Field Generated by Finitely Many Algebraic Elements is Algebraic

Theorem

Let \(\mathbb{F}\) be a field and \(\mathbb{K} = \mathbb{F}(\alpha_1, \dots, \alpha_n)\) be a field extension where \(\alpha_1, \dots, \alpha_n\) are algebraic over \(\mathbb{F}\). Then \(\mathbb{K}/\mathbb{F}\) as finite.

Proof

First note that \(\mathbb{F}(\alpha_1)/\mathbb{F}\) is finite due to the power basis.

Then, if \(\mathbb{F}(\alpha_1, \dots, \alpha_k)\) is finite so is \(\mathbb{F}(\alpha_1, \dots, \alpha_k)(\alpha_{k + 1}) = \mathbb{F}(\alpha_1, \dots, \alpha_{k + 1})\) since \(\alpha_{k + 1}\) is algebraic over \(\mathbb{F}\) which is a subfield of \(\mathbb{F}(\alpha_1, \dots, \alpha_k)\).

Thus the result follows by induction on \(k\).