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

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

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:

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

Prime ideals of $\mathbb{Z}$ decompose as follows in $\mathcal{O}_{K}$ :

Let $p\in\mathbb{Z}$ be a prime.

1.

If $p\mid d$ (divides), then $p\mathcal{O}_{K}=(p,\sqrt{d})^{2}$ ;
2.

If $d$ is odd, then

$2\mathcal{O}_{K}=\begin{cases}(2,1+\sqrt{d})^{2},&\text{ if }d\equiv 3\ % \operatorname{mod}\ 4,\\ \left(2,\frac{1+\sqrt{d}}{2}\right)\left(2,\frac{1-\sqrt{d}}{2}\right),&\text{% if }d\equiv 1\ \operatorname{mod}\ 8,\\ \text{prime},&\text{ if }d\equiv 5\ \operatorname{mod}\ 8.\end{cases}$
3.

If $p\neq 2$ , $p$ does not divide $d$ , then

$p\mathcal{O}_{K}=\begin{cases}(p,n+\sqrt{d})(p,n-\sqrt{d}),&\text{ if }d\equiv n% ^{2}\ \operatorname{mod}\ p,\\ \text{prime},&\text{ if $d$ is not a square }\operatorname{mod}\ p.\end{cases}$

References

Title	prime ideal decomposition in quadratic extensions of $\mathbb{Q}$
Canonical name	PrimeIdealDecompositionInQuadraticExtensionsOfmathbbQ
Date of creation	2013-03-22 13:53:46
Last modified on	2013-03-22 13:53:46
Owner	alozano (2414)
Last modified by	alozano (2414)
Numerical id	7
Author	alozano (2414)
Entry type	Theorem
Classification	msc 11R11
Related topic	CalculatingTheSplittingOfPrimes
Related topic	ExamplesOfPrimeIdealDecompositionInNumberFields
Related topic	PrimeIdealDecompositionInCyclotomicExtensionsOfMathbbQ
Related topic	NumberField
Related topic	SplittingAndRamificationInNumberFieldsAndGaloisExtensions