# number theory Q

by ElDavidas
Tags: number, theory
 P: 80 1. The problem statement, all variables and given/known data Let $\zeta$ be a primative 6-th root of unity. Set $\omega = \zeta i$ where $i^2 = 1$. Find a square-free integer $m$ such that $Q [\sqrt{m}] = Q[ \zeta ]$ 2. Relevant equations The minimal polynomial of $\zeta$ is $x^2 - x + 1$ 3. The attempt at a solution I was intending to use the theorem that: Take $p$ to be a prime and $\zeta$ to be a $p$-th root of unity. if $$S = \sum_{a =1}^{p-1} \big( \frac{a}{p} \big) \zeta^a$$ then $$S^2 = \Big( \frac{-1}{p} \Big) p$$. This would make $S^2$ 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 $( \frac{-1}{p} \Big)$, I mean the legendre symbol.
 Sci Advisor HW Helper P: 9,398 What are the roots of x^2-x+1?
 P: 80 The roots of that polynomial are the primative roots $\zeta$ and $\zeta^5$. How would I now use this information?
No. What are the roots of that polynomial. You're making it too complicated. If I gave you that polynomial in you Freshman calc course, or whatever, you'd be able to write out the roots without thinking. What are the roots? Or better yet, don't write out the roots using THE QUADRATIC FORMULA, just write down a sixth root of unity using elementary complex numbers. HINT: If I asked for a primitive 4th root of unity, would i be acceptable? Or -i? You know that $\exp(2\pi i/n)$ is a primitive n'th root of unity, and that the others are $\exp(2\pi i m/n)$ for m prime to n.