Give an example of a cubic polynomial over \(\mathbb{Q}\) with Galois group \(S_3\), and prove that its Galois group is indeed \(S_3\).
Solution
Take \[ f(x)=x^3-2. \] It is irreducible over \(\mathbb Q\) by Eisenstein at \(2\). Its discriminant is \[ \Delta(f)=-108, \] which is not a square in \(\mathbb Q\).
For an irreducible cubic, the Galois group is a transitive subgroup of \(S_3\), hence either \(A_3\) or \(S_3\). It is contained in \(A_3\) exactly when the discriminant is a square. Therefore \[ \operatorname{Gal}(f/\mathbb Q)\cong S_3. \]
Equivalently, the splitting field is \[ \mathbb Q(\sqrt[3]2,\zeta_3). \] The real cubic field \(\mathbb Q(\sqrt[3]2)\) has degree \(3\), while adjoining the nonreal element \(\zeta_3\) gives degree \(2\), so the splitting field has degree \(6\).