Find $[\mathbb Q (\xi ^2 ) : \mathbb Q ]$

Question

Where $\xi = e^{\frac{\pi}3i}$ and $L=\mathbb Q (\xi ^2 )$

I want to find the minimal polynomial of $L$ over $\mathbb Q$ and then this will determine the degree of the field.

But I am not sure how to do it. Normally if we have something like $\mathbb Q (\sqrt2)$, we would let $x=\sqrt2 $ and then keep squaring (or other power value) both sides until we have a rational on the RHS. In this case, we just square it once. Then prove that its irreducible which is easy if the degree of the polynomial is $2$ or $3$.

But in our case, I don't think it works like that.

Please help.

Answer 1

Note that if $\gamma=\xi^2$, then $\gamma^3-1=\left(e^{2\frac{\pi}{3}}\right)^3-1=e^{2\pi}-1=1-1=0$. So $\gamma$ is a root of $x^3-1=(x-1)(x^2+x+1)$, and since $\gamma\not\in \mathbb{Q}$, the minimal polynomial is $x^2+x+1$, and finally $[L:\mathbb{Q}]=2$.

Answer 2

Hint $$ (\xi^2)^3=1$$

So $\xi^2$ is a root of $X^3-1$. Can you factor this polynomial? Which factor does have $\xi^2$ as a root?