Let $\alpha=\text{exp}(2\pi i/5)$.
I was trying to find an explicit relation between
$\mathbb{Q}(\alpha)=\{a+b\alpha+c\alpha^2+d\alpha^3 : a,b,c,d\in \mathbb{Q}\}$
and
$\mathbb{Q}(i,\ \sqrt{5})=\{(a+ib)+(c+id)\sqrt{5}:a,b,c,d\in\mathbb{Q}\}$
Why $\sqrt{5}$ and $i$? I was able to find that $2x^2+(1-\sqrt{5})x+2$ is an irreducible minimal polynomial in $\mathbb{Q}(\sqrt{5})$ with root $\alpha$. And $$\alpha=\frac{\sqrt{5}-1}{4}+i\frac{\sqrt{10+2\sqrt{5}}}{4}$$ $\alpha$ has $\sqrt{5}$ as you can see.
But I have trouble dealing with double sqrt in imaginary part, $\sqrt{10+2\sqrt{5}}$. $$\sqrt{10-2\sqrt{5}}=\sqrt{5+2\sqrt{5}}+\sqrt{5-2\sqrt{5}}$$ I found above relation by solving $x^2-10x+5=0$ but this process can't reduce double sqrt into single.
For $X=\sqrt{10+2\sqrt{5}}$, $$X^2-10=-2\sqrt{5}$$ shows $\sqrt{10+2\sqrt{5}}$ is algebraic number of degree 2 in $\mathbb{Q}(\sqrt{5})$. Is this an implication that $\mathbb{Q}(\sqrt{5})$ is not large enough to math up with $\mathbb{Q}(\alpha)$? If it is, what algebraic number $k$ should I use to obtain the following? $$\mathbb{Q}(k,i)\cong\mathbb{Q}(\alpha)$$