Let \(L/K\) be a finite separable field extension. Prove that \(L/K\) is normal if and only if \(L\) is the splitting field of some separable polynomial \(p(x) \in K[x]\).
Solution
Suppose first that \(L/K\) is normal. Since the extension is finite and separable, the primitive element theorem gives \[ L=K(\alpha) \] for some \(\alpha\in L\). Let \(m_\alpha\in K[x]\) be its minimal polynomial. Separability of \(L/K\) makes \(m_\alpha\) separable, and normality implies that \(m_\alpha\) splits completely in \(L\). Since one of its roots, namely \(\alpha\), already generates \(L\), the field generated by all its roots is exactly \(L\). Thus \(L\) is the splitting field of the separable polynomial \(m_\alpha\).
Conversely, suppose \(L\) is the splitting field over \(K\) of a separable polynomial \(p\in K[x]\). Let \(\Omega\) be an algebraic closure of \(K\), and let \[ \sigma:L\hookrightarrow\Omega \] be a \(K\)-embedding. Every root \(\alpha\) of \(p\) in \(L\) satisfies \[ p(\sigma(\alpha))=\sigma(p(\alpha))=0, \] so \(\sigma\) permutes the roots of \(p\). Since those roots generate \(L\), \[ \sigma(L)=L. \] Hence every \(K\)-embedding of \(L\) into an algebraic closure has image \(L\), so \(L/K\) is normal.