Alternative proof that a simple field extension just has a finite number of intermediate fields

I wanted to prove:

Let $L/K$ be a simple algebraic field extension. Then $L/K$ has just a finite number of intermediate field.

I worked out a proof and then looked into the solution, which is totally different so I wanted to know if my proof is valid.

The official solution goes like this, let $L = K(a)$ and consider the minimal polynomial $P$ of $a$ over $K$. Let $B$ be an intermediate field, and let $Q$ be the minimal polynomial of $a$ over $B$. Then $Q | P$ in $B[x]$, but $P$ just contains a finite numbers of irreducible factors, and because every intermediate field corresponds to just one factor, it follows that there just could be a finite number of intermediate fields.

My solution: Suppose there exists an infinite number of intermediate fields, then one could construct an infinite tower of field extensions $$K_1 \le K_2 \le K_3 \le K_4 \le \ldots$$ with $[K_{i+1} : K_i] > 1$. And then with the multiplicity and because of $K_i \le L$ for each $i$ it follows that $L/K$ could not be finite. First for $K_1$ set $K_1 := K$, and then proceed inductively. Suppose $K_1 \le \ldots \le K_n$ are choosen, then because there are an infinite number of intermediate fields, there exists some intermediate field $B$ which contains an element $\beta \in B$ which is in none of the of the field $K_i$, i.e. $\beta \notin K_i$ for $i \in \{ 1, \ldots, n \}$. Set $K_{n+1} = K_n(\beta)$, then $K_n \le K_{n+1}$ and $[K_{n+1} : K_n] > 1$ because the inclusion is proper. q.e.d.

EDIT: Added the supposition that $L/K$ has to be algebraic.

• Are there only finitely many vector subspaces between $\{0\}$ and $\mathbb{R}^2$? – Daniel Fischer Jul 28 '13 at 12:42
• no, for example there are an infinite numbers of lines through the origin. – StefanH Jul 28 '13 at 12:45
• Right. But the intermediate fields between $K$ and $L$ are $K$-vector spaces between $K$ and $L$. We just saw an example that having infinitely many intermediate vector spaces does not imply that you have an infinite ascending tower of vector subspaces. Arguing that for intermediate fields it would be so, essentially requires proving that in the given situation, there are only finitely many intermediate fields. – Daniel Fischer Jul 28 '13 at 12:50
• ok, so my proof is wrong. thank you – StefanH Jul 28 '13 at 13:04

This doesn't work. You'd have to prove that $a\notin K_n$ for each $n$.
Of course, I can't provide a counterexample with infinitely many intermediate fields as the theorem is true, but it's not hard to find a counterexample to your reasoning: consider $K={\bf Q}$, $a=\sqrt 2+\sqrt 3$. Then you could pick $K_2={\bf Q}[\sqrt 2]$ and then $B={\bf Q}[\sqrt 3]$ and $\beta=\sqrt 3$ are an intermediate subfield and its element which is not in $K_2$, and then you'd have $K_3=L$.