# 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 |