# “Intersection” of separable subfields [duplicate]

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?

## marked as duplicate by Arnaud D., José Carlos Santos, clathratus, Lee David Chung Lin, zz20sMar 22 at 2:44

• I think your solution is fine, and it is how I would do it. I think the key idea is "get a Galois group containing the groups for $K$ and $L$". – Steve D Oct 10 '17 at 1:15
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 $E$.
• I think you need to say more at the beginning ($E$ is not necessarily Galois over $F$). – Steve D Oct 10 '17 at 1:14
• @Steve D: It's never stated that the extension $E/F$ is Galois, I am just denoting the group of automorphisms by $Gal(E/F)$. The only detail that we need in fact is $[E\colon K\cap L]< \infty$. The important thing ( that unfortunately is covered in details in the statement of the exercise ) is : a field is a Galois extension of a field of invariants of a finite group. This plus: the group of invariants of a finite extension is finite. – Orest Bucicovschi Oct 10 '17 at 11:33
• Yes, I know it's not assumed $E$ is Galois over $F$. But you really need your "fact" from the end of your answer, at the beginning. You need to know $E$ is Galois over the fixed field of $Gal(E/F)$. The fact that $K$ and $L$ contain this fixed field is then enough to finish. – Steve D Oct 10 '17 at 15:01