## Cyclotomic 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 $n \lt m$. 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 $h^2 | f g = x^m - 1$. Thus $h$ divides $\frac{d}{d x} x^m - 1 = m x^{m-1}$. Since $p$ does not divide $m$, $h$ must be a monomial which contradicts $h | x^m - 1$.∎

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

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

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 $\phi(r) \le \phi(m)$, which is a contradiction unless $r = m$ (since $m$ is even). Hence $k | m$.∎

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