QuadraticOrder

The quadratic order of discriminant $D = f^2 D_K$: $\mathbb{Z} + f\,\mathcal{O}_K$, the subring of conductor $f$ in the ring of integers of $\mathbb{Q}(\sqrt D)$. $\mathbb{Z}[\sqrt{d}]$ is $\mathrm{QuadraticOrder}(4d)$.

SageMathQuadraticField(d).order_of_conductor(f)↗

QuadraticOrder(D)the order of discriminant $D$, for $D \equiv 0$ or $1 \pmod 4$ not a square; the field's own discriminant gives its ring of integers

Typed signatures

Type	Package	Applies
(integer) -> set<quadratic_integer>	number-theory	always

Domain: Number theory

Details

• A value for the Over option, as QuadraticIntegers is: IsPrime(x, Over -> QuadraticOrder(-12)) asks whether $x$ is prime in $\mathbb{Z}[\sqrt{-3}]$. An element is written as any number of the field is, and belongs to the order when its coordinate on $\omega_K$ is a multiple of $f$
• Orders of conductor $f > 1$ have fewer units ($\mathbb{Z}[\sqrt{-3}]$ has only $\pm 1$, against six in $\mathbb{Z}[\omega]$) and never factor uniquely, since they are not integrally closed: $4 = 2 \cdot 2 = -(1 + \sqrt{-3})(-1 + \sqrt{-3})$ in $\mathbb{Z}[\sqrt{-3}]$, so FactorInteger stays unevaluated
• At a prime $p$ dividing $f$ the ideals are not all invertible, and an element is prime exactly when $\mathcal{O}/(x)$ is a field, irreducible when nothing of a smaller norm divides it
• QuadraticOrder(D) for the field's discriminant ($D_K = d$ or $4d$) is QuadraticIntegers(d)

Examples

Basic #

2 is not prime in ℤ[√−3]: it divides (1 + √−3)(1 − √−3) = 4 without dividing either, though it is prime in ℤ[ω]

live ↗

In$\mathrm{IsPrime}(2, \operatorname{KeyValuePair}(\mathrm{Over}, \mathrm{QuadraticOrder}(-12)))$Out$\mathrm{False}$

in the maximal order ℤ[ω], 2 is inert and prime

live ↗

In$\mathrm{IsPrime}(2, \operatorname{KeyValuePair}(\mathrm{Over}, \mathrm{QuadraticIntegers}(-3)))$Out$\mathrm{True}$

4 = 2 · 2 = −(1 + √−3)(−1 + √−3): an order of conductor 2 has no unique factorization

live ↗

In$\mathrm{FactorInteger}(4, \operatorname{KeyValuePair}(\mathrm{Over}, \mathrm{QuadraticOrder}(-12)))$Out$\mathrm{FactorInteger}(4, (\mathrm{Over},\mathrm{QuadraticOrder}(-12)))$

7 = (1 + 2√2)(−1 + 2√2) still splits in ℤ[√8]

live ↗

In$\mathrm{IsPrime}(7, \operatorname{KeyValuePair}(\mathrm{Over}, \mathrm{QuadraticOrder}(32)))$Out$\mathrm{False}$

About

Examples	4
Crosswalk	SageMath

Implementation

mappedwolframpackages/oracle/src/mappings.ts

Wolfram has no quadratic orders: its number-field heads take an algebraic number, and work in the maximal order.

See also: QuadraticIntegers, QuadraticInteger, IsPrime