Let \(E/F\) be a finite field extension. Prove that \(E/F\) is algebraic and that there is a polynomial \(g\in F[x]\) whose splitting field \(L/F\) is finite and contains \(E\).
Solution
1. Every element of \(E\) is algebraic over \(F\).
Let \(\alpha\in E\). Since \(E\) is finite-dimensional over \(F\), the vectors \[ 1,\alpha,\alpha^2,\ldots,\alpha^{[E:F]} \] are linearly dependent over \(F\). Thus some nonzero polynomial in \(F[x]\) vanishes at \(\alpha\), so \(\alpha\) is algebraic over \(F\).
2. The extension \(E/F\) is generated by finitely many algebraic elements.
Choose an \(F\)-basis \(e_1,\ldots,e_n\) of \(E\). Then \[ E=F(e_1,\ldots,e_n), \] and by 1 every \(e_i\) is algebraic over \(F\).
3. Let \(f_i\in F[x]\) be the minimal polynomial of \(e_i\) and set\[ g=f_1\cdots f_n\in F[x]. \]If \(L\) is the splitting field of \(g\) over \(F\), then \(E\subseteq L\).
Since \(e_i\) is a root of \(f_i\), every \(e_i\) lies in the splitting field \(L\) of \(g\). Hence \[ E=F(e_1,\ldots,e_n)\subseteq L. \]
4. The splitting field \(L/F\) is finite.
The polynomial \(g\) has finite degree, so it has finitely many roots in an algebraic closure. The splitting field \(L\) is generated over \(F\) by those finitely many roots, each algebraic over \(F\). A field extension generated by finitely many algebraic elements is finite.
5. Therefore every finite extension is algebraic and is contained in a finite splitting field.
1. The field \(L\) constructed in 3 is a splitting field over \(F\), contains \(E\), and is finite over \(F\) by
4.