TITLE: How to show that a prime degree separable field extension containing a nontrivial conjugate of a primitive element is Galois and cyclic
QUESTION [10 upvotes]: Let $F$ be a field and let $E/F$ be a separable extension, with $[E:F]=p$, a prime. Given a primitive $\alpha_1\in E$ with $F(\alpha_1)=E$. Let $\alpha_2\neq \alpha_1$ denote one of the $p$ conjugates of $\alpha$ (in an algebraic closure). If $\alpha_2\in F(\alpha_1)$, I want to show that $E$ is Galois and that $Gal(E/F)$ is cyclic.
Does this just follow by noting that there is an automorphism of $F$ that sends $\alpha_1$ to $\alpha_2$ and deducing that this extends to an automorphism of the closure, which then must send $\alpha_2$ to $\alpha_3$, and so on? And this automorphism must have degree $p$ and so must be cyclic? I feel like I'm missing something.

REPLY [6 votes]: Let $f(x)$ be the minimal polynomial of $\alpha_1$, and let $K$ be the splitting field of $f(x)$. Since $f(x)$ is separable (as we are assuming that $E$ is separable), then $K$ is Galois over $F$ and contains $E$.
The subgroup $H$ of $G=\mathrm{Gal}(K/F)$ that fixes $E$ is a subgroup that satisfies $[G:H]=[F:E]=p$. Hence, $H$ is maximal in $G$. Thus, $H$ is either normal, or $N_G(H) = H$.
If $H$ is normal, then $E$ is Galois over $F$; since $\mathrm{Gal}(E/F)$ has order $p$, it is cyclic of order $p$.
If $H$ were not normal, then $N_G(H)=H$, which means that $H$ has $p$ distinct conjugates in $G$; the conjugates corresponds to $p$ distinct intermediate extensions $E_{i}/F$, $i=1,\ldots,p$. If $\mathrm{id}=\sigma_1,\ldots,\sigma_p$ are coset representatives for $H$ in $G$, then $E_i = \sigma_i(E) = \sigma_i(F(\alpha_1)) = F(\sigma_i(\alpha_1))$. But we know that there are at least two $i$s for which $\sigma_i(E_1)=E_1$, namely the identity, and the $\sigma$ that maps $\alpha_1$ to $\alpha_2$ (no other coset representative can map $\alpha_1$ to $\alpha_1$, because then it would fix $E$ pointwise and thus lie in $H$). This contradicts our assumption that there are $p$ distinct $E_i$. The contradiction arises from the assumption that $H$ is not normal. Thus, $H$ is normal, and $E$ is Galois over $F$. At this point we actually recognize that $E=K$.