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Let $\mathfrak{F}$ be the set of binary quadratic forms over $\mathbb{Z}$. Let $F = ax^2 + bxy + cy^2 \in \mathfrak{F}$. We say $D = b^2 - 4ac$ is the discriminant of $F$. It is easy to see that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$).

Let $\Gamma =SL_2(\mathbb{Z})$.
Let $\alpha = \left( \begin{array}{ccc} p & q \\ r & s \end{array} \right)$ be an element of $\Gamma$. We write $f^\alpha(x, y) = f(px + qy, rx + sy)$. Since $(f^\alpha)^\beta$ = $f^{\alpha\beta}$, $\Gamma$ acts on $\mathfrak{F}$ from right.

Let $D$ be a non-square integer such that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$). We denote the set of binary quadratic forms of discriminant $D$ by $\mathfrak{F}(D)$. It is easy to see that $\Gamma$ acts on $\mathfrak{F}(D)$ from right. Let $\sigma = \left( \begin{array}{ccc} 1 & 1 \\ 0 & 1 \end{array} \right) \in \Gamma$. Let $\Gamma_{\infty} = \{\sigma^n; n \in \mathbb{Z}\}$. $\Gamma_{\infty}$ also acts on $\mathfrak{F}(D)$ from right. We denote the set of $\Gamma_{\infty}$-orbits on $\mathfrak{F}(D)$ by $\mathfrak{F}(D)/\Gamma_{\infty}$.

Let $K$ be a quadratic number field. Let $R$ be an order of $K$. Let $D$ be its discriminant. It is easy to see that $D$ is not not a square integer and $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$).

We denote the group of invertible elements of the ring $\mathbb{Z}$ by $\mathbb{Z}^\times$. Namely $\mathbb{Z}^\times = \{-1, 1\}$.

Let $F = ax^2 + bxy + cy^2 \in \mathfrak{F}(D)$. By this question, $I = \mathbb{Z}a + \mathbb{Z}\frac{(-b + \sqrt{D})}{2}$ is an ideal of $R$. Hence we get a map $\psi_0\colon \mathfrak{F}(D) \rightarrow \mathfrak{i}(R)$, where $\mathfrak{i}(R)$ is the set of fractional ideals of $R$. We define a map $\psi_1\colon \mathfrak{F}(D) \rightarrow \mathfrak{i}(R)\times\mathbb{Z}^\times$ by $\psi_1(F) = (\psi_0(F), sgn(a))$, where $sgn(a)$ is the sign of $a$. It is easy to see that $\psi_1$ induces a map $\mathfrak{F}(D)/\Gamma_{\infty} \rightarrow \mathfrak{i}(R)\times\mathbb{Z}^\times$. Hence this map induces a map $$\psi\colon\mathfrak{F}(D)/\Gamma_{\infty} \rightarrow (\mathfrak{i}(R)/\mathbb{Q}^\times)\times\mathbb{Z}^\times$$

Next we would like to define a map $(\mathfrak{i}(R)/\mathbb{Q}^\times)\times\mathbb{Z}^\times \rightarrow \mathfrak{F}(D)/\Gamma_{\infty}$

Let $I \in \mathfrak{i}(R)$. $I$ can be written as $I = J/\lambda$, where $J$ is an ideal of $R$ and $\lambda \in R$. We define the norm of $I$ as $N(I) = N(J)/N(\lambda R)$. It is easy to see that this is well defined.

Let $\alpha, \beta \in K$. We denote $\alpha\beta' - \alpha'\beta$ by $\Delta(\alpha, \beta)$, where $\alpha'$(resp. $\beta'$) is the conjugate of $\alpha$(resp. $\beta$). $\Delta(\alpha, \beta) \neq 0$ if and only if $\alpha, \beta$ are linearly independent over $\mathbb{Q}$. If $D < 0$, we define $\sqrt{D}$ as i$\sqrt{|D|}$. Let $\{\alpha, \beta\}$ be $\mathbb{Z}$-basis of $I \in \mathfrak{i}(R)$. If $\Delta(-\alpha, \beta)/\sqrt{D} > 0$, we say the basis $\{\alpha, \beta\}$ is positively oriented. If $\Delta(-\alpha, \beta)/\sqrt{D} < 0$, we say the basis $\{\alpha, \beta\}$ is negatively oriented.

Let $\{\alpha, \beta\}$ be positively oriented basis of $I \in \mathfrak{i}(R)$. We can assume that $\alpha \in \mathbb{Q}$. Let $x, y$ be indeterminates. Let $s \in \mathbb{Z}^\times$. We write $f(\alpha, \beta, s; x, y) = sN_{K/\mathbb{Q}}(x\alpha - sy\beta)/N(I)$. Namely $f(\alpha, \beta, s; x, y) = s(x\alpha - sy\beta)(x\alpha' - sy\beta')/N(I)$. It is easy to see that $f(\alpha, \beta, s; x, y)$ is a binary quadratic form of discriminant $D$.

My question Are the following propositions true? If yes, how do we prove them?

Proposition 1 The class of $\mathfrak{F}(D)/\Gamma_{\infty}$ represented by $f(\alpha, \beta, s; x, y)$ is determined only by the class of $(\mathfrak{i}(R)/\mathbb{Q}^\times)$ represented by $I$ and $s$.

Proposition 2 By proposition 1, we can define a map $$\phi\colon (\mathfrak{i}(R)/\mathbb{Q}^\times)\times\mathbb{Z}^\times \rightarrow \mathfrak{F}(D)/\Gamma_{\infty}$$ by $\phi(([I], s)) = [f(\alpha, \beta, s; x, y)]$, where $[I]$(resp. $[f(\alpha, \beta, s; x, y)]$) denotes the class represented by $I$(resp. $f(\alpha, \beta, s; x, y)$).

Then $\psi$ and $\phi$ are inverses of each other.

Corollary Let $\mathfrak{F}_0(D)$ be the set of primitive binary quadratic forms of discriminant $D$. Let $\mathfrak{I}(R)$ be the group of invertible fractional ideals of $R$. Then the map $\psi$ induces a bijection:

$$\mathfrak{F}_0(D)/\Gamma_{\infty} \rightarrow (\mathfrak{I}(R)/\mathbb{Q}^\times)\times\mathbb{Z}^\times$$

Proof: This follows immediately from proposition 2 and the proposition of this question.

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@All Please refrain from discussing meta-matters on the main site. There is already at least one meta-discussion on this question. Please take any meta discussion there, or to another meta thread if need be. –  Bill Dubuque Oct 16 '12 at 4:48
    
Have you seen Henri Cohen's exposition? (excerpted in may answer here) –  Bill Dubuque Oct 18 '12 at 13:53
    
@BillDubuque Actually yes. But it's long time ago and I don't have the book right now as I said. –  Makoto Kato Oct 18 '12 at 18:28
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1 Answer

Notations We denote a binary quadrtic form $ax^2 + bxy + cy^2$ also by $(a, b, c)$. Let $x_1,\cdots,x_n$ be elements of $K$. We denote by $[x_1,\cdots,x_n]$ the subgroup of $K$ generated by the set $\{x_1,\cdots,x_n\}$.

First we prove that the map $\psi\colon\mathfrak{F}(D)/\Gamma_{\infty} \rightarrow (\mathfrak{i}(R)/\mathbb{Q}^\times)\times\mathbb{Z}^\times$ is well-defined. It suffices to prove the following lemma.

Lemma 1 Let $\psi_0\colon \mathfrak{F}(D) \rightarrow \mathfrak{i}(R)$ be the map defined in the question. Let $(a, b, c) \in \mathfrak{F}(D)$. Then $\psi_0((a,b,c)^{\sigma^n}) = \psi_0((a, b, c))$, where $\sigma = \left( \begin{array}{ccc} 1 & 1 \\ 0 & 1 \end{array} \right) \in \Gamma$ and $n \in \mathbb{Z}$.

Proof: Note that $\sigma^n = \left( \begin{array}{ccc} 1 & n \\ 0 & 1 \end{array} \right) \in \Gamma$. $a(x + ny)^2 + b(x + ny)y + cy^2 = ax^2 + (2an + b)xy + an^2 + bn + c$. Hence $(a,b,c)^{\sigma^n} = (a, 2an + b, an^2 + bn + c)$. Hence $\psi_0((a,b,c)^{\sigma^n}) = [a, (-2an - b + \sqrt D)/2] = [a, -an + (-b + \sqrt D)/2] = [a, (-b + \sqrt D)/2] = \psi_0((a, b, c)).$

QED

Definition 1 Let $I$ be a fractional ideal of $R$. Let $\alpha, \beta$ be $\mathbb{Z}$-basis of $I$. It is easy to see that $\Delta(\alpha, \beta)^2$ is a non-zero rational number and independent of the choice of the $\mathbb{Z}$-basis(see Lemma 7 of my answer to this question). We denote it by $d(I)$.

Lemma 2 Let $I$ be a fractional ideal of $R$. Let $\gamma \in I$. Then $N(\gamma)/N(I)$ is a rational integer.

Proof: We may suppose $\gamma \ne 0$. By Proposition 13 of my answer to this question, $N(\gamma R)/N(I)$ is a rational integer. On the other hand, by Proposition 11 of the same answer, $N(\gamma R) = |N(\gamma)|$. QED

Definition 2 Let $I = [\alpha, \beta]$ be a fractional ideal of $R$. Let $x, y$ be indeterminates. Let $s \in \mathbb{Z}^\times$. We write $f(\alpha, \beta, s; x, y) = sN_{K/\mathbb{Q}}(x\alpha - sy\beta)/N(I)$. Namely $f(\alpha, \beta, s; x, y) = s(x\alpha - sy\beta)(x\alpha' - sy\beta')/N(I)$.

Lemma 3 Let $I = [\alpha, \beta]$ be a fractional ideal of $R$. Then $f(\alpha, \beta, s; x, y)$ is an integral binary quadratic form of discriminant $D$, i.e. $f(\alpha, \beta, s; x, y) \in \mathfrak{F}(D)$.

Proof: We first prove that $f(\alpha, \beta, s; x, y)$ has integral coefficients. $N(x\alpha - sy\beta) = (x\alpha - sy\beta)(x\alpha' - sy\beta') = (\alpha \alpha')x^2 - s(\alpha\beta' + \beta\alpha')xy + (\beta\beta')y^2$.

Hence $f(\alpha, \beta, s; x, y) = ax^2 + bxy + cy^2$, where

$a = s(\alpha \alpha')/N(I)$

$b = -(\alpha\beta' + \beta\alpha')/N(I)$

$c = s(\beta\beta')/N(I).$

By Lemma 2, $a, c \in \mathbb{Z}$. $N(\alpha + \beta) = (\alpha + \beta)(\alpha' + \beta') = \alpha \alpha' + (\alpha\beta' + \beta\alpha') + \beta\beta'$. Hence $(\alpha\beta' + \beta\alpha')/N(I) = N(\alpha + \beta)/N(I) - N(\alpha) /N(I) - N(\beta)/N(I)$. Hence, by Lemma 2, $b \in \mathbb{Z}$.

It remains to prove that the discriminant of $f(\alpha, \beta, s; x, y)$ is $D$. Since $(\alpha\beta' + \beta\alpha')^2 - 4\alpha \alpha'\beta\alpha' = (\alpha\beta' - \beta\alpha')^2 = \Delta(\alpha, \beta)^2 = d(I), b^2 - 4ac = d(I)/N(I)^2$. Hence, by the corollary of Proposition 12 of my answer to this question, $b^2 - 4ac = D$. QED

Lemma 4 Let $I = [\alpha, \beta]$ be a fractional ideal of $R$. Let $r \ne 0$ be a rational number. Then $f(r\alpha, r\beta, s; x, y) = f(\alpha, \beta, s; x, y)$.

Proof: $f(r\alpha, r\beta, s; x, y) = sN(xr\alpha - syr\beta)/N(rI) = sN(r)N(x\alpha - sy\beta)/|N(r)|N(I) = sr^2N(x\alpha - sy\beta)/r^2N(I) = f(\alpha, \beta, s; x, y).$

QED

Lemma 5 Let $I$ be a fractional ideal of $R$. Then there exists $r \in \mathbb{Q}$ and a primitive ideal(see here) $I_0$ such that $I = rI_0$.

Proof: There exists $\gamma \ne 0 \in K$ such that $\gamma I \subset R$. We may suppose $\gamma \in R$. Let $c = N(\gamma)$. Then $cI = \gamma'\gamma I \subset \gamma'R \subset R$. There exist $d \in \mathbb{Z}$ and a primitive ideal $I_0$ such that $cI = dI_0$. Hence $I = rI_0$, where $r = d/c$. QED

Lemma 6 Let $I$ be a primitive ideal of $R$. Suppose $I = [c, \beta]$, where $c$ is a rational integer. Then $\beta$ can be written as $\beta = d \pm \omega$, where $d \in \mathbb{Z}, \omega = (D + \sqrt D)/2$.

Proof: Let $a, b + \omega$ be the canonical basis of $I$(see here). Since $I \cap \mathbb{Z} = a\mathbb{Z} = c\mathbb{Z}$, $c = \pm a$. Since $R = [1, \omega]$, $\beta$ can be written as $\beta = d + e\omega, d, e \in \mathbb{Z}$. Then $N(I) = a = |ce| = a|e|$. Hence $e = \pm 1$. QED

Lemma 7 Let $I$ be a primitive ideal of $R$. Suppose $I = [a, \beta] = [a, \gamma]$, where $a \gt 0$ is a rational integer and both $\{a, \beta\}$ and $\{a, \gamma\}$ are positively oriented. Then $\beta - \gamma \in a\mathbb{Z}$.

Proof: By Lemma 6, there exists $d \in \mathbb{Z}$ such that $\beta = d \pm \omega$. If $\beta = d + \omega, \Delta(-a, \beta) = -a(d + \omega') + a(d + \omega) = a(\omega - \omega') = a\sqrt D$. If $\beta = d - \omega, \Delta(-a, \beta) = -a(d - \omega') + a(d - \omega) = a(\omega' - \omega) = -a\sqrt D$. Since $\{a, \beta\}$ is positively oriented, $\beta$ must be $d + \omega$. Similarly $\gamma = c + \omega$. Hence $\beta - \gamma = d - c \in I \cap \mathbb{Z} = a\mathbb{Z}$. QED

Lemma 8 Let $I$ be a fractional ideal of $R$. Suppose $\{\alpha, \beta\}$ is a positively oriented $\mathbb{Z}$-basis of $I$. Then $\{\alpha, \beta+n\alpha\}$ is also a positively oriented $\mathbb{Z}$-basis of $I$ for any $n \in \mathbb{Z}$ and $f(\alpha, \beta+n\alpha, s; x, y) = f(\alpha, \beta, s; x, y)^{\sigma^{-sn}}$.

Proof: Clearly $I = [\alpha, \beta+n\alpha]$. $\Delta(-\alpha, \beta + n\alpha) = -\alpha(\beta' + n\alpha') + \alpha'(\beta + n\alpha) = \Delta(\alpha, \beta)$. Hence $\{\alpha, \beta+n\alpha\}$ is positively oriented. Suppose $f(\alpha, \beta, s; x, y) = (a, b, c)$.

$a = s(\alpha \alpha')/N(I)$

$b = -(\alpha\beta' + \beta\alpha')/N(I)$

$c = s(\beta\beta')/N(I).$

Suppose $f(\alpha, \beta+n\alpha, s; x, y) = (k, l, m)$.

$k = s(\alpha \alpha')/N(I) = a$

$l = -(\alpha(\beta+ n\alpha)' + (\beta + n\alpha)\alpha')/N(I) = b - 2sna$

$m = s(\beta+n\alpha)(\beta'+n\alpha')/N(I) = s(\beta\beta' + n(\beta\alpha' + \alpha\beta') + n^2\alpha\alpha')/N(I) = c - snb + an^2$

Hence $(k, l, m) = (a, b, c)^{\sigma^{-sn}}$. QED

Proof of Proposition 1 Let $I, J$ be fractional ideals of $R$. Suppose $I = [\alpha, \beta], J = [\gamma, \delta]$, where $\alpha, \gamma \in \mathbb{Q}$ and $\{\alpha, \beta\}$ and$\{\gamma, \delta\}$ are positively oriented bases of $I$ and $J$ respectively. Suppose $J = rI$ for some $r \in \mathbb{Q}$. We need to prove that $f(\gamma, \delta, s; x, y) = f(\alpha, \beta, s; x, y)^{\sigma^n}$ for some $n \in \mathbb{Z}$.

By Lemma 5, there exists $q \in \mathbb{Q}$ and a primitive ideal $I_0$ such that $I = qI_0$. Then $J = rqI_0$. Hence $I_0 = [(1/q)\alpha, (1/q)\beta] = [(1/rq)\gamma, (1/rq)\delta]$. Hence, by Lemma 4, we may suppose that $I$ is a primitive ideal and $I = J$. Since $I = [-\alpha, -\beta]$, we may suppose $\alpha \gt 0$ by Lemma 4. Similarly we may suppose $\gamma \gt 0$. Since $I \cap \mathbb{Z} = \alpha \mathbb{Z} = \gamma \mathbb{Z}$, $\alpha = \gamma$. Hence, by Lemma 7 and Lemma 8, we are done. QED

Proof of Proposition 2 We first prove that $\phi\circ \psi = 1$. Let $(a, b, c) \in \mathfrak{F}(D)$. Let $I = [a, (-b + \sqrt D)/2] = [a, - (D + b)/2 + \omega]$, where $\omega = (D + \sqrt D)/2$. Note that $I$ is a primitive ideal.

Suppose $a \gt 0$. Note that $N(I) = a$. Let $\alpha = a, \beta = (-b + \sqrt D)/2$. $\{\alpha, \beta\}$ is positively oriented. Suppose $f(\alpha, \beta, 1; x, y) = (k, l, m)$.

$k = (\alpha \alpha')/N(I) = a^2/a = a$

$l = -(\alpha\beta' + \beta\alpha')/N(I) = ab/a = b$

$m = (\beta\beta')/N(I) = (b^2 - D)/4a = 4ac/4a = c$

Suppose $a \lt 0$. Note that $I = [-a, (-b + \sqrt D)/2$ and $N(I) = -a$. Let $\alpha = -a, \beta = (-b + \sqrt D)/2$. $\{\alpha, \beta\}$ is positively oriented. Suppose $f(\alpha, \beta, -1; x, y) = (k, l, m)$.

$k = -(\alpha \alpha')/N(I) = a^2/a = a$

$l = -(\alpha\beta' + \beta\alpha')/N(I) = ab/a = b$

$m = -(\beta\beta')/N(I) = (b^2 - D)/4a = 4ac/4a = c$

Hence $\phi\circ \psi = 1$ in either case.

It remains to prove that $\psi\circ \phi = 1$. Let $I = [a, b + \omega]$ be a primitive ideal of $R$, where $a \gt 0, b \in \mathbb{Z}, \omega = (D + \sqrt D)/2$. Let $\alpha = a, \beta = b + \omega$. Let $s \in \mathbb{Z}^\times$. Let us compute $f(\alpha, \beta, s; x, y)$. Let $f(\alpha, \beta, s; x, y) = (k, l, m)$.

$k = s(\alpha \alpha')/N(I) = sa$

$l = -(\alpha\beta' + \beta\alpha')/N(I) = -(2b + D)$

$m = s(\beta\beta')/N(I) = s(b^2 + bD + (d^2 - d)/4)/a.$

Hence $\psi\circ\phi([I], s) = ([[sa, b + (D + \sqrt D)/2]], s) = ([I], s)$. QED

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