A simple finite field extension only has finitely many intermediate fields

Let $$L/K$$ be a finite and simple field extension, so $$L = K(\alpha)$$ and $$[L:K] = n$$. Show that $$L/K$$ only has finitely many intermediate fields $$M$$.

I'd like to know if my following proof is correct or how I could improve it:

Suppose $$L = K(\alpha)$$. For every intermediate field $$M$$ let $$f_M \in M[X]$$ be the minimal polynomial of $$\alpha$$ over $$M$$. Then for every $$M, M'$$ we have $$M \subseteq M' \iff f_M \in M'[X]$$. Now I wanna show that every $$f_M$$ uniquely defines the corresponding intermediate field $$M$$:

Let $$M'$$ be another intermediate field s.t $$f_M \in M'[X]$$. Then by the above we have $$M \subseteq M'$$ and by the definition of $$f_M$$ and $$f_{M'}$$ it follows that $$\deg(f_{M'}) \leq \deg(f_M)$$. Since every root of $$f_{M'}$$ in $$M'$$ is also a root of $$f_M$$ (I'm not sure about this one so in case this isn't necessarily true some clarification would be great!), we get $$f_{M'} \mid f_M$$. However, since both polynomials are irreducible and monic, we have $$f_M = f_{M'}$$ which gives us $$M = M'$$. Similarly we have that $$f_M \mid f_K$$ in $$L[X]$$ for every $$M$$.

Now, since $$L[X]$$ is a UFD, we get: $$f_K = \prod_{K \subseteq M \subseteq L}f_M.$$ Since this factorization is unique and all factors are monic and irreducible there are no units in the product so we get: $$\deg(f_K) = \sum_{K \subseteq M \subseteq L}\deg(f_M) = n$$ which means there are only finitely many intermediate fields.

I see a couple of problems with your proof. But we will manage to salvage the idea of your proof:

"Problem": The statement that $$M\subset M'\Leftrightarrow f_M\in M'[X]$$ needs a proof (unless you know this from somewhere). The implication "$$\Rightarrow$$" is immediate, the other direction is not. The problem is, that it could happen that $$M\not\subset M'$$ and $$M'\not\subset M$$.

Solution: If $$f_M\in M'[X]$$, then $$f_M = f_{M\cap M'}$$. Let $$F:=M\cap M'$$ Then we can conclude by a degree argument: $$[L:K] =[L:F]\cdot[F:K]= [L:M]\cdot[M:F]\cdot[F:K]$$ But $$[L:M]=[L:F]$$ (since the minimal polynomials are the same) and hence $$[M:F]=1$$. Thus $$M=F=M\cap M'$$ and $$M\subset M'$$.

Problem: From your assumption that $$f_M\in M'[X]$$ you can't deduce that $$f_M=f_{M'}$$. As a counterexample look at any nontrivial extension $$K(\alpha)/K$$. $$f_K\neq f_{K(\alpha)}$$ since $$f_{K(\alpha)}=X-\alpha$$. The problem in your argument is the sentence: "However, since both polynomials are irreducible and monic...". Both are only irreducible as polynomials over their respective base fields.

Solution: Strike this paragraph. As you mention in the first paragraph: "Now I want to show that any $$f_M$$ uniquely determines the corresponding intermediate field $$M$$." Do just that! Assume that $$f_M=f_{M'}$$ and you can skip a couple of sentences in your proof.

Problem: Your claim $$f_K = \prod_{K \subseteq M \subseteq L}f_M$$ doesn't hold. I assume that again you are making the mistake of assuming that all of those $$f_M$$ are irreducible over $$L[X]$$? You can check that this is wrong with any non-trivial example. If $$f_K = f_K\cdot g$$, then $$g=1$$, so all the $$f_M$$ in your product would have to be units, thus constant).

Solution: You don't need this. It will suffice that in a UFD any element only has (up to associates) only finitely many divisors (why?). Any $$f_M$$ divides $$f_K$$, so there can only be finitely many such $$f_M$$. Thus there can only be finitely many intermediate fields.

Bonus: Why does any $$f_M$$ divide $$f_K$$? We know that $$f_K\in M[X]$$ and by definition (or a basic lemma, depending on your source) $$f_M$$ divides any polynomial $$g\in M[X]$$ that satisfies $$g(\alpha)=0$$. We know that $$f_K(\alpha)=0$$.

• First of all thank you for your very detailed answer! As to your quesiton why every element in a UFD only has finitely many divisors (up to associates): Let $R$ be a UFD and $a \in R$ s.t $a = \varepsilon\prod_{i \in I}p_i^{v_i(a)}$ where $v_i(a) = 0$ for almost every $i \in I$, $\varepsilon \in R^{\times}$ and $p_i$ are irreducible elements. Does this already show that any element in $R$ only has finitely many divisors, since almost every $p_i^{v_i(a)} = 1$? Commented Dec 24, 2020 at 20:22
• More or less, yes! You now have to argue that any divisor of $a$ can only be composed of irreducible factors (with small enough exponents) of $a$ and there are only finitely many such combinations
– CPCH
Commented Dec 24, 2020 at 20:26
• If $d$ divides $a$ then $d\cdot r = a = \varepsilon \prod_{i \in I}p_i^{v_i(a)}$ for $r \in R$. Since this factorization on the RHS is unique, d is composed of irreducible factors of $a$. Is this correct? Commented Dec 24, 2020 at 20:58