# $K(u,v)$ is a simple extension of fields if $u$ is separable

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.

• However it is the exercise 7 at page 59, from the Kaplansky's Book "Fields and Rings". This is the hint: prove that there are only finitely many intermediate fields. It can be assumed that $K(u)$ is the maximal separable subfield. Analyze an intermediate field $L$ by showing that it lies between its maximal separable subfield $L_0$ and $L_0(v)$. – Dubious Jul 1 '13 at 21:07