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Z ); : 2 KRISTYNA´ ZEMKOVA´ the most natural choice seems to be to consider any equivalence given by a , determinant of which is a totally positive unit. Indeed, this approach was taken by Mastropietro in his thesis [16]; he described a construction of the correspondence between the and the equivalence classes of binary quadratic forms, but only for the case when the base field is a real quadratic number field of class number one, and the discriminant is totally negative. The aim of this article is to take into account an equivalence of binary quadratic forms given by all totally positive units of the base number field and to develop, in such settings, a Dedekind-like correspondence between classes of forms and an ideal class group. As the base field, we consider an arbitrary number field of narrow class number one; this is equivalent to having class number one together with the existence of units of all signs. These conditions are necessary for our approach: First, the class number has to be one, for a free module basis of any fractional ideal to exist, and second, units of all signs are needed in order for such a basis to be able to have any orientation. This paper is based on the first part of the author’s Master thesis [23]. The results of this paper are further used in [24] to generalize the composition of Bhargava’s cubes to rings of integers of number fields of narrow class number one. Refined definitions of the equivalence of quadratic forms and of the ideal class group are given in Subsections 2.2 and 2.3; then Section 3 deals with the correspondence itself. The main results of the article are Theorem 3.6 together with Corollary 3.7. At the end of the article, Section 4 is devoted to totally negative discriminants, which allow investigating positive definiteness of quadratic forms.

2. Preliminaries Throughout the whole paper, we fix a number field K of narrow class number one; this is equivalent to K being of class number one and having units of all signs (for reference, see [10, Ch. V, (1.12)]). We write K for the ring of algebraic integers of O + K; the group of units of this ring is denoted by K , and K stands for its of totally positive units. Assume that K has exactlyU r embeddingsU into real numbers, and let σ1,...,σr be these embeddings. Furthermore, we fix a relative quadratic extension L of the number field K. Note that the Galois group Gal(L/K) has two elements; if τ is the nontrivial element of this group, then, for an α L, we write α instead of τ(α). ∈ 2.1. Ideals. Since K has the narrow class number one, the ring is a free -module; OL OK hence L = [1, Ω] K for a suitable Ω L, and every (fractional) L-ideal has a module basis ofO the form [Oα, β] for some α,∈ β O L (see [18, Cor. p. 388]O and [17, Prop. 2.24]). OK ∈ In both cases, the index K will be usually omitted. As an algebraic over K , O 2 O Ω is a root of a monic quadratic polynomial x + wx + z for some w,z K ; the other root is Ω. Put D = w2 4z, and note that, without loss of generality,∈ O Ω − w + √D w √D (2.1) Ω = − Ω , Ω= − − Ω . 2 2

Hence, L = K √DΩ , and 2 (2.2)  D = Ω Ω . Ω − Also note that a + b √D a,b .  OL ⊆ 2 2 Ω ∈ OK One may expect the element D to be square-free (i.e., not divisible by q2 for any  Ω q K K ), but since DΩ is the discriminant of a binary , it may not∈ always O \U be the case. Hence, instead of that, we introduce the following definition of fundamental element: an element, which is “almost square-free” and a quadratic residue modulo 4 at the same time. COMPOSITIONOFQUADRATICFORMS 3

Definition 2.1. An element d of K is called fundamental if d is a quadratic residue modulo 4 in , and O OK either d is square-free, • 2 d or for every p such that p d the following holds: p 2 and 2 is not • ∈ OK \UK | | p a quadratic residue modulo 4 in . OK In the case K = Q, this definition agrees with the one of the fundamental discrimi- nant. The following lemma shows that, from this point of view, DΩ is “a over K”.

Lemma 2.2. DΩ is fundamental. 2 Proof. Obviously, DΩ = w 4z is a quadratic residue modulo 4. − 2 Assume that there exists p K which is not a unit, and such that p DΩ; set DΩ ∈ O 2 | D′ = 2 . Since √D is a root of the polynomial x D′, there exist a,b p ′ ∈ OL − ∈ OK √ √ w+p D′ √ such that D′ = a + bΩ where Ω = − 2 . Comparing the coefficients at D′, we get that p must be a divisor of 2 in K . O 2 For contradiction, suppose that there exists t K such that D′ t (mod 4). We 2 ∈ O ≡ 2 can find m K such that D′ = t 4m; then the quadratic polynomial x + tx + m ∈ O − t+√D has the discriminant equal to D′ and a root κ = − ′ , which is an element of . 2 OL Hence, there exist a′,b′ such that κ = a′ + b′Ω, i.e., ∈ OK t + √D′ w + p√D′ − = a′ + b′ − . 2 2

Comparing the coefficients at √D′, we obtain that b′p = 1. But that is not possible, because b′ , and p is not a unit. Hence, we have found the desired contradiction.  ∈ OK 2 On the other hand, if we take D K such that K √D = L, then clearly p D = 2 ∈ O p q DΩ for some p, q K . Furthermore, if D is fundamental, then q has to be a unit, 2 ∈ O 2  p D q DΩ because both q2 and p2 are quadratic residues modulo 4. We have proved the following lemma. Lemma 2.3. Let D be a fundamental element of and K √D = L. Then there OK exists u such that D = u2D . ∈UK Ω  In the following, the word “ideal” will generally stand for a fractional ideal while to the usual meaning will be referred as to the “integral ideal”. Consider an ideal I = [α, β] in . Then there exists a 2 2 matrix M consisting of elements of K such that OL × α 1 = M . β · Ω     Then also α α 1 1 (2.3) = M , β β · Ω Ω     and thus αβ αβ (2.4) det M = − . Ω Ω − The proofs of the following two lemmas are just direct computations. Lemma 2.4. Let I = [α, β] be an ideal, M the same matrix as above, [pα + rβ,qα + sβ] another -module basis of I, and M the matrix corresponding to this basis. Then OK det M = (ps qr) det M. − f Lemma 2.5. det M K, and if α, β , then det M . f ∈ ∈ OL ∈ OK 4 KRISTYNA´ ZEMKOVA´

The relative norm of an element α is defined as (α)= αα. If I is a fractional NL/K -ideal, then the ideal (I)= (α): α I is the relative norm of the ideal OL NL/K NL/K ∈ I. Note that if I, J are two -ideals such that I J, then (I) (J). L  L/K L/K Since (I) is an -idealO and h(K) = 1 by the⊆ assumption,N this⊆ ideal N has to NL/K OK be principal. For a principal ideal, it clearly holds that L/K ((α)) = L/K (α) . Generally, the generator of the ideal (I) can be writtenN explicitly in termsN of the L/K  -module basis of I, as the followingN lemma shows. OK Lemma 2.6. Let I = [α, β] be an ideal, and M the same matrix as above. Then det M generates the -ideal (I). OK NL/K Proof. This is a well-known result holding for any finite Galois extension E/F such that h(F ) = 1. The proof can be found, e.g., in [15, Th. 1].  Let us now determine what will be later recognized as the inverse class of an ideal. We need to start with a technical lemma, which will also turn out to be useful later in the proof of Proposition 3.1, as the elements here will be exactly the coefficients of the quadratic form obtained from the ideal [α, β]. Lemma 2.7. Let [α, β] be an ideal, and M the same matrix as above. Then αα ββ αβ + αβ , , det M det M det M are coprime elements of . OK αα ββ αβ+αβ Proof. We start by proving that det M , det M , det M are elements of K . First, assume that α, β . Then det M by Lemma 2.5; hence we needO to show that the ∈ OL ∈ OK elements αα,ββ, αβ + αβ are divisible by det M in K . Since α [α, β], there is (α) [α, β]. It follows from Lemma 2.6 that (αα) = O (α) ∈ ([α, β]) = (det M⊂), NL/K ⊂ NL/K and therefore αα is divisible by det M in . By the same argument, ββ is divisible by K  det M in . Similarly, (α+β) [α, β] impliesO that (α + β) is divisible by det M. OK ⊂ NL/K Since L/K (α + β)= αα+ββ+αβ+αβ, we see that αβ+αβ = L/K (α + β) αα ββ is divisibleN by det M in as well. N − − OK In the general case, we can find k K such that kα,kβ L. Hence, we may apply the first part of the proof to the ideal∈ O [kα,kβ]. Since the corresponding∈ O determinant is k2 det M; the terms k2 cancel out in the fractions. αα αβ+αβ ββ Denote a = det M , b = det M , c = det M . To prove that a,b,c are coprime, first note 2 2 that b 4ac = Ω Ω = DΩ. Therefore, if a,b,c were divisible by an element p − DΩ− in , then 2 would be a quadratic residue modulo 4, which is not possible by OK \UK p  Lemma 2.2.  Proposition 2.8. Let [α, β] be an ideal. Then [α, β] α, β = (det M) · − as -ideals. OL   Proof. Denote α β I = [α, β], J = , − . det M det M   We will prove that IJ = [1, Ω], which is equivalent to the statement of the lemma. Note that 1 1 (J)= [α, β] = ; NL/K (det M)2 NL/K − det M   therefore, L/K (IJ) = (1) by the multiplicativity of the norm. Since we have also ([1, Ω])N = (1), to prove that [1, Ω] = IJ, we only need to show that 1, Ω IJ NL/K ∈ COMPOSITIONOFQUADRATICFORMS 5

(see [15, Cor. to Th. 1]). As IJ is an L-ideal as well, it even suffices to prove that 1 IJ. O ∈Clearly, αα αβ αβ ββ IJ = , − , , − . det M det M det M det M  OK By Lemma 2.7, αα αβ + αβ ββ gcd , , = 1; det M det M det M   therefore, 1 IJ.  ∈ 2.2. Quadratic forms. By binary quadratic forms over K we understand homogeneous polynomials of degree 2 with coefficients in , i.e., Q(x, y) = ax2 + bxy + cy2 with OK a,b,c K . For abbreviation, we will refer to them as quadratic forms. By Disc(Q) we denote∈ the O discriminant of the quadratic form Q, i.e., Disc(Q)= b2 4ac. Comparing to the case of quadratic forms over Q, we need to slightly redefine the− . Definition 2.9. Two quadratic forms Q(x, y) and Q(x, y) are equivalent, denoted by + Q Q, if there exist elements p,q,r,s K satisfying ps qr K and a totally ∼ + ∈ O e − ∈ U positive unit u K such that Q(x, y)= uQ(px + qy,rx + sy). e ∈U 2 2 2 2 Let Q(x, y)= ax + bxy + cy eand Q(x, y)= uQ(px + qy,rx + sy)= ax + bxy + cy be equivalent quadratic forms. Then a = u(ap2e+ bpr + cr2), e e e (2.5) b = u(2apq + b(ps + qr)+2crs), ec = u(aq2 + bqs + cs2), and e e Disc(Q) = b2 4ac = u2(ps qr)2(b2 4ac) (2.6) = u2(−ps qr)2 Disc(−Q). − − On the other hand, there ise e ee

1 2 2 a = u(ps qr)2 (as brs + cr ), −1 − (2.7) b = 2 ( 2aqs + b(ps + qr) 2cpr), u(ps qr) − − −1 e 2 e e 2 c = u(ps qr)2 (aq bpq + cp ). − e− e e We say that a quadratic form Q(x, y) represents m K if there exist x0,y0 K such e e e∈2 O 2 ∈ O that Q(x0,y0) = m. A quadratic form Q(x, y) = ax + bxy + cy is called primitive if gcd(a,b,c) K . We state two lemmas classical in the case of quadratic forms over Q; we omit the∈U proofs because they are analogous to that case.

Lemma 2.10. Equivalent quadratic forms represent the same elements of K up to the multiplication by a totally positive unit. O Lemma 2.11. Let Q be a primitive quadratic form, and Q Q. Then Q is also primitive. ∼ e e We are interested in quadratic forms of given discriminant, namely of discriminant 2 DΩ = Ω Ω . But from (2.6) we can see that equivalent quadratic forms do not always have− the same discriminant; their discriminants may differ from each other by a square of a totally positive unit. Therefore, we will consider all quadratic forms of discriminants belonging to the set 2 = u2 Ω Ω u + . D − ∈UK n  o

6 KRISTYNA´ ZEMKOVA´

Note that all the elements of are fundamental (by Lemma 2.2). We will denote by the set of all primitive quadraticD forms of discriminant in modulo the equivalence relationQD described above: D Q(x, y)= ax2 + bxy + cy2 a,b,c , gcd(a,b,c) , Disc(Q) = ∈ OK ∈UK ∈ D . QD  .∼ Remark 2.12. Note that if Disc(Q)= u2D for u +, then Disc 1 Q = D . Hence, Ω ∈UK u Ω in every class of , there is a quadratic form of discriminant exactly DΩ. D  If K is totallyQ real (and of narrow class number one), then every totally positive unit is square of a unit (for a reference, see [9, Prop. 2.4]). Consider equivalent quadratic forms Q and Q′, such that Disc(Q) = Disc(Q′), i.e., 1 Q′(x, y)= Q(px + qy,rx + sy) ps qr − 2 for some p,q,r,s K . If u K is such that ps qr = u , then ∈ O ∈Up q r s − Q x + y, x + y = Q′(x, y) u u u u   gives the equivalence of Q and Q′ with determinant 1. Therefore, in this case, our notion could be simplified to quadratic forms of the discriminant exactly DΩ, and to the equivalence by the matrices of determinant 1. On the other hand, if K is not totally real, then the units other than 1 become important. For example, take K = Q(i), and consider the quadratic forms Q(x, y) = 2 2 2 2 x +4xy +2y and Q′(x, y) = ix +4xy 2iy . Clearly Q′(x, y)= iQ(ix,y), but there − − is no matrix of determinant 1 which would provide the equivalence between Q and Q′. That can be seen as follows: by the first equality of (2.5), we need to find p, r Z[i] such that p2 +4pr +2r2 = i. But that is not possible, because the imaginary part∈ of p2 +4pr +2r2 is divisible by 2. Let us prove two lemmas, which will be useful later. Lemma 2.13. Let µ , and let Q(x, y) be a quadratic form with Disc(Q) . ∈ UL ∈ D Then there exist some elements p0, q0, r0,s0 K such that p0s0 q0r0 = µµ and Q(x, y)= 1 Q(p x + q y, r x + s y). ∈ O − p0s0 q0r0 0 0 0 0 − 2 2 Proof. Let Q(x, y)= ax + bxy + cy , and assume that Disc(Q)= DΩ (multiply Q by a u v suitable totally positive unit if needed). There exist u, v K such that µ = 2 +√DΩ 2 ; 2 2 ∈ O hence µµ = u + D v . From the condition p s q r = µµ and by comparing 2 Ω 2 0 0 − 0 0 the coefficients of the quadratic forms (p0s0 q0r0)Q(x, y) and Q(p0x + q0y, r0x + s0y), we get the following  system  of equations: −

2 2 p s q r = u + D v , 0 0 − 0 0 2 Ω 2 2 2 (p0s0 q0r0)a = ap + bp0r0+cr , (2.8) − 0 0 (p s q r )b = 2ap q + b(p s + q r )+2cr s , 0 0 − 0 0 0 0 0 0 0 0 0 0 (p s q r )c = aq2 + bq s + cs2. 0 0 − 0 0 0 0 0 0 One can check that u bv u + bv p = − , q = cv, r = av, s = 0 2 0 − 0 0 2 fulfill the system of equations (2.8). It remains to show that p0, q0, r0,s0 are elements of . This is obvious for q and r ; for p and s it follows from the computation OK 0 0 0 0 u 2 v 2 u 2 v 2 u2 b2v2 µµ = + D = + (b2 4ac) = − + acv2, 2 Ω 2 2 − 2 4 and from the fact that both acv 2 and µµ are elements of .  OK COMPOSITIONOFQUADRATICFORMS 7

Lemma 2.14. Let Q(x, y)= ax2+bxy+cy2 be a quadratic form, p,q,r,s such that ∈ OK ps qr +. Consider the quadratic form Q(x, y)= Q(px+qy,rx+sy)= ax2+bxy+cy2 − ∈UK equivalent to Q(x, y). Denote D = Disc(Q) and D = Disc(Q). Then e e e e b+√D p − 2a + q be+ √D e e e = − . b+e√D 2a r − 2a + s e e b+√D e 2 Proof. As − 2a is a root of the quadratic polynomial Q(x, 1) = ax + bx + c, we can write e e e e e e e b+√D b+√D − 2a − 1 Q e e =0 2a 1 !  e e  e e e p r p q b+√D b+√D − 2a − 1 Q e e =0 2a q s r s 1 !  e e      e e b+√D p − 2a + q b+√D b+√D Q e e =0 p − 2a + q r − 2a + s e e e e  b+e√D    r − 2a + s e e  e e  b+√e D p − 2a +q b+√D e f p − 2a +q b+√D e f 1 Q r − e +s =0 b+√D 2a e  e f  r − 2a +s  e f  e1 e   b+√D   p − 2a +q b+√D In particular, e f has to be one of the roots of Q(x, 1); either − or b+√D 2a e r − 2a +s √ e f b D e − −2a . One can check by direct computation (using the expressions (2.5) and (2.6)) that the former case holds. 

2.3. Relative Oriented Class Group. In the traditional correspondence, there are binary quadratic forms on one side, and the class group ℓL = IL/ L, or the narrow class + C P group ℓ = L/ +, on the other side. Since we are working with number fields of higher L I PL degrees,C the situation is a bit more complicated. Inspired by Bhargava’s definition of the class group, which considers the orientation of the bases of the ideals, we define the relative oriented class group with respect to the extension L/K. Compared to the rational numbers, our base field K has r real embeddings; therefore, instead of one sign, we consider r signs: one for every real embedding. Later, in Section 4, we will see that these signs are closely connected to the positive definiteness of the corresponding quadratic forms. Definition 2.15. For a K write sgn(a) = (sgn(σ (a)),..., sgn(σ (a))), and set ∈ 1 r o = (I; ε ,...,ε ) I a fractional -ideal,ε 1 ,i =1,...,r , IL/K { 1 r | OL i ∈ {± } } o = (γ);sgn (γ) γ L ; PL/K NL/K | ∈ }   here, (I; ε1,...,εr) is called the oriented ideal. Then the relative oriented class group of the field extension L/K is defined as

o o ℓL/K = IL/K o . C L/K .P The multiplication on o is defined componentwise as IL/K (I; ε ,...,ε ) (J; δ ,...,δ ) = (IJ; ε δ ,...,ε δ ); 1 r · 1 r 1 1 r r 8 KRISTYNA´ ZEMKOVA´ thus, o is clearly an , and o is its subgroup. Therefore, the group IL/K PL/K ℓo is well defined. C L/K If two oriented ideals (I; ε ,...,ε ) and (J; δ ,...,δ ) lie in the same class of ℓo , 1 r 1 r C L/K we say that they are equivalent, and we write (I; ε1,...,εr) (J; δ1,...,δr). Let us first compare the relative oriented class group ℓo ∼ with the class group ℓ C L/K C L of the number field L. In the following, by is understood the principal -ideal OL OL generated by a unit (hence the in the group ℓL), and by L the one-element group. C {O } Proposition 2.16. Denote H = sgn (µ) µ . Then NL/K | ∈UL o ℓL/K  ℓL C 1 r . C ≃ L h± i /H .{O }× Proof. Denote ε = (ε1,...,εr). Define maps f and g: f : 1 r o {OL}×h± i −→ IL/K ( ; ε) ( ; ε) OL 7−→ OL g : o IL/K −→ IL (I; ε) I 7−→ Then f is injective, g surjective, and Ker g = ( ; ε) ε 1 r = Im f. Consider { OL | ∈ h± i } restrictions f ′ and g′ of these two maps: o f ′ : H {OL}× −→ PL/K ;sgn (µ) (µ); sgn (µ) OL NL/K 7−→ NL/K o g′ :   PL/K −→ PL (γ);sgn (γ) (γ) NL/K 7−→ Note that f ′ is indeed a restricton of f, as (µ) = L for any µ L. Again, f ′ O ∈ U is injective, g′ is surjective, and Ker g′ = (γ);sgn L/K (γ) γ L = Im f ′. Hence, we obtain the following commutative diagram: N | ∈U   f g 1 1 r o 1 {OL}×h± i IL/K IL

i1 i2 i3

f ′ g′ 1 H o 1 {OL}× PL/K PL

r o Note that Coker i = 1 /H, Coker i = ℓ , Coker i = ℓ , and Ker i = 1 {OL}× h± i 2 C L/K 3 C L/K 3 1. Hence, by snake lemma, there is a short exact sequence:

r o 1 1 /H ℓ ℓ 1 {OL}× h± i C L/K C L This sequence gives us the required ℓo ℓ L/K r . L C 1 /H C ≃ {OL}× h± i . 

Remark 2.17. The relative oriented class group could be defined for any finite Galois extension E/F such that h+(F ) = 1; Proposition 2.16 would remain true without any modifications. For an -ideal I = [α, β], we define its orientation as OL [α, β];sgn (det M) = ([α, β]; sgn(σ1(det M)),..., sgn(σr(det M)))  COMPOSITIONOFQUADRATICFORMS 9 where α α 1 1 = M , β β · Ω Ω     αβ αβ i.e., det M = − . Ω Ω − Lemma 2.18. Let (I; ε1,...,εr) be an oriented ideal. Then there exists a basis [α, β] αβ αβ of I such that sgn (det M) = (ε ,...,ε ), where det M = − . 1 r Ω Ω − Proof. Consider any basis [α, β] of the ideal I, and multiply α by a unit u K with the appropriate sgn (u); such a unit exists by the assumption that the narrow∈U class number of K is one. 

If [α, β];sgn (det M) and [α′,β′];sgn (det M ′) are two oriented ideals, then their multiple is the ideal [α, β] [α′,β′] with the orientation sgn (det M det M ′). The ex-  ·  · istence of a basis of the ideal [α, β] [α′,β′] with such an orientation is guaranteed by the previous lemma. Note that here· we need the condition h+(K) = 1; otherwise, there may not exist any basis of the multiple of two ideals with the required orientation. The orientation of a principal ideal is given directly by the definition as

(γ);sgn (γγ) = ((γ); sgn(σ1(γγ)),..., sgn(σr(γγ))) . Note that if  γα γα 1 1 = M , γβ γβ · Ω Ω     then det M = γγ det M, and therefore, f (γ);sgn (γγ) [α, β];sgn (det M) = [γα,γβ];sgn det M . f · o   Lemma 2.19. The identity element of the group ℓ is ([1, Ω]; +1,..., +1), and the C L/K f inverse to [α, β];sgn (det M) is α, β ;sgn (det M) (taking all of the oriented ideals − as representatives of classes in ℓo ).  C L/K   Proof. The orientation of the ideal [1, Ω] is (+1,..., +1), because M is in this case the unit matrix. Hence, the oriented ideal ([1, Ω]; +1,..., +1) is a representative of the identity element of the group ℓo . C L/K We know from Proposition 2.8 that [α, β] α, β = (det M); since the orientation · − of the product [α, β] α, β is sgn (det M)2 = sgn det Mdet M , we even have that · −   [α, β];sgn (det M)  α, β ;sgn(det M) = (det M);sgn det Mdet M o . · − ∈PL/K Thus, the oriented ideals  [α, β];sgn (det M) and α, β ;sgn (det M) represent the − inverse classes in the group ℓo .  C L/K     The following lemma states an easy but very useful observation; the proof is imme- diate.

Lemma 2.20. Two oriented ideals (I; ε1,...,εr) and (J; δ1,...,δr) are equivalent if and only if there exists γ L such that γI = J and sgn (γγ) = (ε1δ1,...,εrδr). Moreover, if I = J, then γ has to∈ be a unit.

3. Correspondence between Ideals and Quadratic Forms 3.1. From Ideals to Quadratic Forms. Let us define a map attaching a quadratic form to an oriented ideal. o Φ: ℓL/K C −→ QD 2 2 1 ααx (αβ+αβ)xy+ββy [α, β];sgn (det M) (αx βy)= − 7−→ det M NL/K − det M  10 KRISTYNA´ ZEMKOVA´

To be accurate, the map Φ goes between the classes of oriented ideals and classes of quadratic forms; we omit the classes to relax the notation. We have to check that this map is well defined. First, let us see that the image of this map lies in . QD Proposition 3.1. Let [α, β];sgn (det M) be a representative of a class in ℓo , and C L/K denote by Q its image under the map Φ. Then Q is a representative of a class of α,β  α,β . QD Proof. We have ααx2 (αβ + αβ)xy + ββy2 Q (x, y)= − . α,β det M

Lemma 2.7 ensures both that the coefficients of Qα,β are elements of K , and that this O 2 quadratic form is primitive. One easily computes that Disc(Qα,β)= Ω Ω , which is an element of . −  D  To prove that the map Φ does not depend on the choice of the representative of the class in ℓo , we start with a lemma, which connects units from the quadratic extension C L/K with the equivalence of quadratic forms:

Lemma 3.2. Let Q be a primitive quadratic form, and p,q,r,s be such that ∈ OK ps qr K . If there exists a unit µ L such that sgn (µµ) = sgn (ps qr), then − ∈ U 1 ∈ U − Q(x, y) ps qr Q(px + qy,rx + sy). ∼ − Proof. Use the elements p , q , r ,s from Lemma 2.13: p s q r = µµ, and 0 0 0 0 0 0 − 0 0 1 Q(x, y)= Q(p x + q y, r x + s y). p s q r 0 0 0 0 0 0 − 0 0 ps qr + Note that sgn (p s q r ) = sgn (µµ) = sgn (ps qr), and thus − . We 0 0 0 0 p0s0 q0r0 K − − 1 − ∈ U can find the equivalence between the quadratic forms ps qr Q(px qy, rx + sy) and − − − 1 Q(p x+q y, r x+s y); this equivalence is obtained by the change of coordinates p0s0 q0r0 0 0 0 0 given− by the matrix 1 p q p q − 0 0 , r s r− s  0 0 −  ps qr and by the multiplication by the totally positive unit − . Finally, p0s0 q0r0 − 1 1 Q(px qy, rx + sy) Q(p x + q y, r x + s y)= Q(x, y). ps qr − − ∼ p s q r 0 0 0 0 − 0 0 − 0 0 

Proposition 3.3. The map Φ does not depend on the choice of the representative [α, β];sgn (det M) .

Proof. First, we would like to show that the definition of Φ is independent on the choice of the basis [α, β] of an ideal, i.e., that the quadratic form arising from the basis [pα + rβ,qα + sβ], such that p,q,r,s K , ps qr K , is equivalent to the one obtained from the basis [α, β]; but this∈ is O not true− in general,∈ U since the change of the basis may change the orientation as well. Thus, we have to add an assumption that the oriented ideals obtained from these two bases are equivalent in ℓo : Consider two C L/K oriented ideals

I = [α, β];sgn (det M) , I = [pα + rβ,qα + sβ];sgn det M ,     e f COMPOSITIONOFQUADRATICFORMS 11 such that p,q,r,s , ps qr , and assume I I. Denote Q(x, y) = Φ(I), ∈ OK − ∈ UK ∼ Q(x, y)=Φ(I); we need to prove that Q Q. We have ∼ e L/K ((pα + rβ)x (qα + sβ)y) eQ(x, y)= Ne − e det M p2αα+pr(αβ+αβ)+r2ββ x2 2pqαα+(ps+qr)(αβ+αβ)+2rsββ xy+ q2αα+qs(αβ+αβ)+s2ββ y2 e= − f (ps qr) det M  −   αα(px qy)2 (αβ+αβ)(px qy)( rx+sy)+ββ( rx+sy)2 1 = − − − − − = Q(px qy, rx + sy), (ps qr) det M ps qr − − − − where det M = (ps qr) det M by Lemma 2.4. Since I I and [α, β] = [pα+rβ,qα+sβ], there exists µ − by Lemma 2.20, such that sgn ∼(µµ) = sgn (ps qr). Thus, the ∈ UL − quadraticf forms Q(x, y) and Q(x, y) are equivalent by Lemmae 3.2. Now, let us consider any two equivalent oriented ideals; the equivalence is given by multiplication by a principal orientede ideal (γ);sgn (γγ) , i.e., we have a pair of oriented ideals [α, β];sgn (det M) and [γα,γβ]; sgn (γγ det M) . The situation in this case is  much easier, because the image of the oriented ideal [γα,γβ]; sgn (γγ det M) under   the map Φ is  1 γαγαx2 (γαγβ + γαγβ)xy + γβγβy2 (γαx γβy)= − γγ det M NL/K − γγ det M ααx2 (αβ + αβ)xy + ββy2 = − , det M which is identical to the quadratic form obtained from [α, β];sgn (det M) .  3.2. From Quadratic Forms to Ideals. To get an oriented ideal from a quadratic form, define a map o Ψ: ℓL/K QD −→ C b+√Disc(Q) Q(x, y)= ax2 + bxy + cy2 a, − ;sgn (a) 7−→ 2    Again, we omit to write the classes, and we have to show that this map is well defined. Proposition 3.4. Let Q(x, y)= ax2 + bxy + cy2 be a representative of a class in . Then its image under the map Ψ is an element of o . QD IL/K Proof. Set D = Disc(Q), and note that b uw a b + √D aΩ= − a + − , 2u · u · 2 b + √D c b + uw b + √D Ω= a − , 2 u · − 2u · 2 2 + w+√DΩ where D = u D for a totally positive unit u , andΩ= − by (2.1). Hence, Ω ∈UK 2 b+√D a, − is indeed an -ideal, and we only need to compute the orientation of the 2 OL hbasis. Let iM be a matrix such that a a 1 1 = M . b √D b+√D · Ω Ω  − −2 − 2    We need to prove that sgn (det M) = sgn (a). There is

b+√D b √D a − a − − a√D det M = 2 − 2 = = ua. Ω Ω Ω Ω − − Since u is totally positive, we have sgn (det M) = sgn (a).  12 KRISTYNA´ ZEMKOVA´

Proposition 3.5. The map Ψ does not depend on the choice of the representative Q(x, y).

Proof. Let Q(x, y)= ax2 + bxy + cy2 be a quadratic form of discriminant D , and + ∈ D u . Then ∈UK

ub + √u2D Ψ(uQ(x, y)) = ua, − ;sgn (ua) " 2 # ! b + √D = (u);sgn (uu) a, − ;sgn (a) , · " 2 # !  because both u and uu are totally positive. Hence,

b + √D Ψ(uQ(x, y)) a, − ;sgn (a) = Ψ(Q(x, y)). ∼ " 2 # !

Now, let p,q,r,s be such that ps qr +, and consider the quadratic form ∈ OK − ∈UK Q(x, y)= Q(px + qy,rx + sy)= ax2 + bxy + cy2. We have e e e b +e √D Ψ(Q(x, y)) = a, − ;sgn (a) , " 2 # ! b + D Ψ(Q(x, y)) = a, − ;sgn (a) ; " 2p # ! e e e e e we need to show that these two oriented ideals are equivalent. b+√D Let us first examine only how to come from the basis a, − 2 to the basis

b+√D h i a, − 2 ; we will deal with the orientations afterwards.  e e  e q p 1 s r b+√D · a b+√D b+√D b+√D a, − 2 1, − 2a p − 2a + q, r − 2a + s e e −−−−−−−−→e e e −−−−−−−−→ e e e e   1     b+√D − e e e r − 2 +s √ · a  b+ D (Lemma 2.14) e e f p − 2a +q b+√D e e f , 1 = − , 1 b+√D 2a e −−−−−−−−→ r − 2a +s  e f    0 1 e 1 0   b+√D a b+√D 1, − · a, − −−−−−−−−→ 2a −−−−−−−−→ 2     Recall that the multiplication of the basis by an element γ change the orientation by sgn (γγ), and the transformation of the basis by a matrix M change the orientation by sgn (det M). Since a K, there is aa = a2, which is totally positive. Thus, the ∈ 1 multiplication by a does not change the orientation; the same holds for a K. Also, both (ps qr) and 1 are totally negative, and hence both the transformations∈ by the e − −q p − 0 1 matrices and change the orientation to the opposite one; together they s r 1 0 does not affect the orientation.  Therefore, the only impact might have the multiplication COMPOSITIONOFQUADRATICFORMS 13

1 b+√D − by r − 2a + s :  e e  e 1 1 b + D − b + D − sgn  r − + s r − + s  2ap ! 2ap !  e e e e    be+ D b e D = sgn r − + s r − − + s 2ap ! 2ap !! e e e2 e = sgn rb +2sa r2D = sgn r2b2 4rsab +4s2a2 r2(b2 4ac) − e − e − − −      2 2 (2.7) 2 = sgn 4a(ase brse + cr )e = sgn 4ae(ps qre)ea = sgne (aa) e ee − −  2  2  (we used the factse e thate 4a ande 4(ps qr) aree totally positive). Alle the transformations changed the orientation from sgn (a)− to sgn (a) sgn (aa) = sgn a2a = sgn (a), and that · e b+√D is exactly the desired orientation. Hence, the oriented ideals a, − 2 ;sgn (a) and e e e e b+√D h i   a, − 2 ;sgn (a) are equivalent.  e e   3.3.e Conclusion. Wee are ready to prove that the maps Φ and Ψ defined in the two previous subsections are mutually inverse bijections. Theorem 3.6. Let K be a number field of narrow class number one, and let D be a 2 + fundamental element of K . Set L = K √D , and = u D u K . We have a bijection O D | ∈U   1:1 o ℓL/K QD ←→ C Ψ b+√Disc(Q) Q(x, y)= ax2 + bxy + cy2 a, − ;sgn (a) 7−→ 2    ααx2 (αβ+αβ)xy+ββy2 Φ [ αβ αβ − αβ αβ [α, β];sgn − − √D √D ←−    Proof. Let Ω be such that L = [1, Ω]. By Lemma 2.3, there is a unit u K (not necessarily totally positive)O such that D = u2D ; then we have = [1∈,u UΩ] and Ω OL u2D = (uΩ uΩ)2. If we begin with the canonical basis [1,uΩ] of instead of [1, Ω], Ω − OL we get the same results for D in the place of DΩ. Therefore, without loss of generality, we may assume that D = DΩ. Note that under this assumption αβ αβ − = det M. √D Let I = [α, β];sgn (det M) o be a representative of a class in ℓ , and  C L/K ααx2 (αβ + αβ)xy + ββy2 Q (x, y)= − α,β det M its image under the map Φ. If we use now the map Ψ, we obtain the oriented ideal αα αβ + αβ Ω Ω αα I′ = , + − ;sgn . det M 2 det M 2 det M     14 KRISTYNA´ ZEMKOVA´

αβ αβ αβ+αβ Ω Ω (αβ+αβ)+(αβ αβ) αβ Since det M = − , there is + − = − = . Thus, Ω Ω 2det M 2 2 det M det M − αα αβ αα I′ = , ;sgn . det M det M det M     I det M det M det M If we multiply ′ by the principal oriented ideal α ;sgn α α , we get exactly the ideal I. Therefore, I I′ = ΨΦ(I), and Ψ Φ = id ℓo . ∼ ◦  C L/K  On the other hand, consider Q(x, y)= ax2 + bxy + cy2, a representative of a class in . Its image under the map Ψ is the oriented ideal QD b + Disc(Q) a, − ;sgn (a) . " p2 # ! Using the map Φ, we get a quadratic form 2 2 2 b Disc(Q) 2 a x + abxy + − 4 y Q′(x, y)= , det M where a a 1 1 b √Disc(Q) b+√Disc(Q) = M , − − − · Ω Ω 2 2 !   and det M = ua for a unit u +. Hence, ∈UK 1 2 2 Q′(x, y)= (ax + bxy + cy ), u and Q Q′ = ΦΨ(Q). Therefore, Φ Ψ = id .  ∼ ◦ QD Corollary 3.7. carries a group structure arising from the multiplication of ideals QD in K √D . The identity element of this group is represented by the quadratic form x2 (Ω + Ω)xy +ΩΩy2, and the inverse element to ax2 + bxy + cy2 is the quadratic  form− ax2 bxy + cy2. − o Proof. The group structure of is given by the bijection with ℓL/K from Theorem QD C 3.6. The identity element is given as the image under the map Φ of the oriented ideal ([1, Ω]; +1,..., +1) (which represents the identity element in ℓo ); thus, the identity C L/K element is x2 (Ω + Ω)xy +ΩΩy2. Consider the− quadratic forms ax2 +bxy+cy2 and ax2 bxy+cy2, and set D = b2 4ac. The images of these two quadratic forms under the map− Φ are the oriented ideals− b + √D b + √D a, − ;sgn (a) and a, ;sgn (a) , " 2 # ! " 2 # ! which represent the mutually inverse classes of ℓo by Lemma 2.19. Hence, the C L/K quadratic forms ax2 + bxy + cy2 and ax2 bxy + cy2 are inverse to each other.  − 3.4. Quadratic fields. At the end of this section, let us look at the case of quadratic fields: Assume K = Q, and L = Q √D . In this case, there exists only one real embedding of K, and that is the identity. Thus, Proposition 2.16 says that o  ℓL/K ℓL C 1 , C ≃ L h± i/H .{O }× where H = sgn L/K (µ) µ L . To give more precise results, we need to distin- guish three possibleN cases according| ∈U to the sign of D and the existence of a negative unit:  1. If D < 0, then there is L/K (γ) > 0 for every γ L; therefore, the oriented ideals ([α, β]; sgn(det M)) andN ([ α, β]; sgn(det M)) cannot∈ be equivalent, and it is − − COMPOSITIONOFQUADRATICFORMS 15 easy to see that ℓo ℓ 1 . Moreover, the quadratic forms ax2 +bxy +cy2 and C L/K ≃C L ×h± i ax2 +bxy cy2 cannot be equivalent either; if one of them is positive definite, then the other− one is− negative definite. This explains the factor 1 in the relative oriented class group because in the usual correspondence only positiveh± i definite forms are considered whenever D< 0. 2. Let D > 0, and assume that every unit has positive norm, i.e., for every µ L there is (µ) = +1. Consider the following surjective homomorphism: ∈ U NL/K f : o ℓ+ IL/K −→ C L (I;+1) I + 7−→ PL (I; 1) √DI + − 7−→ PL Since Ker f = o , it follows that ℓo ℓ+. Furthermore, it is well known that in PL/K C L/K ≃C L this case is ℓ+ ℓ 1 ; hence, we have as well that ℓo ℓ 1 . C L ≃C L × h± i C L/K ≃C L × h± i 3. Finally, assume that D> 0, and that there exists a unit of negative norm, i.e., there exists µ0 L such that L/K (µ0) = 1; then H = (+1), ( 1) = 1 . Therefore, we get directly∈ U from PropositionN 2.16 that− ℓo ℓ{ . Since− it is} wellh± knowni that in C L/K ≃ C L this case the groups ℓ and ℓ+ are isomorphic, we have as well that ℓo ℓ+. C L C L C L/K ≃C L Let us summarize our observations into the following proposition:

Proposition 3.8. Let K = Q, and let L = Q √DΩ for DΩ a fundamental element of Z.  1. If D < 0, then ℓo ℓ 1 . Ω C L/Q ≃C L × h± i 2. If D > 0 and µµ =1 for every µ , then ℓo ℓ 1 ℓ+. Ω ∈UL C L/Q ≃C L × h± i≃C L 3. If D > 0 and there exists µ such that µµ = 1, then ℓo ℓ ℓ+. Ω ∈UL − C L/Q ≃C L ≃C L Remark 3.9. The proposition shows us that the relative oriented class group ℓo C L/K (and hence the correspondence between oriented ideals and quadratic forms) is a gener- alization of Bhargava’s view to the classical correspondence; see [1, Section 3.2].

4. Totally positive definite quadratic forms A very interesting and well-studied class of quadratic forms are the totally positive definite ones, which form a natural generalization of sums of squares. As such, they have been studied for example in the context of representations of totally positive integers, e.g., in [2, 3, 6–8, 11, 14, 20]. Of course, a binary quadratic form can never be universal; nevertheless, our results may prove to be useful also in the study of quadratic forms of higher ranks. If DΩ is totally negative (i.e., σi(DΩ) < 0 for all i = 1,...,r; this fact will be denoted by DΩ 0), then we can study the totally positive definite quadratic forms. Recall that a quadratic≺ form Q(x, y) = ax2 + bxy + cy2 is totally positive definite if 2 2 σi(Q(x, y)) = σi(a)x + σi(b)xy + σi(c)y is positive definite for all i = 1,...,r. It is clear from the matrix notation that Q(x, y)= ax2 + bxy + cy2 is totally positive definite if and only if a 0 (i.e., a is totally positive). Hence, the image under the map Ψ of a totally positive≻ definite quadratic form Q(x, y) = ax2 + bxy + cy2 is the oriented ideal a, 1 b + Disc(Q) ;+1,..., +1 . 2 − h One may askp if it is possiblei to describe the totally positive definite quadratic forms in terms of the oriented ideals. We start with the following lemma, which claims that if D 0, then all ideals in the same class of ℓo have the same orientation. Ω ≺ C L/K Lemma 4.1. If DΩ 0 and (I; ε1,...,εr) (J; δ1,...,δr), then εi = δi for all i = 1,...,r. ≺ ∼ 16 KRISTYNA´ ZEMKOVA´

Proof. By Lemma 2.20, if (I; ε ,...,ε ) (J; δ ,...,δ ), then there exists γ L such 1 r ∼ 1 r ∈ that J = γI and δi = sgn(σi(γγ))εi for all i =1,...,r. Since DΩ 0, there is γγ 0 for every γ L. Therefore, sgn(σ (γγ)) = +1, and δ = ε for all i ≺=1,...,r. ≻ ∈ i i i

Proposition 4.2. Let DΩ 0, let Q , and let i 1,...,r . Then the following are equivalent: ≺ ∈QD ∈{ }

(i) σi(Q) is positive definite, (ii) Q is the image under the map Φ of an oriented ideal [α, β];sgn (det M) such that σ (det M) > 0, i  (iii) Q is the image under the map Φ of an oriented ideal [α, β];sgn (det M) such that σ β > 0, where c + c √D = c for any c ,c K. i ℑ α ℑ 1 2 Ω 2 1 2 ∈     2 2  Proof. Let Q(x, y)= ax + bxy + cy . Since DΩ 0, the positive definiteness of σi(Q) is given by the sign of σ (a). First, we will prove≺ that (i) (ii): Recall that i ⇔ b + Disc(Q) Ψ(Q)= a, − ;sgn (a) . " p2 # ! If I = [α, β];sgn (det M) is an oriented ideal such that Φ(I)= Q(x, y), then ΨΦ(I)= Ψ(Q). Hence, since I ΨΦ(I), there is ∼  b + Disc(Q) [α, β];sgn (det M) a, − ;sgn (a) . ∼ " p2 # !  By the previous lemma, sgn (det M) = sgn (a). On the other hand, consider an oriented ideal I = [α, β];sgn (det M) . Then the first coefficient of the quadratic form Φ(I) is equal to αα , and sgn αα = sgn (det M), because αα is totally positive. Therefore, det M det M σ (Q) is positive definite if and only if σ (det M) > 0. i  i Let us prove (ii) (iii). Assume that Q is the image under the map Φ of an ⇔ w+√DΩ oriented ideal [α, β];sgn (det M) . Recall that Ω = − 2 by (2.1), and write α = a + a √D , β = b + b √D for some a ,a ,b ,b K. One can easily compute that 1 2 Ω 1 2 Ω  1 2 1 2 ∈ αβ αβ det M = − = 2(a1b2 a2b1), Ω Ω − − and β a b a b D + (a b a b )√D = 1 1 − 2 2 Ω 1 2 − 2 1 Ω . α a2 a2D 1 − 2 Ω Thus, β a b a b = 1 2 − 2 1 , ℑ α a2 a2D   1 − 2 Ω where a2 a2D = a2 + a2 D is totally positive. Hence, 1 − 2 Ω 1 2 | Ω| β sgn = sgn (a b a b ) = sgn (det M) . ℑ α 1 2 − 2 1    

The result about totally positive quadratic forms follows immediately from Proposi- tion 4.2.

Corollary 4.3. Let DΩ 0. A quadratic form is totally positive definite if and only if it is the image under the≺ map Φ of an oriented ideal [α, β];sgn (det M) such that β is totally positive. ℑ α    COMPOSITIONOFQUADRATICFORMS 17

Acknowledgment I wish to thank V´ıtˇezslav Kala for his excellent guidance and useful suggestions. I also want to thank Professor Rainer Schulze-Pillot for pointing out references [13,21]. References

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Charles University, Faculty of Mathematics and Physics, Department of Algebra, Sokolovska´ 83, 18600 Praha 8, Czech Republic

Fakultat¨ fur¨ Mathematik, Technische Universitat¨ Dortmund, D-44221 Dortmund, Ger- many E-mail address: [email protected]