# multiples of an algebraic number

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

 $x^{n}\!+\!r_{1}x^{n-1}\!+\!r_{2}x^{n-2}\!+\cdots+\!r_{n}=0,$

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

 $0=l^{n}(\alpha^{n}\!+\!r_{1}\alpha^{n-1}\!+\!r_{2}\alpha^{n-2}\!+\cdots+\!r_{n% })=(l\alpha)^{n}\!+\!lr_{1}(l\alpha)^{n-1}\!+\!l^{2}r_{2}(l\alpha)^{n-2}\!+% \cdots+\!l^{n}r_{n},$

i.e. the the algebraic equation

 $x^{n}\!+\!lr_{1}x^{n-1}\!+\!l^{2}r_{2}x^{n-2}\!+\cdots+\!l^{n}r_{n}=0$

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 MultiplesOfAnAlgebraicNumber 2014-05-16 19:58:36 2014-05-16 19:58:36 pahio (2872) pahio (2872) 10 pahio (2872) Theorem msc 11R04 TheoryOfAlgebraicNumbers AlgebraicSinesAndCosines SomethingRelatedToAlgebraicInteger RationalAlgebraicIntegers