Quadratic forms are functions $f(X,Y)$ that can be written as $aX^2+bXY+cY^2$ with integer coeficients. Quadratic orders are subrings of finite index of the ring of integers in a quadratic extension of $\mathbb{Q}$.
The 1986 article "Nonsingular Plane Cubic Curves over Finite Fields" by Schoof mentions that the number of equivalence classes for binary quadratic forms and of quadratic orders.
The definition of $\Delta(\mathcal{O})$ doesn't appear in the text. What is $\Delta(\mathcal{O})$, and how does it relate to the discriminant of a binary form?
Best Answer
A binary quadratic form, with integer coefficients, is some $$ f(x,y) = A x^2 + B xy + C y^2. $$ The discriminant is $$ \Delta = B^2 - 4 A C. $$ We will abbreviate this by $$ \langle A,B,C \rangle. $$ It is primitive if $\gcd(A,B,C)=1. $ Standard fact, hard to discover but easy to check: $$ (A x^2 + B x y + C D y^2 ) (C z^2 + B z w + A D w^2 ) = A C X^2 + B X Y + D Y^2,$$ where $ X = x z - D yw, \; Y = A xw + C yz + B yw. $ This gives us Dirichlet's definition of "composition" of quadratic forms of the same discriminant, $$ \langle A,B,CD \rangle \circ \langle C,B,AD \rangle = \langle AC,B,D \rangle. $$ In particular, if this $D=1,$ the result represents $1$ and is ($SL_2 \mathbf Z$) equivalent to the "principal" form for this discriminant. Oh, duplication or squaring in the group; if $\gcd(A,B)=1,$ $$ \langle A,B,AD \rangle^2 = \langle A^2,B,D \rangle. $$ This comes up with positive forms: $ \langle A,B,C \rangle \circ \langle A,-B,C \rangle = \langle 1,B,AC \rangle $ is principal, the group identity. Probably should display some $SL_2 \mathbf Z$ equivalence rules, these are how we calculate when things are not quite right for Dirichlet's rule: $$ \langle A,B,C \rangle \cong \langle C,-B,A \rangle, $$ $$ \langle A,B,C \rangle \cong \langle A, B + 2 A, A + B +C \rangle, $$ $$ \langle A,B,C \rangle \cong \langle A, B - 2 A, A - B +C \rangle. $$
Imaginary first. Suppose we want to know about $\mathbf Q(\sqrt {-47}).$ Reduced positive forms $ \langle A,B,C \rangle $ obey $|B| \leq A \leq C$ and $B \neq -A,$ also whenever $A=C$ we have $B \geq 0.$ Our group of binary forms is
This is an abelian group in any case, so it is cyclic of order 5. These are also the five elements in the ring of integers of $\mathbf Q(\sqrt {-47}).$ Here is the mapping from forms to ideals: given $ \langle A,B,C \rangle, $ drop the letter $C.$ That's it. $$ \langle A,B,C \rangle \mapsto \left[ A, \frac{B + \sqrt \Delta}{2} \right]. $$ Oh, why is this an ideal, rather than just some $\mathbf Z$-lattice? Because, given $\alpha,\beta$ rational integers, $$ \left[ \alpha, \frac{\beta + \sqrt \Delta}{2} \right] $$ is an ideal if and only if $$ 4 \alpha | ( \Delta - \beta^2 ). $$
Group: we already see how to do $$ \langle 2,1,6 \rangle^2 \cong \langle 4,1,3 \rangle \cong \langle 3,-1,4 \rangle; $$ $$ \langle 2,1,6 \rangle \circ \langle 3,-1,4 \rangle \cong \langle 2,5,9 \rangle \circ \langle 3,5,6 \rangle \cong \langle 6,5,3 \rangle \cong \langle 3,-5,6 \rangle \cong \langle 3,1,4 \rangle; $$ $$ \langle 2,1,6 \rangle \circ \langle 3,1,4 \rangle \cong \langle 6,1,2 \rangle \cong \langle 2,-1,6 \rangle. $$ $$ \langle 2,1,6 \rangle \circ \langle 2,-1,6 \rangle \cong \langle 1,1,12 \rangle $$ in any case.