TITLE: If $u$ is separable over $K$ and $v$ is purely inseparble over $K$, then $K(u,v)=K(u+v)$.
QUESTION [7 upvotes]: I've been having problems with this exercise of Hungerford's book, Algebra. I have been studying the section of separability by myself, and I already have a solution of the excercise, but it's very lengthy and cumbersome. I'm here looking for a better solution:
Let $F$ be a field extention of $K$, and $u\in F$ a separable element over K, $v\in F$ a purely inseparable element over $K$. Show that $K(u,v)=K(u+v)$.
My proof consist in show that $K(u+v)\subseteq K(u,v)$ and then show that $K(u,v)$ is both, separable and purely inseparable over $K(u+v)$, leading to the conclusion desired, but my method is not a very elegant one.
I would be grateful to see another solutions, thanks.

REPLY [8 votes]: We can suppose char$(K)=p>0$. We have $v^{p^m}=a\in K$ for some $m\geq0$, so $u^{p^m}=\bigl[(u+v)-v\bigr]^{p^m}=(u+v)^{p^m}-a\in K(u+v)\,.$ Therefore the element $u$ is purely inseparable over $K(u+v)$; but since $u$ is separable over $K$, then $u$ is also separable over $K(u+v)$. Consequently $u\in K(u+v)$, which in turn implies $v=(u+v)-u\in K(u+v)$, as desired.