Let $f(x)\in \mathbb Q[x]$ irreducible of degree $n$ and $K$ its splitting field over $\mathbb Q$. Prove that if $\operatorname{Gal}(K/\mathbb Q)$ is abelian, then $|\operatorname{Gal}(K/\mathbb Q)|=n$.

How can I prove this?

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    $\begingroup$ Of course there is nothing special to Q here! It could be any perfect field. In the case of finite fields (where every Galois extension is abelian) this is in particular useful. $\endgroup$ – Martin Brandenburg Feb 11 '14 at 15:27
  • $\begingroup$ It doesn't even need to be a perfect field. Only assuming $f$ is also separable, the theorem would work. $\endgroup$ – MaudPieTheRocktorate May 27 '17 at 18:19

I wonder if this works out:

Since $K/ℚ$ is abelian, every intermediate extension is normal and so is $ℚ(α)$ for some zero $α ∈ K$ of $f$. This must mean that $ℚ(α)$ is a splitting field of $f$ and so $K = ℚ(α)$.

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  • $\begingroup$ this is how I tried to prove it in the begining!(to prove that K=Q(a)) but I m not familiar with the term of normal extensions. I guess I'll have to do some searching $\endgroup$ – lea Feb 8 '14 at 13:16
  • $\begingroup$ Yes. It does work. Delightful! $\endgroup$ – Jyrki Lahtonen Feb 8 '14 at 13:21
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    $\begingroup$ lea: For an extension $F/K$, being normal with respect to the base field $K$ means that every galois automorphism from an extension $E/F/K$ restricts to an automorphism of the extension $F/K$. This corresponds to saying that $\mathrm{Gal}(E/F)$ is a normal subgroup of $\mathrm{Gal}(E/K)$. If $E/K$ is abelian, this true for any intermediate field $F$. It’s part of the main theorem of Galois theory. $\endgroup$ – k.stm Feb 8 '14 at 13:28
  • $\begingroup$ I understand what you say. It' s what i was looking for! $\endgroup$ – lea Feb 8 '14 at 14:28

I would try the following line of argument. Only outlining it now (think of it as extended hints):

  1. Identify the Galois group $G$ as a subgroup of the permutation of the $n$ roots. Why is $G$ transitive?
  2. Let $G_1, G_2, \ldots, G_n$ be the point stabilizers. That is, if $x_i$, $i=1,2,\ldots,n$, is one of the roots, then $G_i=\{\sigma\in G\mid \sigma(x_i)=x_i\}$. Show that all the groups $G_i$ are conjugate to each other and hence equal to each other.
  3. Show that the intersection $\cap_iG_i$ is trivial, and conclude that all the subgroups $G_i$ are trivial.
  4. Show that $|G|=n|G_i|$ for any $i$. Enjoy the happiness that comes from having completed this task.
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  • $\begingroup$ A great set of hints for this. $\endgroup$ – Tobias Kildetoft Feb 8 '14 at 12:37
  • $\begingroup$ @lea: I'm assuming that basics about groups acting on sets have been covered prior to this. If not, then ask for clarifications at the troubling points. $\endgroup$ – Jyrki Lahtonen Feb 8 '14 at 12:48
  • $\begingroup$ I didn't even think to solve it with actions and group theory. It's nice. I m ok with 1 and 2 (1 because f(x) is irreducible and 2 because the action is transitive hence the orbits match). I'm having a problem with 3 and 4 and where to use the fact that G is abelian. $\endgroup$ – lea Feb 8 '14 at 13:10
  • $\begingroup$ @lea: Conjugation is trivial in an abelian group, and the automorphisms in the intersection $G_i\cap G_j$ fix both $x_i$ and $x_j$, hence the automorphisms in $\cap_iG_i$ fix ... $\endgroup$ – Jyrki Lahtonen Feb 8 '14 at 13:14

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