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.

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Depending on what order you're doing things in, this might be an easy application of the fundamental theorem of Galois theory (clearly a finite group has finitely many subgroups). –  Chris Eagle Jan 20 '12 at 15:33
The key words here are "primitive element theorem". This is shown in most introductory books on abstract algebra, algebraic number theory... Have you had a look at, say Dummit and Foote, Section 14.4, Theorem 25? –  Álvaro Lozano-Robledo Jan 20 '12 at 15:38
@ÁlvaroLozano-Robledo: It looks like he already has Artin's version of the primitive element theorem. He wants to know why a number field has finitely many subfields. –  Cam McLeman Jan 20 '12 at 15:40
I'm happy to convince myself of this using Galois correspondence. However, I always seem to forget about the primitive element theorem; it is necessary that all polynomials over $\mathbb Q$ are separable, but this is easy to prove. Thanks guys –  Matt Jan 20 '12 at 15:50
@Matt: I guess the point of my comment is that with the work you've already done (the hard part of the primitive element theorem, that it suffices to verify that there are finitely many intermediate subfields), it seems a waste to finish the problem by just citing the more-frequently-stated version of the result. –  Cam McLeman Jan 20 '12 at 15:58

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.

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Cam, this is exactly the proof that I alluded to in Dummit and Foote, Thm 25 of Section 14.4... –  Álvaro Lozano-Robledo Jan 20 '12 at 15:44
Thanks. I've never come across the idea of "Galois closure"; I can guess what it is (the smallest extension of $L$ that is Galois over $K$), but how do I convince myself of its existence? –  Matt Jan 20 '12 at 15:48
Okay, sure, it's not an altogether unsurprising proof. –  Cam McLeman Jan 20 '12 at 15:51
But he's trying to show the existence of a primitive element... –  Cam McLeman Jan 20 '12 at 15:58
Yes, problems certainly arise. In fact, the argument above works precisely for the separable case. The wikipedia page on the primitive root theorem gives the standard counter-example in the non-separable case. –  Cam McLeman Jan 20 '12 at 16:14

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.

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This is of course not my proof - it's folklore, but I can't recall where I first saw it. –  vgty6h7uij May 20 '12 at 16:39
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).