# cyclotomic field

A *cyclotomic field ^{}* (or

*cyclotomic number field*) is a cyclotomic extension of $\mathbb{Q}$. These are all of the form $\mathbb{Q}({\omega}_{n})$, where ${\omega}_{n}$ is a primitive $n$th root of unity

^{}(http://planetmath.org/PrimitiveNthRootOfUnity).

The ring of integers of a cyclotomic field always has a power basis over $\mathbb{Z}$ (http://planetmath.org/PowerBasisOverMathbbZ). Specifically, the ring of integers of $\mathbb{Q}({\omega}_{n})$ is $\mathbb{Z}[{\omega}_{n}]$.

Given a ${\omega}_{n}$, its minimal polynomial over $\mathbb{Q}$ is the cyclotomic polynomial^{} ${\mathrm{\Phi}}_{n}(x)$. Thus, $[\mathbb{Q}({\omega}_{n}):\mathbb{Q}]=\phi (n)$, where $\phi $ denotes the Euler phi function.

If $n$ is odd, then $\mathbb{Q}({\omega}_{2n})=\mathbb{Q}({\omega}_{n})$. There are many ways to prove this, but the following is a relatively short proof: Since ${\omega}_{n}=\omega _{2n}{}^{2}\in \mathbb{Q}({\omega}_{2n})$, we have that $\mathbb{Q}({\omega}_{n})\subseteq \mathbb{Q}({\omega}_{2n})$. We also have that $[\mathbb{Q}({\omega}_{2n}):\mathbb{Q}]=\phi (2n)=\phi (2)\phi (n)=\phi (n)=[\mathbb{Q}({\omega}_{n}):\mathbb{Q}]$. Thus, $[\mathbb{Q}({\omega}_{2n}):\mathbb{Q}({\omega}_{n})]=1$. It follows that $\mathbb{Q}({\omega}_{2n})=\mathbb{Q}({\omega}_{n})$.

Note. If $n$ is a positive integer and $m$ is an integer such that $\mathrm{gcd}(m,n)=1$, then ${\omega}_{n}$ and $\omega _{n}{}^{m}$ are the same cyclotomic field.

Title | cyclotomic field |
---|---|

Canonical name | CyclotomicField |

Date of creation | 2013-03-22 17:10:25 |

Last modified on | 2013-03-22 17:10:25 |

Owner | Wkbj79 (1863) |

Last modified by | Wkbj79 (1863) |

Numerical id | 9 |

Author | Wkbj79 (1863) |

Entry type | Definition |

Classification | msc 11R18 |

Classification | msc 11-00 |

Synonym | cyclotomic number field |

Related topic | CyclotomicExtension |

Related topic | CyclotomicPolynomial |