Show that the Galois group over \(\QQ\) of \[ f(x)=x^3-3x-3 \] is isomorphic to \(S_3\).
Solution
1. The polynomial \(f\) is irreducible over \(\QQ\).
Apply Eisenstein’s criterion with the prime \(3\). The leading coefficient is \(1\), so it is not divisible by \(3\). Every lower coefficient is divisible by \(3\), and the constant term \(-3\) is not divisible by \(9\). Hence \(f\) is irreducible over \(\QQ\).
2. The discriminant of \(f\) is\[ \Delta=-135. \]
For a depressed cubic \[ x^3+ax+b, \] the discriminant is \[ -4a^3-27b^2. \] Here \(a=-3\) and \(b=-3\), so \[ \Delta=-4(-3)^3-27(-3)^2 =108-243 =-135. \]
3. The Galois group is \(S_3\).
Because \(f\) is an irreducible cubic, its Galois group acts transitively on the three roots. Thus it is either \(A_3\) or \(S_3\).
For an irreducible cubic over a field of characteristic different from \(2\), the Galois group lies in \(A_3\) exactly when the discriminant is a square in the base field. Here \(-135\) is not a square in \(\QQ\). Therefore the Galois group is not contained in \(A_3\), so it must be \(S_3\).