abstract algebra - Is the algebraic subextension of a finitely generated field extension finitely generated?

This question is motivated by this other question (and its answer).

Suppose we have a field $F$, possibly imperfect. Consider the finitely generated field extension $F(a_1,\ldots,a_n)$. Is it always true that $K=F(a_1,\ldots,a_n)\cap F^{\textrm{alg}}$ is finitely generated?

The proof from 1 generalises to case when $F$ is perfect (or more generally when $F(a_1,\ldots,a_n)$ is separable over $F$, I guess), thanks to the primitive element theorem.

But what about the general case? What if the initial extension is inseparable?

Answer

Given an arbitrary field extension $F\subset G$, if $G$ is finitely generated (as a field) over $F$, then any intermediate field extension $F\subset K\subset G$ is also finitely generated over $F$.
Applying this to your situation (with $G=F(a_1,\cdots,a_n))$ you see immediately that your $K$ is both algebraic and finitely generated over $F$ so that actually it is even a finite-dimensional vector space over $F$.

The powerful theorem mentioned in the first sentence is unfortunately not as well-known as it should.
As often the best reference is Bourbaki: Algebra, Chapter 5, §15, Corollary 3, page 118.