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Let $K$ be a quadratic number field. Let $R$ be an order of $K$, i.e. a subring of $K$ which is a free $\mathbb{Z}$-module of rank $2$. Let $D$ be the discriminant of $R$. Let $x_1,\cdots, x_n$ be a sequence of elements of $R$. We denote by $[x_1,\cdots,x_n]$ the $\mathbb{Z}$-submodule of $R$ generated by $x_1,\cdots, x_n$. By this question, $R = [1, \omega]$, where $\omega = (D + \sqrt D)/2$. Let us consider the following problem.

Problem Let $I, J$ be non-zero ideals of $R$. Under what condition does there exist $\alpha \in K$ such that $I = \alpha J$?

This problem is clearly important in computing the ideal class group of $R$(see this question for the definition of the ideal class group of $R$). Clearly we can assume that $I$ and $J$ are primitive ideals(see this question for the definition of a primitive ideal).

My question Is the following proposition correct? If yes, how do you prove it?

Proposition Let $I = [a, k + \omega], J = [b, l + \omega]$ be primitive ideals of $R$, where $a,k, b, l \in \mathbb{Z}, a \gt 0, b \gt 0$. Let $\theta = (k + \omega)/a, \tau = (l + \omega)/b$. Then there exists $\alpha \in K$ such that $I = \alpha J$ if and only if there exist integers $p, q, r, s$, such that $ps - qr = \pm 1$ and $\theta = (p\tau + q)/(r\tau + s)$. Moreover, if we require an additional condition $N(\alpha) \gt 0$, then we must replace $ps - qr = \pm 1$ by $ps - qr = 1$. Note that when $D \lt 0$, this condition is automatically satisfied, i.e. $N(\alpha) \gt 0$ for all non-zero $\alpha \in K$.

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Makoto Kato
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Notation Let $K$ be a quadratic number field. Let $\sigma$ be the unique non-identity automorphism of $K/\mathbb{Q}$. Let $\alpha \in K$. We denote $\sigma(\alpha)$ by $\alpha'$.

Let $\alpha, \beta \in K$. We denote the determinant of the matrix $\left( \begin{array}{ccc} \alpha & \alpha' \\ \beta & \beta' \end{array} \right)$ by $\Delta(\alpha, \beta)$. Namely $\Delta(\alpha, \beta) = \alpha\beta' - \alpha'\beta$.

Lemma 1 Let $K$ be a quadratic number field. Let $I$ be a free $\mathbb{Z}$-submodules of $K$ of rank $2$. Let $\mu, \nu$ be its basis over $\mathbb{Z}$. Let $\alpha, \beta \in I$. There exist $p, q, r, s \in \mathbb{Z}$ such that $\alpha = p\mu + q\nu, \beta = r\mu + s\nu$. Then $\Delta(\alpha, \beta) = (ps - qr)\Delta(\mu, \nu)$.

Proof: This follows immediately from the following equation.

$\left( \begin{array}{ccc} \alpha & \alpha' \\ \beta & \beta' \end{array} \right)$ = $\left( \begin{array}{ccc} p & q \\ r & s \end{array} \right)$ $\left( \begin{array}{ccc} \mu & \mu' \\ \nu & \nu' \end{array} \right)$.

Lemma 2 Let $K$ be a quadratic number field. Let $\alpha, \beta, \gamma \in I$. Then $\Delta(\gamma\alpha, \gamma\beta) = N(\gamma)\Delta(\alpha, \beta)$.

Proof: $\Delta(\gamma\alpha, \gamma\beta) = \gamma\alpha\gamma'\beta' - \gamma'\alpha'\gamma\beta = \gamma\gamma'(\alpha\beta' - \alpha'\beta) = N(\gamma)\Delta(\alpha, \beta)$.

Proof of the proposition

Suppose there exists $\alpha \in K$ such that $I = \alpha J$. Then $[a, k + \omega] = \alpha[b, l + \omega] = [\alpha b, \alpha (l + \omega)]$. Hence there exist $p, q, r, s \in \mathbb{Z}$ such that $ps - qr = \pm 1$ and

$k + \omega = p\alpha (l + \omega) + q\alpha b = \alpha(p(l + \omega) + qb)$,

$a = r\alpha (l + \omega) + s\alpha b = \alpha(r(l + \omega) + sb)$.

Hence $\theta = (p(l + \omega) + qb)/(r(l + \omega) + sb) = (p\tau + q)/(r\tau + s)$.

Suppose $N(\alpha) \gt 0$. Let $P = ps - qr$. By Lemma 1 and Lemma 2, $\Delta(a, k + \omega) = P\Delta(\alpha b, \alpha (l + \omega)) = PN(\alpha)\Delta(b, l + \omega)$. Since $\Delta(a, k + \omega) = -a\sqrt D, \Delta(b, l + \omega) = -b\sqrt D$, $a = PN(\alpha)b$. Hence $P$ must be positive, i.e. $P = 1$.

Conversely suppose $\theta = (p\tau + q)/(r\tau + s)$, where $p, q, r, s \in \mathbb{Z}$ and $ps - qr = \pm 1$. Let $\alpha = a/(b(r\tau + s))$. Then $a = \alpha b(r\tau + s)$. $a\theta = \alpha b(r\tau + s)\theta = \alpha b(p\tau + q)$. Hence $[a, a\theta] = [\alpha b(r\tau + s), \alpha b(p\tau + q)] = \alpha b[r\tau + s, p\tau + q] = \alpha b[1, \tau] = \alpha [b, b\tau]$. Hence $I = \alpha J$.

Let $P = ps - qr$. By the same argument as above, $a = PN(\alpha)b$. Hence, if $P = 1$, $N(\alpha) \gt 0$. QED

Makoto Kato
  • 42,602