Why this extension is purely inseparable? A field $F$ is separably closed if whenever $\alpha\in\bar{F}$ is separable over $F$ we have $\alpha\in F$. A separable closure of $F$ is a field $E\supset F$ such that $E$ is separably closed and $E/F$ is separable. 
Now suppose that $E/F$ is normal and $K$ is the separable closure of $F$, how can I prove that $E/K$ is purely inseparable?
I was trying to prove that if we take $\alpha\in E$ and an homomorphism $\tau: K(\alpha)\rightarrow \bar{K}$ that fixes $K$ then it is the identity, but I got stuck, any help?
Another thing that is not so clear is how do we know that $K\subset E$?
 A: This only makes sense if $K$ is the separable closure of $F$ in $E$, that is, the set of elements of $E$ which are separable over $F$. Now, you may as well assume the characteristic of $F$ is $p>0$, since otherwise $K=E$. I claim that for any $\alpha\in E$, $\alpha^{p^n}$ is separable over $F$ for some $n\geq 0$. You can assume the minimal polynomial $f$ of $\alpha$ over $F$ it not separable, since otherwise we can take $n=0$. You can also assume $f$ has positive degree. Because $f^\prime=0$ (inseparability), $f(X)=g(X^p)$ for some $g\in F[X]$. Thus $\alpha^p$ is algebraic over $K$ of degree less than $\alpha$, so by induction, for some $n\geq 0$, $\alpha^{p^{n+1}}=(\alpha^p)^{p^n})$ is separable over $F$. So you're done. It follows that the minimal polynomial for $\alpha$ over $K$ looks like $(X-\alpha)^{p^m}$ for some $m$, which is to say that $\alpha$ is purely inseparable over $K$.
Note this doesn't use the fact that $E/F$ is normal. It does use the assumption that $E/F$ is algebraic. 
