Ring of integers (${\mathcal{O}}_K$)

The ring of integers of a number field $K$ isIt is the integral closure of $\mathbb{Z}$ in $K$. Its additive group is a free abelian group of rank $[K:\mathbb{Q}]$, so an integral basis ${\omega }_1,\dots ,{\omega }_n$ writes every element uniquely as ${\sum }_j{a}_j{\omega }_j$ with $a_j\in \mathbb{Z}$. Its field of fractions is $K$.^[ref]

For $K=\mathbb{Q}$, this gives ${\mathcal{O}}_K=\mathbb{Z}$; for $K=\mathbb{Q}(\sqrt{2})$, it gives $\mathbb{Z}[\sqrt{2}]$. The Gaussian integers give the corresponding example inside $\mathbb{Q}(i)$. The ring of integers is determined by the field, and need not equal $\mathbb{Z}[\theta ]$ for a chosen algebraic integer generator: in $\mathbb{Q}(\sqrt{5})$, the larger ring $\mathbb{Z}[(1+\sqrt{5})/2]$ contains $\mathbb{Z}[\sqrt{5}]$ with index two.

This is the setting for divisibility, global units, and prime ideal factorization. A ring of integers is a Dedekind domain, so every nonzero proper ideal factors uniquely into nonzero prime ideals, even when unique factorization of elements fails.^[ref]

A global unit and its inverse are algebraic integers in the number field. Equivalently its valuation is zero at every nonarchimedean prime. Global units embed in local unit groups; their p-adic closures can be smaller than the full local groups. This discrepancy is measured by the class-field unit sequence.

An order in a number field $K$ is a subring $A\subseteq {\mathcal{O}}_K$ containing $1$ and having full rank $[K:\mathbb{Q}]$ as an additive abelian group. Its finite index in the ring of integers of a number field is related to its discriminant by the discriminant-index formula for an integral lattice.

Every nonzero ideal of the ring of integers of a number field factors uniquely as a finite product of nonzero prime ideals. Thus each $x\in K^{\times }$ has integer valuations ${\mathrm{ord}}_{\mathfrak{p}}(x)$, almost all zero, determined by the factorization of its principal ideal.

If $\mathfrak{p}$ is a prime ideal of ${\mathcal{O}}_K$ above a rational prime $p$, its ramification index is the exponent $e_{\mathfrak{p}}$ of $\mathfrak{p}$ in the factorization of $(p)$. Equivalently, for the normalized ideal valuation, $e_{\mathfrak{p}}=v_{\mathfrak{p}}(p)$.

Let $B$ be the integral closure of a Dedekind domain $R$ in a finite separable extension $L/K$ of degree $n$. For a nonzero prime ideal $\mathfrak{p}$ of $R$, writeand let $f_{\mathfrak{q}/\mathfrak{p}}=[B/\mathfrak{q}:R/\mathfrak{p}]$. Then

For a prime number $p$ and $\alpha =\sqrt[p]{p}$, the polynomial $X^p-p$ is Eisenstein at $p$, andThe discriminant-index formula for an integral lattice reduces the proof to the prime $p$. At the unique prime ideal above $p$, one has $v(\alpha )=1$ and $v(p)=p$, which rules out an integral element $g(\alpha )/p$ with $g\notin p\mathbb{Z}[X]$.

Let $d\ne 0,1$ be a square-free integer, and let $K=\mathbb{Q}(\sqrt{d})$ be the associated quadratic field. ThenThe pair $1,{\omega }_d$ is an integral basis. The field discriminant is $d$ in the first case and $4d$ in the second. This statement requires $d$ to be a square-free integer: for example, $\mathbb{Q}(\sqrt{8})=\mathbb{Q}(\sqrt{2})$ but $\mathbb{Z}[\sqrt{8}]$ omits $\sqrt{2}$.^[ref]

For $d=2$, the integral basis is $1,\sqrt{2}$, and the field discriminant is $8$. For $d=5$, it is $1,\varphi$ with $\varphi =(1+\sqrt{5})/2$, and the field discriminant is $5$. The identity ${\varphi }^2=\varphi +1$ makes $\mathbb{Z}[\varphi ]$ closed under multiplication. Since $\sqrt{5}=2\varphi -1$, the smaller order in a number field $\mathbb{Z}[\sqrt{5}]$ consists precisely of $a+b\varphi$ with even $b$. This computes its index two and shows concretely why adjoining a square root does not always supply every algebraic integer.

In this second integral basis, the field norm isIn particular, $N(\varphi )=-1$ and ${\varphi }^{-1}=\varphi -1$: the extra algebraic integer is also a unit. The units of the quadratic field Q square root of five describe all its powers.

Since $2\not\equiv 1(\mathrm{mod}4)$, the ring of integers of $\mathbb{Q}(\sqrt{2})$ iswith integral basis $1,\sqrt{2}$ and field discriminant eight. Multiplication is ordinary polynomial multiplication followed by $(\sqrt{2}{)}^2=2$:For example, $(1+\sqrt{2}{)}^2=3+2\sqrt{2}$ and $(1+\sqrt{2}{)}^3=7+5\sqrt{2}$. The integer coefficients stay integral, as required for a ring.

Under the two field embeddings, the element $a+b\sqrt{2}$ becomes the point $(a+b\sqrt{2},a-b\sqrt{2})$. The integral basis vectors become $(1,1)$ and $(\sqrt{2},-\sqrt{2})$, giving a Minkowski embedding of a real quadratic field. Their determinant has absolute value $2\sqrt{2}=\sqrt{8}$, consistent with the field discriminant.

The ring $\mathbb{Z}[\sqrt{2}]$ is a Euclidean domain for the absolute field norm. Approximate both rational coefficients of a quotient by nearest integers. If the errors are $u,w\in [-1/2,1/2]$, then $|u^2-2w^2|\le 1/2<1$, giving a remainder of strictly smaller absolute norm. In particular this ring is a principal ideal domain and has trivial ideal class group.