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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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    @David: They live in the latter. I got a linear system as suggested in your answer below, and had SAGE do the computation, which gave a good answer. Thanks David!2012-03-01

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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.