Simple Field Extensions from a Separable Element and an Arbitrary Element 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.
 A: 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)
