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

Let $K$ be a quadratic number field, i.e. $K=\\mathbb{Q}(\\sqrt{d})$ forsome 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 $p\eq 2$ , $p$ does not divide $d$ , then
$p\\mathcal{O}_{K}=\\begin{cases}(p,n+\\sqrt{d})(p,n-\\sqrt{d}),&\\text{ if }d\\equivn%^{2}\\ \\operatorname{mod}\\ p,\\\\\\text{prime},&\\text{ if $d$ is not a square }\\operatorname{mod}\\ p.\\end{cases}$

References