cyclotomic fieldCyclotomicField2007-05-30 02:58:092007-06-24 12:35:03Definitionroot of unity\usepackage{amssymb}
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A \emph{cyclotomic field} (or \emph{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 \PMlinkname{primitive $n$th root of unity}{PrimitiveNthRootOfUnity}.
The ring of integers of a cyclotomic field always has a \PMlinkname{power basis over $\mathbb{Z}$}{PowerBasisOverMathbbZ}. Specifically, the ring of integers of $\mathbb{Q}(\omega_n)$ is $\mathbb{Z}[\omega_n]$.
Given a \PMlinkescapetext{primitive $n$th root of unity} $\omega_n$, its minimal polynomial over $\mathbb{Q}$ is the cyclotomic polynomial $\Phi_n(x)$. Thus, $[\mathbb{Q}(\omega_n)\!:\!\mathbb{Q}]=\varphi(n)$, where $\varphi$ 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 $[\mathbb{Q}(\omega_{2n})\!:\!\mathbb{Q}]=\varphi(2n)=\varphi(2)\varphi(n)=\varphi(n)=[\mathbb{Q}(\omega_n)\!:\!\mathbb{Q}]$. Thus, $[\mathbb{Q}(\omega_{2n})\!:\!\mathbb{Q}(\omega_n)]=1$. It follows that $\mathbb{Q}(\omega_{2n})=\mathbb{Q}(\omega_n)$.
\textbf{Note.}\, If $n$ is a positive integer and $m$ is an integer such that $\gcd(m,n)=1$, then\, $\omega_n$\, and\, ${\omega_n}^m$\, are \PMlinkescapetext{primitive $n$th roots of unity and generate} the same cyclotomic field.