Let \(K\) be a field, \(f\in K[x]\) an irreducible polynomial of degree \(n\geq 5\) and \(F\) a splitting field of \(f\) over \(K\). Assume that \(Aut_{k}(F)\simeq S_{n}\). Let \(u\) be a root of \(f\) in \(F\). Then,
\(K(u)\) is not Galois over \(K\); \([K(u):K]=n\) and \(Aut_{K}(K(u))=1\) (and hence solvable).
Every normal closure over \(K\) that contains \(u\) also contains an isomorphic copy of \(F\).
There is no radical extension field \(E\) of \(K\) such that \(K\subset K(u)\subset E\).
Solution
Let \[ G=\operatorname{Aut}_K(F)\cong S_n. \]
1. One has\[ [F:K]=n!, \]and \(F/K\) is Galois.
Because \(F\) is a splitting field of a polynomial of degree \(n\), adjoining its roots one at a time gives \[ [F:K]\le n!. \] For every finite extension, the number of \(K\)-automorphisms is at most its degree, so \[ n!=|\operatorname{Aut}_K(F)|\le [F:K]\le n!. \] Hence equality holds throughout. A finite extension has as many \(K\)-automorphisms as its degree exactly when it is Galois. Thus \(F/K\) is Galois and \[ \operatorname{Gal}(F/K)=G\cong S_n. \]
2. The action of \(G\) on the \(n\) roots of \(f\) is the natural faithfultransitive action of \(S_n\).
Since \(f\) is irreducible, \(G\) acts transitively on its roots. Since \(F\) is generated by those roots, an automorphism fixing every root is the identity, so the action is faithful. A faithful action of a group of order \(n!\) on an \(n\)-element set identifies \(G\) with the full symmetric group on that set.
3. Let\[ H=\operatorname{Gal}(F/K(u)). \]Then \(H\) is the stabilizer of the root \(u\), hence \(H\cong S_{n-1}\), and\[ [K(u):K]=[G:H]=n. \]
By the Galois correspondence, \(H\) consists exactly of the automorphisms of \(F\) fixing \(u\). Under the action in 2, this is the stabilizer of one point, so it is isomorphic to \(S_{n-1}\). Therefore \[ [K(u):K]=[G:H]=\frac{n!}{(n-1)!}=n. \]
4. \(K(u)/K\) is not Galois and\[ \operatorname{Aut}_K(K(u))=1. \]
The extension \(K(u)/K\) is Galois if and only if its corresponding subgroup \(H\) is normal in \(G\). A point stabilizer \(S_{n-1}\) is not normal in \(S_n\) for \(n\ge3\), so \(K(u)/K\) is not Galois.
For an intermediate field \(F^H\) of a finite Galois extension, restriction induces \[ \operatorname{Aut}_K(F^H)\cong N_G(H)/H. \] Here \(H\) is a point stabilizer. If \(g\in N_G(H)\), then \[ gHg^{-1}=H. \] But \(gHg^{-1}\) is the stabilizer of \(g(u)\). Distinct points have distinct stabilizers in the natural \(S_n\)-action, so \(g(u)=u\), hence \(g\in H\). Thus \(N_G(H)=H\), and therefore \[ \operatorname{Aut}_K(K(u))\cong H/H=1. \]
5. Every normal extension of \(K\) containing \(u\) contains a \(K\)-copyof \(F\); inside a fixed algebraic closure it contains \(F\) itself.
Let \(N/K\) be normal and suppose \(u\in N\). Since \(f\) is the minimal polynomial of \(u\) over \(K\), normality implies that every \(K\)-conjugate of \(u\), equivalently every root of \(f\), lies in \(N\). The field generated by all roots of \(f\) is its splitting field. Hence \(N\) contains the splitting field of \(f\), which is \(K\)-isomorphic to \(F\), and equals \(F\) after choosing both inside the same algebraic closure.
6. There is no radical extension \(E/K\) satisfying\[ K\subset K(u)\subset E. \]
By 1, \(K(u)/K\) is finite separable, and by 5 its normal closure is \(F\). The solvability-by-radicals theorem says that a finite separable extension is contained in a radical extension only if the Galois group of its normal closure is solvable. Here that group is \[ \operatorname{Gal}(F/K)\cong S_n. \] For \(n\ge5\), \(S_n\) is not solvable because it contains the nonabelian simple normal subgroup \(A_n\). Therefore \(K(u)\) cannot be contained in any radical extension of \(K\).