# number field that is not norm-Euclidean

Proposition. The real quadratic field^{} $\mathbb{Q}(\sqrt{14})$ is not norm-Euclidean.

Proof. We take the number $\gamma =\frac{1}{2}+\frac{1}{2}\sqrt{14}$ which is not integer of the field ($14\equiv 2\phantom{\rule{veryverythickmathspace}{0ex}}(mod4)$). Antithesis: $\gamma =\varkappa +\delta $ where $\varkappa =a+b\sqrt{14}$ is an integer of the field ($a,b\in \mathbb{Z}$) and

$$ |

Thus we would have

$$ |

And since ${(2a-1)}^{2}=4(a-1)a+1\equiv 1\phantom{\rule{veryverythickmathspace}{0ex}}(mod8)$, it follows $E\equiv 1-14\cdot 1\equiv 3\phantom{\rule{veryverythickmathspace}{0ex}}(mod8)$, i.e. $E=3$. So we must have

${(2a-1)}^{2}\equiv {(2a-1)}^{2}-14{(2b-1)}^{2}\equiv 3\phantom{\rule{veryverythickmathspace}{0ex}}(mod7).$ | (1) |

But $\{0,\pm 1,\pm 2,\pm 3\}$ is a complete residue system^{} modulo 7, giving the set $\{1,\mathrm{\hspace{0.17em}2},\mathrm{\hspace{0.17em}4}\}$ of possible quadratic residues^{} modulo 7. Therefore (1) is impossible. The antithesis is wrong, whence the theorem 1 of the parent entry (http://planetmath.org/EuclideanNumberField) says that the number field^{} is not norm-Euclidean.

Note. The function N used in the proof is the usual

$$\text{N}:r+s\sqrt{14}\mapsto {r}^{2}-14{s}^{2}\mathit{\hspace{1em}}(r,s\in \mathbb{Q})$$ |

defined in the field $\mathbb{Q}(\sqrt{14})$. The notion of norm-Euclidean number field is based on the norm (http://planetmath.org/NormAndTraceOfAlgebraicNumber). There exists a fainter function, the so-called Euclidean valuation, which can be defined in the maximal orders^{} of some algebraic number fields (http://planetmath.org/NumberField); such a maximal order, i.e. the ring of integers^{} of the number field, is then a Euclidean domain^{}. The existence of a Euclidean valuation guarantees that the maximal order is a UFD and thus a PID. Recently it has been shown the existence of the Euclidean domain $\mathbb{Z}[\frac{1+\sqrt{69}}{2}]$ in the field $\mathbb{Q}(\sqrt{69})$ but the field is not norm-Euclidean.

The maximal order $\mathbb{Z}[\sqrt{14}]$ of $\mathbb{Q}(\sqrt{14})$ has also been proven to be a Euclidean domain (Malcolm Harper 2004 in Canadian Journal of Mathematics).

Title | number field that is not norm-Euclidean |
---|---|

Canonical name | NumberFieldThatIsNotNormEuclidean |

Date of creation | 2013-03-22 16:56:56 |

Last modified on | 2013-03-22 16:56:56 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 15 |

Author | pahio (2872) |

Entry type | Example |

Classification | msc 13F07 |

Classification | msc 11R21 |

Classification | msc 11R04 |

Related topic | UniqueFactorizationAndIdealsInRingOfIntegers |