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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?

user26857
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lea
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2 Answers2

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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 $K = ℚ(α)$.

k.stm
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  • 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 – lea Feb 08 '14 at 13:16
  • Yes. It does work. Delightful! – Jyrki Lahtonen Feb 08 '14 at 13:21
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    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. – k.stm Feb 08 '14 at 13:28
  • I understand what you say. It' s what i was looking for! – lea Feb 08 '14 at 14:28
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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.
Jyrki Lahtonen
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    A great set of hints for this. – Tobias Kildetoft Feb 08 '14 at 12:37
  • @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. – Jyrki Lahtonen Feb 08 '14 at 12:48
  • 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. – lea Feb 08 '14 at 13:10
  • @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 ... – Jyrki Lahtonen Feb 08 '14 at 13:14
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    That's a really nice one! Alternatively, from 3. we could also deduce that any root of $f$ is fixed only by the trivial automorphism, so any root generates the whole splitting field. – ალექსანდრე ჭაღალიძე Feb 01 '25 at 16:57