- If \(F\) is some (not necessarily Galois) field extension over \(K\) such that \([F: K]=6\) and Aut \(_{K} F \simeq S_{3}\), then \(F\) is the splitting field of an irreducible cubic over \(K[x]\).
Solution
Let \[ G=\operatorname{Aut}_K(F)\cong S_3. \]
1. The extension \(F/K\) is Galois and \(G=\Gal(F/K)\).
For every finite extension, \[ |\operatorname{Aut}_K(F)|\le [F:K], \] with equality if and only if the extension is Galois. Here \[ |\operatorname{Aut}_K(F)|=|S_3|=6=[F:K]. \] Hence \(F/K\) is Galois.
2. Let \(H\le G\) be a subgroup of order \(2\), and put \(E=F^H\). Then \([E:K]=3\).
By the Galois correspondence, \[ [E:K]=[G:H]=6/2=3. \] Since \(F/K\) is finite Galois, the intermediate extension \(E/K\) is finite separable.
3. There exists \(\alpha\in E\) such that \(E=K(\alpha)\), and its minimal polynomial \(f\in K[x]\) is irreducible of degree \(3\).
By the primitive element theorem, the finite separable extension \(E/K\) is simple, say \(E=K(\alpha)\). Then \[ \deg f=[K(\alpha):K]=[E:K]=3. \] Thus \(f\) is an irreducible cubic.
4. The polynomial \(f\) splits over \(F\).
Because \(F/K\) is normal, every \(K\)-embedding of \(E=K(\alpha)\) into an algebraic closure extends to an element of \(G\). Hence every \(K\)-conjugate of \(\alpha\) is of the form \(\sigma(\alpha)\) for some \(\sigma\in G\), and therefore lies in \(F\). Thus every root of the separable polynomial \(f\) lies in \(F\).
5. The splitting field of \(f\) inside \(F\) is all of \(F\).
Let \(L\) be the field generated over \(K\) by all roots of \(f\). Since those roots are the \(G\)-orbit of \(\alpha\), \[ L=K(\sigma(\alpha):\sigma\in G). \] An element \(\tau\in G\) fixes \(L\) pointwise exactly when it fixes every \(\sigma(\alpha)\). Since \[ \operatorname{Stab}_G(\alpha)=\Gal(F/K(\alpha))=\Gal(F/E)=H, \] this pointwise stabilizer is \[ \bigcap_{\sigma\in G}\sigma H\sigma^{-1}, \] the core of \(H\) in \(G\).
Under \(G\cong S_3\), a subgroup of order \(2\) is generated by a transposition. Its three conjugates are the three distinct order-\(2\) subgroups, whose intersection is trivial. Hence \[ \Gal(F/L)=1. \] By the Galois correspondence, \(L=F\).
6. Therefore \(F\) is the splitting field over \(K\) of the irreducible cubic \(f\).
By 3, \(f\) is irreducible cubic, and by <1>4–<1>5 its splitting field is exactly \(F\).