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# 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.

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

## Mathematics Subject Classification

11R18*no label found*11-00

*no label found*

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