E Field Of A Ring

Complex number that solves a monic polynomial with integer coefficients

In algebraic number theory, an algebraic integer is a complex number which is integral over the integers. That is, an algebraic integer is a circuitous root of some monic polynomial (a polynomial whose leading coefficient is 1) whose coefficients are integers. The set of all algebraic integers $A$ is airtight under addition, subtraction and multiplication and therefore is a commutative subring of the complex numbers.

The ring of integers of a number field $One thousand$ , denoted by $O K$ , is the intersection of $G$ and $A$ : information technology can also be characterised as the maximal guild 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 $\mathbb {Z} [\blastoff ]$ is finitely generated as an abelian group, which is to say, as a $\mathbb {Z}$ -module.

Definitions [edit]

The following are equivalent definitions of an algebraic integer. Permit $K$ exist a number field (i.east., 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.

Algebraic integers are a special case of integral elements of a ring extension. In detail, an algebraic integer is an integral element of a finite extension $K/\mathbb {Q}$ .

Examples [edit]

The merely 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 $a$ . Annotation that the leading coefficient of the polynomial $bx - a$ is the integer $b$ . As some other special case, the foursquare root ${\sqrt {n}}$ of a nonnegative integer $northward$ is an algebraic integer, but is irrational unless $n$ is a perfect foursquare.
If $d$ is a foursquare-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 $ten 2 - d$ . Moreover, if $d \equiv 1 mod 4$ , and so the element ${\textstyle {\frac {1}{2}}(1+{\sqrt {d}}\,)}$ is also an algebraic integer. It satisfies the polynomial $x ii - x + 1 / four (1 - d)$ where the abiding term $1 / iv (1 - d)$ is an integer. The full ring of integers is generated by ${\sqrt {d}}$ or ${\textstyle {\frac {i}{2}}(i+{\sqrt {d}}\,)}$ respectively. Run across Quadratic integer for more.
The band of integers of the field $F=\mathbb {Q} [\alpha ]$ , $α = 3 \sqrt 1000$ , has the following integral footing, writing $thou = hk ii$ for 2 square-free coprime integers $h$ and $k$ :^[1] ${\begin{cases}one,\alpha ,{\dfrac {\alpha ^{2}\pm yard^{two}\alpha +thousand^{two}}{3k}}&one thousand\equiv \pm 1{\bmod {9}}\\1,\blastoff ,{\dfrac {\alpha ^{2}}{k}}&{\text{otherwise}}\end{cases}}$
If $ζ northward$ is a primitive $due north$ th root of unity, then the band of integers of the cyclotomic field $\mathbb {Q} (\zeta _{due north})$ 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 past substituting $x n$ in the polynomial for $α$ .

Not-case [edit]

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 (simply are algebraic numbers). Hither primitive is used in the sense that the highest mutual factor of the coefficients of $P$ is 1; this is weaker than requiring the coefficients to be pairwise relatively prime.

Facts [edit]

The sum, difference and product of two algebraic integers is an algebraic integer. In general their quotient is non. The monic polynomial involved is more often than not of higher degree than those of the original algebraic integers, and tin be found by taking resultants and factoring. For instance, if $x 2 - x - 1 = 0$ , $y three - y - one = 0$ and $z = xy$ , then eliminating $x$ and $y$ from $z - xy = 0$ and the polynomials satisfied by $10$ and $y$ using the resultant gives $z 6 - three z 4 - 4 z 3 + z 2 + z - i = 0$ , which is irreducible, and is the monic equation satisfied past the product. (To see that the $xy$ is a root of the $x$ -resultant of $z - xy$ and $102 - x - one$ , ane might use the fact that the resultant is contained in the platonic 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 and so constructible: in a naïve sense, near 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 primary platonic theorem.
If the monic polynomial associated with an algebraic integer has abiding term 1 or −1, and so the reciprocal of that algebraic integer is also an algebraic integer, and is a unit, an element of the group of units of the ring of algebraic integers.
Every algebraic number tin be written every bit the ratio of an algebraic integer to a non-zero algebraic integer. In fact, the denominator can always be chosen to exist a positive integer. Specifically, if $ten$ is an algebraic number that is a root of the polynomial $p (x)$ with integer coefficients and leading term $a due north x due north$ for $a northward > 0$ so $a due north x / a n$ is the promised ratio. In particular, $y = a n x$ is an algebraic integer considering it is a root of $a north - 1 northward p (y / a n)$ , which is a monic polynomial in $y$ with integer coefficients.