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ElDavidas
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Homework Statement
Let [itex] \zeta [/itex] be a primative 6-th root of unity. Set [itex] \omega = \zeta i [/itex] where [itex] i^2 = 1[/itex].
Find a square-free integer [itex]m[/itex] such that [itex] Q [\sqrt{m}] = Q[ \zeta ] [/itex]
Homework Equations
The minimal polynomial of [itex] \zeta [/itex] is [itex] x^2 - x + 1 [/itex]
The Attempt at a Solution
I was intending to use the theorem that:
Take [itex] p [/itex] to be a prime and [itex] \zeta [/itex] to be a [itex] p [/itex]-th root of unity. if
[tex]S = \sum_{a =1}^{p-1} \big( \frac{a}{p} \big) \zeta^a [/tex]
then
[tex] S^2 = \Big( \frac{-1}{p} \Big) p [/tex].
This would make [itex]S^2[/itex] an integer. However, 6 is not a prime though. I'm really stumped in what to do. Any help would be greatly appreciated.
Oh, and by [itex]( \frac{-1}{p} \Big)[/itex], I mean the legendre symbol.
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