Quadratic integers

The quadratic integers within a quadratic field $𝐾 = ℚ (\sqrt 𝑑)$ where $𝑑$ is a squarefree integer are #m/thm/ring

O 𝐾 = {ℤ [\sqrt 𝑑] 𝑑 \equiv 2, 3 (m o d 4) ℤ [1 + \sqrt 𝑑 2] 𝑑 \equiv 1 (m o d 4)

Proof

Let $𝐾 = ℚ (\sqrt 𝑑)$ Clearly an element of $𝐾$ of degree 1 is an algebraic integer iff it is an integer. Let $𝛼 = 𝑎 + 𝑏 \sqrt 𝑑 \in 𝐾$ be a degree 2 element. Then the minimal polynomial of $𝛼$ is

𝑚 𝛼 (𝑥) = 𝑥 2 - 2 𝑎 𝑥 + 𝑎 2 - 𝑏 2 𝑑

By ^P1, we have $𝛼 \in O 𝐾$ iff $𝑚 𝛼 (𝑥) \in ℤ [𝑥]$ , which is precisely the case when $2 𝑎, 𝑎 2 - 𝑏 2 𝑑 \in ℤ$ . It follows $(2 𝑏) 2 𝑑 \in ℤ$ , so since $𝑑$ is squarefree $2 𝑏 \in ℤ$ . Letting $𝑟 = 2 𝑎$ , $𝑠 = 2 𝑏$ , we have $𝛼 \in O 𝐾$ iff $𝑟, 𝑠 \in ℤ$ and $𝑟 2 - 𝑑 𝑠 2 \equiv 4 0$ . Since $𝑑$ is squarefree it follows $𝑑 ≢ 4 0$ , so we need only consider the cases

If $𝑑 \equiv 4 1$ then $0 \equiv 4 𝑟 2 - 𝑑 𝑠 2 \equiv 4 𝑟 2 - 𝑠 2$ which holds iff $𝑟 \equiv 2 𝑠$ ;
If $𝑑 \equiv 4 2$ then $0 \equiv 4 𝑟 2 - 𝑑 𝑠 2 \equiv 4 𝑟 + 2 𝑠 2$ which holds iff $𝑟, 𝑠 \equiv 2 0$ ;
If $𝑑 \equiv 4 3$ then $0 \equiv 4 𝑟 2 - 𝑑 𝑠 2 \equiv 4 𝑟 2 + 𝑠 2$ which holds iff $𝑟, 𝑠 \equiv 2 0$ .

It follows that the general expression for an algebraic integer $𝛼 \in O 𝐾$ is

$𝛼 = 𝑝 + 𝑞 \sqrt 𝑑$ if $𝑑 ≢ 4 1$
$𝛼 = 𝑝 + 𝑞 1 + \sqrt 𝑑 2$ if $𝑑 \equiv 4 1$

where $𝑝, 𝑞 \in ℤ$ , whence the above.

In general, a quadratic integer is the solution to some monic quadratic with integer coëfficients.

Properties

Let $𝛼 \in O 𝐾$ be a (proper) quadratic integer with minimal polynomial $𝑥 2 + 𝑎 𝑥 + 𝑏$

The discriminant is $Δ 𝐾 : ℚ (𝛼) = 𝑎 2 - 4 𝑏$ .
It follows that

Δ 𝐾 = {4 𝑑 𝑑 \equiv 4 2, 3 𝑑 𝑑 \equiv 4 1

Proof of 1

See Discriminant of an algebraic integer.

Prime ideals

Let $𝑝$ be an odd prime and $(𝑑 𝑝)$ be the corresponding Legendre symbol.

If $(𝑑 𝑝) = 1$ then $𝐾 : ℚ$ is unramified at $⟨ 𝑝 ⟩ = ⟨ 𝑝, 𝑎 + \sqrt 𝑑 ⟩ ⟨ 𝑝, 𝑎 - \sqrt 𝑑 ⟩$ , where $𝑎 2 \equiv 𝑝 𝑑$ .
If $(𝑑 𝑝) = - 1$ then $𝐾 : ℚ$ is inert at $𝑝$ .¹

Proof

First suppose $𝑎 2 \equiv 𝑝 𝑑$ for $𝑎 \neq 0$ . Then

𝔞 = ⟨ 𝑝, 𝑎 + \sqrt 𝑑 ⟩ ⟨ 𝑝, 𝑎 - \sqrt 𝑑 ⟩ = ⟨ 𝑝 2, 𝑝 (𝑎 \pm \sqrt 𝑑), 𝑎 2 - 𝑑 2 ⟩ \subseteq ⟨ 𝑝 ⟩

and on the other hand $𝔞$ contains both $𝑝 2$ and $𝑝 (𝑎 + \sqrt 𝑑) + 𝑝 (𝑎 - \sqrt 𝑑) = 2 𝑝 𝑎$ . Thus by Bézout's lemma we have $𝑝 = g c d {𝑝 2, 2 𝑝 𝑎} \in 𝔞$ , so $𝔞 = ⟨ 𝑝 ⟩$ .

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2022. Algebraic number theory course notes, ¶2.12, pp. 38–39. ↩