$x^{n} - 1$ [1505C]

The splitting field of $x^{n} - 1 \in \mathbb{Q}[x]$ is $\mathbb{Q}(\zeta_{n})$.

The polynomial splits completely in $\mathbb{C}$. Let $\zeta_{n} = e^{ 2\pi i / n } = \cos \left( \frac{2\pi}{n} \right) + i \sin \left( \frac{2\pi}{n} \right) \in \mathbb{C}$. Then $\zeta_{n}$ is a root of $x^{n} - 1$ and $\zeta_{n}^{k}, 0 \leq k \leq n -1$ are distinct roots of $x^{n} - 1$. It follows that $$ (x - \zeta_{n}^{0})(x - \zeta_{n}^{1}) \cdots (x - \zeta_{n}^{n-1}) \mid x^{n} - 1 $$ in $\mathbb{C}[x]$. By comparing degrees, we know that $$ x^{n} - 1 = (x - \zeta_{n}^{0})(x - \zeta_{n}^{1}) \cdots (x - \zeta_{n}^{n-1}). $$ Therefore, a splitting field of $x^{n} - 1$ is $\mathbb{Q}(\zeta_{n}, \dots, \zeta_{n}^{n-1}) = \mathbb{Q}(\zeta_{n}) = \mathbb{Q}[\zeta_{n}]$.

If $k$ is coprime to $n$, then $\mathbb{Q}(\zeta_{n}) = \mathbb{Q}(\zeta_{n}^{k})$.

When $n = p$ is prime, we have shown that $\Phi_{p}(x) = \frac{x^{p} - 1}{x - 1} = x^{p-1} + \cdots + x + 1$ is irreducible in $\mathbb{Q}[x]$ in MATH 111B. And $\Phi_{p}(\zeta_{p}) = 0$ and $\Phi_{p}(x)$ is monic, so $m_{\zeta_{p}, \mathbb{Q}}(x) = \Phi_{p}(x)$. Therefore, $[\mathbb{Q}(\zeta_{p}) : \mathbb{Q}] = \deg \Phi_{p}(x) = p-1$.

Later, we will show that $[\mathbb{Q}(\zeta_{n}) : \mathbb{Q}] = \lvert (\mathbb{Z} / n\mathbb{Z})^{\times} \rvert$.