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A cyclotomic field (or cyclotomic number field) is a cyclotomic extension of
 . These are all of the form
 , where  is a primitive  th root of unity.
The ring of integers of a cyclotomic field always has a power basis over
 . Specifically, the ring of integers of
 is
![$ \mathbb{Z}[\omega_n]$](http://images.planetmath.org:8080/cache/objects/9486/l2h/img7.png "$ \mathbb{Z}[\omega_n]$") .
Given a primitive  th root of unity  , its minimal polynomial over
 is the cyclotomic polynomial  . Thus,
![$ [\mathbb{Q}(\omega_n)\!:\!\mathbb{Q}]=\varphi(n)$](http://images.planetmath.org:8080/cache/objects/9486/l2h/img12.png "$ [\mathbb{Q}(\omega_n)\!:\!\mathbb{Q}]=\varphi(n)$") , where  denotes the Euler phi function.
If  is odd, then
 . There are many ways to prove this, but the following is a relatively short proof: Since
 , we have that
 . We also have that
![$ [\mathbb{Q}(\omega_{2n})\!:\!\mathbb{Q}]=\varphi(2n)=\varphi(2)\varphi(n)=\varphi(n)=[\mathbb{Q}(\omega_n)\!:\!\mathbb{Q}]$](http://images.planetmath.org:8080/cache/objects/9486/l2h/img18.png "$ [\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$](http://images.planetmath.org:8080/cache/objects/9486/l2h/img19.png "$ [\mathbb{Q}(\omega_{2n})\!:\!\mathbb{Q}(\omega_n)]=1$") . It follows that
 .
Note. If  is a positive integer and  is an integer such that
 , then  and
 are primitive  th roots of unity and generate the same cyclotomic field.
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