Number Fields

Let $\omega =e^{2\pi i/m}$. Then every conjugate of $\omega$ must be of the form ${\omega }^k$ for some $1\le k\le m$ coprime to $m$ (since every conjugate must also be a $m$ root of unity, and not an $n$th root for any $n<m$. The converse is also true:

Theorem: The conjugates of $\omega$ are ${\omega }^k$ for $1\le k\le m$ coprime to $m$.

Proof: Let $\theta ={\omega }^k$ for some $k\le m$ coprime to $m$. We show ${\theta }^p$ is a conjugate of $\theta$ for all primes $p$ not dividing $m$. (Applying this fact repeatedly proves the theorem.)

Let $f(x)$ be the minimal polynomial for $\theta$ over $\mathbb{Q}$. Then $x^m-1=f(x)g(x)$ for some monic $g\in \mathbb{Q}[x]$, and we must have in fact $f,g\in \mathbb{Z}[x]$.

Now ${\theta }^p$ is a root of $x^m-1$, so it must be a root of $f$ or $g$. If $g({\theta }^p)=0$ then $\theta$ is a root of $g(x^p)$, thus $g(x^p)$ must be divisible by $f(x)$.

For the remainder of the proof we work in ${\mathbb{Z}}_p[x]$. Then $g(x^p)=g(x{)}^p$, and since ${\mathbb{Z}}_p[x]$ is a UFD, it follows $f,g$ have a common nonunit divisor $h$ (any prime factor of $f$ divides $g(x{)}^p$ hence divides $g(x)$) and hence $h^2\mid fg=x^m-1$. Thus $h$ divides $\frac{d}{dx}{x}^m-1=m{x}^{m-1}$. Since $p$ does not divide $m$, $h$ must be a monomial which contradicts $h\mid x^m-1$. $\blacksquare$

Corollary: $[\mathbb{Q}[\omega ]:\mathbb{Q}]=\varphi (m)$

Corollary: $\mathrm{Gal}(\mathbb{Q}[\omega ]/\mathbb{Q})\cong {\mathbb{Z}}_m^{*}$

This corollary implies that the subfields of $\mathbb{Q}[\omega ]$ correspond to the subgroups of ${\mathbb{Z}}_m^{*}$. For $p$ prime, the $p$th cyclotomic field contains a unique subfield of order $d$ for every divisor $d$ of $p-1$. In particular the $p$th cyclotomic field contains a unique quadratic field. It turns out to be $\mathbb{Q}[\sqrt{\pm p}]$ where the sign is determined by $p$.

Corollary: Let $\omega =e^{2\pi i/m}$. For even $m$, the only roots of unity in $\mathbb{Q}[\omega ]$ are the $m$th roots of unity, and for odd $m$, the only roots of unity are the $2m$th roots of unity.

Proof: For $m$ odd, we know that the $m$th cyclotomic field is the same as the $2m$th cyclotomic field. Hence assume $m$ is even. Then if $\theta$ is a primitive $k$th root of unity, then $\mathbb{Q}[\omega ]$ must also contain a primitive $r$th root of unity where $r$ is the least common multiple of $m$ and $k$. But then we must have $\varphi (r)\le \varphi (m)$, which is a contradiction unless $r=m$ (since $m$ is even). Hence $k\mid m$. $\blacksquare$

Corollary: For even $m$, the $m$ cyclotomic fields are all distinct and pairwise nonisomorphic.