Composition of two simple field extension with coprime degree is simple?

Question

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.

Answer 1

We know that if $F(a)/F$ is separable and $F(b)/F$ is simple then $F(a,b)/F$ is simple.

Next in your notations write $E=F(u),L=E(v)=F(u,v)$.

• If $F(u)/F$ is separable then apply the previous point to get $F(u,v)=F(w)$.

• If $F(u)/F$ is not separable then $p| [F(u):F]$ so $p\nmid [E(v):E]$ and hence $E(v)/E$ is separable. For $n$ large enough $v^{p^n}$ is separable over $F$.

  $E(v)/E$ being separable gives that $v\in E(v^{p^n})$ ie. $F(u,v)=F(u,v^{p^n})$, and you can apply again the previous point to get $F(u,v^{p^n})=F(z)$.