The field of algebraic numbers in $\mathbb Q (a_1,\ldots, a_l)$ is finite over $\mathbb Q$

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.

• Doesn't it has to do more with the fact that an extension is finite iff is algebraic? I mean, if $a_1,...,a_l$ were already in $\Bbb Q$. – Ana Galois Apr 10 '14 at 12:06
• @Ana, it is not true in general that an algebraic extension is finite. – Gerry Myerson Apr 10 '14 at 12:14
• @GerryMyerson well maybe I'm interpretting wrongly this theorem that I have in my notes: $E/F$ is finite iff $E/F$ is algebraic and there exists $\alpha_1,\dots,\alpha_n\in E$ such that $E=F(\alpha_1,\dots,\alpha_n)$ – Ana Galois Apr 10 '14 at 12:57
• But there are infinite algebraic extensions. – Lubin Apr 10 '14 at 14:10
• @AnaGalois: The second part "and the exists..." is essential, you can't skip it. – Najib Idrissi Apr 10 '14 at 14:12

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 $Q$.

• How does it follow that $K$ has dimension at most $m$? (I see how it does, but I don't think this is sufficiently trivial so as to be entirely omitted...) – tomasz Apr 10 '14 at 19:35
• @tomasz My last edit actually removed the argument :-( I've restored it. – Magdiragdag Apr 10 '14 at 19:45
• I still don't think it's complete. Any vector space is the union of its $1$-dimensional subspaces. – tomasz Apr 10 '14 at 19:48
• @tomasz in this case, all finite dimensional subspaces have dimension at most m – Magdiragdag Apr 10 '14 at 20:26
• But let me add a single sentence clarifying a small point anyway. – Magdiragdag Apr 10 '14 at 20:53