Imaginary quadratic field

An imaginary quadratic field $𝐾 = ℚ (\sqrt 𝑑)$ is a quadratic field with $𝑑 < 0$ , #m/def/num/alg and hence signature $(𝑟 1, 𝑟 2) = (0, 1)$ .

Properties

The group of units $O \times 𝐾$ is ${1, - 1}$ except for $𝑑 = - 1$ , giving ring of integers $ℤ [𝑖]$ , or $𝑑 = - 3$ , giving $ℚ (\sqrt - 3)$ .

Proof of 1.

First consider the monogenic case, i.e. $𝑑 ≢ 4 1$ and hence $O 𝐾 = ℤ [\sqrt 𝑑]$ . Since the field norm of $𝛼 = 𝑎 + 𝑏 \sqrt 𝑑$

N 𝐾 : ℚ (𝛼) = 𝑎 2 - 𝑏 2 𝑑

where both terms are positive, the only ways to get $N 𝐾 : ℚ (𝛼) = 1$ are if

This exceptional case is $ℤ [𝑖]$ .

For $𝑑 \equiv 4 1$ , we have $O 𝐾 = ℤ [1 + \sqrt 𝑑 2]$ . The field norm of a generic $𝛼 = 𝑎 + 𝑏 1 + \sqrt 𝑑 2$ we have

N 𝐾 : ℚ (𝛼) = (𝑎 + 𝑏 2) 2 - 𝑏 2 𝑑 4

where both terms are positive, the only ways to get $N 𝐾 : ℚ (𝛼) = 1$ are if

This exceptional case is $ℚ (\sqrt - 3)$ .

#state/tidy | #lang/en | #SemBr