Proper ideals of quadratic orders are invertible #
In an order O of a quadratic number field K, a fractional ideal is proper (its multiplier ring
is exactly O) if and only if it is invertible. In higher degree only the implication from
invertible to proper survives; the cubic order ℤ + 2ℤ∛2 + 2ℤ∛4 has a proper noninvertible ideal.
So for quadratic orders the Picard group Pic O sees every proper ideal class, which is what
lets the ring class group of a quadratic order be described by proper ideals or, equivalently,
by primitive binary quadratic forms.
The proof is Cox's. Write a nonzero fractional ideal as a lattice I = ℤα + ℤβ and put
τ = β / α, a root of a primitive integral quadratic a τ² - b τ - e = 0 (gcd (a, b, e) = 1).
Then a τ preserves I, so it lies in O when I is proper. With τ' = b / a - τ the
conjugate of τ, the elements a / α and a τ' / α carry I into O, and the products of
α, β ∈ I with them are a, b and -e up to integer combinations. Since a, b and e
generate the unit ideal of ℤ, the product of I with 1 / I = (O : I) contains 1, so it is
the unit ideal.
Main results #
In the namespace TauCeti.GlobalNumberFields.NumberFieldOrder:
IsProperFractionalIdeal.isUnit_of_finrank_eq_two: a proper fractional ideal of an order in a quadratic field is invertible.isProperFractionalIdeal_iff_isUnit_of_finrank_eq_two: for an order in a quadratic field, proper and invertible fractional ideals coincide.
References #
- D. A. Cox, Primes of the Form x² + ny², §7, Lemma 7.5.
- G. S. Kopp and J. C. Lagarias, Class Field Theory for Orders of Number Fields, §2.
A proper fractional ideal of a quadratic order is invertible (Cox, Lemma 7.5).
Proper and invertible fractional ideals coincide in a quadratic order. In higher degree only invertible ideals are guaranteed to be proper.