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 💡