Let \(L/K\) be a Galois extension of fields, with Galois group \(G = \{\sigma_1, \dots, \sigma_n\}\), and let \(\alpha \in L\). Prove that \(L = K(\alpha)\) iff \(\sigma_1(\alpha), \dots, \sigma_n(\alpha)\) are distinct.
Solution
The finite Galois group in the hypothesis gives \([L:K]=|G|=n\) [@DF04]. Let \(m_\alpha\in K[x]\) be the minimal polynomial of \(\alpha\) and put \(d=\deg m_\alpha=[K(\alpha):K]\).
1. If \(L=K(\alpha)\), the \(n\) displayed conjugates are distinct.
Suppose \(\sigma_i(\alpha)=\sigma_j(\alpha)\). Every element of \(K(\alpha)\) is a rational expression in \(\alpha\) with coefficients in \(K\). Both automorphisms fix those coefficients, and their agreement on \(\alpha\) therefore makes them agree on every element of \(L=K(\alpha)\). Thus \(\sigma_i=\sigma_j\), and the enumeration of \(G\) gives \(i=j\).
2. If the \(n\) displayed conjugates are distinct, then \(L=K(\alpha)\).
For each \(i\), fixing the coefficients of \(m_\alpha\) gives \[ m_\alpha(\sigma_i(\alpha)) =\sigma_i(m_\alpha(\alpha))=0. \] Thus \(m_\alpha\) has at least \(n\) distinct roots, so \(d\geq n\). On the other hand, the tower law in \(K\subseteq K(\alpha)\subseteq L\) gives \(n=[L:K(\alpha)]d\), and hence \(d\leq n\). It follows that \(d=n\) and \([L:K(\alpha)]=1\), proving the conclusion.