosly follows [Cox(1989), p 8]. First w ar going to show that th thr givn condition can b rarrangd to giv thr quivalnt conditions. If an intgr B satiss th rst two congruncs thn B (b + b )B + bb (B b)(b b ) 0 mod aa. Hnc rarranging and dividing by, which w can, sinc (b + b ) and (b(b + b ) ac) = (bb + D), w gt that b+b B bb +D mod aa. Multiplying th rst two congruncs by a and a rspctivly w hav that out initial conditions bcom: a B a aa b mod, a B a b mod aa, b + b B bb + D mod aa. (.1) W can just work backward to s th implication th othr-way, as th rst two congrunc ar quivalnt to B b mod a and B b mod a. Thn th argumnt works in rvrs. So w nd to nd B which satisfy th lattr thr conditions, which is asir to nd as th quations ar all modulo aa and all linar (w rmovd th B trm). Sinc gcd( a, a, b+b ) = 1 by Euclid's a algorithm w know thr xists n 1, n, n 3 Z such that n 1 + n a + n 3 b+b = 1. Fix such n 1, n, n 3 and lt B b th uniqu numbr btwn 0 and aa ab such that B n 1 Not that sinc D = b a c = b ac thn a D a b mod aa furthrmor sinc b b mod w hav aa b aa b mod aa a B n a ab a a b a bb + a ( D a 1 + n + n 3 n 1 + n a + n 3 a B n aab aa b abb + a D 1 + n + n 3 ( a n 1 + n a + n 3 + n a b + n 3 bb +D mod aa and ad ab mod aa. Now w can s that b + b ) a b + b b a aa b mod ) a b a b mod aa b + b B n abb + ab a b + a b b bb + D b + b 1 + n + n 3 bb + D mod aa Hnc w hav constructd a B which satiss th congruncs (.1). Suppos that thr is a scond B which satiss th thr rlations, thn aftr som rarrangmnt w hav a (B B ) a (B B ) b+b (B B ) 0 mod aa. Hnc w hav aa a (B B ), aa a (B B ) and aa b+b (B B ), this implis that a, a (B B ). Using th fact that gcd( a, a, b+b ) = 1 and th last divisibility condition w s that aa (B B ), i.., B B mod aa. Hnc w hav found a uniqu (modulo ) B which satiss th conditions. aa Dnition.10. Lt f(x, y) = (a, b, c), g(x, y) = (a, b, c ) b two binary quadratic forms of discriminant D. Thn th Dirichlt composition of f(x, y) and g(x, y) is th form F (x, y) = aa x + Bxy + (B D) aa y whr B is th (uniqu up to modulo aa ) intgr B of Lmma.9. Rmark. Th proof of.9 gav us an xplicit formula to calculat th intgr B that w can us for ab Dirichlt composition, namly B = n 1 + n a b + n 3 bb +D whr n 1, n, n 3 ar such that n 1 a + n a b+b + n 3 =. At this point th radr might wondr why w wnt to prov a non-trivial lmma instad of dning B for Dirichlt composition as in th prvious sntnc, which is a nicr and mor straightforward dnition. On of th rasons is that way w stablishd som of th congruncs that B satisfy, which w will us in othr proofs of this papr, and, in a fw cass, th congrunc givs an asir B to work with than th xplicit formula, which still involv running Euclid's algorithm. Naturally w nd to chck a fw proprtis of Dirichlt composition to s that thy it is usful whn considring quivalnc classs of binary quadratic forms, spcially th implid claim that th Dirichlt composition of two binary form is uniqu, in fact on can asily s that it is not. Th following thorm will clarify ths issus.., 6

7 Thorm.11. Lt f(x, y) = (a, b, c) and g(x, y) = (a, b, c ) b two binary quadratic forms of discriminant D that ar both positiv dnit or both indnit and F (x, y), F (x, y) b two Dirichlt compositions. Thn F (x, y) is a primitiv binary quadratic form with discriminant D. positiv dnit. If D < 0 thn F (x, y) is [F (x, y)] = [F (x, y)]. Proof. For as of notation w will assum gcd(a, a, b+b ) = 1 (on can chck that rplacing a, a, b+b by a, a, b+b rspctivly in th right placs will still mak th argumnt hold). W prov that F is primitiv. Lt C = B D aa so that F (x, y) = aa x + Bxy + Cy. Rcall that (a, b, c) is quivalnt to (a, b + an, an + bn + c) (s proof of.6) and by dnition B = b + an for som n Z. Notic that a C = B D a = a n +ban+b b +ac a = an +bn+c, hnc (a, b, c) is quivalnt to (a, B, a C) similarly (a, b, c ) is quivalnt to (a, B, ac). Notic that if w lt X = xz Cyw and Y = axw + a yz + Byw thn w can s that F (X, Y ) = aa x z + abx zw + a Cx w + aa C y w +a Cy z + a BCy zw + B xyzw + abcxyw = (ax + Bxy + a Cy )(a z + Bzw + acw ) Sinc (a, B, a C) and (a, B, ac) rprsnt th sam numbrs as (a, b, c) and (a, b, c ) rspctivly w s that F (x, y) rprsnts all th numbrs of th form f(x 1, y 1 )g(z 1, w 1 ). Suppos that F (x, y) is not primitiv, that is, thr is a prim p that divids th cocints of F (x, y), thn p divids all th numbrs F (x, y) rprsnts. Hnc w hav that p divids all th numbrs of th form f(x 1, y 1 )g(z 1, w 1 ), so p aa. Suppos that p a, thn p a. But p ac implis p c and p a(a +b +c ) hnc p b contradicting th primitivity of g(x, y). So w hav that p a and p a, but th sam argumnt can b rpatd on p cc implying that p divids c and c. Furthrmor p (a + b + c)(a + b + c ) implis p bb so ithr all th cocints of f(x, y) or all th cocints of g(x, y) ar divisibl by p again, contradicting primitivity of f(x, y) or g(x, y). Hnc thr is no prim p that divids th cocints of F (x, y). Th discriminant of F (x, y) is B (aa ) B D aa = D. If D < 0 thn both a, a > 0, hnc so is aa, so F (x, y) is positiv dnit. Lt F (x, y) = (aa, B, C) and F (x, y) = (aa, B, C ) whr B = B + aa n for som n Z and C = B D aa. Thn applying T n to (aa, B, C) w gt (aa, B, C ). Hnc [F (x, y)] = [F (x, y)]. Th group of positiv dnit binary quadratic forms. W ar now on our way to show that th st of (primitiv) binary quadratic forms of discriminant D, which for this sction w will dnot C(D), is a nit ablian group. W will split th proof in two cass: whn D < 0 and whn D > 0 and squar-fr. Th scond cas is a slight gnralisation of th rst cas and th proof is fairly similar, in fact it covrs th rst cas. W split it into two cass so to introduc nw concpts slowly. W ar going to look at quadratic lds, lt us dnot such a ld as Q( N) whr N is a squarfr intgr. So if α K = Q( N) thn α = a + b N for som a, b Q, w dnot th conjugat of α, i.., th non-trivial automorphism acting on α, by α = a b N. Also in this papr by minimal polynomial of an lmnt α w will man th last dgr polynomial f Z[x] with coprim cocints such that f(α) = 0 and th lading cocint of f is positiv, not th last dgr monic polynomial f Q[x] such that f(α) = 0. Dnition.1. Lt K = Q( N) b a quadratic ld. W dnd an ordr O in K to b a subst of K such that 1. O is a subring of K,. O is a nitly gnratd Z-modul, 3. O contains a Q-basis of K. 7

8 W not that O has rank two ovr Z and hnc w can dn it by its basis α, β. W us th notation [α, β] to man th Z-modul with α, β as its basis. Lt O K dnot th maximal ordr and dn th discriminant of K to b as follow: { N if N 1 mod, d K = N othrwis. Th conductor of O is th indx [O K : O] = f and dn th discriminant of th ordr O = [α, β] to b ( ( )) α α D = dt. β β On can show that O K = [1, ω K ] whr ω K = d K+ d K (a fact from MA3A6 Algbraic Numbr Thory) and if f is th conductor of O thn O = [1, fω K ]. Sinc th discriminant dos not dpnd on th choic of basis w can s that D = f d K using th prvious basis. W can s from th this that if D is th discriminant of O th quadratic ld, K, which O is an ordr of is K = Q( D), furthrmor D satiss D 1, 0 mod. Dnition.13. Lt K b a quadratic ld and lt I b an idal of an ordr O. W say that I is a propr idal if O = {β K : βi I}. A fractional idal I of O is a O-modul such that ωi O for som ω O. Givn a fractional idal I, w say that it is invrtibl if thr xists a fractional idal J such that IJ = O. Not that th dnition of propr xtnds to fractional idals, also by th dnition of propr idal, vry idal I is a propr idal of a uniqu ordr O. Rcall that th norm of an lmnt α K is N(α) = αα, whil th norm of an fractional idal I = [α, β] O = [1, τ], is ( N(I) = abs α α ) 1 β β, D whr D is th discriminant of O (or K), or quivalntly if α = a 1 + a τ, β = b 1 + b τ, thn ( ) a N(I) = abs 1 a b 1 b. This is oftn thought as th indx of I in O, that is, N(I) = [L:I] [L:O] whr L is lattic containing both O and I, that is, L is a Z-modul of rank which contains both O and I. Not that w hav to tak th absolut valu so that th dnition is indpndnt of th choic of th basis for I. W will us Tr to dnot th trac functions which maps α to α + α. Thorm.1. Lt K = Q(τ) b a quadratic ld and ax + bx + c th minimal polynomial of τ. Thn [1, τ] is a propr fractional idal for th ordr [1, aτ] of K. Proof. W can asily s that [1, τ] is an [1, aτ]-modul as 1, τ, aτ and aτ = bτ c ar all in [1, τ], furthrmor a[1, τ] [1, aτ], hnc [1, τ] is a fractional idal of [1, aτ]. From this it follows that [1, aτ] {β K : β[1, τ] [1, τ]}. Th condition {lt β K b such that β[1, τ] [1, τ]} is quivalnt to th condition {lt β K b such that β [1, τ] and βτ [1, τ]}. Now β [1, τ] mans that thr xists m, n Z such that β = m+nτ. Hnc βτ = mτ +nτ = mτ + n cn a ( bτ c) = a +( bn a + m) τ. So βτ [1, τ] if and only if a n (sinc gcd(a, b, c) = 1 as thy ar th cocints of th minimal polynomial). Hnc β = m+an τ, implying that {β L : β[1, τ] [1, τ]} [1, aτ]. So {β K : β[1, τ] [1, τ]} = [1, aτ] In particular th prvious thorm stats that [1, τ] is a propr idal of only [1, aτ]. Th nxt thorm shows us a proprty of quadratic lds, which in gnral dos not hold for highr dgr numbr lds. Thorm.15. Lt O b an ordr in a quadratic ld K and lt I b a fractional O-idal. Thn I is propr if and only if I is invrtibl. 8

9 Proof. Suppos I is invrtibl, that is, thr is anothr fractional idal J such that IJ = O. If β K and βi I thn βo = βij IJ = O. Sinc βo O w hav β O, hnc I is propr. Suppos I = [α, β] (α, β K) is a propr fractional idal. Ltting τ = β α w hav I = α[1, τ]. Lt ax + bx + c b th minimal polynomial of τ. Thn by Thorm.1, w know that O = [1, aτ]. Sinc τ is also a root of th minimal polynomial of τ w know that [1, aτ] = [1, aτ] = O and by Thorm.1 I = α[1, τ] is propr fractional idal of O. Nxt considr aii = aαα[1, τ][1, τ] = N(α)[a, aτ, aτ, aττ]. Furthrmor sinc w hav an xplicit quadratic polynomial which has τ and τ as its only roots w know τ + τ = b a and ττ = c a, so aiī = N(α)[a, aτ, b, c] = N(α)[1, aτ] = N(α)O (whr th scond to last quality follows from th fact gcd(a, b, c) = 1). Hnc thr xists a fractional idal, namly J = I, such that IJ = O a N(α) Dnition.16. Lt O b an ordr of a quadratic ld K. Lt I(O) dnot th st of invrtibl fractional idals of O, and P (O) b th st of non-zro principal idals of O. It is quit clar, sinc vry invrtibl idal has an invrs and th multiplication of any two invrtibl idal is an invrtibl idal, that I(O) forms an ablian group undr multiplication. Notic also that P (O) I(O), which allows us to dn our objct of intrst. Dnition.17. Lt O b an ordr of a quadratic ld K. W dn th idal class group of O to b C(O) = I(O)/P (O). W can nally prov our claim of C(D), with D < 0, forming a group by rlating C(O) to C(D). Thorm.18. Lt D < 0 and O = [1, fω K ] an ordr with discriminant D of th quadratic ld K = Q( D), whr f is th conductor of O and ω K = d K+ d K with d K th discriminant of K. [ ] 1. If f(x, y) = (a, b, c) is a binary quadratic form with discriminant D thn a, b+ D is a propr idal of O, [ ]. C(D) is a group and th map snding f(x, y) to inducs an isomorphism btwn a, b+ D C(D) and C(O). Hnc th ordr of C(O) is th class numbr h(d). Proof. This proof follows [Cox(1989), p 137]. Lt f(x, y) = (a, b, c) b a binary quadratic form with discriminant D < 0. Sinc D < 0, th complx roots of f(x, 1) = ax + bx + c ar not ral, so lt τ b th uniqu root with positiv imaginary part, w say τ is th root of f(x, y). Sinc a > 0, it follows that τ = b+ D a, hnc [ a, b+ D ] = [a, aτ] = a[1, τ]. Rcall that if D is th discriminant of O thn K = Q( D) hnc τ K. Now by Lmma.1 w know that a[1, τ] is a propr idal of th ordr [1, aτ] so w want to prov that O = [1, aτ]. First rcall that f d K = D = b ac hnc fd K and b hav th sam parity, so b+fd k Z. W us this fact to show: aτ = b + D = b + f d K = b + fd K = b + fd K ( dk + ) d K + f + fw K Hnc [1, aτ] = [1, fω K ], so w hav nishd proving th rst part. Bfor w prov th scond part, w claim that if τ C and r, s, t, u Z thn ( ) ( ) rτ + s r s I(τ) I = dt (.) tτ + u t u tτ + u 9

10 To prov this, lt τ = x + iy thn ( ) rτ + s I tτ + u ( ) (rx + riy + s)(tx tiy + u) = I (u + tx) + t y 1 = I(rtixy + ruiy rtixy stiy) tτ + u ( ) 1 r s = tτ + u dt I(iy) t u W ar going to show th map is wll-dnd. Lt f(x, y) and f (x, y) b binary quadratic forms of discriminant D and τ, τ b thir rspctiv root. Suppos that thy ar quivalnt, i.., lt f(x, y) = f (rx + sy, tx + uy) with ru st = 1. Thn 0 = f(τ, 1) = f (rτ + s, tτ + u) = (tτ + u) f ( rτ + s tτ + u, 1 ), (.3) but using quation (.) w hav that rτ+s tτ+u has positiv imaginary part, hnc by uniqunss of τ w hav τ = rτ+s tτ+u. Lt λ = tτ + u K thn λ[1, τ ] = (tτ + u)[1, rτ+s tτ+u ] = [tτ + u, rτ + s] = [1, τ]. To show that th map is injctiv w work backward. Convrsly suppos [1, τ] = λ[1, τ ] for som λ K thn [1, τ] = [λ, λτ ], in othr words λτ = rτ + s and λ = tτ + u for som r, s, t, u Z, such that th matrix ( ) r s S = t u is invrtibl. Rarranging w gt τ = rτ+s tτ+u, but sinc τ and τ both hav positiv imaginary parts by quation (.), w hav that th dtrminant of S is positiv, i.., ru st = 1. Thn by quation (.3) w hav that f (rx + sy, tx + uy) and f(x, y) hav th sam roots, so it follows that thy ar qual hnc f(x, y) is quivalnt to f (x, y). W hav just shown that two binary quadratic forms ar quivalnt if and only if [1, τ] = λ[1, τ ], but th last quality is th sam as saying that [1, τ] and [1, τ ] ar in th sam quotint. So w hav provd that th map snding f(x, y) to a[1, τ] inducs an injction C(D) C(O) Now w show th map is surjctiv. Lt I b a fractional idal of O and lt I = [α, β] for som α, β K. Switching α and β if ncssary w can assum that τ = β α has a positiv imaginary part and lt ax + bx + c b th minimal polynomial of τ with a > 0. Lt f(x, y) = ax + bxy + cy, not that sinc τ is a root and is not a ral numbr th form f(x, y) must hav a ngativ discriminant and hnc is positiv dnit. Furthrmor sinc th discriminant of O = [1, aτ] is D = (aτ aτ) = b ac w hav that th discriminant of f(x, y) is D. So f(x, y) is a binary quadratic form (it is primitiv as it ax + bx + c is a minimal polynomial) that maps to a[1, τ]. But a[1, τ] lis in th class of I = [α, β] = α[1, τ] in C(O) so th map is surjctiv. W hav a bijction btwn th sts C(D) C(O). Lt f(x, y) = (a, b, c) and g(x, y) = (a, b, c ) b two binary quadratic forms of discriminant D and F (x, y) thir Dirichlt composition, thn thir imags ar [a, b+ D ], [a, b + D ], [ aa, B+ D ] rspctivly whr = gcd(a, a, b+b ) and B = 1 (n 1ab +n a bb b+n +D 3 ) for som n 1, n, n 3 Z such that n 1 a + n a b+b + n 3 =. To prov our claim that Dirichlt composition corrsponds to multiplication of idals (up to quivalnc classs in both cass) w nd to show that [a, b+ D ][a, b + D ] = [aa, a b + D, a b+ D, 1 (bb +D) 1 (b+b ) D ] is quivalnt to [ aa, B+ D ]. W claim that in fact th product is qual to [ aa, B+ D ]. To s this, if w rcall that th norm is multiplicativ, w s, 1 (bb +D) 1 (b+b ) D that th N([aa, a b + D, a b+ D ]) = aa = N([ aa, B+ D w can us our congruncs (.1) and th fact a, a, b+b to show that [a, b+ D ][a, b + D ]), furthrmor w ] +m B+ D, whr m = a and n is such that a B = a b + [ aa, B+ D ] (for xampl a b + D = n aa n aa ). Putting ths two facts togthr togthr w hav [a, b+ D ][a, b + D ] = [ aa, B+ D ], but [ aa, B+ D ] = 1 [ aa, B+ D ], hnc thy ar in th sam idal class, complting th proof. Rmark. W mak a rmark about th map w dscribd whn proving surjctivity of th bijction. Not that if τ has for minimal polynomial ax + bx + c thn N(τ) = c a and Tr(τ) = b a. Furthrmor, w hav that N([1, τ]) = 1 a, so th binary quadratic form associatd to [1, τ] was x +Tr(τ)xy+N(τ)y N([1,τ]). 10

11 Now if τ = β α, notic that N([1, τ]) = β α β α = βα αβ αα w hav that th binary quadratic form associatd to I = [α, β] is = N([α,β]) N(α) and Tr( β α ) = Tr( βα N(α) ) = Tr(αβ) N(α). So x + Tr(αβ) N(α) xy + N(β) N(α) y N(I) N(α) = N(α)x + Tr(αβ)xy + N(β)y N(I) = N(αx + βy). N(I).5 Gnral cas W ar now going to considr th quivalnc classs of primitiv indnit binary quadratic forms. Th rason th prvious proof would fail if D > 0 is that th roots of f(x, 1) ar ral, and hnc our invrs map is not uniqu as it dpnds on th choic of D. This can b rctid by considring th narrow class group instad of th class group. Bfor w can dn this w nd a fw dnitions. Dnition.19. An ordr O of discriminant D is said to b orintd onc a choic of D has bn mad. Onc this choic has bn mad w can dn a map π : O Z by π(τ) = τ τ D, which has th particularity that π(x + y D) = y With th ida of orintd ordr coms th ida of orintating fractional idal. Dnition.0. An orintd fractional idal of an orintd ordr O is a pair (I, ɛ), whr I is a fractional idal of O and ɛ {±1} is th orintation. W say (I, ɛ) is positivly orintd if ɛ = 1, othrwis it is ngativly orintd. A basis α, β of an orintd fractional idal (I, ɛ) is said to b corrctly orintd if π(αβ) = αβ αβ D has th sam sign as ɛ. W dn th product of two orintd fractional idal (I, ɛ) and (I, ɛ ), of an ordr O of discriminant D, to b (II, ɛɛ ). This can asily b xtndd to th product of an orintd fractional idal and an lmnt of Q( D) to b α(i, ɛ) = (αi, sgn(n(α))ɛ), whr sgn(α) is th usual sign function. W will from now on assum that all fractional idals of an orintd ordr ar thmslvs orintd, hnc w will writ I to man th pair (I, ɛ). Furthrmor, whn talking about a basis of an orintd idal, w will amus that th basis is corrctly orintd. W just nd on mor dnition, a rnmnt of quivalnt idals for our nw stting, and thn w can dn th narrow class group. Dnition.1. Two orintd fractional idals, I, J of an ordr O of a discriminant D, ar quivalnt if thr xists α Q( D) such that αi = J. W dnd th narrow class group, C + (O), of an orintd ordr O, to b th st of quivalnc classs of orintd fractional invrtibl idals. Rmark. In many litratur, th narrow class group is dnd dirntly as follows: Lt P + (O) b th st of principal fractional idal of O with a gnrator of positiv norm. Th narrow class group is C + (O) = I(O)/P + (O). Whil th usual narrow class groups has th proprty that C + (O) = C(O) whn D < 0, it has th problm that you nd to dal with positiv dnit binary quadratic forms sparatly from ngativ dnit binary quadratic forms. Our dnition has th advantag that it givs th corrct notion whn D < 0 as it dos not distinguish btwn positiv dnit and ngativ dnit, a distinction that w can not mak in sction 3. Anothr sid rmark is that w can rlat C + (O) to C(O) by considring th following short xact squnc: 1 {±1}/N(O ) C + (O) C(O) 1 ɛ (O, ɛ) (I, ɛ) I If th discriminant D of O is ngativ, thn vry lmnt of O, including units, hav a positiv norm, hnc {±1}/N(O } = {±1}. Furthrmor th short xact squnc split, sinc th map dnd by I (I, 1) is a sction of th map (I, ɛ) I, it is only wll dnd sinc idals ar only quivalnt to idals with th sam orintation. Sinc th short xact squnc split, w hav by th splitting lmma C + (O) = {±1} C(O). If D > 0 and O has a unit with ngativ norm, thn {±1}/N(O ) = 1, so C + (O) = C(O). Finally if D > 0 and O dos not hav any unit with ngativ norm, thn all w can say is that C + (O) C(O). 11

12 With th abov rmark w rdn C(D) to b th st of quivalnc classs of primitiv binary quadratic forms of discriminant D. In particular whn D < 0 thn C(D) includs both positiv dnit and ngativ dnit form, hnc it is twic th siz of our prvious C(D). With orintd fractional idals, w nd to mak a slight modication to th dnition of th norm of an idal. Rcall that whn w dnd it, it was pointd out that w had to tak th absolut valu so that th dnition is indpndnt of th choic of basis. It turns out that with orintd fractional w do not nd to tak th absolut valu anymor, or quivalntly th norm of an orintd idal is ɛn(i). Tha is if α, β is a corrctly orintd basis of I, thn ( ) α α 1 N(I) = β β D Thorm.. Lt D b a non-squar intgr and O an orintd ordr with discriminant D of th quadratic ld K = Q( D). Thn thr is a bijction btwn th C(D) and C + (O) dnd by [(a, b, c)] maps to th quivalnc class of [a, b+ D ]. This bijction is an isomorphism of groups. Proof. This proof will follow th sam structur as th proof of Thorm.18 with a fw slight altration. W rst show that [a, b+ D ] is a propr idal of O. Lt f(x, y) = (a, b, c) b a binary quadratic form. Lt τ b th root of f(x, 1) = ax + bx + c such that π(τ) has th sam sign as a. Explicitly, using th quadratic formula, w hav τ = b+ D a. Sinc N(a) > 0, w hav [a, b+ D ] = a[1, τ] is quivalnt to [1, τ] in C + (O). By Thorm.1 w know [1, τ] is a propr idal. W now chck th map is wll dnd. Suppos that f(x, y) = (a, b, c) and f (x, y) = (a, b, c ) such that f(x, y) = f (rx + sy, tx + uy) with ru st = 1, this mans that a = au but + ct. Lt τ b th root of f(x, 1) with π(τ) having th sam sign as a, and τ th root of f (x, 1) with π(τ ) having th sam sign as a, w show how to rlat τ and τ. Considr and w calculat that 0 = f(τ, 1) = f (rτ + s, tτ + u) = (tτ + u) f ( rτ + s tτ + u, 1 π ( ) rτ + s tτ + u ( ) (rτ + s)(tτ + u) = π N(tτ + u) ( ) rtττ + su + ruτ + stτ = π N(tτ + u) (ru st)(τ τ) = N(tτ + u) D π(τ) = N(tτ + u) But w hav that N(tτ + u) = (tτ + u)(tτ + u) = u + Tr(τ)tu + N(τ)t = u b a tu + c a t = a a. So w hav two cass: Cas 1. Cas. ) (.) (.5) N(tτ + u) > 0: in which cas a and a hav both th sam sign and τ = rτ+s tτ+u. Thn if w lt λ = tτ + u Q( D), w hav N(λ) > 0 and λ[1, τ ] = [1, τ]. Hnc our two idals ar quivalnt, as both [1, τ ] and [1, τ] hav th sam orintation. N(tτ + u) < 0: in which cas a and a hav dirnt sign and τ = rτ+s tτ+u. Thn if w lt λ = tτ + u Q( D), w hav N(λ) < 0 and λ[1, τ ] = [1, τ]. But sinc [1, τ ] and [1, τ] hav dirnt orintation, w hav that thy ar in fact quivalnt. Convrsly, to show injctivity, suppos that [1, τ] = λ[1, τ ] for som λ = tτ + u Q( D). Thn w hav λτ = rτ + s and λ = tτ + u for som r, s, t, u Z such that th matrix ( ) r s S = t u is invrtibl. Rarranging w gt τ = rτ+s tτ+u, and by quation (.5), w hav π(τ ) N(λ) = ru st. π(τ) 1

13 But sinc if N(λ) > 0 thn π(τ) and π(τ ) hav th sam sign, whil if N(λ) < 0 thn π(τ) and π(τ ) hav dirnt sign, w hav that ru st = 1. So S is in SL (Z) and f(x, 1) and f (x, 1) hav th sam roots, hnc thy ar quivalnt. W hav shown that two binary quadratic forms ar quivalnt if and only if [1, τ] = λ[1, τ ] for som λ. Hnc w hav provd th map is injctiv. W nally show th map is surjctiv. Lt I b a fractional idal of O and lt I = [α, β] for som α, β K. Without loss of gnrality w can assum N(α) > 0, as vry idal has a basis with N(α) > 0 which can b achivd by taking th smallst α Q 0 I. Lt τ = β α, so I = α[1, τ]. Considr th quadratic form f(x, y) = N(x + τy) N([1, τ]) = x + Tr(τ)xy + N(τ)y. N([1, τ]) Lt ax + bxy + cy b th minimal polynomial of τ with sgn(a) = sgn(π(τ)), hnc gcd(a, b, c) = 1. Thn by Thorm.1 w s that that [1, τ] is a propr fractional idal for th ordr O, with N([1, τ]) = 1 a. Thrfor f(x, y) = ax a Tr(τ)xy + N(τ)y = ax + bxy + cy. This shows that th map is surjctiv sinc f(x, y) [a, b+ D ] and I = α[1, τ] = α[1, b+ D a ] ar both in th sam idal class as N(α) > 0 Now that w hav shown th bijction btwn C(D) and C + (O), w claim that th work w hav don in Thorm.18 shows how th bijction inducs th isomorphism. W know that, for D < 0, C(D) is nit and sinc Dirichlt composition is symmtric w hav that C(D) is a nit ablian group, w can prov th sam rsults for D > 0, but this rquirs mor background in th rduction thory of indnit forms. W not that th invrs of th bijction is givn in th abov thorm, that is I = [α, β] N(αx+βy) N(I). W nish this sction by nding th idntity and th invrs of vry lmnt in C(D), as this is oftn ssntial to group calculations. Thorm.3. Lt D 0, 1 mod. Th idntity lmnt of C(D) is th class containing th principal form { x D y if D 0 mod, x xy + 1 D y if D 1 mod. Th invrs of th class containing th form ax + bxy + cy is th class containing ax bxy + cy. Proof. Lt f(x, y) = (a, b, c) and I(x, y) b th principal form. Lt ɛ {0, 1} b such that ɛ D mod. First w not that gcd(a, 1, b+ɛ ) = 1. W show that in this cas B = b, whr B is th intgr ndd in Dirichlt composition. Sinc D = b ac w asily s that b D mod a, furthrmor b b mod a. So if D thn b so b 0 mod. If on th othr hand D thn b so b 1 mod. Hnc b satisfy th thr congrunc rlation ndd to nd B in th Dirichlt composition. Hnc th Dirichlt composition of f(x, y) and I(x, y) is ax + bxy + b (b ac) a y = f(x, y) as rquird. Lt f(x, y) = (a, b, c) and f(x, y) = (a, b, c). Rcall that (a, b, c) is quivalnt to (c, b, a) = g(x, y) and gcd(a, b, c) = 1. It is asy to s that b satiss th thr congrunc rlations ndd to nd B, hnc th Dirichlt composition of f(x, y) and g(x, y) is acx + bxy + b D ac y = acx + bxy + y = (ac, b, 1). Whil this is not th principal form, rcall that (a, b, c) is quivalnt to (a, b+an, an +bn+c) for any n Z. If D 0 mod thn b so lt n = b thn, if w lt dnot quivalnc, notic that (ac, b, 1) (1, b, ac) (1, b + n, n bn + ac) (1, 0, b b + ac ) (1, 0, D ) which is th principal form whn D 0 mod. Similarly if D 1 mod w can show that (ac, b, 1) is quivalnt to th principal form by noticing that b so if w lt n = b 1 thn (ac, b, 1) (1, b+n, n bn+ac) (1, 1, b +b+1 b b+ac ) = (1, 1, 1 D ). Hnc in both cass th Dirichlt composition is in th class of th principal form, proving that th invrs of (a, b, c) is (a, b, c). 13

14 3 Bhargava's cubs 3.1 A nw look at Gauss composition In this sction w will rfr to svral typs of quivalnc, so w will say two objcts A, B ar G- quivalnt if thr xists g G such that A = B g, whr G is a group and B g th imag of B aftr bing actd on by g. Not that by this dnition, in Sction w talkd about SL (Z)-quivalnt binary quadratic forms. W will also switch from multiplicativ notation to additiv notation for th group of binary quadratic forms, sinc w know th group is ablian. Furthrmor throughout this sction w will assum all our spacs ar non-dgnrat, that is, whnvr w talk about th discriminant of an objct; b it a cub of intgrs, a quadratic ring or othrwis, w will assum that it is non-zro. As notd bfor, thr is a link btwn binary quadratic forms and matrics, so on might wondr if w can nd anything of intrst using cub of intgrs. This is what Bhargava dos [Bhargava(00)]: h considrs cubs of intgrs and xplors six dirnt ways ths cubs of intgrs can rprsnt forms; with ach of ths forms h xplors th qustion of whthr a composition law can b associatd to th forms so to crat a group. W shall xplor v dirnt composition laws complt with proofs and xampls. Considr th spac ( Z ) 3, this has a basis {i j k : i, j, k {1, }} whr 1, ar th standard Z-basis of Z. So using this basis w hav that ach lmnt of (Z ) 3 is uniquly dtrmind by th ight intgrs a ij,k, that is x = i,j,k a ijk( i j k ). If w lt (a, b, c, d,, f, g, h) = (a 1,1,1, a 1,1,, a 1,,1, a 1,,, a,1,1, a,1,, a,,1, a,, ) thn w can can rprsnt vry lmnt of (Z ) 3 by a cub of intgrs: f (3.1) a b c g h, So (Z ) 3 can b idntid with th spac of all cubs of intgrs. If w lt Γ = (SL (Z)) 3, w hav a right natural action on (Z ) 3, that is γ = γ 1 γ γ 3 Γ givs a map γ 1 γ γ 3 : Z Z Z Z Z Z, with th map γ i : Z Z bing th usual lft action. W want to s what this corrsponds to in trms of cubs of intgrs. Lt A (Z ) 3 and not that A has thr dirnt ways to b rprsntd by two matrics namly ( ) ( ) a b f M 1 =, N c d 1 = ; g h ( ) ( ) a c b d M =, N g = ; f h ( ) ( ) a c g M 3 =, N b f 3 =. d h Lt us considr γ id id acting on our cub, and for as of us lt us tmporarily go back to a i,j,k notation. With this in mind w s that M 1 can b put in corrspondnc with a 1,j,k ( 1 j k ), N with a,j,k ( j k ), M with a i,1,k ( i 1 k ), ans so on. Lt ( ) r s γ = SL t u (Z), thn th map γ : Z Z snds ( 1, ) (r 1 + s, t 1 + u ), from this w s that th imag of a1,j,k ( 1 j k ) + a,j,k ( j k ) undr γ id id is a 1,j,k ((r 1 + t ) j k ) + a,j,k ((s 1 + u ) j k ). Rgrouping trms togthr, w gt (ra 1,j,k + sa,j,k )( 1 j k ) + (ta1,j,k + ua,j,k )( j k ). Whn putting this back in trms of matrics, w s that γ id id d 1

15 snds (M 1, N 1 ) (rm 1 + sn 1, tm 1 + un 1 ). If w rpat this argumnt with id γ id and id id γ, thn w s that in gnral, th natural action is dnd as follows: w tak an lmnt ( ) ( ) ( ) r1 s γ = 1 r s r3 s 3 Γ t 1 u 1 t u t 3 u 3 thn w dn th action of γ on A, dnotd A γ, by rplacing, in any ordr, (M i, N i ) by (r i M i + t i N i, s i M i +u i N i ). Notic that at ach stag all th M i, N i ar actd but from our abov discussion w s that th ordr in which you apply ach individual γ i dos not mattr. Exampl. Lt A b th cub which has ( ) 1 M 1 = N 1 =, γ = 3 ( ) ( ) ( ) W calculat A γ : at th rst stag w rplac (M 1, N 1 ) by (M 1 + N 1, M 1 + N 1 ) so w gt ( ) ( ) 3 6,, hnc A = (,, 6, 8, 3, 6, 9, 1). Nxt w nd to rplac our nw M and N, bing rspctivly ( ) ( ) 6 8,, by (M +3N, M +N ) which givs A = (16, 10, 36,,, 15, 5, 33). Finally w apply th last componnt of γ, which just ngats vry ntry to nd that A γ = ( 16, 10, 36,,, 15, 5, 33). So w hav that 15 A γ = , 36 Dnition 3.1. Lt A (Z ) 3 b a cub and M i, N i for i {1,, 3} b as abov, thn w construct thr binary quadratic forms Q A i, dnotd Q i if A is obvious, by stting Q i (x, y) = dt(m i x N i y) Explicitly Q 1 = ((ad bc)x + (cf + bg d ah)xy + (h fg)y ) = (dt M 1 x + (cf + bg d ah)xy + dt N 1 y ). W rcall from Sction that th discriminant of a binary quadratic form ax +bxy+cy is D = b ac. An xplicit calculation shows that, givn an arbitrary A (Z ) 3, th discriminant of Q 1, Q and Q 3 is th sam, lading to th following dnition. Dnition 3.. W dn th discriminant of A to b D = Disc(Q i ) = a h + b g + c f + d (abgh + cdf + acfh + bdg + adh + bfcg) + (adfg + bch) Lmma 3.3. Lt A (Z ) 3 giv ris to Q 1, Q, Q 3 and γ = γ 1 γ γ 3 thn A γ givs riss to th thr binary quadratic forms Q γ 1 1, Qγ and Q γ3 3. That is ) Q(Aγ i = (Q A i )γ i. Proof. W will only prov this for Q 1 as th xact sam argumnt can b usd for Q and Q 3. Lt γ = id γ γ 3, whr id is th idntity matrix. By dnition id dos not chang our original M 1, N 1 whil γ (rspctivly γ 3 ) just do column (rspctivly row) oprations on M 1, N 1 simultanously. Mor prcisly if ( ) r s γ = t u 15

16 and w dnot th columns of M 1 x N 1 y by c 1, c, thn w hav that γ acts by th column opration c 1 = rc 1 + tc and c = sc1+( st+ru)c r. So th dtrminant of M 1 x N 1 y is, using linar algbra, scald by a factor of r st+ru r = dt(γ 1 ) = 1. Hnc dt(m 1 x N 1 y) is unchangd undr γ, so Q 1 is unactd by γ. Nxt considr γ = γ 1 id id. In this cas M 1, N 1 ar only actd by ( ) r s γ 1 = t u and an xplicit calculation shows that dt((rm 1 + tn 1 )x (sm 1 + un 1 )y) = Q 1 (rx + sy, tx + uy) = Q γ1 1 So a cub A (Z ) 3 givs ris to thr binary quadratic form with th sam discriminant. W want to stablish an opration on thos thr binary quadratic forms so to nd a group. Barghava inspird himslf from th group law on lliptic curvs: if thr points P 1, P, P 3 on a an lliptic curv ar collinar thn th sum of P 1, P, P 3 is 0 Axiom (Th cub law). For all Q A 1, Q A, Q A 3 of Q A 1, Q A, Q A 3 is zro. that aris from som A (Z ) 3 w hav that th sum A usful consqunc of this axiom is that it lads to an idntication of two quivalnt binary quadratic forms. By that w man if γ = γ 1 id id thn, as in th proof, w hav two sts of thr binary quadratic cubs Q 1, Q, Q 3 and Q γ1 1, Q, Q 3, but w know th sums of ths two sts ar zro, hnc Q 1 bcoms idntid with Q γ1 1. As in Sction givn a binary quadratic form Q w will dnot by [Q] th st of SL (Z)-quivalnc classs of Q. W dnot by C( ( Sym Z ) ; D) th st of quivalnc classs of primitiv binary quadratic forms with discriminant D. Not that C( ( Sym Z ) ; D) is th sam as C(D) in sction, but w now spcify (Sym Z ) to mak it xplicit that w ar considring binary quadratic forms. W us th notation Sym Z to man that w ar looking at th spac of -variabl (i.., binary) symmtric (if w look at th associatd matrix) forms of dgr (i.., quadratic). Furthrmor w us (... ) to man that th associatd matrix dos not ncssarily hav ntris in Z, in othr words, if our quadratic form is (a, b, c), w do not rquir b to b vn. Thorm 3. (Gauss composition). Lt D 0, 1 mod and lt { x D Q id,d = y D 0 mod x xy + 1 D y D 1 mod Thn thr xists a uniqu binary opration to turn C( ( Sym Z ) ; D) into an additiv group with: 1. [Q id,d ] is th idntity,. for any cub A of discriminant D such that Q A 1, Q A, Q A 3 ar primitiv w hav [QA 1 ] + [Q A ] + [Q A 3 ] = [Q id,d ]. (Part of th cub law) Givn Q 1, Q, Q 3 with [Q 1 ] + [Q ] + [Q 3 ] = [Q id,d ] thn thr xists a uniqu, up to Γ-quivalnc, cub A (Z ) 3 of discriminant D such that A givs ris to Q 1, Q, Q 3. Bfor w prov this, w ar going to show that th cub law axiom is quivalnt to Dirichlt composition. Hnc this thorm is a rstatmnt of th fact that Dirichlt composition turns C(D) into a group, which w hav provd in th cass whn D is not a squar. Lt A id,d b th following cub A id,d = D, 1 0 A id,d = D dpnding on whthr D 0 mod or D 1 mod rspctivly. Thn w not that Q id,d, what w hav calld th principl form in sction, is such that A id,d givs ris to Q 1 = Q = Q 3 = Q id,d. 16

17 Dnition 3.5. W say that A is primitiv if th thr binary quadratic forms it givs ris to ar primitiv Using this dnition w can now show how th cub law axiom and Gauss composition ar quivalnt. If w start with a primitiv cub, as in gur (3.1), w not that sinc th cub is primitiv its cocints hav gcd 1. To s this lt G = gcd(a, b, c, d,, f, g, h) this implis that G gcd(bc ad, d + ah bg fc, fg h) = 1, whr th quality is du to th fact that Q 1 is primitiv, which implis that G = 1. W us that fact to nd γ Γ such that A γ is of th form (1, 0, 0, d, 0, f, g, h) for som nw d, f, g, h by applying th following stps: rst lt ( ) ( ) 0 1 U =, T n 1 n = SL (Z). By applying appropriat copis of U id id, id U id and id id U, w can assum without loss of gnrality that a is th smallst non-zro absolut ntry of A. If a is coprim with ithr b, c or, w can us Euclid's algorithm to nd a matrix in SL (Z) changing a into 1. If not thn apply appropriat copis of T n id id, id T n id and id id T n to rduc b, c and modulo a. W can thn rplac a with th smallst non-zro absolut ntry and rpat th procss. W ithr stop as soon as w hav a = 1, or if th rst cas dos not occurs whn w nd up with A bing of th form (a, 0, 0, d, 0, f, g, h). In th lattr cas, notic that gcd(a, f) = 1 for primitivity to hold, and w can apply T to A to gt (a, 0, d, d, f, f, g + h, h), but thn w ar back in th cas of gcd(a, ) = 1. Hnc w can nd γ Γ such that A γ is of th form (1, b, c, d,, f, g, h), thn us th 1 to rduc b, c and to 0, i.., w hav 0 f 1 0 g h. 0 d Th thr binary quadratic forms this cub givs riss to ar Q 1 = x fy gy dx hy = dx + hxy + fgy Q = x dy fy gx hy = gx + hxy + dfy Q 3 = x gy dy fx hy = fx + hxy + dgy Not that by construction Q A i [Q i ], and th cub law stats that [Q 1 ] + [Q ] = [Q 3 ]. In trms of Dirichlt composition, noting that gcd(d, h, fg) [ = 1 implis gcd(d, g, h) = ] 1, w hav [Q 1 ] + [Q ] = [ dx + hxy + fgy ] + [ gx + hxy + dfy ] = dgx + Bxy + B (h +dfg) dg y for som B that satiss B h mod d, B h mod g, B h + dfg mod dg. W can asily s that B = h works hnc [Q 1 ] + [Q ] = [ dgx + hxy fy ], but rcall that ax + bxy + cy cy bxy + ax, hnc [Q 1 ] + [Q ] = [ fx hxy + dgy ] which in trms of Dirichlt composition w know is [Q 3 ]. So Dirichlt composition corrsponds to th Cub Law. Proof of rst part of Thorm 3. in th cas D is not a squar. Th prcding paragraph shows how th cub law is quivalnt to Dirichlt composition. So, in th cas D is not a squar, Dirichlt composition is a binary opration that satisfy th givn condition sinc: w know it is an additiv binary opration dnd on C((Sym Z ) ; D); its idntity is th quivalnc class of th principal form, now dnotd [Q id,d ]; it is quivalnt to th cub law. Furthrmor, th prcding paragraph shows that if a binary opration satisfy th cub law, thn it is Dirichlt composition, proving th uniqunss part of th statmnt. W will not nish th proof of Thorm 3. as such, but instad it will b provn as parts of othr discussion and provs of latr sction. W will covr th cas of whn D is a squar as part of th scond proof of Thorm 3.6, which w do on pag. As for th scond statmnt of th thorm, w will show how w can xtnd Thorm 3.7 to prov this. 17