Construct an explicit polynomial over \(\mathbb{Q}\) whose Galois group is the cyclic group of order 3: \[\operatorname{Gal}(f/\mathbb{Q}) \cong \mathbb{Z}/3\mathbb{Z}.\] Prove that your polynomial satisfies this condition.
Solution
Take \[ f(x)=x^3-3x+1. \] By the rational root test, its only possible rational roots are \(\pm1\), and \[ f(1)=-1, \qquad f(-1)=3. \] Thus \(f\) is irreducible over \(\mathbb Q\).
For a depressed cubic \(x^3+px+q\), the discriminant is \[ \Delta=-4p^3-27q^2. \] Here \(p=-3\) and \(q=1\), so \[ \Delta=108-27=81=9^2. \] An irreducible cubic has transitive Galois group, hence either \(A_3\) or \(S_3\), and it lies in \(A_3\) exactly when its discriminant is a square. Therefore \[ \boxed{\operatorname{Gal}(f/\mathbb Q)\cong A_3\cong C_3}. \]