(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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]

2. Relevant equations

The minimal polynomial of [itex] \zeta [/itex] is [itex] x^2 - x + 1 [/itex]

3. 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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# Number theory Q

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