TITLE: "Intersection" of separable subfields
QUESTION [3 upvotes]: I have the following question, from Isaacs' Algebra book.
Suppose $F\le E$ is a finite-degree normal field extension, and that $K$ and $L$ are intermediate subfields (between $F$ and $E$).  If $E$ is separable over $K$ and $L$, prove $E$ is separable over $K\cap L$.
I've tried to prove this by recognizing that $E$ is Galois over $K$ and $L$, and then showing $E$ is Galois over $K\cap L$.  Here is my attempt:
Let $M$ be the fixed field of $Gal(E/K\cap L)$, so that $K\cap L\subset M$. Now every element of $Gal(E/K)$ fixes $K\cap L$, and the same is true of $Gal(E/L)$. So $Gal(E/K)$ and $Gal(E/L)$ are subgroups of $Gal(E/K\cap L)$. By the fundamental theorem of Galois theory, $K$ (as the fixed field of the subgroup of $Gal(E/K)$) and $L$ (as the fixed field of the subgroup of $Gal(E/L)$) are fields containing $M$.  Then $M\subset K\cap L$, so $M=K\cap L$ and thus $E$ is Galois over $K\cap L$.
Can someone let me know if this proof works? Or if there is a different way to solve this exercise?

REPLY [2 votes]: The group $G=Gal(E/F)$ is finite. By Galois theory $K$ is the fixed field of the subgroup $G_1 = Gal(E/K)$ of $G$, and $L$ is the fixed field of the subgroup $G_2 = Gal(E/L)$ of $G$. So the fixed field of $(G_1, G_2)$ ( the subgroup of $G$ generated by $G_1$, $G_2$) is $K\cap L$.  Now we use the following fact: If $H$ is a finite group of automorphisms of a field $E$ and $M$ is fixed subfield then $E/M$ is a Galois extension. Indeed, let $e\in E$ and $e_1=e, e_2, \ldots, e_s$ the orbit of $e$ under $H$. The polynomial $\prod_i(X-e_i)$ has coefficients in $M$, and distinct roots, all in $E$.
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