Give a polynomial with \(S_3\) as its Galois group over \(\mathbb{Q}\), and prove it.
Solution
Take \[ f(x)=x^3-2. \] It is irreducible over \(\mathbb Q\) by Eisenstein at \(2\). Let \(\alpha=\sqrt[3]{2}\) and let \(\omega\) be a primitive cube root of unity. The roots are \[ \alpha,\quad \alpha\omega,\quad \alpha\omega^2, \] so the splitting field is \[ K=\mathbb Q(\alpha,\omega). \] Now \[ [\mathbb Q(\alpha):\mathbb Q]=3, \] and \(\mathbb Q(\alpha)\subset\mathbb R\) while \(\omega\notin\mathbb R\), so \[ [K:\mathbb Q(\alpha)]=2. \] Hence \[ [K:\mathbb Q]=6. \] Since \(K/\mathbb Q\) is the splitting field of a separable cubic, its Galois group embeds in \(S_3\), and its order is \(6\). Therefore \[ \operatorname{Gal}(K/\mathbb Q)\cong S_3. \]