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 $\mathcal{O}_{K}$, 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, $$K = \Q(\theta)$$ for some algebraic number $$\theta \in \Complex$$ by the primitive element theorem.


 * $α ∈ K$ is an algebraic integer if there exists a monic polynomial $$f(x) \in \Z[x]$$ such that $f(α) = 0$.
 * $α ∈ K$ is an algebraic integer if the minimal monic polynomial of $α$ over $$\mathbb{Q}$$ is in $$\Z[x]$$.
 * $α ∈ K$ is an algebraic integer if $$\Z[\alpha]$$ is a finitely generated $$\Z$$-module.
 * $α ∈ K$ is an algebraic integer if there exists a non-zero finitely generated $$\Z$$-submodule $$M \subset \Complex$$ 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
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}$$
 * The only algebraic integers that are found in the set of rational numbers are the integers. In other words, the intersection of $$\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}\,)$$ is a quadratic field of rational numbers. The ring of algebraic integers $\mathcal{O}_{K}$ contains $$\sqrt{d}$$ since this is a root of the monic polynomial $x^{2} − d$. Moreover, if $d ≡ 1 mod 4$, then the element $\frac{1}{2}(1 + \sqrt{d}\,)$ is also an algebraic integer. It satisfies the polynomial $x^{2} − x + 1⁄4(1 − d)$ where the constant term $1⁄4(1 − d)$ is an integer. The full ring of integers is generated by $$\sqrt{d}$$ or $\frac{1}{2}(1 + \sqrt{d}\,)$  respectively. See Quadratic integer for more.
 * The ring of integers of the field $$F = \Q[\alpha]$$, $3}$, has the following integral basis, writing $m = hk^{2}$ for two square-free coprime integers $h$ and $k$: $$\begin{cases}
 * If $ζ_{n}$ is a primitive $n$th root of unity, then the ring of integers of the cyclotomic field $$\Q(\zeta_n)$$ is precisely $$\Z[\zeta_n]$$.
 * If $α$ is an algebraic integer then $n}$ 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 $&alpha;$, the ring extension (in the sense that is equivalent to field extension) of the integers by $&alpha;$, denoted by $$\Z(\alpha) \equiv \{\sum_{i=0}^n \alpha^i z_i | z_i\in \Z, n\in \Z\}$$, is finitely generated if and only if $&alpha;$ is an algebraic integer.

The proof is analogous to that of the corresponding fact regarding algebraic numbers, with $$\Q$$ there replaced by $$\Z$$ here, and the notion of field extension degree replaced by finite generation (using the fact that $$\Z$$ is finitely generated itself); the only required change is that only non-negative powers of $&alpha;$ 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 $$\Z$$ or $$\Q$$, 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 $$\Z$$ instead of the rationals $$\Q$$.

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 $q(y) = an −&thinsp;1 n&thinsp;p(y&hairsp;/a_{n})$, 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 $&alpha;$ is an algebraic integers and $$\alpha\in\Q$$, then $$\alpha\in\Z$$. This is a direct result of the rational root theorem for the case of a monic polynomial.