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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 $p$.2012-03-09
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    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 $E$.2012-03-09
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    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).2012-03-09
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    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.2012-03-09
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    I thought any automorphism that fixes the ground field has to permute the roots of the minimal polynomial.2012-03-09
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    "Permute" does not mean that it has to cycle through them; it just means it has to sends the set of roots to the set of roots.2012-03-09

2 Answers 2

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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 $F$.2012-03-10
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    @Misty: Yes, your answer is more general; which is why I voted it up.2012-03-10