Compute the Galois group of \(x^3 - 2\) over the rationals.
Solution
Let \(\alpha=\sqrt[3]{2}\) and let \(\zeta_3\) be a primitive cube root of unity.
1. The polynomial \(x^3-2\) is irreducible over \(\QQ\).
It is Eisenstein at the prime \(2\).
2. Its splitting field is\[ L=\QQ(\alpha,\zeta_3). \]
The three roots are \[ \alpha,\ \zeta_3\alpha,\ \zeta_3^2\alpha. \]
3. One has \([L:\QQ]=6\).
By 1, \([\QQ(\alpha):\QQ]=3\). The field \(\QQ(\alpha)\) is real, while \(\zeta_3\) is nonreal, so \(\zeta_3\notin\QQ(\alpha)\). Since \(\zeta_3\) has degree \(2\) over \(\QQ\), adjoining it doubles the degree: \[ [L:\QQ]=6. \]
4. Therefore\[ \operatorname{Gal}(L/\QQ)\cong S_3. \]
The Galois group acts faithfully on the three roots, so it embeds in \(S_3\). Since \(L/\QQ\) is Galois of degree \(6\), the Galois group has order \(6\), hence equals \(S_3\).