In the book 'Algebra IV: Infinite Groups, Linear Groups' by Kostrikin and Shafarevich, there is a sketch of a proof (on page 84) of a theorem by Schur. I'm struggling to understand the line:
Since the field $\mathbb Q (a_1,\ldots, a_l)$ is finitely generated, the field $K$ of algebraic numbers in it is finite over $\mathbb Q$.
Why is this so? I can't find a proof for it anywhere. My best guess is that $K$ is a finitely generated algebra over $\mathbb Q$, and hence by the weak Nullstellensatz is of finite degree, but I can't see why $K$ would be finitely generated.
Well, there ought to be an easy answer to this, but here's a hard work version.
Split the extension ${\mathbb Q}(a_1,\dots,a_n):{\mathbb Q}$ in a purely transcendental extension ${\mathbb Q}(t_1,\dots,t_k):{\mathbb Q}$ and an algebraic extension ${\mathbb Q}(a_1,\dots,a_n) :{\mathbb Q}(t_1,\dots,t_k)$. This last one, being algebraic and finitely generated, has finite degree, say $m$.
Now consider a single element of $K$. Its minimum polynomial over ${\mathbb Q}$ is also its minimum polynomial over ${\mathbb Q}(t_1,\dots,t_k)$, so has degree at most $m$. Since every finitely generated subfield of $K$ can be generated by a single element, every such finitely generated subfield has dimension at most $m$ over ${\mathbb Q}$. So every finite dimensional subspace of $K$ over $Q$, being contained in the subfield generated by a basis, has dimension at most $m$ over $Q$. Therefore $K$ itself has dimension at most $m$ over $Q$.