Consider $\mathbb{Q}(\zeta_8) / \mathbb{Q}$. I know that this has a Galois group isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_2$. Therefore, our automorphisms better have the structure that they square to give the identity. Furthermore, let $\{\alpha_1, \alpha_2,\alpha_3,\alpha_4\} := \{\zeta_8, \zeta_8^3, \zeta_8^5,\zeta_8^7\}$ be an enumeration of the roots

This gives us the automorphisms

\begin{align*} \sigma_1 = ID,\hspace{0.2cm} \sigma_2: \begin{cases} \alpha_1 \mapsto \alpha_2\\ \alpha_2 \mapsto \alpha_1\\ \alpha_3 \mapsto \alpha_4\\ \alpha_4 \mapsto \alpha_3 \end{cases} \sigma_3: \begin{cases} \alpha_1 \mapsto \alpha_1\\ \alpha_2 \mapsto \alpha_2\\ \alpha_3 \mapsto \alpha_4\\ \alpha_4 \mapsto \alpha_3 \end{cases} \sigma_4: \begin{cases} \alpha_1 \mapsto \alpha_2\\ \alpha_2 \mapsto \alpha_1\\ \alpha_3 \mapsto \alpha_3\\ \alpha_4 \mapsto \alpha_4 \end{cases} \end{align*}

We have found the correct automorphisms since they square to give the identity, preserving the structure of the Klein four group.

The next thing I want to know is the fixed fields corresponding to subgroups of the Galois group. As far as I know, there are 3 non-trivial subgroups \begin{align*} H_1=\langle\sigma_2\rangle,\hspace{0.2cm} H_2=\langle\sigma_3\rangle,\hspace{0.2cm} H_3=\langle\sigma_4\rangle \end{align*} Which infact are all isomorphic to $\mathbb{Z}/2$.

As far as I know finding these fixed fields means taking the extension field $E:= \mathbb{Q}(\zeta_8)$ and running the permutations over elements in it and checking if they're preserved. The only ones which stick out are the sums in each permutation. These are the elements

\begin{align*} E^{H_1} &= \{\alpha \in E \hspace{0.2cm}\mid\sigma\alpha = \alpha \hspace{0.2cm}\forall\sigma\in H_1\}\\ &= \mathbb{Q}(\zeta_8 + \zeta_8^3, \zeta_8^5 + \zeta_8^7) \end{align*}

And I gather that for $H_2$, $H_3$ the process is similar. These fixed fields are the elements in bijection with the subgroups of the Galois group.

My questions are, have I fully determined the fixed field? How do I know? More over, does this reduce and give us something more familiar?


EDIT: Due to the help of several posters below, I was able to determine that the automorphisms are actually given by

\begin{align*} \sigma_1 = ID,\hspace{0.2cm} \sigma_2: \zeta_8^k \mapsto (\zeta_8^3)^k,\hspace{0.2cm} \sigma_3: \zeta_8^k \mapsto (\zeta_8^5)^k,\hspace{0.2cm} \sigma_3: \zeta_8^k \mapsto (\zeta_8^7)^k \end{align*}

Furthermore that the subgroups of the Galois group are still given as above.

And the fixed fields \begin{align*} \mathbb{Q}(\zeta_8)^{H_1} &= \mathbb{Q}(\zeta_8 + \zeta_8^3, \zeta_8^5 + \zeta_8^7) = \mathbb{Q}(i\sqrt{2})\\ \mathbb{Q}(\zeta_8)^{H_2} &= \mathbb{Q}(\zeta_8 + \zeta_8^5, \zeta_8^5 + \zeta_8^7) = \mathbb{Q}\\ \mathbb{Q}(\zeta_8)^{H_3} &= \mathbb{Q}(\zeta_8 + \zeta_8^7, \zeta_8^3 + \zeta_8^5) = \mathbb{Q}(\sqrt{2})\\ \end{align*}

These were found by actually drawing the 8 roots of unity in the complex plane and using the fact that $$\zeta_8 = \frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}$$

  • $\begingroup$ I am not sure the $\sigma_3$ and $\sigma_4$ you gave are automorphisms. Since $\sigma_3(\zeta_8) = \zeta_8$ it must be the identity automorphism, because they are defined by their effect on the primitive element of $\mathbb{Q}(\zeta_8)$. $\endgroup$
    – Tob Ernack
    Commented Jun 4, 2017 at 1:15
  • $\begingroup$ What about $\sigma_k$ defined by left multiplication of the roots by $\zeta_8^{2k}$? Then, we can get 3 automorphisms from powers of $\sigma_k$ and the fourth power being the identity. This would make the Galois group cyclic, isomorphic to $\mathbb{Z}/4$. I think these are all automorphisms for the reason you mentioned above. However, I am still unsure. $\endgroup$
    – user311475
    Commented Jun 4, 2017 at 1:21
  • $\begingroup$ Are you saying the map $\sigma_k(\zeta_8^i) = \zeta_8^{2k}\zeta_8^i = \zeta_8^{2k + i}$? That would also not be an automorphism because it would need to satisfy $\sigma_k(\zeta_8^i) = \sigma_k(\zeta_8)^i = (\zeta_8^{2k}\zeta_8)^i = \zeta_8^{2ki + i} \neq \zeta_8^{2k + i}$ in general. $\endgroup$
    – Tob Ernack
    Commented Jun 4, 2017 at 1:30
  • 1
    $\begingroup$ Your Galois group needs to act transitively on the roots. Try $\zeta\to \zeta^3$, $\zeta\to \zeta^5$, $\zeta\to \zeta^7$ Your fixed fields should be generated by one element and can be identified as quadratics over $\Bbb{Q}$ $\endgroup$
    – sharding4
    Commented Jun 4, 2017 at 1:30

2 Answers 2


These are not the correct automorphisms. Instead they should be $$\sigma_1 = \mathrm{id}, \; \; \sigma_2 : \zeta^k \mapsto \zeta^{3k}, \; \; \sigma_3 : \zeta^k \mapsto \zeta^{5k}, \; \; \sigma_4 : \zeta^k \mapsto \zeta^{7k}.$$

For any subfield $K \subseteq \mathbb{Q}(\zeta_8)$, the trace $$\mathrm{Tr} : \alpha \mapsto \sum_{\sigma \in \mathrm{Gal}(\mathbb{Q}(\zeta_8) / K)} \sigma(\alpha)$$ is a surjective linear map. In particular, if you choose a vector space basis of $\mathbb{Q}(\zeta_8)$ and apply $\mathrm{Tr}$, you will get a $\mathbb{Q}$-basis of $K$ (which in particular generates $K$ as a field.) This seems to be more or less what you are doing anyway.

For example, the fixed field of $\sigma_2$ is $$K = \mathbb{Q}\Big( 1 + \sigma(1), \zeta + \sigma(\zeta), \zeta^2 + \sigma(\zeta^2),..., \zeta^7 + \sigma(\zeta^7) \Big).$$ It is not hard to see that $\zeta + \sigma(\zeta)$ already generates $K$ as a field, since $[K: \mathbb{Q}] = 2.$

  • $\begingroup$ How did you pick those automorphisms? As far as I can see it doesn't preserve the structure of the Klein 4 group. $\endgroup$
    – user311475
    Commented Jun 4, 2017 at 1:59
  • 1
    $\begingroup$ @jamesmartini In general the Galois group of $\mathbb{Q}(\zeta_n) / \mathbb{Q}$ is naturally $(\mathbb{Z}/n\mathbb{Z})^{\times}$, where $a \in (\mathbb{Z}/n \mathbb{Z})^{\times}$ acts by $\zeta \mapsto \zeta^a$ $\endgroup$
    – user452093
    Commented Jun 4, 2017 at 2:01
  • $\begingroup$ In our case $(\mathbb{Z}/8\mathbb{Z})^{\times} = \{1,3,5,7\}$, each element in this group has $1^2 = 3^2 = 5^2 = 7^2 =1 mod 8$, therefore $(\mathbb{Z}/8\mathbb{Z})^{\times}\cong \mathbb{Z}_2 \times \mathbb{Z}_2$, which is the Klein four group. $\endgroup$
    – user311475
    Commented Jun 4, 2017 at 2:03
  • $\begingroup$ @jamesmartini Yes, that's what I mean $\endgroup$
    – user452093
    Commented Jun 4, 2017 at 2:05
  • $\begingroup$ @u452093 From what I can see, when you square those automorphisms you don't get the identity? Sorry for labouring the point, but this is why I can't understand the problem $\endgroup$
    – user311475
    Commented Jun 4, 2017 at 2:13

You might note that the minimal polynomial of $\zeta_8$ (over $\Bbb Q$) is:

$\Phi_8(x) = \dfrac{x^8 - 1}{\Phi_4(x)\Phi_2(x)\Phi_1(x)} = x^4 + 1$,

that is, the roots are primitive eighth roots of unity.

Note that the eighth roots of unity form a (multiplicative) cyclic group of order $8$, and the primitive roots correspond to the units of $\Bbb Z_8$. It follows that any automorphism of $\Bbb Q(\zeta_8)$ preserving $\Bbb Q$ must correspond to an automorphism of $\Bbb Z_8$, so:

$\text{Gal}(\Bbb Q(\zeta_8)/\Bbb Q)$ is isomorphic to a subgroup of $\text{Aut}(\Bbb Z_8) \cong U(8) \cong \Bbb Z_2 \times \Bbb Z_2$.

If you know as well, that $\Phi_8(x)$ is irreducible over $\Bbb Q$ (a non-trivial fact, but not that hard to show in this case), then we know that:

$|\text{Gal}(\Bbb Q(\zeta_8)/\Bbb Q)| = 4$, so this is isomorphic to $\Bbb Z_2 \times \Bbb Z_2$.

As these roots of $x^4 + 1$ are all complex, we know that complex-conjugation is one of our automorphisms, which then has a fixed field which is a sub-field of the reals, which is algebraic of degree $2$.

It might also be helpful to know an explicit form for $\zeta_8$, namely:

$\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}$.

I leave it to you to show that $\Bbb Q(\zeta_8) = \Bbb Q(\sqrt{2},i)$, one inclusion is obvious. This shows that $\Bbb Q(\sqrt{2})$ is a subfield of $\Bbb Q(\zeta_8)$, and it corresponds to (is the fixed field of) the automorphism:

$\zeta_8 \mapsto (\zeta_8)^7$ (this is just the automorphism $g \mapsto g^{-1}$ that every abelian group has).

There are two other non-trivial automorphisms:

$\zeta_8 \mapsto (\zeta_8)^3\\ \zeta_8 \mapsto (\zeta_8)^5.$

In your terms, these are the automorphisms:

$\alpha_1 \mapsto \alpha_2\\ \alpha_2 \mapsto \alpha_1\\ \alpha_3 \mapsto \alpha_4\\ \alpha_4 \mapsto \alpha_3$


$\alpha_1 \mapsto \alpha_3\\ \alpha_2 \mapsto \alpha_4\\ \alpha_3 \mapsto \alpha_1\\ \alpha_4 \mapsto \alpha_2.$

You may find it rewarding to show that the latter automorphism fixes $\Bbb Q(\zeta_8 + (\zeta_8)^3) = \Bbb Q(i\sqrt{2})$, and the former fixes $\Bbb Q((\zeta_8)^3(\zeta_8)^7) = \Bbb Q(i)$.

  • $\begingroup$ It seems as though there are two sums which are fixed in each automorphism, are both sums equal or perhaps simply one is rational? e.g. $\zeta_8^k \mapsto (\zeta_8^k)^3$ gives the two fixed sums $\zeta_8 + \zeta_8^3$, and $\zeta_8^5 + \zeta_8^7$. But you didnt mention the latter in the first fixed field $\mathbb{Q}(i\sqrt{2})$. $\endgroup$
    – user311475
    Commented Jun 4, 2017 at 4:20
  • $\begingroup$ I figured it out, I drew all the 8th roots of unity in the complex plane and determined what numbers the sums produced quite easily, thx again! $\endgroup$
    – user311475
    Commented Jun 4, 2017 at 5:01

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