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Let $\zeta = e^{2\pi i/110}$, and set $K = \mathbb{Q}(\zeta)$. There is an $\alpha$ in $\mathbb{Q}(\zeta^{11})$ of absolute value 1, which I'm trying to find.

Consider $\sigma$, the Galois automorphism of $K$ which sends $\zeta \mapsto \zeta^{7}$. Note that $\sigma^2$ sends $\alpha$ to its complex conjugate $\bar{\alpha}$.

I have two different equations that $\alpha$ must satisfy, of the form

$$-10 = A\alpha + B\alpha^\sigma + C\bar{\alpha} + D\alpha^{\sigma^3}$$ $$-10 = E\alpha + F\alpha^\sigma + G\bar{\alpha} + H\alpha^{\sigma^3}$$

where $A,\ldots,H \in K$ are explicit known constants (none of which are zero).

Is there any way I can get my hands on $\alpha$ based on this information? I could simultaneously eliminate $\alpha^\sigma$ and $\alpha^{\sigma^3}$ if $B/D = F/H$, and then I'd have a quadratic in $\alpha$; but unfortunately $B/D \neq F/H$.

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$\sigma^2$ sends $\zeta$ to $\zeta^{49}=\zeta^5$, which is not the complex conjugate of $\zeta$. If $\sigma^2$ sends $\alpha$ to its complex conjugate, that is a very strong condition on $\alpha$. –  Gerry Myerson Mar 1 '12 at 1:55
    
@Alvaro: The 110 in my original question was not meant to be 11. I've changed it back. –  Barinder Banwait Mar 1 '12 at 11:49
    
@GerryMyerson: $\sigma^2$ sends $\zeta^{11}$ to $\zeta^{-11}$ which is the complex conjugate of $\zeta^{11}$, and hence $\alpha$ maps to $\bar{\alpha}$. Maybe you saw the question when $\zeta = e^{2 \pi i/11}$, and not $e^{2 \pi i/110}$. –  Barinder Banwait Mar 1 '12 at 11:52
    
This question is all about $\mathbb{Q}(e^{2\pi i / 10})$, surely? Only $\zeta^{11} = e^{2\pi i / 10}$ enters, so why mention $e^{2\pi i / 110}$? –  David Loeffler Mar 1 '12 at 14:26
    
Hang on: where do $A ... H$ live? In $\mathbb{Q}(\zeta^{11})$ or $\mathbb{Q}(\zeta)$? If it's the latter, we can reduce to the former by taking the trace down to $\mathbb{Q}(\zeta^{11})$. –  David Loeffler Mar 1 '12 at 14:49

1 Answer 1

up vote 2 down vote accepted

Write $\xi = \zeta^{11} = e^{2 \pi i / 10}$, which clearly satisfies a quartic over $\mathbb{Q}$. Write $\alpha$ as a linear combination of $1, \xi, \xi^2, \xi^3$. Then your two equations become eight linear equations in four variables, and any computer algebra system will tell you instantly if a solution exists and if so what it is.

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