Finitely generated extension implies finitely generated field?

Question

Let $K/F$ be a field extension. If $K$ is a finitely generated extension, namely $K=F[u_1,~\cdots,~u_n]$ for some $u_i\in K$. Then is $K$ also a finitely generated field(i.e. $K=\langle a_1,~\cdots,~a_m\rangle$ for some $u_i\in K$, where $\langle.\rangle$ means the subfield generated by the things inside)?

Answer 1

If $K/F$ is a field extension where $K$ is finitely-generated as an $F$-algebra, then in fact $K/F$ is algebraic and finite-dimensional as an $F$-vector space. This is Zariski's Lemma. In particular $K/F$ is finitely-generated as a field extension.

Be careful: the converse is far from being true though.

For example, say $K=\mathbb{Q}(\pi)$ where $\pi=3.14...$. Since $\pi$ is transcendental, $K/\mathbb{Q}$ is not finitely-generated as a $\mathbb{Q}$-algebra.