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Question is to :

Determine the Subfields of $\mathbb{Q}(\zeta_8)$ generated by the periods of $\zeta_8$ and in particular show that not every subfield has such a period as primitive element.

What I have done so far is :

I could see that $Gal(\mathbb{Q}(\zeta_8)/\mathbb{Q})\cong (\mathbb{Z}/8\mathbb{Z})^*\equiv \mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$

I do not understand the question properly but then I would first of all find all sub fields of $\mathbb{Q}(\zeta_8)/\mathbb{Q}$

For that i would use fundemental theorem of galois theory which gives bijection between subfields of $\mathbb{Q}(\zeta_8)/\mathbb{Q}$ and subgroups of $Gal(\mathbb{Q}(\zeta_8)/\mathbb{Q})$ i.e., subgroups of $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$

I thought it would be helpful to write explicitly what are all the elements of the galois group...

$Gal(\mathbb{Q}(\zeta_8)/\mathbb{Q})=\{Id,\sigma_3,\sigma_5,\sigma_7\}$ where :

  • $\sigma_3(\zeta)=\zeta^3$
  • $\sigma_5(\zeta)=\zeta^5=-\zeta$
  • $\sigma_7(\zeta)=\zeta^7=-\zeta^3$

Subfields of Galois extension are fixed fields of the subgroups :

  • $H_3=\{Id,\sigma_3\}$
  • $H_5=\{Id,\sigma_5\}$
  • $H_7=\{Id,\sigma_7\}$

By doing similar calculations that are prescribed in examples, I could see that :

  • Fixed field of $H_3=\{Id,\sigma_3\}$ is $\mathbb{Q}(\zeta+\zeta^3)$

$\sigma_3(\zeta+\zeta^3)=\sigma_3(\zeta)+\sigma_3(\zeta^3)=\zeta^3+\zeta$

  • Fixed field of $H_5=\{Id,\sigma_5\}$ is $\mathbb{Q}(\zeta+\zeta^5)$

$\sigma_5(\zeta+\zeta^5)=\sigma_5(\zeta)+\sigma_5(\zeta^5)=\zeta^5+\zeta$

EDIT : Derel Holt reminded me that $\zeta+\zeta^5=0$ that clearly says that fixed field is just zero field.. As $\sigma_5(\zeta)=-\zeta$ It is obvious that only zero field is fixed field of $\zeta$

I am not able to make any sense out of this... Please help me to see this clearly..

Can a non trivial subgroup give trivial fixed field.. :O

  • Fixed field of $H_7=\{Id,\sigma_7\}$ is $\mathbb{Q}(\zeta+\zeta^7)$

$\sigma_7(\zeta+\zeta^7)=\sigma_7(\zeta)+\sigma_7(\zeta^7)=\zeta^7+\zeta$

Thus, I have found all sub fields but then I see that

  • "Each subfield has a period as primitive element"

I do not understand where did i go wrong...

Definition of period is :

Let $H$ be any subgroup of Galois group of $\mathbb{Q}(\zeta_p)$ over $\mathbb{Q}$ and let $$\alpha_H=\sum_{\sigma\in H}\sigma(\zeta_p)$$ The elements constructed in above equation and their conjugates are called the periods of $\zeta$.

Please help me to see where did i go wrong..

Thank you

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    $\begingroup$ $\zeta+\zeta^5=0$. $\endgroup$ – Derek Holt Feb 27 '14 at 20:40
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    $\begingroup$ You have drawn the wrong conclusion from Derek's comment. The correct conclusion is that your method fails to find the fixed field of $H_5$. After all, on general principles, you know this field will be a quadratic extension of the rationals. In fact, you have stumbled on the answer --- you have found a period that doesn't generate a subfield. $\endgroup$ – Gerry Myerson Feb 28 '14 at 12:08
  • $\begingroup$ @GerryMyerson : Yes Yes... I should get a quadratic extension but then as $\sigma_5=\sigma_3\circ \sigma_7$ I thought fixed fields also dependent and tried in that way and ended up with that zero field case :( I do not understand your " In fact, you have stumbled on the answer --- you have found a period that doesn't generate a subfield" :O $\endgroup$ – user87543 Feb 28 '14 at 12:11
  • $\begingroup$ Hint: what is $\sigma_5(i)$? $\endgroup$ – Gerry Myerson Feb 28 '14 at 12:13
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    $\begingroup$ @GerryMyerson : As $i=\zeta^2$ I see that $\sigma_5(i)=\sigma_5(\zeta^2)=(\sigma_5(\zeta))^2=\zeta^2=i$ so, Our fixed field contains $\mathbb{Q}(i)$ but then our field is of degree $2$ over $\mathbb{Q}$ so this should be the fixed field... Is my justification sufficient :O $\endgroup$ – user87543 Feb 28 '14 at 12:18
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To clarify, looking at the fields themselves:

  • the three quadratic fields in $\Bbb Q(\zeta_8)$ are $\Bbb Q(i),\Bbb Q(\sqrt 2)$ and the less obvious $\Bbb Q(i\sqrt 2)$.

  • two are generated with periods, the third is not.

  • two Galois automorphisms are `classical' conjugacies: $i\mapsto -i$ and $\sqrt 2 \mapsto -\sqrt 2$. Look up what the third does.

  • this goes to show that despite the fundamental theorem, field extensions can still be somewhat messy :)

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  • $\begingroup$ Can you explain why $Gal(\mathbb{Q}(\zeta_8)/\mathbb{Q}) \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ rather than $\mathbb{Z}/4Z$? This is where I am getting stuck on understanding this whole thing. I think the Galois group here should be cyclic. $\endgroup$ – jamesmartini Jun 3 '17 at 15:37
  • $\begingroup$ Sure. Each of these automorphisms repeated yields identity. Try it with the worst: $i\sqrt 2 \mapsto -i\sqrt 2 \mapsto i\sqrt 2$. Since no element has order 4 the 4-element group is not cyclic. $\endgroup$ – Emmanuel Amiot Jan 1 '18 at 21:55

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