# 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  $\Phi_{n}(x)$. 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 the same cyclotomic field.

Title cyclotomic field CyclotomicField 2013-03-22 17:10:25 2013-03-22 17:10:25 Wkbj79 (1863) Wkbj79 (1863) 9 Wkbj79 (1863) Definition msc 11R18 msc 11-00 cyclotomic number field CyclotomicExtension CyclotomicPolynomial