So the problem is,
"Prove that $FL/K$ is separable, given that $F/K$ and $L/K$ are separable extension."
I'm a newcomer in this field, so I don't know much results of separability. This is the definition I learn:
"An algebraic extension $L/K$ is separable iff $min(K,\alpha)$ is separable for all $\alpha \in L$."
Here's what I do so far, let $\alpha\in FL$, then I have to prove that $min(K,\alpha)$ is separable. We have $\alpha \in K(\beta_1,...,\beta_n,\gamma_1,...,\gamma_m)$ in which $\beta_i\in F, \gamma_i\in L$. I want to prove that $K(\beta_1,...,\beta_n,\gamma_1,...,\gamma_m)/K$ is a separable extension. This is where I'm stuck, am I on the right track?
THank you for your help
This may be proven more easily by using an alternative characterization of separability: A finite extension $ L/K $ of degree $ [L : K] = n $ is separable if a given embedding $ K \to \bar K $ can be extended to an embedding $ L \to K $ in exactly $ n $ different ways. (This number is less than or equal to $ n $ for any finite extension $ L/K $.) The number of such extensions is the separability degree of the extension $ L/K $. An infinite algebraic extension $ L/K $ is separable if and only if every finite subextension $ F/K $ is separable.
Then, we have the following claims:
Equivalence of definitions. An algebraic extension $ L/K $ is separable in the above sense if and only if each element $ \alpha \in L $ has minimal polynomial over $ K $ which is separable in $ K[X] $. Indeed, the number of distinct extensions of a fixed embedding $ K \to \bar K $ to an embedding $ K(\alpha) \to \bar K $ is equal to the number of distinct roots of the minimal polynomial of $ \alpha $ in $ \bar K $. If $ \alpha $ were inseparable over $ K $, this number would be strictly less than $ [K(\alpha) : K] $. Conversely, if every $ \alpha \in L $ has separable minimal polynomial, then given a finite subextension $ F/K $, we may pick a primitive element such that $ F = K(\beta) $. Since $ \beta $ is separable over $ K $, we have $ [K(\beta) : K] $ distinct extensions of a given embedding $ K \to \bar K $, by sending $ \beta $ to its distinct $ K $-conjugates.
Separability is transitive. Indeed, given a tower $ K \subset L \subset M $, we fix an embedding $ K \to \bar K = \bar L $. If both $ L/K $ and $ M/L $ are finite and separable, then this embedding extends to an embedding $ L \to \bar K $ in $ [L : K] $ distinct ways; and each of those embeddings extend to an embedding $ M \to \bar K = \bar L $ in $ [M : L] $ distinct ways. This gives a total of $ [M : L][L : K] = [M : K] $ embeddings for the extension $ M/K $, thus by the above characterization, the extension $ M/K $ is separable. The infinite case is left as an exercise.
Finite extensions generated by separable elements are separable. That is, if a finite extension $ L/K $ may be written as $ L = K(\beta_1, \beta_2, \ldots, \beta_m) $ where each $ \beta_i $ is separable over $ K $, then the extension $ L/K $ is separable. By the above transitivity result, it suffices to prove this for $ m = 1 $; in which case the claim is trivial by the above characterization. This extends readily to infinite algebraic extensions $ L/K $. From this, it follows that sums and products of separable elements are separable, and thus we have the claim:
Compositums of separable extensions are separable. Follows immediately from the above, since $ FL/K $ in a larger field $ M $ is exactly the subfield of $ M $ containing elements of the form $ f_1 l_1 + f_2 l_2 + \ldots + f_n l_n $ where $ f_i \in F $ and $ l_i \in L $. These clearly form a ring, and they form a field since the inverse of such an element is a polynomial function of itself with coefficients in $ K $, thus lies in the same collection. These elements are sums of products of separable elements, thus they are separable; hence the extension $ FL/K $ is separable as well.