Let $K=\mathbb Q$ and $L=\mathbb Q(\zeta_8)$, where $\zeta_8$ is a primitve $8th$ root of unity.

I have to determine the Galois group of this extension, the subgroups of it and the associated intermediate fields.

It was not hard to determine the Galois group. We have $Gal(L/K)\cong\mathbb Z_8^{*}\cong\mathbb Z_2\times \mathbb Z_2$

So the Galois group has exactly $4$ subgroups, namely the generated groups of the following automorphisms:





So the non trivial subgroups are $<\sigma_2>,<\sigma_3>\text{ and }<\sigma_4>$.

Now its not hard to see that $\mathbb Q(\zeta^1+\zeta^3)\subseteq L^{<\sigma_2>}$

But if we want to show equality, we have to show that $[\mathbb Q(\zeta^1+\zeta^3):\mathbb Q]=2$

Does the following argument work?: $\zeta^1+\zeta^3$ is real if and only if $\overline{\zeta^1}=\zeta^3$, but $\overline{\zeta^1}=\zeta^7$, so $\zeta^1+\zeta^3\in \mathbb C/\mathbb R$, hence the degree over $\mathbb Q$ must be greater than $1$, so it must be $2$.

Is this okay?

However, what we should do with the fixed field of $\sigma_4$?. We know that $\mathbb Q(\zeta^1+\zeta^7)\subseteq L^{<\sigma_4>}$ But the generator is real. How we can show equality?


1 Answer 1


You have made a good start, so I give some extended hints. Ask for more if needed.

For handling $\zeta^1+\zeta^7$ it may be best to use the explicit formula (and Moivre) $$\zeta=e^{2\pi i/8}=\cos\frac\pi4+i\sin\frac\pi4=\frac{1+i}{\sqrt2}.$$

Actually using Moivre will give you another useful viewpoint to the fixed field of $\langle\sigma_2\rangle$ as well :-)

With $\zeta^5$ (or $\langle \sigma_3\rangle$) the sum $\zeta^1+\zeta^5$ won't help you because it vanishes. What about the product $\zeta^5\zeta^1$?

  • $\begingroup$ Thanks for the answer. Shouldn't it be $\zeta^7$ instead of $\zeta^5$ in the last lines? Showing that the $\mathbb Q(\zeta^1+\zeta^5)$ is the fixed field of $\sigma_3$ was not so hard for me. (Very similar to the first fixed field). $\endgroup$ Feb 19, 2015 at 11:44
  • $\begingroup$ Something must have gone wrong then, @Mathlearner. $\Bbb{Q}(\zeta^1+\zeta^5)$ is NOT the fixed field of $\langle\sigma_3\rangle$. This is because $\zeta^4=-1$, so $\zeta^1+\zeta^5=0$. OTOH Galois theory tells us that the fixed field is quadratic over $\Bbb{Q}$. $\endgroup$ Feb 19, 2015 at 11:52
  • $\begingroup$ Okay, I see. But I don't know then how to determine the fixed field. Until now I did always the following: If I want to determine the fixed field of $\sigma_3$, then I took $\zeta$ and looked at the image of $\zeta$ under $\sigma_3$. The image is $\zeta^5$. And the image of $\zeta^5$ under $\sigma_3$ is $\zeta$, hence their sum is the fixed field. (If the degree is correct). This method worked at least for $\sigma_2$..can you help me? $\endgroup$ Feb 19, 2015 at 12:03
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    $\begingroup$ Okay, if $\sigma_3$ maps $\zeta$ to $\zeta^5=-\zeta$, then the fixed field is maybe $\mathbb Q(\zeta^2)$? $\endgroup$ Feb 19, 2015 at 12:18
  • $\begingroup$ Correct. One way to see that is to observe that $\zeta\cdot\zeta^5=\zeta^6=-\zeta^2$ is fixed under $\sigma_3$. Your teacher may appreciate it if you identify $\zeta^2=i$. $\endgroup$ Feb 19, 2015 at 13:43

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