I have problems to prove the following statement.
Let $K$ be a field and let $K(u,v)$ be an algebraic extension of $K$. If $u$ is separable over $K$ then $K(u,v)$ is a simple extension.
(My incomplete) proof: We can suppose that $char K=p$, because in characteristic $0$ all finite extensions are simple. Let's divide the problem in $3$ cases (the third is problematic):
1) $v$ is separable. In this case $K(u,v)$ is a separable extension of $K$ and so it is simple.
2) $v$ is purely inseparable. We prove that $K(u,v)=K(u+v)$; by definition exists $m\in\mathbb N$ such that $v^{p^m}\in K$, so $(u+v)^{p^m}=u^{p^m}+v^{p^m}\in K(u^{p^m})$. So $K(u^{p^m})\subseteq K(u+v)$, but $u$ is separable on $K$ and follows that $K(u^{p^m})=K(u)$. Moreover $v=(u+v)-u\in K(u+v)$ and we can conclude that $K(u+v)=K(u,v)$.
3) $v$ is neither separable nor purely inseparable. ???
I would appreciate any hint about the point 3), but also a more organic proof without any subdivision.
Thanks in advance.
Look at the usual proof in the separable case. The only place they use separability is to show the gcd of the minimal polynomials has distinct roots. This only really requires one of the polynomials to be separable.