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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 OK, 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 is finitely generated as an abelian group, which is to say, as a -module.

Definitions

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

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 .

Examples

Non-example

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 , 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 there replaced by here, and the notion of field extension degree replaced by finite generation (using the fact that 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 or , 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 instead of the rationals .

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 x2x − 1 = 0, y3y − 1 = 0 and z = xy, then eliminating x and y from zxy = 0 and the polynomials satisfied by x and y using the resultant gives z6 − 3z4 − 4z3 + z2 + 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 zxy and x2x − 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

See also

References

  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.