Cyclotomic extension

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

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 ).

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

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

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

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

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