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Why is it that the subfield fixed by the subgroup of this Galois group is $\mathbb{Q}(\sqrt5)$. Can someone explain it without using the cyclotomic extension of $\mathbb{Q}$?

Thank you

edit:

Using the definition used by Don in his response we find that the galois grp is $\{id, \omega, \omega^2, \omega^3 \}$. This grp has one subgroup $K = \{id, \omega^2\}$ I am trying to find the fixed subfield of $\mathbb{Q}(\zeta) \ (\zeta$ being the root of unity of $x^5-1)$ that is fixed by $K$

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  • $\begingroup$ Since $x^5 - 1$ is a cyclotomic extension of $\mathbb{Q}$, one has to use it. But perhaps you mean certain facts about cyclotomic extensions that you do not know. Also, when you say "the subgroup", what subgroup are you referring to? $\endgroup$
    – RghtHndSd
    Apr 18, 2014 at 14:31
  • $\begingroup$ The subgroup of automorphisms $\{id, \sigma^2\}$ that is isomorphic to $\mathbb{Z}_2$. This is a subgroup of the galois group $Gal_{\mathbb{K}/\mathbb{Q}}$ of the said polynomial. $\mathbb{K}$ being the splitting field of $x^5 - 1$ $\endgroup$
    – Quester
    Apr 18, 2014 at 14:47
  • $\begingroup$ What is $\sigma$? Please add these details to the question itself by editing it. $\endgroup$
    – RghtHndSd
    Apr 18, 2014 at 14:57
  • $\begingroup$ "The" subfield of that Galois group? Perhaps you meant the real (quadratic or whatever) subfield of ...? $\endgroup$
    – DonAntonio
    Apr 18, 2014 at 14:59
  • $\begingroup$ Using the definition used by Don in his response we find that the galois grp is $\{id, \omega, \omega^2, \omega^3 \}$. This grp has one subgroup $K = \{id, \omega^2\}$ I am trying to find the fixed subfield of $\mathbb{Q}(\zeta) \ (\zeta$ being the root of unity of $x^5-1)$ that is fixed by $K$ $\endgroup$
    – Quester
    Apr 18, 2014 at 18:36

1 Answer 1

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Highlights

Put $\;\zeta:=e^{\frac{2\pi i}5}$, and let $\;\omega\in Gal(\Bbb Q(\zeta)/\Bbb Q)\;$ be complex conjugation.

Observe that for$\;z\in\Bbb C\;\;,\;\;\overline z=z^{-1}\iff |z|=1\;$ , and thus:

$$\omega(\zeta+\zeta^{-1})=\omega(\zeta)+\omega(\zeta)^{-1}=\zeta^{-1}+\zeta\implies \zeta+\zeta^{-1}\in\Bbb R$$

(or, of course, simpler: $\;\zeta+\zeta^{-1}=\zeta+\overline\zeta=2\,\text{Re}\,\zeta\in\Bbb R)\;$

which means

$$\zeta+\zeta^{-1}\in\Bbb Q^{\langle\omega\rangle}:=\;\text{the fixed field of}\;\;\omega$$

and since the order of $\;\omega\;$ in the Galois group is two and clearly $\;\zeta+\zeta^{-1}\notin\Bbb Q\;$ , we're done.

You can also check try to find the irreducible rational polynomial of $\;\zeta+\zeta^{-1}\;$ ...and it is a quadratic.

Finally, just check that

$$2\cos\frac{2\pi}5=\frac12(\sqrt5-1)\;\ldots$$

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  • $\begingroup$ Let $Gal(\mathbb{Q}(\zeta)/\mathbb{Q}) = \{id, \rho, \rho^2, \rho^3\}$ where $\rho(\zeta) = \zeta^2$ and $\rho^2(\zeta) = \zeta^3$ and $\rho^2(\zeta) = \zeta^4$ Could you write this complex conjugation $\omega$ in terms of the automorphism $\rho$ $\endgroup$
    – Quester
    Apr 23, 2014 at 23:12
  • $\begingroup$ @Quester: $$\omega(\zeta)=\overline z=z^{-1}=z^4=\rho^2(\zeta)\;\ldots$$ $\endgroup$
    – DonAntonio
    Apr 24, 2014 at 3:28
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    $\begingroup$ Thank you Don. I figured out the error I was making before when I asked this question. Now all is clear. :) $\endgroup$
    – Quester
    Apr 25, 2014 at 23:42

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