I came across this question: Let K be a quadratic number field and let $\alpha \in \mathcal{O}_K$. Suppose there exist caprice coprime positive integers $a$ and $b$ such that $|N_K(\alpha)| = ab$. Prove that $(a,\alpha)(b,\alpha) = (\alpha).$

What I know is that: Quadratic number field is $\mathbb{Q}(\sqrt(d))$, where $d$ is a square free integer, basis are $\{1,\sqrt(d)\}$.

Because $\alpha \in \mathcal{O}_{Q_{\sqrt(d)}}$ the coefficients of the minimal polynomial are integers and $N_K (\alpha)\in Z$.

$\alpha$ is of the form $p +q\sqrt(d)$ with $p,q \in \mathbb{Q}$ and since $\mathbb{Q}(\sqrt(d))$ is a degree two extension the conjugate of $\alpha$ is $p - q\sqrt(d)$

$|N_K(\alpha)| = |p^2 - q^2d|$

How to show that $(a,\alpha)(b,\alpha) = (\alpha)$?

• Does "Caprice positive numbers" = "arbitrary positive numbers", or is that word defining something. And what does $\;(a,\alpha)\;$ mean in this context? – DonAntonio Nov 13 '16 at 13:41
• @Don Some software autocorrects "coprime" to "caprice". Coprimes are a bit eccentric... – Bill Dubuque Nov 13 '16 at 14:48

If you mean "coprime positive integers $a$ and $b$ ...", then I have a solution. Otherwise tell me what caprice means.
Since $a,b$ are coprime there are integers $r,s$ with $ar + bs = 1$. Now $(a, \alpha)(b, \alpha) = (ab, a\alpha, b\alpha, \alpha^2)$. Thus $ra \alpha + s b \alpha = (ra + sb) \alpha = \alpha \in (a, \alpha)(b, \alpha)$. The other direction is obvious, since $\sigma(\alpha) \in R$ and thus $ab = \alpha \sigma(\alpha), a \alpha, b \alpha, \alpha^2 \in (\alpha)$, where $\alpha$ is the non-trivial field automorphism of the quadratic number field.
$\color{#c00}{ab=\bar\alpha\alpha}\,\Rightarrow\ (a,\alpha)(b,\alpha) = (\color{#c00}{ab},a\alpha,b\alpha,\alpha\alpha) = (\color{#c00}{\bar\alpha},\color{#0a0}{a,b},\alpha)\color{#c00}\alpha\ [ = (\alpha)\$ if $\ (\color{#0a0}{a,b}) = 1 ]$