Real quadratic field

ℚ(√223)

Consider the monogenic Real quadratic field 𝐾 =ℚ(𝛼) where 𝛼 =√223. #m/thm/num/alg

K.<α> = QuadraticField(223)

Discriminant

By Discriminant of an algebraic integer, we have

Group of units

Take the reduced element with simple continued fraction

whence

is the fundamental unit, and we have

Class group

Minkowski's bound is given by

so applying Kummer's factorization theorem:

Some algebraic integers of small field norm are

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

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

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

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.

#state/develop | #lang/en | #SemBr