multiples of an algebraic number

Theorem.  If $\alpha$ is an algebraic number, then there exists a non-zero multiple (http://planetmath.org/GeneralAssociativity) of $\alpha$ which is an algebraic integer.

Proof.  Let $\alpha$ be a of the equation

$$x^n+r_1{x}^{n-1}+r_2{x}^{n-2}+\mathrm{\cdots }+r_n=0,$$

where $r_1$, $r_2$, …, $r_n$ are rational numbers ($n>0$).  Let $l$ be the least common multiple of the denominators of the $r_j$’s.  Then we have

$$0=l^n({\alpha }^n+r_1{\alpha }^{n-1}+r_2{\alpha }^{n-2}+\mathrm{\cdots }+r_n)={(l\alpha )}^n+l{r}_1{(l\alpha )}^{n-1}+l^2{r}_2{(l\alpha )}^{n-2}+\mathrm{\cdots }+l^n{r}_n,$$

i.e. the the algebraic equation

$$x^n+l{r}_1{x}^{n-1}+l^2{r}_2{x}^{n-2}+\mathrm{\cdots }+l^n{r}_n=0$$

with rational integer coefficients.

According to the theorem, any algebraic number $\xi$ is a quotient (http://planetmath.org/Division) of an algebraic integer (of the field $\mathbb{Q}(\xi )$) and a rational integer.

Title	multiples of an algebraic number
Canonical name	MultiplesOfAnAlgebraicNumber
Date of creation	2014-05-16 19:58:36
Last modified on	2014-05-16 19:58:36
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	10
Author	pahio (2872)
Entry type	Theorem
Classification	msc 11R04
Related topic	TheoryOfAlgebraicNumbers
Related topic	AlgebraicSinesAndCosines
Related topic	SomethingRelatedToAlgebraicInteger
Related topic	RationalAlgebraicIntegers