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Let us consider a primitive $7^{\text{th}}$ root of unity $\eta$. Then the minimal polynomial of $ \eta $ over $ \mathbb{Q} $ is $1 + \eta +.....+ \eta^{6}$. So the dimension of the $\mathbb{Q}$-space $ \mathbb{Q}(\eta) = L$ (say) is $ 6 $. Now the group $G=\operatorname{Gal}(\mathbb{Q}(\eta)/\mathbb{Q}) $ is a cyclic group of order $ 6 $ and let $ \sigma $ be a generator. Let $ a \in L^{\times} $ and $ b,c \in L_{3} $, where $ L_3 $ is the fixed field of $ \sigma^{3} $. Consider the equation $$ \frac{\sigma^{2}(a)}{a} = \frac{b}{c} - 1 .$$ What is the maximum dimension of a subspace in $ L_3 \oplus L_3 $ such that this equation has no solution?

Note that the maximum dimension is greater than $3$ since if $ b=0 $ or $ c=0 $ or $ b=c $ then it has no solution, in all cases the dimension of the corresponding subspaces is $ 3 $. Also the eigenvalues of $ \sigma^{2} $ are $ 1,\omega ,\omega^2 $.

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  • $\begingroup$ You know all the ways $z\in L_3$ can be written as $b/c-1$ with $b$ and $c$ in $L_3$. So the question seems to be the intersection of $L_3$ with the image of $a\mapsto \sigma^2(a)/a$ from $L^\times$ to itself. $\endgroup$
    – Chris Wuthrich
    Commented Oct 3, 2021 at 9:42
  • $\begingroup$ So @Chris that means you want to say when the intersection of the image with $L_3$ is $\phi $ $\endgroup$
    – Sky
    Commented Oct 3, 2021 at 17:46

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