I just was looking at an exercise which asks the reader to show that for $F \subset L$, if $L = F(\theta)$ for some $\theta \in L$ then there exists only finitely many subfields $K$ of $L$ containing $F$. Can I just prove this by the following argument:
The tower extension theorem says $[L:F] = [L:K][K:F] $
This means: $[L:F] > [K:F] > 0$ , since $[L:K] = 1 \iff L = K$, which is clear when viewing the extensions as vector spaces.
Continuing to find distinct intermediate subfields $F \subset K' \subset K \subset L $, we see that the degree of the field extension decreases at each step and is bounded below by 0, so the process will eventually terminate.
So there can be at most $[L:K] - 1$ distinct intermediate subfields, which is finite.
Is this a valid argument or is there some subtlety I'm missing here.
Thank you in advance.
Your solution is wrong because $K,K'$ being two intermediate fields does not imply that one of them is contained in the other.
Here is a possible solution. Let $K$ be an intermediate field. Then $L/K$ is a finite extension, hence algebraic. Let $g=\sum_{i=0}^n b_ix^i\in K[x]$ be the minimal polynomial of $\theta$ over $K$. Then $[L:K]=n$. Now let's write $K'=F(b_0,b_1,...,b_n)$. Then $K'\subseteq K$, and hence $K'(\theta)\subseteq K(\theta)=L$. On the other hand $L=F(\theta)\subseteq K'(\theta)$ and combining these results we conclude that $L=K'(\theta)$. Also note that $g\in K'[x]$ and is irreducible over $K'$. (because if it was reducible then it would be reducible over $K$ as well). Hence $g$ is also the minimal polynomial of $\theta$ over $K'$, so $[L:K']=n$ as well. This of course tells us that $K=K'$. And the last thing we should take in note is that if $m$ is the minimal polynomial of $\theta$ over $F$ then obviously $m\in K[x]$ and hence $g|m$.
Alright, so what did we prove? We showed that any intermediate field between $F$ and $L$ is generated by the coefficients of a fixed polynomial in $L[x]$ which divides $m$. There are only finitely many such polynomials (to see this write $m$ in the form $(x-\alpha_1)(x-\alpha_2)...(x-\alpha_k)$ and see which fixed polynomials might divide it) and hence there are finitely many intermediate fields.