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Let $\omega=e^{i\pi/4}$ be the $8$-th root of the unity. I know that $\mathbb{Q}(\omega)/\mathbb{Q}$ is Galois, as it is the splitting field of the separable polynomial $p(x)=x^8-1$. However, I ran into trouble calculating the subfields of $\mathbb{Q}(\omega)$. The $4$ automorphisms in $G = \operatorname{Gal}(\mathbb{Q}(\omega)/\mathbb{Q})$ are: \begin{equation} \operatorname{id}, \quad \sigma_3(\omega):=\omega^3, \quad \sigma_5(\omega):=\omega^5, \quad \sigma_7(\omega):=\omega^7. \end{equation}

But when finding the fixed field of $H = \{\operatorname{id},\sigma_3\}$, I found that $\omega^k=\sigma_3(\omega^k)$ if and only if $2k \equiv 0 \!\pmod{\!\!8}$, in which case $k=0,4$. But $\omega^4=-1$, so the fixed field of $H$ is $\mathbb{Q}(-1)=\mathbb{Q}$.

However, since $H$ is a non-trivial subgroup of $G$, and the extension $\mathbb{Q}(\omega)/\mathbb{Q}$ is Galois, I thought the fixed field of $H$ should correspond to a nontrivial subfield of $\mathbb{Q}(\omega)$. Any insight as to where my calculations or assumptions may be off would be much appreciated.

azif00
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ChrisWong
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    You're definitely on the right track! But remember, you're looking for all elements of $\Bbb Q(\omega)$ that are invariant under $\sigma_3$ (for example), not just whether the individual roots of unity are invariant. What about $\omega+\omega^3$ for example? – Greg Martin Jan 03 '21 at 17:41
  • I see now, $\omega+\omega^3$ is fixed under $\sigma_3$. Similarly, $\omega+\omega^7$ is fixed under $\sigma_7$. I suppose writing out where each root is sent under the transformation helps, which is doable for $n=8$. Thanks! – ChrisWong Jan 03 '21 at 17:54
  • $Tr_{K/F}$ is surjective $K\to F$ and $Tr_{K/F}(a)=\sum_{h\in Gal(K/F)} h(a)$ – reuns Jan 03 '21 at 19:09
  • Yup, the sum (or the trace as reuns correctly points out) works here. Excpet with $\sigma_5$ given that $\omega+\omega^5=0$. Try with $\omega\cdot \omega^5$ instead in that case. – Jyrki Lahtonen Jan 03 '21 at 20:29
  • Also, see this. Close to being a duplicate actually. What do others think? I have many fitting dupehammers, but want to exercise a bit of caution. – Jyrki Lahtonen Jan 03 '21 at 20:31

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