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 $\gcd (m,n)=1$, then  ${\omega }_n$  and  $\omega _n{}^m$  are the same cyclotomic field.