Let \(u\) be algebraic over a field \(K\), let \(f\in K[x]\) be the minimal polynomial of \(u\), and let \(L\) be the normal closure of \(K(u)/K\). Let \(F\) be the splitting field of \(f\) over \(K\). Prove that \(L=F\).
Solution
1. The splitting field \(F\) is a normal extension of \(K\) containing \(K(u)\).
By definition, \(u\) is a root of \(f\), so \(u\in F\) and therefore \(K(u)\subseteq F\). A splitting field of a polynomial over \(K\) is normal over \(K\).
2. Hence \(L\subseteq F\).
The normal closure \(L\) is the smallest normal extension of \(K\) containing \(K(u)\). By 1, \(F/K\) is normal and contains \(K(u)\). Minimality of \(L\) therefore gives \[ L\subseteq F. \]
3. Every root of \(f\) lies in \(L\).
The field \(L/K\) is normal and contains the root \(u\) of the irreducible polynomial \(f\in K[x]\). For a normal extension, every irreducible polynomial over the base that has one root in the extension splits completely there. Hence all roots of \(f\) lie in \(L\).
4. Hence \(F\subseteq L\).
The splitting field \(F\) is generated over \(K\) by all roots of \(f\). By 3, all these roots lie in \(L\), so \[ F\subseteq L. \]
5. Therefore \(L=F\).
Combine 2 and 4.