Prove a certain cyclic extension with prime power order is simple

Question

Let $E/F$ be a cyclic extension of degree $p^n$, where p is prime. Let $L$ be an intermediate field such that $[E:L] = p$. If $E = L(\alpha)$, prove that $E = F(\alpha)$.

I've tried to work it out, but my line of reasoning hasn't gotten me anywhere yet. We have $$ p = [E:L] = [E:L(\alpha)]\cdot[L(\alpha):L] $$

By assumption $[E:L(\alpha)] = 1$, so the order of $\alpha \in E$ must be $p$.

I'm not sure why adding an element of order $p$ to $F$ will give me $E$, but it must make use of the fact that $E/F$ is a cyclic extension (which I take to mean its Galois group is in fact cyclic).

Answer 1

Since E/F is cyclic of degree $p^n$, for any integer m less or equal to n, there is a unique subextension $F_m$/F of degree $p^m$ , and E is the union of the ascending chain of the $F_m$ ' s . Your hypothesis is that E = $F_{n-1}$(a) . If F(a) were not equal to E, a would belong to a certain $F_m$ with m < n , hence a would be in $F_{n-1}$ : contradiction .

Answer 2

Let $\sigma$ be a generator of $Aut_F E$. By the fundamental theorem of Galois theory we see that $\sigma|_L$ generates $Aut_F L$ with $\left( \sigma|_L\right)^{p^{n-1}}=1_L$. As $E=L(\alpha)$, for the order of $\sigma$ to be $p^n$ there must be $\sigma^{p^{n-1}}(\alpha)\neq \alpha$. As $\sigma|_{F(\alpha)}$ generates $Aut_F F(\alpha)$, the order of $Aut_F F(\alpha)$ must be strictly larger than $p^{n-1}$ which forces $[F(u):F]=|Aut_F F(\alpha)|=p^n=[E:F]=[E:F(\alpha)][F(\alpha):F]$, hence $[E:F(\alpha)]=1$ and $E=F(\alpha)$ follows.