Algebraic integer

In Algebraic number theory, an algebraic integer is a complex number that is a root of some monic polynomial (a polynomial whose leading coefficient is 1) with coefficients in $\mathbb {Z}$ (the set of integers). The set of all algebraic integers, $A$ , is closed under addition and multiplication and therefore is a commutative subring of the complex numbers. The ring $A$ is the integral closure of regular integers $\mathbb {Z}$ in 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 $x$ is an algebraic integer if and only if the ring $\mathbb {Z} [x]$ 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 set of rational numbers), in other words, $K=\mathbb {Q} (\theta )$ for some algebraic number $\theta \in \mathbb {C}$ by the primitive element theorem.

$\alpha \in K$ is an algebraic integer if there exists a monic polynomial $f(x)\in \mathbb {Z} [x]$ such that $f(\alpha )=0$ .
$\alpha \in K$ is an algebraic integer if the minimal monic polynomial of $\alpha$ over $\mathbb {Q}$ is in $\mathbb {Z} [x]$ .
$\alpha \in K$ is an algebraic integer if $\mathbb {Z} [\alpha ]$ is a finitely generated $\mathbb {Z}$ -module.
$\alpha \in K$ is an algebraic integer if there exists a finitely generated $\mathbb {Z}$ -submodule $M\subset \mathbb {C}$ such that $\alpha M\subseteq M$ .

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 which 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 $a$ . Note that the leading coefficient of the polynomial $bx - a$ is the integer $b$ . As another special case, the square root $\sqrt n$ of a non-negative integer $n$ is an algebraic integer, and so 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 $d \equiv 1 (mod 4)$ the element $(1 + \sqrt d)/2$ is also an algebraic integer. It satisfies the polynomial $x 2 - x + (1 - d)/4$ where the constant term $(1 - d)/4$ is an integer. The full ring of integers is generated by $\sqrt d$ or $(1 + \sqrt d)/2$ respectively. See quadratic integers for more.
The ring of integers of the field $F=\mathbb {Q} [\alpha ],\alpha ={\sqrt[{3}]{m}}$ has the following integral basis, writing $m=hk^{2}$ for two square-free coprime integers $h$ and $k$ :^[1]

{\begin{cases}1,\alpha ,{\frac {\alpha ^{2}\pm k^{2}\alpha +k^{2}}{3k}}&m\equiv \pm 1\mod 9\\1,\alpha ,{\frac {\alpha ^{2}}{k}}&\mathrm {else} \end{cases}}

If $ζ n$ is a primitive $n$ -th 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 $\beta ={\sqrt[{n}]{\alpha }}$ 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 which has integer coefficients but is not monic, and $P$ is irreducible over $\mathbb {Q}$ , then none of the roots of $P$ are algebraic integers. (Here primitive is used in the sense that the highest common factor of the set of coefficients of $P$ is 1; this is weaker than requiring the coefficients to be pairwise relatively prime.)

Facts

The sum, difference and product of two algebraic integers is an algebraic integer. In general their quotient is not. The monic polynomial involved is generally of higher degree than those of the original algebraic integers, and can be found 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 $z - xy$ and the polynomials satisfied by $x$ and $y$ using the resultant gives $z 6 - 3 z 4 - 4 z 3 + z 2 + z - 1$ , which is irreducible, and is the monic polynomial 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.)
Any number constructible out of the integers with roots, addition, and multiplication is therefore 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.
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 which is integrally closed in any of its extensions.
The ring of algebraic integers is a Bézout domain, as a consequence of the principal ideal theorem.

References

^ Marcus, Daniel A. (1977), Number fields, Berlin, New York: Springer-Verlag, ISBN 978-0-387-90279-1, chapter 2, p. 38 and exercise 41.

Daniel A. Marcus, Number Fields, third edition, Springer-Verlag, 1977

[1] Marcus, Daniel A. (1977), Number fields, Berlin, New York: Springer-Verlag, ISBN 978-0-387-90279-1, chapter 2, p. 38 and exercise 41.

[1]

Definitions

Examples

Non-example

Facts

See also

References