Real quadratic field

ℚ(√223)

Consider the monogenic Real quadratic field 𝐾 =ℚ(𝛼) where 𝛼 =√223. alg

Sage

K.<α> = QuadraticField(223)

Discriminant

By Discriminant of an algebraic integer, we have

Δ𝐾=22⋅223

Group of units

Take the reduced element with simple continued fraction

𝜗=1𝛼−14=[――――――1;13,1,28]

whence

𝜖=𝑞2+𝑞3𝜗=14+405𝜗=15𝛼+224

is the fundamental unit, and we have

O×𝐾={±𝜖𝑚:𝑚∈ℤ}

Class group

Minkowski’s bound is given by

𝑀𝐾=√233<15,

so applying Kummer’s factorization theorem:

𝑝	𝑥2 −223mod𝑝	⟨𝑝⟩	norms
2	(𝑥 −1)2	𝔭22	2
3	(𝑥 +1)(𝑥 −1)	𝔭3𝔭′3	3,3
5	𝑥2 −3	⟨5⟩	52
7	𝑥2 −6	⟨7⟩	72
11	(𝑥 +5)(𝑥 −5)	𝔭11𝔭′11	11,11

Some algebraic integers of small field norm are

𝑡	N𝐾:ℚ⁡(𝛼 +𝑡)
±14	−33
±15	2
±16	3 ⋅11

so

• from 𝑡 =15, we see 𝔭2 =⟨𝛼 +15⟩ ∼⟨1⟩;
• from 𝑡 =16, we see 𝔭3𝔭11 =⟨𝛼 +16⟩ ∼⟨1⟩;
• from 𝑡 = −14, wee see 𝔭33 =⟨𝛼 −14⟩ ∼⟨1⟩.

Therefore the ideal class group Cl⁡𝐾 =⟨[𝔭3]⟩ is cyclic of order 1 or 3. We show 𝔭3 cannot be principal, whence Cl⁡𝐾 ≅C3.

Suppose towards contradiction 𝔭3 =⟨𝛽⟩ for some 𝛽 ∈O𝐾, so |N𝐾:ℚ⁡(𝛽)| =3. Then ⟨𝛽3⟩ =⟨𝛼 −14⟩, so 𝛽3 =𝑢(𝛼 −14) for some 𝑢 ∈O×𝐾. It follows

𝛽3=±𝜖𝑘(𝛼−14),𝑘∈{0,1,2},

Thus 𝛽3 =𝜖𝑘(𝛼 −14) for 𝑘 ∈{0,1,2}, where by direct calculation

𝛽3∈{𝛼−14,−434𝛼+6481,−14𝛼−209}.

Now suppose 𝛽 =𝑎 +𝑏𝛼, where N𝐾:ℚ⁡(𝛽) =∣𝑎2−223𝑏2∣ =3, so both 𝑎,𝑏 ≠0. We have

𝛽3=𝑎(𝑎2+669𝑏2)+𝑏(3𝑎2+223𝑏2)𝛼

where the absolute value of the coëfficient of 𝛼 must be at least 3 +223 =226, leaving only the case 𝛽3 = −434𝛼 +6481. Thus 𝑎(𝑎2 +669𝑏2) =6481, a prime number. Thus 𝑎, which must be positive, must equal 1, so 669𝑏2 =6480 for some 𝑏 ∈ℤ, which is impossible.

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