TITLE: Intermediate fields of cyclotomic field $\mathbb{Q}(\zeta_8)$ - Dummit Foote $14.5.2$
QUESTION [6 upvotes]: 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

REPLY [5 votes]: 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 :)