# How to show that a prime degree separable field extension containing a nontrivial conjugate of a primitive element is Galois and cyclic

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.

#### Solutions Collecting From Web of "How to show that a prime degree separable field extension containing a nontrivial conjugate of a primitive element is Galois and cyclic"

Yes, the argument (essentially) works. Let me phrase it in a more careful way,
without any assumption on the degree.
Let $F_i:=E(\alpha_i)$ and view the $E_i$ as subfields of their Galois closure.
There is an equivalence relation on the roots, namely, $\alpha_i \sim \alpha_j$ if
$F_i = F_j$. The Galois group $G \subseteq S_n$ acts on the roots, and preserves this equivalence. In particular, either

1. No two roots generate the same field

2. All the roots generate the same field (and thus $F$ is Galois of degree $n$).

3. $G$ is an imprimitive subgroup of $S_n$ (i.e. preserves a non-trivial decomposition of $1,\ldots n$). This is clear, because the decomposition is
given explicitly by the equivalence relation. In particular, this implies
that $G$ is a subgroup of $S_A$ wreath $S_B$ for some $A,B \ne 1$ with $n = AB$.

If $n$ is prime, then, since $G$ acts transitively, 3 cannot occur, so
“not 1” implies 2.

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 $E=K$.