Orders and Algebraic Integers

Quadratic Integers works in one ring per field: $\mathcal{O}_d$, all of $\mathbb{Q}(\sqrt d)$'s integers. A field has other rings with the same fraction field, its orders, and fields of higher degree have rings of integers that no single generator reaches. This page goes through both.

Orders

An order of a number field $K$ is a subring that is a lattice of full rank: finitely generated over $\mathbb{Z}$, spanning $K$ over $\mathbb{Q}$. Every order lies inside the ring of integers $\mathcal{O}_K$, the maximal order. In a quadratic field each order is $\mathbb{Z} + f\,\mathcal{O}_K$ for one conductor $f$, and is named by its discriminant $D = f^2 D_K$: QuadraticOrder(D). The ring $\mathbb{Z}[\sqrt{-3}]$ is the order of conductor 2 in $\mathbb{Z}[\omega]$:

An order shares its field's arithmetic away from the conductor, and loses it at the conductor. In $\mathbb{Z}[\omega]$ the prime 2 is inert, and so prime:

In $\mathbb{Z}[\sqrt{-3}]$ it is not, though nothing smaller divides it: it divides $(1 + \sqrt{-3})(1 - \sqrt{-3}) = 4$ without dividing either factor.

So 4 has two factorizations into irreducibles, $2 \cdot 2$ and $(1 + \sqrt{-3})(1 - \sqrt{-3})$. An order of conductor $f > 1$ is never integrally closed, and never factors uniquely. The units shrink too: $\mathbb{Z}[\omega]$ has six, $\mathbb{Z}[\sqrt{-3}]$ only $\pm 1$.

Past degree two

A number field of degree $n$ is $\mathbb{Q}(\theta)$ for $\theta$ a root of an irreducible polynomial of degree $n$, the minimal polynomial. The number $\theta$ can be a radical, $i$, the golden ratio, or a root named by its polynomial, PolynomialRoot(p, k) (Wolfram's Root):

The field's real and complex embeddings are its signature $(r_1, r_2)$, with $r_1 + 2r_2 = n$. $\mathbb{Q}(\sqrt[3]{2})$ has one real cube root of 2 and a complex pair:

By Dirichlet's unit theorem the units of $\mathcal{O}_K$ have rank $r_1 + r_2 - 1$. That rank is 1 here, as in a real quadratic field.

When ℤ[θ] is not enough

$\mathbb{Z}[\theta]$ is always an order, AlgebraicOrder(θ). It need not be the maximal one. In $\mathbb{Q}(\sqrt[3]{19})$, the number

$$\frac{1 + \theta + \theta^2}{3}, \qquad \theta = \sqrt[3]{19},$$

is an algebraic integer, though its coordinates aren't integers:

The ring of integers comes with a basis, written on $\theta$'s powers as Wolfram writes it:

The index $[\mathcal{O}_K : \mathbb{Z}[\theta]]$ is 3 here. The discriminants differ by its square: $\mathrm{disc}(x^3 - 19) = -27 \cdot 19^2$, and the field's is a ninth of that:

How the ring of integers is found: an order $\mathcal{O}$ fails to be maximal at a prime $p$ exactly when some element of $\frac1p\mathcal{O}$ outside $\mathcal{O}$ is integral. Only primes with $p^2$ dividing the discriminant can fail. So $\mathbb{Z}[\theta]$ is enlarged at each such $p$, one integral element at a time, until none is left.

Dedekind's cubic

Some fields have no $\theta$ with $\mathbb{Z}[\theta] = \mathcal{O}_K$ at all. Dedekind's example is the field of a root of $x^3 - x^2 - 2x - 8$. Its ring of integers is $\mathbb{Z} + \mathbb{Z}\theta + \mathbb{Z}\frac{\theta + \theta^2}{2}$:

The reason is the prime 2. It splits into three distinct primes, each of norm 2, so $\mathcal{O}_K/2\mathcal{O}_K \cong \mathbb{F}_2^3$. In a ring $\mathbb{Z}[\alpha]$ that would need $\alpha$'s minimal polynomial to have three distinct roots mod 2, and $\mathbb{F}_2$ has only two elements. So 2 is far from prime:

while 3, where $x^3 - x^2 - 2x - 8$ stays irreducible, is inert:

Primality here is that of the ideal: $(\alpha)$ is prime when $\mathcal{O}/(\alpha)$ is a field. With $|N(\alpha)| = p^f$ that comes down to linear algebra over $\mathbb{F}_p$. $p$ must lie in $(\alpha)$, and Frobenius $x \mapsto x^p$ must be injective on the quotient and fix only $\mathbb{F}_p$. The same test works in any order, at the conductor too.

What's not here yet

Class groups and units in degree 3 and up, and fields given by two generators ($\sqrt 2 + \sqrt 3$), stay unevaluated for now. So does any polynomial whose irreducibility no prime certifies: $x^4 + 1$ splits mod every prime, though not over $\mathbb{Q}$.