PlanetMath: prime ideal decomposition in quadratic extensions of $\mathbb{Q}$

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prime ideal decomposition in quadratic extensions of $\mathbb{Q}$

(Theorem)

Let $K$ be a quadratic number field, i.e. $K=\Rats(\sqrt{d})$ for some square-free integer $d$ . The discriminant of the extension is

$\displaystyle D_{K/\mathbb{Q}}=\begin{cases}d, & \text{ if } d\equiv 1 \ \operatorname{mod}\ 4,\\ 4d, & \text{ if } d\equiv 2,3 \operatorname{mod}\ 4.\end{cases}$

Let $\mathcal{O}_K$ denote the ring of integers of $K$ . We have:

$\displaystyle \mathcal{O}_K\cong \begin{cases}\mathbb{Z}\oplus \frac{1+\sqrt{d}... ...sqrt{d}\mathbb{Z}, & \text{ if } d\equiv 2,3 \operatorname{mod}\ 4. \end{cases}$

Prime ideals of $\Ints$ decompose as follows in $\mathcal{O}_K$ :

Theorem 1 Let $p\in \Ints$ be a prime.

If $p\mid d$ (divides), then $p\mathcal{O}_K=(p,\sqrt{d})^2$ ;
If $d$ is odd, then

$\displaystyle 2\mathcal{O}_K=\begin{cases}(2,1+\sqrt{d})^2, & \text{ if } d\equ... ... 8,\\ \text{prime}, & \text{ if } d\equiv 5\ \operatorname{mod}\ 8. \end{cases}$
If $p\neq 2$ , $p$ does not divide $d$ , then

$\displaystyle p\mathcal{O}_K=\begin{cases}(p,n+\sqrt{d})(p,n-\sqrt{d}), & \text... ...t{prime}, & \text{ if $d$\ is not a square } \operatorname{mod}\ p. \end{cases}$