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7.3 Binary Quadratic Forms

Gauss's approach to ideals (which were not defined in his time!) was to represent elements of the class group (groups were also not defined in his time!) by equivalence classes of quadratic forms. The idea is to make use of the fact that for each ideal I we are actually interested in objects like Nm($ \alpha$)/Nm(I) for some element $ \alpha$ in I. As seen above, the ideal class is represented by some ideal J with Nm(J) = Nm($ \alpha$)/Nm(I).

To fix notation, let the quadratic order R be given as $ \mathbb {Z}$ + $ \mathbb {Z}$ . $ \omega$, where $ \omega$ = (D + $ \sqrt{D}$)/2 with D = DR the discriminant of the order R; then $ \omega$ satisfies the equation

$\displaystyle \omega^{2}_{}$ - D . $\displaystyle \omega$ + % latex2html id marker 16899
$\displaystyle {\frac{D^2-D}{4}}$ = 0

Any non-zero ideal I in R is then of the form $ \mathbb {Z}$ . a + $ \mathbb {Z}$ . (b + c$ \omega$), where I $ \cap$ $ \mathbb {Z}$ = $ \mathbb {Z}$ . a is the restriction of I to $ \mathbb {Z}$; we can assume that a > 0. Moreover, by Euclidean division we can subtract a multiple of a from b to ensure that 0 $ \leq$ b < a. Now the fact that I is an ideal gives us
a . $\displaystyle \omega$ = p . a + q . (b + c$\displaystyle \omega$)  
(b + c$\displaystyle \omega$) . $\displaystyle \omega$ = r . a + s . (b + c$\displaystyle \omega$)  

For some integers p, q, r and s. From this we deduce

a = qc    and  0 = pa + qb    
b + cD = sc    and  - % latex2html id marker 16935
$\displaystyle {\frac{D^2-D}{4}}$ = ra + sb    

Hence a = qc and b = - pc are multiples of c. Moreover, the quadratic expression b2 + bcD + c2% latex2html id marker 16940
$ {\frac{D^2-D}{4}}$ = - rac is divisible by ac. Now, it is clear that c is determined by the condition that (I + $ \mathbb {Z}$)/$ \mathbb {Z}$ is the subgroup $ \mathbb {Z}$ . (c$ \omega$) of R/$ \mathbb {Z}$ $ \cong$ $ \mathbb {Z}$ . $ \omega$; or equivalently that Nm(I) = ac, with a determined as before. It follows that any tuple (a, b, c) that satisfies the above conditions uniquely determines an ideal and vice versa.

Now, it is clear that the ideal c-1I = $ \mathbb {Z}$ . q + $ \mathbb {Z}$(- p + $ \omega$) is equivalent to I in the class group. Thus, we say the ideal is primitive if the representative tuple (a, b, c) satisfies c = 1. Clearly, we only need to look at primitive ideals for the purpose of computing the class group; but there are more equivalence relations.

We write a general element of I as ax + (b + c$ \omega$)y; its norm is a multiple of Nm(I) = ac. Thus,

QI(x, y) = % latex2html id marker 16969
$\displaystyle {\frac{\operatorname{\rm Nm}(ax+(b+c\omega)y)}{\operatorname{\rm Nm}(ac)}}$ = qx2 + sxy - ry2

(with notation as above) is a form with integer coefficients. Moreover, it is invariant (by construction) under the replacement of I by a rational multiple. We easily check the identity s2 + 4qr = D. Conversely, given any form Q(x, y) = qx2 + sxy - ry2 satisfying this identity, we note that s $ \equiv$ D(mod 2). Thus we can consider the ideal $ \mathbb {Z}$ . q + $ \mathbb {Z}$(s + $ \sqrt{D}$)/2. When Q(x, y) = QI(x, y), s = - 2p + D so that (s + $ \sqrt{D}$)/2 = - p + $ \omega$; hence we recover the primitive ideal associated with I. However, we not that for a general quadratic form Q(x, y) the integers q and (s - D)/2 need not be positive unless we impose this as an additional requirement on the quadratic forms under consideration.

Now, if I = $ \mathbb {Z}$ . u1 + $ \mathbb {Z}$ . u2, for some elements u1, u2 in R, then the quadratic form Qu1, u2(x, y) = Nm(xu1 + yu2)/Nm(I) is (in general) different from QI(x, y). However, it is obtained from QI(x, y) by a substitution (x, y) $ \mapsto$ (Ax + By, Cx + Dy) where $ \left(\vphantom{
\begin{smallmatrix}A & B   C & D
\end{smallmatrix} }\right.$$ \begin{smallmatrix}A & B   C & D
\end{smallmatrix}$$ \left.\vphantom{
\begin{smallmatrix}A & B   C & D
\end{smallmatrix} }\right)$ is an integer matrix with integer inverse. One way to obtain a new basis is to consider I = $ \alpha$ . J for some ideal J in R and some $ \alpha$ in K. Then, we write J = $ \mathbb {Z}$ . a' + $ \mathbb {Z}$ . (b' + c'$ \omega$) as before. Clearly u1 = a'$ \alpha$ and u2 = (b' + c'$ \omega$)$ \alpha$ is another basis of I.

Conversely, given a basis u1 and u2 of the ideal I, let d be a denominator of u2/u1; i. e. d is a positive integer so that du2/d1 lies in R. Consider the ideal J = (d /u1) . I, we see that J = $ \mathbb {Z}$ . d + $ \mathbb {Z}$ . (du2/u1) and J $ \cap$ $ \mathbb {Z}$ = $ \mathbb {Z}$ . d. Thus, as above we can find e and f so that 0 $ \leq$ e < d and (du2/u1) = nd±(e + f$ \omega$) for some integer n. Thus J = $ \mathbb {Z}$ . d + $ \mathbb {Z}$ . (e + f$ \omega$). Putting $ \alpha$ = u1/d we see that u1 = d$ \alpha$ and u2 = (nd±(e + f$ \omega$))$ \alpha$; in particular, I = $ \alpha$ . J. Moreover, we have

Qu1, u2(x, y) = Qd,(du2/u1)(x, y) = QJ(x, nx±y)

the latter form being clearly equivalent to QJ.

Thus we have shown that QI(x, y) and QJ(x, y) are equivalent under an integer change of co-ordinates for the variables (x, y) if and only if the corresponding ideals are equivalent in the class group. The problem of finding representatives of ideal classes can be replaced by the problems of finding quadratic forms that represent equivalence classes.

We now separate the cases D < 0 and D > 0. In the first case, we restrict our attention to quadratic forms Q(x, y) = qx2 + sxy - ry2 (continuing the above notation) such that q > 0. Since D = s2 + 4qr < 0, we see that r < 0. In fact Q(x, y) > 0 for all (x, y) $ \neq$ (0, 0). Pictorially, the region Q(x, y) $ \leq$ r is bounded by an ellipse. Thus, among lattice points we can choose u1 to be an element where the Q(u1) takes its minimum (non-zero) value. Now, we can complete u1 to a basis by picking a suitable vector u2. The only possible alternative choices for u2 are nu1±u2 for some integer n. Let u2 be so chosen that the value Q(u2) is minimum in this collection. It is not too difficult to show that the expression for Q in this basis is independent of the finitely many choices available. (In fact for D| > 4 the choices of u1 and u2 are unique upto sign). Now, in this basis we get Q(x, y) = Ax2 + Bxy + Cy2 with A $ \leq$ C and | B| $ \leq$ A. Moreover, if one of these is an equality (which can only happen if | D| $ \leq$ 4), we have B $ \geq$ 0 as well. A quadratic form with negative discriminant is said to be reduced if it has this special form. Clearly, there are only finitely many such forms for a given D; one for each equivalence class of quadratic forms. Thus we have found representatives for the class group.

When D > 0, the quadratic forms are indefinite. The locus Q(x, y) = r represents a hyperbola. Now the value 0 is not attained at non-zero (x, y) (else D would have a square root in integers) and the values are all integers. Thus, the absolute value of Q attains a minimum at some point u1 on the lattice. But this u1 is far from unique (in fact there are infinitely many points where Q takes this value. One can show that upto a finite number of choices these are related by an integer change of co-ordinates. Now, as before, u1 can be completed to a basis by a choice of u2. The alternatives for this choice are nu1±u2 as earlier. Again, there are only finitely many of these with sign opposite to that of Q(u1) (since the term n2Q(u1) in the expansion of the quadratic form will dominate for n large). Among this finite set we choose u2 so that the absolute value of Q is minimum (again with only finitely many options for this choice). Thus, each equivalence class of quadratic forms has been represented upto a finite ambiguity. Moreover, one can bound the ambiguity depending on DR.

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Next: 8 Algebraic Schemes for Up: 7 Quadratic fields Previous: 7.2 Naive computation of
Kapil Hari Paranjape 2002-10-20