Find a Square-Free Int. m such that Q[√m] = Q[ζ]

[ElDavidas]

Homework Statement

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 $Q [\sqrt{m}] = Q[ \zeta ]$

Homework Equations

The minimal polynomial of $\zeta$ is $x^2 - x + 1$

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.

 

[matt grime]

What are the roots of x^2-x+1?

 

[ElDavidas]

The roots of that polynomial are the primative roots $\zeta$ and $\zeta^5$. How would I now use this information?

 

[matt grime]

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.