Let $E/F$ be a simple field extension of degree $m$ and $L/E$ be a simple field extension of degree $n$, where $\gcd(m,n)=1$. Is it necessary that $L/F$ is simple?
In the setting of characterstic $0$ there actually isn't much to say, so I will suppose that $F$ is of characterstic $p\neq 0$. The following result may be useful and a proof can be found here:
Proposition. Suppose that $\alpha,\beta$ are algebraic over $F$ and at least one of $\alpha$ and $\beta$ is separable over a field $F$, then $F(\alpha,\beta)/F$ is simple.
And here are my thoughts. In the original question, write $E = F(\alpha)$ and $L = F(\alpha,\beta)$.
$\bullet$ If $\beta$ has degree $mn$ over $F$ then $L = F(\beta)$, done.
$\bullet$ If $p\nmid m$ then the result follows immediately since $\alpha$ is separable (it is well-known that the degree of an inseparable extension of a field must divide the characteristic).
$\bullet$ If there exists $\gamma\in L$ such that $[F(\gamma):F] = n$, then we also have the result. In this case $L = F(\alpha,\gamma)$ and at least one of $\alpha$ and $\gamma$ is separable since $m$ and $n$ cannot both be multiples of $p$.
But it occurs to me that in a general setting, such $\gamma\in L$ need not exist (essentially the same idea as here). Consider (in characteristic $0$) $F = \mathbb{Q}$, $L$ be a field such that $[F:L] = 6$ and the Galois group of the splitting field is $A_4$. Then $L$ is the fixing field of some order-$2$ subgroup $H$ of $A_4$. Let $E$ be the fixing field of the Klein-four subgroup of $A_4$, then $[E:\mathbb{Q}] = 3$ and $[L:E] = 2$, but $\gamma\in L$ such that $\mathbb{Q}(\gamma)$ has degree $2$ does not exist since $A_4$ has no subgroup of order $6$. Here is an explicit such example.
So is the state true in general? Thank you in advance for any help.
