Which of the following subfields of $\mathbb{C}$ are isomorphic?

Question

Which of the following subfields of $\mathbb{C}$ are isomorphic?

A: $\mathbb{Q}[\sqrt[3]{2}]$ and $\mathbb{Q}[-\sqrt{2}]$
B: $\mathbb{Q}[e^{-2i\pi/3}]$ and $\mathbb{Q}[e^{2i\pi/3}]$
C: $\mathbb{Q}[e^{2i\pi/5}]$ and $\mathbb{Q}[e^{2i\pi/3}]$
D: $\mathbb{Q}[e^{2i\pi/3}]$ and $\mathbb{Q}[-i]$

I know that $\mathbb{Q}[-\sqrt{2}]$ is isomorphic since there exists a polynomial where $$\Bbb Q(\sqrt2)\cong\mathbb{Q}[t]/(t^2-2)\cong\Bbb Q(-\sqrt2). $$

However I'm not sure when it comes to the exponential ones.

Answer 1

Note that $\mathbb{Q} [ e^{2\pi\iota/3} ]$ is the smallest subfield of $\mathbb{C}$ containing the union $\mathbb{Q} \cup \left\{ e^{2\pi\iota/3} \right\}$, and since $$ e^{2\pi\iota/3} = \cos \frac{2\pi}{3} + \iota \sin \frac{2\pi}{3} = \frac{1}{2} + \frac{\iota \sqrt{3}}{2}, $$ therefore we can write $$ \mathbb{Q} [ e^{2\pi\iota/3} ] = \{ p + q \iota \sqrt{3} : p, q \in \mathbb{Q} \} $$

Similarly, as $$ e^{-2\pi\iota/3} = \frac{1}{2} - \frac{\iota \sqrt{3}}{2}, $$ so we can also write $$ \mathbb{Q} [ e^{-2\pi\iota/3} ] = \{ p - q \iota \sqrt{3} : p, q \in \mathbb{Q} \} = \{ p + q \iota \sqrt{3} : p, q \in \mathbb{Q} \}. $$ Thus we have the identity $$ \mathbb{Q} [ e^{2\pi\iota/3} ] = \mathbb{Q} [ e^{-2\pi\iota/3} ]. $$

Alternatively, note that $$ e^{-2\pi\iota/3} = e^{-2\pi\iota/3 + 2 \pi \iota } = e^{4\pi \iota/3} = \left( e^{2 \pi \iota/3} \right)^2, $$ which implies that $$ e^{-2\pi\iota/3} \in \mathbb{Q} [ e^{2\pi\iota/3} ]; $$ similarly, we have $$ e^{2\pi\iota/3} = e^{2\pi\iota/3 - 2 \pi \iota } = e^{- 4\pi \iota/3} = \left( e^{-2 \pi \iota/3} \right)^2, $$ which implies that $$ e^{2\pi\iota/3} \in \mathbb{Q} [ e^{-2\pi\iota/3} ]. $$