# 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.

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A priori, it is not immediately clear to me that the image of $\alpha_2$ under the automorphism must be different from $\alpha_1$ (if $p\gt 2$, of course); it just must be some root in the extension different from $\alpha_2$, but why can't the automorphism just swap $\alpha_1$ and $\alpha_2$ if $p\gt 2$?. If you do show that the extension is Galois, then the fact that the Galois group must be cyclic is immediate, since $|\mathrm{Gal}(E/F)| = [E:F] = p$, and there is one and only one group of prime order: the cyclic group of order $p$. – Arturo Magidin Mar 9 '12 at 4:39
Should I consider the minimal polynomial of $\alpha_1$ and consider its splitting field? I was thinking that in the splitting field the automorphisms must permute the roots and that the nontrivial autormphisms of $E$ must be these automorphisms on the splitting field that fix $E$. – Lepidopterist Mar 9 '12 at 4:50
If $E$ is really Galois over $F$, then it will be the splitting field of the minimal polynomial of $\alpha_1$. And the automorphism of the splitting field that maps $\alpha_1$ to $\alpha_2$ will map $F$ to itself (since $F(\alpha)_2\subseteq F(\alpha_1)$, but of degree $p$ over $E$, hence equal). What is not clear to me is on what grounds you assert that this automorphism must act transitively on the roots (it will if your conclusion is correct, but I don't see how you justify it). – Arturo Magidin Mar 9 '12 at 4:52
Well I think we know that minimal polynomial of $\alpha_1$ should be irreducible and separable (otherwise the extension couldn't be prime), right? So any automorphism on E is going to send $\alpha_1$ to $\alpha_2$ and the subgroup generated by this map has to divide $p$, no? I may very well be confused. – Lepidopterist Mar 9 '12 at 5:03
I thought any automorphism that fixes the ground field has to permute the roots of the minimal polynomial. – Lepidopterist Mar 9 '12 at 5:05

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.

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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 $E=K$.

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This answer seems to be more complicated and less general than my answer. Also, I thought that the OP wanted to know if their answer was correct or not? Since G acts faithfully on a set of $p$ elements, it contains an element $\sigma$ of order $p$ sending $\alpha_1$ to $\alpha_2$, and thus $F = E(\alpha_1)$ is preserved by $\sigma$. Since $\sigma$ acts transitively, all the roots lie in $F$. – Misty Mar 10 '12 at 0:28
@Misty: Yes, your answer is more general; which is why I voted it up. – Arturo Magidin Mar 10 '12 at 3:57