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.

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

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 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 $.mw-parser-output .sfrac{white-space:nowrap}.mw-parser-output .sfrac.tion,.mw-parser-output .sfrac .tion{display:inline-block;vertical-align:-0.5em;font-size:85%;text-align:center}.mw-parser-output .sfrac .num{display:block;line-height:1em;margin:0.0em 0.1em;border-bottom:1px solid}.mw-parser-output .sfrac .den{display:block;line-height:1em;margin:0.1em 0.1em}.mw-parser-output .sr-only{border:0;clip:rect(0,0,0,0);clip-path:polygon(0px 0px,0px 0px,0px 0px);height:1px;margin:-1px;overflow:hidden;padding:0;position:absolute;width:1px}⁠a/b⁠$ 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 $d \equiv 1 mod 4$ , then the element ${\textstyle {\frac {1}{2}}(1+{\sqrt {d}}\,)}$ 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 ${\textstyle {\frac {1}{2}}(1+{\sqrt {d}}\,)}$ respectively. See Quadratic integer for more.
The ring of integers of the field $F=\mathbb {Q} [\alpha ]$ , $α = 3 \sqrt m$ , has the following integral basis, writing $m = hk 2$ for two square-free coprime integers $h$ and $k$ :^[1] ${\begin{cases}1,\alpha ,{\dfrac {\alpha ^{2}\pm k^{2}\alpha +k^{2}}{3k}}&m\equiv \pm 1{\bmod {9}}\\1,\alpha ,{\dfrac {\alpha ^{2}}{k}}&{\text{otherwise}}\end{cases}}$
If $ζ n$ is a primitive $n$ 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 $β = n \sqrt α$ 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 $z 6 - 3 z 4 - 4 z 3 + z 2 + z - 1 = 0$ , 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.

References

^ Marcus, Daniel A. (1977). Number Fields (3rd ed.). Berlin, New York: Springer-Verlag. ch. 2, p. 38 and ex. 41. ISBN 978-0-387-90279-1.

Stein, William. Algebraic Number Theory: A Computational Approach (PDF). Archived (PDF) from the original on November 2, 2013.

[1] Marcus, Daniel A. (1977). Number Fields (3rd ed.). Berlin, New York: Springer-Verlag. ch. 2, p. 38 and ex. 41. ISBN 978-0-387-90279-1.

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