Let \(K\) be a field of characteristic \(0\), let \(f\in K[x]\) be irreducible of degree \(n\ge5\), and let \(F\) be the splitting field of \(f\) over \(K\). Assume \[ \Gal(F/K)\cong S_n. \] Let \(u\in F\) be a root of \(f\).
Show that there is no radical extension field \(E/K\) such that \[ K\subset K(u)\subset E. \]
Solution
Because \(f\) is irreducible and \(u\) is a root, the minimal polynomial of \(u\) over \(K\) is \(f\). Hence the normal closure of the simple extension \[ K(u)/K \] is exactly the splitting field \(F\) of \(f\).
By Galois’s theorem on solvability by radicals, a finite separable extension \(M/K\) in characteristic \(0\) is contained in a radical extension of \(K\) only if the Galois group of the normal closure of \(M/K\) is solvable.
Apply this with \[ M=K(u). \] If there were a radical extension \(E/K\) with \[ K\subset K(u)\subset E, \] then the Galois group of the normal closure of \(K(u)/K\) would be solvable. But that normal closure is \(F\), so this would force \[ \Gal(F/K)\cong S_n \] to be solvable.
For \(n\ge5\), the symmetric group \(S_n\) is not solvable: its normal subgroup \(A_n\) is nonabelian simple, so the derived series cannot terminate at the identity. This contradiction proves that no such radical extension \(E\) exists.