Fix a field \(F\), a separable polynomial \(f\in F[x]\) of degree \(n \geq 3\), and a splitting field \(L\) for \(f\). Prove that if \([L:F] = n!\) then:
\(f\) is irreducible.
For each root \(r\) of \(f\), \(r\) is the unique root of \(f\) in \(F(r)\).
For every root \(r\) of \(f\), there is no field \(E\) with \(F \subsetneq E \subsetneq F(r)\).
Solution
Let \(X\) be the set of the \(n\) distinct roots of \(f\) in \(L\), and put \(G=\operatorname{Gal}(L/F)\).
1. The action of \(G\) on \(X\) identifies it withthe full symmetric group \(\operatorname{Sym}(X)\).
The extension is Galois because it is the splitting field of a separable polynomial [@DF04]. Its automorphisms permute the roots. If an automorphism fixes every root, it fixes the field they generate over \(F\), namely \(L\), so it is the identity. The action is therefore faithful. Since \(|G|=[L:F]=n!=|\operatorname{Sym}(X)|\), its image is the whole symmetric group.
2. The polynomial \(f\) is irreducible over \(F\).
Choose \(r\in X\) and let \(m_r\) be its monic minimal polynomial over \(F\). This polynomial divides \(f\). For every \(\sigma\in G\), applying \(\sigma\) to \(m_r(r)=0\) gives \(m_r(\sigma(r))=0\). Step 1 says that these images include all \(n\) roots. The root bound gives \(\deg m_r\geq n\), while \(m_r\mid f\) gives \(\deg m_r\leq n\). Thus \(f\) is a nonzero constant multiple of the irreducible polynomial \(m_r\).
3. For each root \(r\), no other root belongs to \(F(r)\).
The subgroup \(H=\operatorname{Gal}(L/F(r))\) consists exactly of the permutations fixing \(r\). Suppose \(s\in X\setminus\{r\}\). Since \(n\geq3\), there is \(t\in X\setminus\{r,s\}\). Step 1 supplies the automorphism acting as the transposition \((s\ t)\). It fixes \(r\) and hence fixes \(F(r)\) pointwise, but does not fix \(s\). Consequently \(s\notin F(r)\).
4. The root field \(F(r)\) has no strict intermediatefield over \(F\).
The point stabilizer \(H\) is a maximal proper subgroup of \(G\). Indeed, if \(H\subsetneq J\leq G\), choose \(g\in J\) moving \(r\). Since \(H\) permutes the remaining roots arbitrarily, the \(J\)-orbit of \(r\) is all of \(X\). Its stabilizer in \(J\) is precisely \(H\). Thus \(|J|=n|H|=n!\), and \(J=G\) by orbit-stabilizer.
For \(F\subseteq E\subseteq F(r)\), Galois correspondence gives \(H\subseteq\operatorname{Gal}(L/E)\subseteq G\) [@DF04]. The subgroup must be \(H\) or \(G\), so its fixed field \(E\) must be \(F(r)\) or \(F\). These are exactly the two allowed endpoints.