# rational integers in ideals

Any non-zero ideal of an algebraic number field^{} $K$, i.e. of the maximal order^{} ${\mathcal{O}}_{K}$ of $K$, contains positive rational integers.

*Proof.* Let $\U0001d51e\ne (0)$ be any ideal of ${\mathcal{O}}_{K}$. Take a nonzero element $\alpha $ of
$\U0001d51e$. The norm (http://planetmath.org/NormInNumberField) of $\alpha $ is the product

$$\text{N}(\alpha )={\alpha}^{(1)}\underset{\gamma}{\underset{\u23df}{{\alpha}^{(2)}\mathrm{\cdots}{\alpha}^{(n)}}}$$ |

where $n$ is the degree of the number field and ${\alpha}^{(1)},{\alpha}^{(2)},\mathrm{\cdots},{\alpha}^{(n)}$ is the set of the http://planetmath.org/node/12046$\mathrm{K}$-conjugates^{} of $\alpha ={\alpha}^{(1)}$. The number

$$\gamma =\frac{\text{N}(\alpha )}{\alpha}$$ |

belongs to the field $K$ and it is an algebraic integer^{}, since ${\alpha}^{(2)},\mathrm{\cdots},{\alpha}^{(n)}$ are, as algebraic conjugates of $\alpha $, also algebraic integers. Thus $\gamma \in {\mathcal{O}}_{K}$. Consequently, the non-zero integer

$$\text{N}(\alpha )=\alpha \gamma $$ |

belongs to the ideal $\U0001d51e$, and similarly its opposite number. So, $\U0001d51e$ contains positive integers, in fact infinitely many.

Title | rational integers in ideals |
---|---|

Canonical name | RationalIntegersInIdeals |

Date of creation | 2013-03-22 19:08:47 |

Last modified on | 2013-03-22 19:08:47 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 6 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 12F05 |

Classification | msc 06B10 |

Classification | msc 11R04 |

Related topic | CharacteristicPolynomialOfAlgebraicNumber |

Related topic | IdealNorm |