Cyclotomic extension - Imported Knowlpedia Content

Let $K$ be a field and fix $n\ge 1$ . Choose a primitive n-th root of unity $\zeta_n$ in an algebraic closure of $K$ (when it exists). The cyclotomic extension of level $n$ is the simple extension

K(\zeta_n)/K.

Assume $\mathrm{char}(K)\nmid n$ . Then $x^n-1$ has distinct roots, so $K(\zeta_n)/K$ is separable. Moreover, $K(\zeta_n)$ contains all $n$ -th roots of unity (since every root is $\zeta_n^k$ ), so it is the splitting field of $x^n-1$ over $K$ . Hence $K(\zeta_n)/K$ is normal and therefore Galois (see separable + normal ⇔ Galois).

Remarks

In the classical case $K=\mathbb{Q}$ , the minimal polynomial of $\zeta_n$ is the cyclotomic polynomial $\Phi_n(x)$ , so

[\mathbb{Q}(\zeta_n):\mathbb{Q}] = \deg(\Phi_n)=\varphi(n).

The Galois group $\mathrm{Gal}(\mathbb{Q}(\zeta_n)/\mathbb{Q})$ identifies with $(\mathbb{Z}/n\mathbb{Z})^\times$ by sending an automorphism $\sigma$ to the unique class $a\bmod n$ with $\sigma(\zeta_n)=\zeta_n^{\,a}$ .

Examples

$n=3$ . $\mathbb{Q}(\zeta_3)=\mathbb{Q}(e^{2\pi i/3})=\mathbb{Q}\!\left(\frac{-1+\sqrt{-3}}{2}\right)=\mathbb{Q}(\sqrt{-3})$ , a quadratic extension of $\mathbb{Q}$ .

$n=4$ . $\mathbb{Q}(\zeta_4)=\mathbb{Q}(i)$ , and $\mathrm{Gal}(\mathbb{Q}(i)/\mathbb{Q})\cong C_2$ generated by complex conjugation (a field automorphism).

$n=5$ . $\Phi_5(x)=x^4+x^3+x^2+x+1$ , so $[\mathbb{Q}(\zeta_5):\mathbb{Q}]=4$ and $\mathrm{Gal}(\mathbb{Q}(\zeta_5)/\mathbb{Q})\cong(\mathbb{Z}/5\mathbb{Z})^\times\cong C_4$ .