Inmathematics, thering of integers of analgebraic number field${\displaystyle K}$ is thering of allalgebraic integers contained in${\displaystyle K}$.^[1] An algebraic integer is aroot of amonic polynomial withintegercoefficients:${\displaystyle x^n+c_{n-1}{x}^{n-1}+\cdots +c_0}$.^[2] This ring is often denoted by${\displaystyle O_K}$ or${\displaystyle {\mathcal{O}}_K}$. Since anyinteger belongs to${\displaystyle K}$ and is anintegral element of${\displaystyle K}$, the ring${\displaystyle \mathbb{Z}}$ is always asubring of${\displaystyle O_K}$.

The ring of integers${\displaystyle \mathbb{Z}}$ is the simplest possible ring of integers.^[a] Namely,${\displaystyle \mathbb{Z}=O_{\mathbb{Q}}}$ where${\displaystyle \mathbb{Q}}$ is thefield ofrational numbers.^[3] And indeed, inalgebraic number theory the elements of${\displaystyle \mathbb{Z}}$ are often called the "rational integers" because of this.

The next simplest example is the ring ofGaussian integers${\displaystyle \mathbb{Z}[i]}$, consisting ofcomplex numbers whosereal and imaginary parts are integers. It is the ring of integers in the number field${\displaystyle \mathbb{Q}(i)}$ ofGaussian rationals, consisting of complex numbers whose real and imaginary parts are rational numbers. Like the rational integers,${\displaystyle \mathbb{Z}[i]}$ is aEuclidean domain.

The ring of integers of an algebraic number field is the unique maximalorder in the field. It is always aDedekind domain.^[4]

The ring of integersO_K is afinitely-generated${\displaystyle \mathbb{Z}}$-module. Indeed, it is afree${\displaystyle \mathbb{Z}}$-module, and thus has anintegral basis, that is abasisb₁, ...,b_n ∈ O_K of the${\displaystyle \mathbb{Q}}$-vector space K such that each element x inO_K can be uniquely represented as

${\displaystyle x=\sum _{i=1}^n{a}_i{b}_i,}$

with${\displaystyle a_i\in \mathbb{Z}}$.^[5] Therank n ofO_K as a free${\displaystyle \mathbb{Z}}$-module is equal to thedegree of K over${\displaystyle \mathbb{Q}}$.

A useful tool for computing the integral closure of the ring of integers in an algebraic field${\displaystyle K/\mathbb{Q}}$ is thediscriminant. IfK is of degreen over${\displaystyle \mathbb{Q}}$, and${\displaystyle {\alpha }_1,\dots ,{\alpha }_n\in {\mathcal{O}}_K}$ form a basis of${\displaystyle K}$ over${\displaystyle \mathbb{Q}}$, set${\displaystyle d={\mathrm{\Delta }}_{K/\mathbb{Q}}({\alpha }_1,\dots ,{\alpha }_n)}$. Then,${\displaystyle {\mathcal{O}}_K}$ is asubmodule of the${\displaystyle \mathbb{Z}}$-module spanned by${\displaystyle {\alpha }_1/d,\dots ,{\alpha }_n/d}$.^{[6]pg. 33} In fact, ifd is square-free, then${\displaystyle {\alpha }_1,\dots ,{\alpha }_n}$ forms an integral basis for${\displaystyle {\mathcal{O}}_K}$.^{[6]pg. 35}

Ifp is aprime,ζ is apthroot of unity and${\displaystyle K=\mathbb{Q}(\zeta )}$ is the correspondingcyclotomic field, then an integral basis of${\displaystyle {\mathcal{O}}_K=\mathbb{Z}[\zeta ]}$ is given by(1, ζ, ζ², ..., ζ^p−2).^[7]

If${\displaystyle d}$ is asquare-free integer and${\displaystyle K=\mathbb{Q}(\sqrt{d})}$ is the correspondingquadratic field, then${\displaystyle {\mathcal{O}}_K}$ is a ring ofquadratic integers and its integral basis is given by${\displaystyle \left(1,\frac{1+\sqrt{d}}{2}\right)}$ ifd ≡ 1 (mod 4) and by${\displaystyle (1,\sqrt{d})}$ ifd ≡ 2, 3 (mod 4).^[8] This can be found by computing theminimal polynomial of an arbitrary element${\displaystyle a+b\sqrt{d}\in \mathbb{Q}(\sqrt{d})}$ where${\displaystyle a,b\in \mathbb{Q}}$.

In a ring of integers, every element has a factorization intoirreducible elements, but the ring need not have the property ofunique factorization: for example, in the ring of integers${\displaystyle \mathbb{Z}[\sqrt{-5}]}$, the element 6 has two essentially different factorizations into irreducibles:^[4][9]

${\displaystyle 6=2\cdot 3=(1+\sqrt{-5})(1-\sqrt{-5}).}$

A ring of integers is always aDedekind domain, and so has unique factorization ofideals intoprime ideals.^[10]

Theunits of a ring of integersO_K is afinitely generated abelian group byDirichlet's unit theorem. Thetorsion subgroup consists of theroots of unity ofK. A set of torsion-free generators is called a set offundamental units.^[11]

One defines the ring of integers of anon-archimedean local fieldF as the set of all elements ofF with absolute value≤ 1; this is a ring because of the strong triangle inequality.^[12] IfF is the completion of an algebraic number field, its ring of integers is the completion of the latter's ring of integers. The ring of integers of an algebraic number field may be characterised as the elements which are integers in every non-archimedean completion.^[3]

For example, thep-adic integers${\displaystyle {\mathbb{Z}}_p}$ are the ring of integers of thep-adic numbers${\displaystyle {\mathbb{Q}}_p}$.

• Minimal polynomial (field theory)
• Integral closure – gives a technique for computing integral closures

• Ring theory
• Algebraic number theory

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