Showing that $\sqrt{2} \in \mathbb{Q}[\varepsilon^{2}]$ If we work in the field of complex numbers and let $\varepsilon$ be a primitive 16th root of unity and set $b=\frac{\varepsilon}{\sqrt{2}}$ and $A=\mathbb{Q}[\varepsilon]$ where $\mathbb{Q}$ represents the rationals, and also set $f(X)=X^{8}+16 \in \mathbb{Q}[X]$ and note that $b$ is a root of $f(X)$, then how can we prove that $\sqrt{2} \in \mathbb{Q}[\varepsilon^{2}]$?
 A: $\epsilon$ is a root of $p(x)=x^8+1$ as it is a primitive root of $x^{16}-1 = (x^8-1)(x^8+1)$.
Therefore $\epsilon^2$ is a root of
$$q(x) = x^4+1= x^2\left(x^2+\frac{1}{x^2}\right)= x^2\left(x+\frac{1}{x} - \sqrt 2\right)\left(x+\frac{1}{x} + \sqrt 2\right)$$
As $\epsilon^4 \neq 0$, $\epsilon^2$ is a zero of
$$r(x) = \left(x+\frac{1}{x} - \sqrt 2\right)\left(x+\frac{1}{x} + \sqrt 2\right)$$
This implies $\epsilon^2 + \frac{1}{\epsilon^2} \in \lbrace \sqrt 2, - \sqrt 2 \rbrace$ and the desired result $\sqrt 2 \in \mathbb Q(\epsilon^2)$
A: Since your question starts with "If we work in the field of complex numbers", let's try to work with complex numbers.
A primitive 16-th root of unity is just
$$\varepsilon = e^{2 \pi i / 16}$$
So that
$$\varepsilon^2 = e^{4 \pi i / 16} = e^{\pi i / 4} = \frac{\sqrt 2}{2}(1+i)$$
If you take its inverse, you will get the conjugate
$$(\varepsilon^2)^{-1} = \frac{\sqrt 2}{2}(1-i)$$
So that
$$\sqrt{2} = 2\varepsilon^2 + 2 (\varepsilon^2)^{-1} \in \Bbb Q [\varepsilon^2] $$
