I was looking for some proof for the transitivity property of separable field extensions. Although this might sound like a very well-known fact and is referred to frequently, I do not seem to find a straightforward proof for the direction of the proof which sounds less obvious. Some books say it requires separable degree and skip it, some leave it as an exercise, and some say the method is like that property of algebraic extensions.
Any suggestions would be greatly appreciated.
On
The separable degree of $K/k$ is equal to the number of distinct $\sigma : K \rightarrow E$ where $E$ is a Galois extension containing $K$. In this sense $K/k$ is separable if there are $[K:k]$ many such maps. Thus transitivity of separability amounts to showing that the number of maps from $[F:k]$ is $[F:K][K:k]$. This in turn can be reduced by induction to showing that the number of maps from $k(\alpha,\beta)$ is equal to $[k(\alpha,\beta):k(\alpha)][k(\alpha):k]$ and this is easy to see by analysing the roots of polynomials. So I guess that was a little sketch of what need to be done.
Let's suppose E/K and K/F are both separable extensions. Let S be the separable closure of F in E.(In fact S consists of all elements of E which are separable over K). Since we have K/F is separable, then S contains K, and therefore E is separable over S. But E is purely inseparable over S, thus E=S and the result follows.