- Show that \(\gal(x^3+4x+2)/\QQ \cong S_3\), a symmetric group.
Solution
Let \[ f(x)=x^3+4x+2\in\QQ[x], \] and let \(L\) be its splitting field.
1. The polynomial \(f\) is irreducible over \(\QQ\).
A reducible cubic over \(\QQ\) has a rational root. Since \(f\) is monic, the rational-root theorem says any rational root must be among \(\pm1,\pm2\). Directly, \[ f(1)=7,\quad f(-1)=-3,\quad f(2)=18,\quad f(-2)=-14, \] so none is a root. Hence \(f\) is irreducible.
2. The polynomial \(f\) has exactly one real root and two nonreal complex-conjugate roots.
Its derivative is \[ f'(x)=3x^2+4>0 \] for every real \(x\). Thus \(f\) is strictly increasing on \(\RR\), so it has at most one real root. Since a real cubic has at least one real root, it has exactly one. The remaining two roots are nonreal and, because the coefficients are real, are complex conjugates.
3. The Galois group \(G=\Gal(L/\QQ)\) is a transitive subgroup of \(S_3\) containing a transposition.
Irreducibility of \(f\) implies that \(G\) acts transitively on its three roots. Complex conjugation fixes the unique real root and exchanges the two nonreal roots, so its action on the roots is a transposition. Therefore the permutation representation of \(G\) is transitive and contains a transposition.
4. Hence\[ \Gal(L/\QQ)\cong S_3. \]
3. Thus \(G=S_3\).