# rational algebraic integers

Theorem. A rational number^{} is an algebraic integer^{} iff it is a rational integer.

*Proof.* ${1}^{\circ}$. Any rational integer $m$ has the minimal polynomial $x-m$, whence it is an algebraic integer.

${2}^{\circ}$. Let the rational number $\alpha =\frac{m}{n}$ be an algebraic integer where $m,n$ are coprime integers and $n>0$. Then there is a polynomial^{}

$$f(x)={x}^{k}+{a}_{1}{x}^{k-1}+\mathrm{\dots}+{a}_{k}$$ |

with ${a}_{1},\mathrm{\dots},{a}_{k}\in \mathbb{Z}$ such that

$$f(\alpha )={\left(\frac{m}{n}\right)}^{k}+{a}_{1}{\left(\frac{m}{n}\right)}^{k-1}+\mathrm{\dots}+{a}_{k}=\mathrm{\hspace{0.33em}0}.$$ |

Multiplying this equation termwise by ${n}^{k}$ implies

$${m}^{k}=-{a}_{1}{m}^{k-1}n-\mathrm{\dots}-{a}_{k}{n}^{k},$$ |

which says that $n\mid {m}^{k}$ (see divisibility in rings). Since $m$ and $n$ are coprime^{} and $n$ positive, it follows that $n=1$. Therefore, $\alpha =m\in \mathbb{Z}$.

Title | rational algebraic integers |
---|---|

Canonical name | RationalAlgebraicIntegers |

Date of creation | 2013-03-22 19:07:34 |

Last modified on | 2013-03-22 19:07:34 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 5 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 11R04 |

Related topic | MultiplesOfAnAlgebraicNumber |