algebraic integers

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 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 , denoted by , is the intersection of and : it can also be characterized as the maximal order of the field . Each algebraic integer belongs to the ring of integers of some number field. A number is an algebraic integer if and only if the ring $\mathbb$alpha/math> is finitely generated as an abelian group, which is to say, as a $\mathbb$- module.

Definitions

The following are equivalent definitions of an algebraic integer. Let be a number field (i.e., a finite extension of $\mathbb$, the field of rational numbers), in other words, $K = \Q(\theta)$ for some algebraic number $\theta \in \Complex$ by the primitive element theorem.

* is an algebraic integer if there exists a monic polynomial f(x) \in \Z /math> such that .
* is an algebraic integer if the minimal monic polynomial of over $\mathbb$ is in \Z /math>.
* is an algebraic integer if $\Z$alpha/math> is a finitely generated $\Z$-module.
* is an algebraic integer if there exists a non-zero finitely generated $\Z$-submodule $M \subset \Complex$ such that .

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 $K / \mathbb$.

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

Examples

* The only algebraic integers that are found in the set of rational numbers are the integers. In other words, the intersection of $\mathbb$ and is exactly $\mathbb$. The rational number is not an algebraic integer unless divides . The leading coefficient of the polynomial is the integer .
* The square root $\sqrt$ of a nonnegative integer is an algebraic integer, but is irrational unless is a perfect square.
*If is a square-free integer then the extension $K = \mathbb(\sqrt\,)$ is a quadratic field of rational numbers. The ring of algebraic integers contains $\sqrt$ since this is a root of the monic polynomial . Moreover, if , then the element $\frac(1 + \sqrt\,)$ is also an algebraic integer. It satisfies the polynomial where the constant term is an integer. The full ring of integers is generated by $\sqrt$ or $\frac(1 + \sqrt\,)$ respectively. See Quadratic integer for more.
*The ring of integers of the field $F = \Q$alpha/math>, , has the following integral basis, writing for two square-free coprime integers and : $\begin 1, \alpha, \dfrac & m \equiv \pm 1 \bmod 9 \\ 1, \alpha, \dfrack & \text \end$
* If is a primitive th root of unity, then the ring of integers of the cyclotomic field $\Q(\zeta_n)$ is precisely \Z zeta_n/math>.
* If is an algebraic integer then is another algebraic integer. A polynomial for is obtained by substituting in the polynomial for .

Finite generation of ring extension

For any , the ring extension (in the sense that is equivalent to field extension) of the integers by , denoted by $\Z$alpha\equiv \left\, is finitely generated if and only if 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 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 , and , then eliminating and from and the polynomials satisfied by and using the resultant gives , which is irreducible, and is the monic equation satisfied by the product. (To see that the is a root of the -resultant of and , 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 is an algebraic number then is an algebraic integer, where satisfies a polynomial with integer coefficients and where is the highest-degree term of . The value is an algebraic integer because it is a root of , where is a monic polynomial with integer coefficients.
* If 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 , where satisfies a polynomial with integer coefficients and where is the highest-degree term of .
* The only rational algebraic integers are the integers. That is, if is an algebraic integer and $x\in\Q$ then $x\in\Z$. This is a direct result of the rational root theorem for the case of a monic polynomial.

See also

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

References

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{{Algebraic numbers

Algebraic numbers
Integers