Algebraic integer

Algebraic integer

"This article deals with the ring of complex numbers integral over" Z. "For the general notion of algebraic integer, see Integrality".

In number theory, an algebraic integer is a complex number which is a root of some monic polynomial (leading coefficient 1) with integer coefficients. The set of all algebraic integers is closed under addition and multiplication so it forms a subring of complex numbers denoted by A. The ring A is the integral closure of regular integers in complex numbers.

The ring of integers of a number field "K", denoted by "OK" , is the intersection of "K" and A: it may 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 Z ["x"] is finitely generated as an abelian group, which is to say, a Z-module.

Examples

* The only algebraic integers in rational numbers are the ordinary integers. In other words, the intersection of Q and A is exactly 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 √"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" = Q(√"d") is a quadratic field extension of rational numbers. The ring of algebraic integers "OK" contains √"d" since this is a root of the monic polynomial "x"² − "d". Moreover, if "d" ≡ 1 (mod 4) the element (1 + √"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 √"d" or (1 + √"d")/2 respectively.
* If $zeta_n$ is a primitive "n"-th root of unity, then the ring of integers of the cyclotomic field $mathbf\left\{Q\right\}\left(zeta\right)$ is precisely $mathbf\left\{Z\right\} \left[zeta\right]$.
* If "α" is an algebraic integer then 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 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: most roots of irreducible quintics are not.

* 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 extension.

* The ring of algebraic integers A is a Bézout domain.

References

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

ee also

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

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