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[parent] multiples of an algebraic number (Theorem)
Theorem 1   If $ \alpha$ is an algebraic number, then there exists a non-zero multiple of $ \alpha$ which is an algebraic integer.

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

$\displaystyle x^n\!+\!r_1x^{n-1}\!+\!r_2x^{n-2}\!+\cdots+\!r_n = 0,$
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
$\displaystyle 0 = l^n(\alpha^n\!+\!r_1\alpha^{n-1}\!+\!r_2\alpha^{n-2}\!+\cdots... ...l\alpha)^n\!+\!lr_1(l\alpha)^{n-1}\!+\!l^2r_2(l\alpha)^{n-2}\!+\cdots+\!l^nr_n,$
i.e. the multiple $ l\alpha$ of $ \alpha$ satisfies the algebraic equation
$\displaystyle x^n\!+\!lr_1x^{n-1}\!+\!l^2r_2x^{n-2}\!+\cdots+\!l^nr_n = 0$
with rational integer coefficients.

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



"multiples of an algebraic number" is owned by pahio.
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See Also: theory of algebraic and transcendental numbers


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Cross-references: field, quotient, coefficients, rational integer, algebraic equation, denominators, least common multiple, rational numbers, equation, proof, algebraic integer, algebraic number
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This is version 5 of multiples of an algebraic number, born on 2005-05-30, modified 2008-05-10.
Object id is 7126, canonical name is MultiplesOfAnAlgebraicNumber.
Accessed 1280 times total.

Classification:
AMS MSC11R04 (Number theory :: Algebraic number theory: global fields :: Algebraic numbers; rings of algebraic integers)

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