What are the Galois groups of irreducible cubics?
Solution
Let \(F\) be a field of characteristic not \(2\), let \(f\in F[x]\) be an irreducible separable cubic with roots \(r_1,r_2,r_3\) in a splitting field \(L\), and let \(G=\Gal(L/F)\) act on the roots. Put \(\delta=(r_1-r_2)(r_1-r_3)(r_2-r_3)\) and \(\Delta(f)=\delta^2\in F\).
1. \(G\) is a transitive subgroup of \(S_3\), so \(G=A_3\cong\ZZ/3\ZZ\) or \(G=S_3\).
\(G\) acts faithfully on the roots, and transitively since \(f\) is irreducible. A transitive subgroup of \(S_3\) has order divisible by \(3\) by orbit-stabilizer, so it is \(A_3\) or \(S_3\).
2. \(G\le A_3\) if and only if \(\Delta(f)\) is a square in \(F\).
Each \(\sigma\in G\) satisfies \(\sigma(\delta)=\operatorname{sgn}(\sigma)\delta\). Since \(f\) is separable, \(\delta\neq0\), and since \(\operatorname{char}F\neq2\), \(-\delta\neq\delta\). So \(\delta\) is fixed by \(G\), that is \(\delta\in F\), exactly when \(G\) consists of even permutations; and \(\Delta(f)\) is a square in \(F\) exactly when \(\delta\in F\).
3. Q.E.D.
By steps 1 and 2, \(G\cong\ZZ/3\ZZ\) when \(\Delta(f)\) is a square in \(F\), and \(G\cong S_3\) otherwise.