$ℚ (\sqrt - 14)$

Consider the monogenic imaginary quadratic field $𝐾 = ℚ (𝛼)$ where $𝛼 = \sqrt - 14$ . #m/thm/num/alg

Sage

K.<α> = QuadraticField(-14)

Discriminant

By Discriminant of an algebraic integer,

Δ 𝐾 = - 56 .

Group of units

By ^P1,

O \times 𝐾 = {1, - 1} .

Class group

Minkowski's bound is given by

𝑀 𝐾 = 4 \sqrt 14 𝜋 < 5,

so applying Kummer's factorization theorem

$𝑝$	$𝑥 2 + 14 m o d 𝑝$	$⟨ 𝑝 ⟩$	norms
$2$	$𝑥 2$	$𝔭 22$	$2$
$3$	$(𝑥 + 1) (𝑥 - 1)$	$𝔭 3 𝔭' 3$	$3, 3$

Clearly no algebraic integers can have these norms, so we can be satisfied that these are not principal. Since $𝔭 - 1 3 = 𝔭' 3$ , the ideal class group is generated by ${[𝔭 2], [𝔭 3]}$ . Some algebraic integers of small field norm are

$𝑡$	$N 𝐾 : ℚ (𝛼 + 𝑡)$
$\pm 1$	$3 \cdot 5$
$\pm 2$	$2 \cdot 32$
$\pm 3$	$23$
from which we derive the relation

𝔭 2 𝔭 23 = ⟨ 2, 𝛼 ⟩ ⟨ 3, 𝛼 + 1 ⟩ 2 = ⟨ 𝜗 - 2 ⟩ \sim ⟨ 1 ⟩

whence $𝔭 23 \sim 𝔭 2$ , so $𝔭 43 \sim ⟨ 1 ⟩$ . Therefore

C l 𝐾 = ⟨ [𝔭 3] ⟩ ≅ C 4 .

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

ℚ(√−14)

Discriminant

Group of units

Class group

$ℚ (\sqrt - 14)$