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.

The ring of integers of a cyclotomic field always has a power basis over $\mathbb{Z}$ . Specifically, the ring of integers of $\mathbb{Q}(\omega_n)$ is $\mathbb{Z}[\omega_n]$ .

Given a 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 $\mathbb{Q}(\omega_{2n})=\mathbb{Q}(\omega_n)$ .

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 primitive $n$ th roots of unity and generate the same cyclotomic field.