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Find the minimal polynomial $\exp\left(\frac{2\pi i}{5}\right)$ over $\mathbb{Q}(i)$
It is clear, that polynomial $\exp\left(\frac{2\pi i}{5}\right)$ over $\mathbb{Q}(i, \sqrt{5})$ is $f(x) = x - \exp\left(\frac{2\pi i}{5}\right)$. I found the minimal polynomial $\exp\left(\frac{2\pi i}{5}\right)$ over $\mathbb{Q}(\sqrt{5})$. It is $f(x) = 16 x^2 -8 x (\sqrt{5} - 1 ) + 16$. And I have no idea what to do next.
Thanks in advance for any help.
Your first claim is wrong, as it's well-known that $\mathbb{Q}(\zeta_5)/\mathbb{Q}$ has a cyclic Galois group, namely $\mathbb{Z}/4\mathbb{Z}$. Thus $\mathbb{Q}(\zeta_5)/\mathbb{Q}$ has a unique field extension of degree $2$. So we must have that $\zeta_5 \notin \mathbb{Q}(i,\sqrt{5})$.
On the other side we have that $\zeta_5 + \zeta_5^{-1} = \frac 12 (\sqrt{5}-1)$ and thus we must have that the unique quadratic extension of $\mathbb{Q}(\zeta_5)/\mathbb{Q}$ is $\mathbb{Q}(\sqrt{5})$. This should help you conclude that $i \notin \mathbb{Q}(\zeta_5)$. Thus we get that $\mathbb{Q}(\zeta_5,i)$ is an extension of degree $8$. Finally:
$$[\mathbb{Q}(\zeta_5,i):\mathbb{Q}] = [\mathbb{Q}(\zeta_5,i):\mathbb{Q}(i)][\mathbb{Q}(i):\mathbb{Q}] \implies [\mathbb{Q}(\zeta_5,i):\mathbb{Q}(i)] = 4$$
From here the minimal polynomial of $\zeta_5$ over $\mathbb{Q}(i)$ is of fourth degree. And as it satisfies $x^4+x^3+x^2+x+1$ we must have that this is the actual minimal polynomial.
The proof of Stefan4024 is correct and completely satisfying. Here’s an argument that’s also correct, but unsatisfying unless you know the requisite algebraic number theory:
The extension $\Bbb Q(i)\supset\Bbb Q$ is totally ramified above $2$, and unramified elsewhere. The extension $\Bbb Q(\zeta_5)\supset\Bbb Q$ is totally ramified above $5$, and unramified elsewhere. Their intersection must be unramified everywhere, and thus is equal to $\Bbb Q$. So the extensions are linearly disjoint: the minimal polynomial for $\zeta_5$ over $\Bbb Q$ is also minimal over $\Bbb Q(i)$.
