Here is the problem:
Suppose $F$ is a field. Let $\mu$ and $\nu$ be such that $F(\mu,\nu)/F$ is a finite extension with $\mu$ the root of a separable polynomial in $F[x]$. Show that there exists $\theta \in F(\mu,\nu)$ such that $F(\mu,\nu) = F(\theta)$; in other words, $F(\mu,\nu)/F$ is a simple extension.
I have spent over two hours on this problem, but have not managed to solve it. My efforts so far are as follows:
My educated guess is that I must make use of the following theorem:
Theorem If $K/F$ is a finite extension, then $K = F(\theta)$ if and only if there exist only finitely many subfields of $K$ containing $F$.
Since $\mu$ is the root of a separable polynomial, its minimal polynomial (call it $f(x)$) must divide this separable polynomial, hence $f$ itself must be separable. Let $K$ be the splitting field of $f$; then $K$ is Galois because $f$ is separable. Since $K/F$ is Galois, it is finite and separable, and hence by the Primitive Element Theorem, we have that $K = F(\omega)$ for some $\omega \in K$.
If I can show that $K(\nu)$ has only finitely many subfields containing $F$, then so will $F(\mu,\nu)$ since $F(\mu,\nu)$ is a subfield of $K(\nu)$. However, I'm not sure how to do this. I have a very vague idea of trying to introduce the splitting field for the minimal polynomial of $\nu$, but I'm not sure if that would work.
In any case, any help you could give me would be appreciated.
This answer is not particularly intuitive, but the technique is well-known in proving versions of the 'separable extensions are simple' theorem. It does not really require Galois theory.
Effectively, the idea is to consider fields of the form $F(\nu+t\mu)$. It turns out that any choice of $t\in F$ (aside from a handful of 'bad choices') makes it so that $F(\nu+t\mu)=F(\mu,\nu)$.
Note though that this proof does not directly work for finite fields (or at least I can think of no easy way to adjust it for finite fields). However, a separate proof for finite fields can be done quite easily, as all finite extensions of finite fields are simple. This can be proved in a few ways, though I think the easiest may be by using the fact that all finite multiplicative subgroups of fields are cyclic.
Anyway, the proof for infinite fields:
Theorem: Let $F$ be an infinite field. Let $\mu$ be any element that is separable over $F$ and $\nu$ be any element that is algebraic over $F$. Then, for all but finitely many $t\in F$, we have $F(\nu,\mu)=F(\nu+t\mu)$.\
Proof:
Let $g(x)$ be the minimal polynomial of $\nu$ over $F$ and let $f(x)$ be the minimal polynomial of $\mu$ over $F$. Let $E$ be a splitting of $g(x)f(x)$ over $F$. Let $\nu_1,\nu_2,\dots,\nu_n$ and $\mu_1,\mu_2,\dots,\mu_m$ be all the roots of $g(x)$ in $E$ and all the roots of $f(x)$ in $E$, respectively. The set $$\left\{\frac{\nu_i-\nu}{\mu-\mu_j}\Big| \mu_j\neq \mu\right\}$$ is clearly finite. We choose any $t\in F$ outside of this set and show that $F(\nu,\mu)=F(\nu+t\mu)$.
If we can show that $\mu\in F(\nu+t\mu)$, then we will be done. Let $h(x)$ be the minimal polynomial of $\mu$ over $F(\nu+t\mu)$. Clearly $h(x)|f(x)$, so all of $h(x)$'s roots belong to $\{\mu_i\}$. Also, by Theorem $2$, $h(x)$ is separable. This means $h(x)$ may be factored as $$\prod_{\mu_i\in K} (x-\mu_i)$$ where $K$ is some set of distinct $\mu_i$s that includes $\mu$.
At the same time, we note that the polynomial $g(\nu +t\mu-tx)$ belongs to $F(\nu+t\mu)[X]$ and clearly has $\mu$ as a root. So we must also have $h(x)|g(\nu +t\mu-tx)$. But because of how we have chosen $t$, $\nu+t\mu-t\mu_j$ does not equal any $\nu_i$ whenever $\mu_j\neq \mu$. Since the $\nu_i$s are all the roots of $g(x)$, this means no $\mu_j$ aside from $\mu$ is a root of $g(\nu +t\mu-tx)$. Using this and the facts derived from $h(x)$'s separability, we conclude $h(x)=x-\mu$. Then, since $h(x)\in F(\nu+t\mu)[X]$, we get $\mu\in F(\nu+t\mu)$. It immediately follows that $F(\nu,\mu)=F(\nu+t\mu)$.
(please comment or edit for any corrections or suggestions)