Suppose \(K/F\) is a Galois extension and that \(\mathrm{Gal}(K/F)\) is isomorphic to \(A_n\), with \(n \geq 5\). Suppose \(\alpha \in K\setminus F\). Prove that the minimal polynomial of \(\alpha\) over \(F\) has degree at least \(n\). (You may use whatever group theoretic facts you know about \(A_n\) without proof.)
Solution
Let \(G=\operatorname{Gal}(K/F)\), let \(m(T)\in F[T]\) be the minimal polynomial of \(\alpha\), and put \(d=\deg m\).
1. The action of \(G\) on the roots of \(m\) defines anontrivial homomorphism \(\rho:G\to S_d\).
Normality and separability of the Galois extension imply that \(m\) splits in \(K\) with \(d\) distinct roots [@DF04]. Every \(\sigma\in G\) fixes its coefficients, so sends each root to a root. These permutations respect composition and give the stated homomorphism.
The fixed field \(K^G\) is \(F\) [@DF04]. Since \(\alpha\notin F\), some \(\sigma\in G\) satisfies \(\sigma(\alpha)\ne\alpha\). Thus the root action is not trivial and \(\ker\rho\ne G\).
2. The homomorphism \(\rho\) is injective.
For \(n\geq5\), the group \(A_n\) is simple [@DF04]. Hence \(G\cong A_n\) has no normal subgroups other than \(1\) and \(G\). The kernel of a homomorphism is normal, and step 1 excludes the latter possibility. Therefore \(\ker\rho=1\).
3. One has \(d\geq n\).
Injectivity gives \(|A_n|=n!/2\leq |S_d|=d!\). If \(d<n\), then \(d\leq n-1\), so \[ d!\leq(n-1)!<\frac n2(n-1)!=\frac{n!}{2}, \] where strictness follows from \(n\geq5>2\). This contradicts \(n!/2\leq d!\). Thus \(d\geq n\), as required.