# Algebraic integer

In algebraic number theory, an algebraic integer is a complex number that is integral over the integers. That is, an algebraic integer is a complex root of some monic polynomial (a polynomial whose leading coefficient is 1) whose coefficients are integers. The set of all algebraic integers A is closed under addition, subtraction and multiplication and therefore is a commutative subring of the complex numbers.
The ring of integers of a number field K, denoted by O$_K$, is the intersection of K and A: it can also be characterised as the maximal order of the field K. Each algebraic integer belongs to the ring of integers of some number field. A number α is an algebraic integer if and only if the ring $\mathbb {Z} [\alpha ]$ is finitely generated as an abelian group, which is to say, as a $\mathbb {Z}$-module.

## Definitions

The following are equivalent definitions of an algebraic integer. Let K be a number field (i.e., a finite extension of $\mathbb {Q}$, the field of rational numbers), in other words, $K=\mathbb {Q} (\theta )$ for some algebraic number $\theta \in \mathbb {C}$ by the primitive element theorem.
* α ∈ K is an algebraic integer if there exists a monic polynomial $f(x)\in \mathbb {Z} [x]$ such that f(α) = 0.
* α ∈ K is an algebraic integer if the minimal monic polynomial of α over $\mathbb {Q}$ is in $\mathbb {Z} [x]$.
* α ∈ K is an algebraic integer if $\mathbb {Z} [\alpha ]$ is a finitely generated $\mathbb {Z}$-module.
* α ∈ K is an algebraic integer if there exists a non-zero finitely generated $\mathbb {Z}$-submodule $M\subset \mathbb {C}$ such that αM ⊆ M.
Algebraic integers are a special case of integral elements of a ring extension. In particular, an algebraic integer is an integral element of a finite extension $K/\mathbb {Q}$.

## Examples

* The only algebraic integers that are found in the set of rational numbers are the integers. In other words, the intersection of $\mathbb {Q}$ and A is exactly $\mathbb {Z}$. The rational number ⁠a/b⁠ is not an algebraic integer unless b divides a. The leading coefficient of the polynomial bx − a is the integer b.
* The square root ${\sqrt {n}}$ of a nonnegative integer n is an algebraic integer, but is irrational unless n is a perfect square.
* If d is a square-free integer then the extension $K=\mathbb {Q} ({\sqrt {d}}\,)$ is a quadratic field of rational numbers. The ring of algebraic integers O$_K$ contains ${\sqrt {d}}$ since this is a root of the monic polynomial x$^2$ − d. Moreover, if d ≡ 1 mod 4, then the element 1/2 (1+√d){\textstyle {\frac {1}{2}}(1+{\sqrt {d}}\, )} is also an algebraic integer. It satisfies the polynomial x$^2$ − x + ⁠1/4⁠(1 − d) where the constant term ⁠1/4⁠(1 − d) is an integer. The full ring of integers is generated by ${\sqrt {d}}$ or 1/2 (1+√d){\textstyle {\frac {1}{2}}(1+{\sqrt {d}}\, )} respectively. See Quadratic integer for more.
* The ring of integers of the field $F=\mathbb {Q} [\alpha ]$, α = $^3$√m, has the following integral basis, writing m = hk$^2$ for two square-free coprime integers h and k: ${\begin{cases}1,\alpha ,{\dfrac {\alpha ^{2}\pm k^{2}\alpha +k^{2}}{3k}}&m\equiv \pm 1{\bmod {9}}\\1,\alpha ,{\dfrac {\alpha ^{2}}{k}}&{\text{otherwise}}\end{cases}}$
* If ζ$_n$ is a primitive nth root of unity, then the ring of integers of the cyclotomic field $\mathbb {Q} (\zeta _{n})$ is precisely $\mathbb {Z} [\zeta _{n}]$.
* If α is an algebraic integer then β = $^n$√α is another algebraic integer. A polynomial for β is obtained by substituting x$^n$ in the polynomial for α.

## Non-example

* If P(x) is a primitive polynomial that has integer coefficients but is not monic, and P is irreducible over $\mathbb {Q}$, then none of the roots of P are algebraic integers (but are algebraic numbers). Here primitive is used in the sense that the highest common factor of the coefficients of P is 1, which is weaker than requiring the coefficients to be pairwise relatively prime.

## Finite generation of ring extension

For any α, the ring extension (in the sense that is equivalent to field extension) of the integers by α, denoted by $\mathbb {Z} (\alpha )\equiv \{\sum _{i=0}^{n}\alpha ^{i}z_{i}|z_{i}\in \mathbb {Z} ,n\in \mathbb {Z} \}$, is finitely generated if and only if α is an algebraic integer.
The proof is analogous to that of the corresponding fact regarding algebraic numbers, with $\mathbb {Q}$ there replaced by $\mathbb {Z}$ here, and the notion of field extension degree replaced by finite generation (using the fact that $\mathbb {Z}$ is finitely generated itself); the only required change is that only non-negative powers of α 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 $\mathbb {Z}$ or $\mathbb {Q}$, respectively.

## Ring

The sum, difference and product of two algebraic integers is an algebraic integer. In general their quotient is not. Thus the algebraic integers form a ring.
This can be shown analogously to the corresponding proof for algebraic numbers, using the integers $\mathbb {Z}$ instead of the rationals $\mathbb {Q}$.
One may also construct explicitly the monic polynomial involved, which is generally of higher degree than those of the original algebraic integers, by taking resultants and factoring. For example, if x$^2$ − x − 1 = 0, y$^3$ − y − 1 = 0 and z = xy, then eliminating x and y from z − xy = 0 and the polynomials satisfied by x and y using the resultant gives z$^6$ − 3z$^4$ − 4z$^3$ + z$^2$ + z − 1 = 0, which is irreducible, and is the monic equation satisfied by the product. (To see that the xy is a root of the x-resultant of z − xy and x$^2$ − x − 1, one might use the fact that the resultant is contained in the ideal generated by its two input polynomials.)

### Integral closure

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 is integrally closed in any of its extensions.
Again, the proof is analogous to the corresponding proof for algebraic numbers being algebraically closed.

## Additional facts

* 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 irreducible quintics are not. This is the Abel–Ruffini theorem.
* The ring of algebraic integers is a Bézout domain, as a consequence of the principal ideal theorem.
* If the monic polynomial associated with an algebraic integer has constant term 1 or −1, then the reciprocal of that algebraic integer is also an algebraic integer, and each is a unit, an element of the group of units of the ring of algebraic integers.
* If x is an algebraic number then a$_n$x is an algebraic integer, where x satisfies a polynomial p(x) with integer coefficients and where a$_n$x$^n$ is the highest-degree term of p(x). The value y = a$_n$x is an algebraic integer because it is a root of q(y) = a$^{n−1}$ $_n$ p(y /a$_n$), where q(y) is a monic polynomial with integer coefficients.
* If x 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$_n$|, where x satisfies a polynomial p(x) with integer coefficients and where a$_n$x$^n$ is the highest-degree term of p(x).
* The only rational algebraic integers are the integers. Thus, if α is an algebraic integers and $\alpha \in \mathbb {Q}$, then $\alpha \in \mathbb {Z}$. This is a direct result of the rational root theorem for the case of a monic polynomial.