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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.

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1 Answer 1

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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 $f_K(\alpha)=0$.

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  • $\begingroup$ 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 $p_i^{v_i(a)} = 1$? $\endgroup$
    – Shuster
    Dec 24, 2020 at 20:22
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    $\begingroup$ 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 $\endgroup$
    – CPCH
    Dec 24, 2020 at 20:26
  • $\begingroup$ 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? $\endgroup$
    – Shuster
    Dec 24, 2020 at 20:58

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