This article is about the ring of complex numbers integral over${\displaystyle \mathbb{Z}}$. For the general notion of algebraic integer, seeIntegrality.

Inalgebraic number theory, analgebraic integer is acomplex number that isintegral over theintegers. That is, an algebraic integer is a complexroot of somemonic polynomial (apolynomial whoseleading coefficient is 1) whose coefficients are integers. The set of all algebraic integersA is closed under addition, subtraction and multiplication and therefore is acommutativesubring of the complex numbers.

Thering of integers of anumber fieldK, denoted byO_K, is theintersection ofK andA: it can also be characterized as the maximalorder of thefieldK. Each algebraic integer belongs to the ring of integers of some number field. A numberα is an algebraic integerif and only if the ring${\displaystyle \mathbb{Z}[\alpha ]}$ isfinitely generated as anabelian group, which is to say, as a${\displaystyle \mathbb{Z}}$-module.

The following are equivalent definitions of an algebraic integer. LetK be anumber field (i.e., afinite extension of${\displaystyle \mathbb{Q}}$, the field ofrational numbers), in other words,${\displaystyle K=\mathbb{Q}(\theta )}$ for somealgebraic number${\displaystyle \theta \in \mathbb{C}}$ by theprimitive element theorem.

• α ∈K is an algebraic integer if there exists a monic polynomial${\displaystyle f(x)\in \mathbb{Z}[x]}$ such thatf(α) = 0.
• α ∈K is an algebraic integer if theminimal monic polynomial ofα over${\displaystyle \mathbb{Q}}$ is in${\displaystyle \mathbb{Z}[x]}$.
• α ∈K is an algebraic integer if${\displaystyle \mathbb{Z}[\alpha ]}$ is a finitely generated${\displaystyle \mathbb{Z}}$-module.
• α ∈K is an algebraic integer if there exists a non-zero finitely generated${\displaystyle \mathbb{Z}}$-submodule${\displaystyle M\subset \mathbb{C}}$ such thatαM ⊆M.

Algebraic integers are a special case ofintegral elements of a ring extension. In particular, an algebraic integer is an integral element of a finite extension${\displaystyle K/\mathbb{Q}}$.

Note that ifP(x) is aprimitive polynomial that has integer coefficients but is not monic, andP isirreducible over${\displaystyle \mathbb{Q}}$, then none of the roots ofP are algebraic integers (butarealgebraic numbers). Hereprimitive is used in the sense that thehighest common factor of the coefficients ofP is 1, which is weaker than requiring the coefficients to be pairwise relatively prime.

• The only algebraic integers that are found in the set of rational numbers are the integers. In other words, the intersection of${\displaystyle \mathbb{Q}}$ andA is exactly${\displaystyle \mathbb{Z}}$. The rational number⁠a/b⁠ is not an algebraic integer unlessbdividesa. The leading coefficient of the polynomialbx −a is the integerb.
• Thesquare root${\displaystyle \sqrt{n}}$ of a nonnegative integern is an algebraic integer, but isirrational unlessn is aperfect square.
• Ifd is asquare-free integer then theextension${\displaystyle K=\mathbb{Q}(\sqrt{d})}$ is aquadratic field of rational numbers. The ring of algebraic integersO_K contains${\displaystyle \sqrt{d}}$ since this is a root of the monic polynomialx² −d. Moreover, ifd ≡ 1mod 4, then the element$\frac{1}{2}(1+\sqrt{d})$ is also an algebraic integer. It satisfies the polynomialx² −x +⁠1/4⁠(1 −d) where theconstant term⁠1/4⁠(1 −d) is an integer. The full ring of integers is generated by${\displaystyle \sqrt{d}}$ or$\frac{1}{2}(1+\sqrt{d})$ respectively. SeeQuadratic integer for more.
• The ring of integers of the field${\displaystyle F=\mathbb{Q}[\alpha ]}$,α =³√m, has the followingintegral basis, writingm =hk² for twosquare-freecoprime integersh andk:^[1]
• Ifζ_n is aprimitiventhroot of unity, then the ring of integers of thecyclotomic field${\displaystyle \mathbb{Q}({\zeta }_n)}$ is precisely${\displaystyle \mathbb{Z}[{\zeta }_n]}$.
• Ifα is an algebraic integer thenβ =ⁿ√α is another algebraic integer. A polynomial forβ is obtained by substitutingxⁿ in the polynomial forα.

For anyα, thering extension (in the sense that is equivalent tofield extension) of the integers byα, denoted by${\displaystyle \mathbb{Z}[\alpha ]\equiv \left\{\sum _{i=0}^n{\alpha }^i{z}_i|z_i\in \mathbb{Z},n\in \mathbb{Z}\right\}}$, isfinitely generated if and only ifα is an algebraic integer.

The proof is analogous to that of thecorresponding fact regardingalgebraic numbers, with${\displaystyle \mathbb{Q}}$ there replaced by${\displaystyle \mathbb{Z}}$ here, and the notion offield extension degree replaced by finite generation (using the fact that${\displaystyle \mathbb{Z}}$ is finitely generated itself); the only required change is that only non-negative powers ofα are involved in the proof.

The analogy is possible because both algebraic integers and algebraic numbers are defined as roots of monic polynomials over either${\displaystyle \mathbb{Z}}$ or${\displaystyle \mathbb{Q}}$, respectively.

The sum, difference and product of two algebraic integers is an algebraic integer. In general their quotient is not. Thus the algebraic integers form aring.

This can be shown analogously tothe corresponding proof foralgebraic numbers, using the integers${\displaystyle \mathbb{Z}}$ instead of the rationals${\displaystyle \mathbb{Q}}$.

One may also construct explicitly the monic polynomial involved, which is generally of higherdegree than those of the original algebraic integers, by takingresultants and factoring. For example, ifx² −x − 1 = 0,y³ −y − 1 = 0 andz =xy, then eliminatingx andy fromz −xy = 0 and the polynomials satisfied byx andy using the resultant givesz⁶ − 3z⁴ − 4z³ +z² +z − 1 = 0, which is irreducible, and is the monic equation satisfied by the product. (To see that thexy is a root of thex-resultant ofz −xy andx² −x − 1, one might use the fact that the resultant is contained in theideal generated by its two input polynomials.)

Every root of a monic polynomial whose coefficients are algebraic integers is itself an algebraic integer. In other words, the algebraic integers form a ring that isintegrally closed in any of its extensions.

Again, the proof is analogous tothe corresponding proof foralgebraic numbers beingalgebraically closed.

• Any number constructible out of the integers with roots, addition, and multiplication is an algebraic integer; but not all algebraic integers are so constructible: in a naïve sense, most roots of irreduciblequintics are not. This is theAbel–Ruffini theorem.
• The ring of algebraic integers is aBézout domain, as a consequence of theprincipal ideal theorem.
• If the monic polynomial associated with an algebraic integer has constant term 1 or −1, then thereciprocal of that algebraic integer is also an algebraic integer, and each is aunit, an element of thegroup of units of the ring of algebraic integers.
• Ifx is an algebraic number thena_nx is an algebraic integer, wherex satisfies a polynomialp(x) with integer coefficients and wherea_nxⁿ is the highest-degree term ofp(x). The valuey =a_nx is an algebraic integer because it is a root ofq(y) =a^{n − 1}
  _np(y/a_n), whereq(y) is a monic polynomial with integer coefficients.
• Ifx is an algebraic number then it can be written as the ratio of an algebraic integer to a non-zero algebraic integer. In fact, the denominator can always be chosen to be a positive integer. The ratio is|a_n|x / |a_n|, wherex satisfies a polynomialp(x) with integer coefficients and wherea_nxⁿ is the highest-degree term ofp(x).
• The only rational algebraic integers are the integers. That is, ifx is an algebraic integer and${\displaystyle x\in \mathbb{Q}}$ then${\displaystyle x\in \mathbb{Z}}$. This is a direct result of therational root theorem for the case of a monic polynomial.

• Gaussian integer
• Eisenstein integer
• Root of unity
• Dirichlet's unit theorem
• Fundamental units

• Algebraic numbers
• Integers

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