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Let $\xi_5$ be a primitive fifth root of unity in $\mathbb{C}$, then we know the extension $\mathbb{Q}(\xi_5)\supseteq\mathbb{Q}$ has degree 4 so the Galois group is of order 4. I am trying to find all intermediate extensions and it turns out the only one is $\mathbb{Q}(\sqrt{5})\supseteq\mathbb{Q}$. It is not hard to find this, but how to show that there are no more intermediate extensions?

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The Galois group is of order $4$ because the degree of the extension is $4$, but more is true. It's canonically isomorphic to $(\mathbf{Z}/5\mathbf{Z})^\times\cong\mathbf{Z}/4\mathbf{Z}$, i.e. it is cyclic of order $4$. Galois theory gives a bijective correspondence between intermediate fields and subgroups of the Galois group, so, since $\mathbf{Z}/4\mathbf{Z}$ only has one proper, non-trivial subgroup, $\mathbf{Q}(\xi_5)/\mathbf{Q}$ has a unique non-trivial intermediate extension.

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  • $\begingroup$ Thank you!.I didn't see the fact that this Galois group is cyclic. Now it is clear. $\endgroup$
    – Frank Lu
    Nov 23, 2013 at 20:17
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    $\begingroup$ finite extensions of finite fields are always Galois, and his Galois group cyclic ;) $\endgroup$
    – Lotus
    Feb 4, 2014 at 1:34
  • $\begingroup$ so sorry, i'm now studying finite fields and i see finite fields everywhere. Should i delete my comment? $\endgroup$
    – Lotus
    Feb 4, 2014 at 1:40

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