Decomposition of Normal Field Extensions

Question

I am trying to prove the following homework problem:

Let $K/k$ be a normal field extension and $K_i$ and $K_s$ be intermediate extensions so that $K_i/k$ and $K/K_s$ are purely inseparable and $K_s/k$ and $K/K_i$ are separable. Show that $K=K_i\cdot{K_s}$ and $K_i \cap K_s=k$.

Showing $K_s \cap K_i =k$ is simple. Proving $K=K_i \cdot{K_s}$ seems a bit harder. My question is the following:

Is the condition $K=K_i \cdot K_s$ equivalent to showing $K_i\setminus k$ and $K_s \setminus k$ form a partition of $K\setminus k$?

I dont think this is correct, because I cannot see how the hypothesis of normality comes into play.

Answer 1

It is very easy to see that $K=K_i\cdot K_s$ and it does not use normality of $K/k$: I'll leave that to you.
However the result cannot be obtained (in general) without assuming $K/k$ to be normal.
$$ ?????$$
The key to the paradox is that normality of $K/k$ is a sufficient condition for the existence of a subfield $K_i\subset K$ such $K/K_i$ is separable and $K_i/K$ is purely inseparable !
For a general algebraic extension $K/k$, no such $K_i$ exists .
Indeed $K_i$ can only be the the perfect closure $k^{p^{-\infty}}$ of $k$ in $K$, but in general $K$ won't be separable over $k^{p^{-\infty}}$.
A counterexample and further explanations can be found here.

Edit: a hint
To prove that $K=K_i\cdot K_s$ (the original question!) it is enough to show that $K$ is both separable and purely inseparable over $K_i\cdot K_s$.