Let \(g\in F[x]\) be a separable polynomial, and let \(E/F\) be its splitting field. Prove that \(E/F\) is Galois.
Solution
1. The extension \(E/F\) is finite and normal.
The field \(E\) is generated over \(F\) by the finitely many roots of \(g\), each of which is algebraic over \(F\), so \(E/F\) is finite. Since \(E\) is the splitting field of a polynomial over \(F\), it is normal over \(F\).
2. Every root of \(g\) is separable over \(F\).
Because \(g\) is separable, it has no repeated roots in an algebraic closure. If \(\alpha\) is a root and \(m_\alpha\in F[x]\) is its minimal polynomial, then \(m_\alpha\) divides \(g\). Hence \(m_\alpha\) also has no repeated roots, so \(\alpha\) is separable over \(F\).
3. The extension \(E/F\) is separable.
The splitting field \(E\) is generated by the roots of \(g\), and by 2 each of those generators is separable over \(F\). A finite extension generated by separable elements is separable.
4. Therefore \(E/F\) is Galois.
By 1, \(E/F\) is finite and normal, and by 3 it is separable. Hence it is Galois.