Let $y=e^{2\pi i/7} $ and let $x =y+y^2+y^4$then,
which of the following given option is correct
1)$x\in Q$
2)$x\in Q(\sqrt D)$ for some $ D>0$
3)$x\in Q(\sqrt D)$ for some $ D<0$
4)$x\in iR$
My thinking:
$x=e^{(2\pi i/7)} +e^{2(2\pi i/7)} +e^{4(2\pi i/7)}$ =$e^{2\pi i/7}(1+e^2+e^3)$
Here, $(1+e^2+e^4)$ is real number and $e^{(2πi/7)}$ is complex number so option $4$ is clearly true.
I don't know about the other option(s). Please help me! Thanks in advance.
Note that $y$ is a root of $y^7-1$, but since $y\neq 1$ it is actually a root of $(y^7-1)/(y-1)=y^6+y^5+y^4+y^3+y^2+y+1$. Check that
\begin{align*} x^2 &= (y+y^2+y^4)^2 \\ &= y^2+2y^3+y^4+2y^5+2y^6+y^8 \\ &= y+y^2+y^4 + 2 (y^6+y^5+y^3) \\ &= x + 2(-1-x) \\ &= -x-2 \end{align*} so $x$ is a root of $x^2+x+2$. Since the discriminant of that quadratic is $-7$, it's irreducible over $\mathbb Q$, and the correct answer must be 3)