I am trying to prove the following statement:
Let $l \subseteq E \subseteq L$ and $l \subseteq F \subseteq L$ be a towers of field extensions, and suppose $E/l$ is a separable extension. $EF/F$ is necessarily separable.
For the sake of completion, I am posting a slightly expanded version of Georeges' solution suggestion:
Note that $EF$ is generated over $F$ by $\alpha \in E$. By hypothesis, $\alpha \in E$ is separable over $k$, and therefore $\alpha$ is separable over $F$ since $m_{\alpha,F}(X)$ divides $m_{\alpha,k}(X)$ and $m_{\alpha,k}$ has no repeated roots. Since $EF/F$ is generated by separable elements, it must be a separable extension.
EDIT: Thanks to a comment by Chris Eagle, the attempt below has no reason to be a workable strategy, since if it were $F/l$ would be separable, which it does not have to be.
Note that we have the following tower of extensions:
$$ l \subseteq E \subseteq EF $$
Thus, if we can show $EF/l$ is separable, $EF/F$ is necessarily a separable extension. Assuming $\alpha \in EF$ is not separable over $l$, we can deduce that the minimal polynomial of $\alpha$ over $l$, $m(x)$, is not separable, hence has repeated roots. We can then conclude that $m(x)=g(x^p)$ for some separable, irreducible polynomial $g(x) \in l(x)$. This leads to the conclusion that $\alpha^p$ is separable over $l$, despite $\alpha$ being inseparable. From here, I am unsure how to derive a contradiction. Maybe it just hasn't hit me yet. Can anyone see a way to continue, or is this attempt doomed to failure?
Best Answer
The field $EF$ is generated over $F$ by the elements $a\in E $ i.e. $ EF=F(E)$
Since each $a\in E$ is separable over $l$ it is a fortiori separable over $F$:
Indeed the minimal polynomial $Irr(a,F)$ of $a$ over $F$ divides $Irr(a, l)$, the minimal polynomial of $a$ over $l$ , and so has simple roots too.
Since an extension generated by separable elements is separable (Lang, Algebra, Ch.V, Thm. 4.4), we conclude that $EF/F$ is separable .