TITLE: Why is a number field always of the form $\mathbb Q(\alpha)$ for $\alpha$ algebraic?
QUESTION [7 upvotes]: My definition of a number field is "a finite extension of $\mathbb Q$". I want to prove that if $L$ is a finite field extension of $\mathbb Q$, then $L = \mathbb Q(\alpha)$ for some $\alpha$ algebraic over $\mathbb Q$.
I can prove things that look helpful. I know that if $L/K$ is a finite extension with $\mathrm{char}K = 0$  and with finitely many intermediate fields, then $L = K(\alpha)$ for some algebraic $\alpha$. I also know that if $\alpha$ is algebraic over $K$, then there are only finitely many intermediate fields between $K$ and $K(\alpha)$ (where those fields are determined by the factors of the minimal polynomial of $\alpha$ over $K$). 
I can't pull a proof together though. I'd obviously be done if I could prove that a finite extension of $\mathbb Q$ has finitely many intermediate extensions. Any hints would be greatly appreciated.

REPLY [2 votes]: Use the primitive element theorem to get the existence of a generator. Then the fact that a number field forms a finite extension of $\mathbb{Q}$ tells you that any such generator has to be algebraic (since finite extensions are automatically algebraic).