My lecturer gave the following proof (I've written it word for word) to show that an extension of finite degree $E/F$ is a simple extension of $F$ if and only if the number of fields $K$ with $F\subset K \subset E$ is finite. I'll add what I think is happening with a * as I go along.
Note: the case where F is finite is rather straight forward so I'll skip that part.
$(\Rightarrow)$ Suppose $E=F(\alpha)$, $\alpha \in E$. Suppose $F\subset K\subset E$. $K=($coefficients of the minimal polynomial).
((1*)I think that from this we can imply that there must be finitely many $K$ as $\alpha$ can only have finitely many different minimal polynomials)
$(\Leftarrow)$ Suppose there are finitely many intermediate fields K , $F\subset K \subset E$.
Choose $\alpha$ s.t. $|F(\alpha):F|$ is max possible.
((2*) this becomes relevant later on in the proof, it's saying that we set $\alpha$ to be such that it's extension field has the largest possible finite degree, but it confuses me as to how the proof still holds for lower degree extensions when we rely on this claim)
Suppose $F(\alpha)$ is properly contained in $E$, $\Rightarrow \exists \beta \in E\setminus F(\alpha)$ in particular $F(\alpha)$ is properly contained in $F(\alpha, \beta)$.
((3*)I don't understand how this doesn't immediately contradict our supposition that $|F(\alpha):F|$ was the ,maximum possible and this is my main question about the formulation of this proof.
Apart from that it's saying If we have a field properly contained In $E$ of course there must be some element in $E$ that's not in the field it contains, although we haven't said that $E=F(\alpha, \beta)$ just that there definitely exists an element in $E$ not in $F(\alpha)$ by our supposition, and so $F(\alpha)$ \subsetneq in $F(\alpha, \beta)$ as clearly $F(\alpha)$ does not contain $\beta$ )
Consider all expressions of the form:
$F(\alpha +\lambda \beta)$ where $\lambda \in F$
$F\subsetneq F(\alpha+\lambda \beta) \subsetneq E$
((4*) We're taking this step judging by the rest of the proof to come , so we can get a contradiction by showing the degree of this extension is greater than the degree of $F(\alpha)/F$, but again I don't understand why that wasn't already shown by adjoing $\beta$)
As $|F|$ is infinite $\exists$ $\lambda \neq \mu \in F$, s.t. $F(\alpha+\lambda \beta)=F(\alpha+\mu \beta)$
((5*) I'm not sure why the size of $F$ being infinite is important here , I would have thought that existence of such elements would be true if they were finite also, just from the fact that $F$ is the ground field I would have thought we could say this arbitrarily for all elements of F as the basis of the field extension would be unchanged simply by putting a coefficient from the ground field in front of $\beta$ (except zero if $F$ were a field containing zero) )
$\therefore$ $(\alpha + \lambda \beta)-(\alpha +\mu \beta)=(\lambda-\mu) \beta \in F(\alpha+\lambda \beta) \Rightarrow \beta \in F(\alpha+\lambda \beta) \Rightarrow (\alpha+ \lambda \beta) - \lambda \beta = \alpha \in F(\alpha + \lambda \beta) $.
So $F(\alpha , \beta)\subseteq F(\alpha + \lambda \beta)$ and clearly $F(\alpha , \beta)\supseteq F(\alpha + \lambda \beta)$
But $|F(\alpha + \lambda \beta):F|>|F(\alpha):F|$ which contradicts that $|F(\alpha):F| $ was of maximum possible degree. $\square$
So all my questions boil down to three things:
1) How can this still be true in general when we had to make the assumption $F(\alpha)/F$ was of max possible degree (what about lower degree?)
2) Why wasn't our supposition immediately contradicted and why did we have to go through all that work to find an extension with a greater degree when clearly $|F(\alpha,\beta):F|$ is greater than $|F(\alpha):F|$?
3) The question I raised in $5*$
This was not really an assumption. Since there are only finitely many inbetween extensions there must be a simple extension of maximal degree. The assumption was that this extension then is proper with respect to the bigger field.
Because it was not clear that $F(\alpha,\beta)/F$ is simple. Notice that $\alpha$ was defined to be an element such that $F(\alpha)/F$ is maximal with respect to degree. To get a contradiction we show that $F(\alpha,\beta)/F$ is simple as done in the proof.
This part is the tricky part of the proof. The existence of such $\mu\in F$ is more involved than I expected (linear algebra proof) and needs $F$ to be infinite indeed. I am thinking about lemma 3.3.4 from 'A Field Guide to Algebra' from Chambert-Loir.