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.

Ben Lynn blynn@cs.stanford.edu 💡