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
Best Answer
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 :)