Quadratic field

Quadratic integers

The quadratic integers within a quadratic field 𝐾 =ℚ(√𝑑) where 𝑑 is a squarefree integer are ring

O𝐾={ℤ[√𝑑]𝑑≡2,3(mod4)ℤ[1+√𝑑2]𝑑≡1(mod4)

Proof

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

𝑚𝛼(𝑥)=𝑥2−2𝑎𝑥+𝑎2−𝑏2𝑑

By ^P1, we have 𝛼 ∈O𝐾 iff 𝑚𝛼(𝑥) ∈ℤ[𝑥], which is precisely the case when 2𝑎,𝑎2 −𝑏2𝑑 ∈ℤ. It follows (2𝑏)2𝑑 ∈ℤ, so since 𝑑 is squarefree 2𝑏 ∈ℤ. Letting 𝑟 =2𝑎, 𝑠 =2𝑏, we have 𝛼 ∈O𝐾 iff 𝑟,𝑠 ∈ℤ and 𝑟2 −𝑑𝑠2 ≡40. Since 𝑑 is squarefree it follows 𝑑 ≢40, so we need only consider the cases

1. If 𝑑 ≡41 then 0 ≡4𝑟2 −𝑑𝑠2 ≡4𝑟2 −𝑠2 which holds iff 𝑟 ≡2𝑠;
2. If 𝑑 ≡42 then 0 ≡4𝑟2 −𝑑𝑠2 ≡4𝑟 +2𝑠2 which holds iff 𝑟,𝑠 ≡20;
3. If 𝑑 ≡43 then 0 ≡4𝑟2 −𝑑𝑠2 ≡4𝑟2 +𝑠2 which holds iff 𝑟,𝑠 ≡20.

It follows that the general expression for an algebraic integer 𝛼 ∈O𝐾 is

• 𝛼 =𝑝 +𝑞√𝑑 if 𝑑 ≢41
• 𝛼 =𝑝 +𝑞1+√𝑑2 if 𝑑 ≡41

where 𝑝,𝑞 ∈ℤ, whence the above.

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

Properties

Let 𝛼 ∈O𝐾 be a (proper) quadratic integer with minimal polynomial 𝑥2 +𝑎𝑥 +𝑏

1. The discriminant is Δ𝐾:ℚ(𝛼) =𝑎2 −4𝑏.
2. It follows that

Δ𝐾={4𝑑𝑑≡42,3𝑑𝑑≡41

Proof of 1

See Discriminant of an algebraic integer.

Prime ideals

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

1. If (𝑑𝑝) =1 then 𝐾 :ℚ is unramified at ⟨𝑝⟩ =⟨𝑝,𝑎 +√𝑑⟩⟨𝑝,𝑎 −√𝑑⟩, where 𝑎2 ≡𝑝𝑑.
2. If (𝑑𝑝) = −1 then 𝐾 :ℚ is inert at 𝑝.¹

Proof

First suppose 𝑎2 ≡𝑝𝑑 for 𝑎 ≠0. Then

𝔞=⟨𝑝,𝑎+√𝑑⟩⟨𝑝,𝑎−√𝑑⟩=⟨𝑝2,𝑝(𝑎±√𝑑),𝑎2−𝑑2⟩⊆⟨𝑝⟩

and on the other hand 𝔞 contains both 𝑝2 and 𝑝(𝑎 +√𝑑) +𝑝(𝑎 −√𝑑) =2𝑝𝑎. Thus by Bézout’s lemma we have 𝑝 =gcd{𝑝2,2𝑝𝑎} ∈𝔞, so 𝔞 =⟨𝑝⟩.

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Footnotes

1. 2022. Algebraic number theory course notes, ¶2.12, pp. 38–39. ↩