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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.

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up vote 3 down vote accepted

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 $K_i\cdot K_s$.

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I'm unsure why the containment $K \subseteq K_i \cdot K_s$ is easy to see. Letting $x \in K$ and assuming $x \not \in K_i \cdot K_s$ seems to pose no immediate problems. Perhaps I am going in the wrong direction and just not seeing the obvious. –  Holdsworth88 May 20 '12 at 12:58
    
Dear Holdsworth88, I have added a hint in an Edit to my answer. Feel free to tell me if it doesn't suffice. –  Georges Elencwajg May 20 '12 at 14:01
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