Why is a number field always of the form $\mathbb Q(\alpha)$ for $\alpha$ algebraic? 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.
 A: The result you're looking for is more general. In fact, it holds for any finite separable extension of fields. In particular, it holds for any finite extension of a perfect field (e.g. of $\mathbf{Q}$). I will sketch a proof below.
Claim: If $L/K$ is a finite separable extension of fields, then there is an element $\gamma\in L$ such that $L=K(\gamma)$.
Notation: If $x\in L$ and if $F$ is a subfield of $L$, then I will let $m_{x,F}$ denote the minimal polynomial of $x$ over $F$. Note that the minimal polynomial satisfies the following property: if $b\in L$ and if $P\in F[X]$ with $P(b)=0$, then $m_{x,F}|P$ in $F[X]$.
Proof: Suppose that $L/K$ is a finite separable extension of an infinite field $K$ (if $K$ is finite, the claim is easy to show) and let $\overline{L}=\overline{K}$ be an algebraic closure (one can quickly show that every finite extension of $\mathbf{Q}$ is separable; see Keith Conrad's notes on perfect fields). Since $L/K$ is finite, there are elements $a_1,\ldots, a_n\in L$ such that $L=K(a_1,\ldots, a_n)$. We can prove that $L=K(\alpha)$ for some $\alpha\in L$ by induction on the number of generators.
Let $F=K(a_1,\ldots, a_{n-2})$, so that $L=F(a_{n-1},a_n)$. Put $\alpha=a_{n-1}$ and $\beta=a_n$. Consider the finite set $$S=\{\lambda\in K | \lambda=\frac{\alpha-\alpha'}{\beta'-\beta}\text{ with } \alpha', \beta'\in\overline{K} \text{ roots of }m_{\alpha,K}\text{ and }m_{\beta,K}\text{ resp., and }\beta'\neq\beta\}$$ Let $\lambda\in K\backslash S$ (which exists since $K$ is infinite). Consider the sub-extension $L/F(\gamma)/K$ where $\gamma=\alpha+\lambda\beta$. Suppose that $L\neq F(\gamma)$; we will search for a contradiction. Note that if $\beta\in F(\gamma)$, then so is $\alpha$ (since $S$ contains $0$, $\lambda\neq 0$). We therefore must have $\deg m_{\beta,F(\gamma)}(X)\geq 2$. Consider the polynomial $m_{\alpha,K}(\gamma - \lambda X)\in F(\gamma)[X]$, which has $\beta$ as a root. By the general fact on minimal polynomials that I stated at the top, we have that $m_{\beta,F(\gamma)}|m_{\alpha,K}(\gamma - \lambda X)$ in $F(\gamma)[X]$. In particular, if $\beta'\neq \beta$ is a root of $m_{\beta,F(\gamma)}$ (which exists since $m_{\beta,F(\gamma)}$ is of degree $>1$), then $\beta'$ is also a root of $m_{\alpha,K}(\gamma - \lambda X)$; that is to say, the element $\alpha':=\gamma - \lambda\beta'$ is a root of $m_{\alpha,K}(X)$, and therefore $\lambda = \frac{\alpha - \alpha'}{\beta'-\beta}$ and $\lambda\in S$ (a contradiction). 
Therefore, $L=F(\gamma)=K(a_1,\ldots,a_{n-2},\gamma)$ and we've reduced the number of generators of $L/K$ by $1$, go by induction now.
A: 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).
A: Here's a proof that picks up with your observations (that it suffices to verify that there are finitely many intermediate subfields): 
Let $L^{gc}$ denote the Galois closure of $L$.  Then every subfield of $L$ is a subfield of $L^{gc}$, and subfields of $L^{gc}$ are, by Galois theory, in 1-1 correspondence with the subgroups of the Galois group $\operatorname{Gal}(L^{gc}/\mathbb{Q})$.  Since the Galois group is finite, there are finitely many subgroups, hence finitely many subfields.
