# Proof of finite subfields for a finite field extension

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.

• Not sure if I understood the question correctly. If you want to show that if $L/F$ is a finite field extension then the number of intermediate fields is finite then the claim is false. See this question: math.stackexchange.com/questions/2944053/… – Mark May 25 '19 at 15:49
• When $K=\Bbb Q$ and $L=\Bbb Q(\sqrt3,\sqrt5,\sqrt7)$, $|L:K|=8$, but there are $14$ fields strictly between $K$ and $L$. – Lord Shark the Unknown May 25 '19 at 15:52
• @Mark I might have butchered the question, here it is verbatim: Let $F \subset L$ be a field extension with [L:F] = d < infinity. Show that if L = F(theta) for some theta in L then there exists only finitely many subfields K of L containing F – Rzmwood May 25 '19 at 16:15
• @Rzmwood, Oh, now it is a very different question. If a finite extension $L/F$ is simple (i.e generated by one element) then it is true that there are only finitely many intermediate fields. Actually it is "if and only if"\$. I will write an answer soon. – Mark May 25 '19 at 16:34
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.