Showing $\sqrt{5} \in \mathbb{Q}(w)$ where $w = e^{2\pi i/5}$ I'm having troubles understanding what the basis for $\mathbb{Q}(w)$ is. Would it just be $\{ 1 , w \}$? What about $w^{-1}?$
 A: Here’s another approach, which generalizes:
Since $\omega$ is a primitive fifth root of unity, it’s a root of $(X^5-1)/(X-1)=X^4+X^3+X^2+X+1$. And so $\omega^2+\omega+1+\omega^{-1}+\omega^{-2}=0$. You can find the maximal real subfield of $\Bbb Q(\omega)$ by looking at $\xi=\omega+\omega^{-1}=2\cos(2\pi/5)$. We have
\begin{align}
\xi^2&=\omega^2+2+\omega^{-2}\\
&=-\omega+1-\omega^{-1}\\
&=-\xi+1\,,
\end{align}
giving the minimal equation $\xi^2+\xi-1=0$, whose roots are $\frac{-1}2\pm\frac{\sqrt5}2$, which proves the desired fact.
A: $\Bbb Q(\omega)$ has degree $4$ so a basis is $1,\omega,\omega^2,\omega^3$.
We know that the only prime that ramifies in $\Bbb Q(\omega)$ is $5$, and since the discriminant of a quadratic field $\Bbb Q(\sqrt{d})$ with $d$ square-free is
$$\Delta=\begin{cases} d & d\equiv 1\mod 4 \\ 4d & d\equiv 2,3\mod 4\end{cases}$$
we see that the quadratic subfield must be $\Bbb Q(\sqrt{5})$ since this is the only one that gives a discriminant divisible by $5$ but not $2$, and we know the primes that ramify are exactly those dividing the discriminant.
