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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.

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

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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.

Let $T$ be a transcendence basis of $F(a_1,\ldots ,a_n)/F$. Then $F(a_1,\ldots ,a_n)/F(T)$ is a finite extension and $F(T)$ and $K$ are linearly disjoint over $F$. Hence the cardinality of any $F$-linearly independent set of elements of $K$ is bounded from above by the degree of $F(a_1,\ldots ,a_n)/F(T)$.

If an algebraic extension $K/F$ of fields is not finite, then within $F$ there exists an infinite tower of proper finite extensions $K\subset K_1\subset K_2\subset \ldots $. Since $K_i$ is a $K$-sub vectorspace of $K_{i+1}$ a basis of $K_i$ can be extended to a basis of $K_{i+1}$. Thus the above infinite tower yields an infinite $K$-linearly independent set within $F$. By linear disjointness this set remains linearly independent over $F(T)$ – a contradiction.

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