Find the minimal polynomial over $\mathbb{Q}(i)$

Question

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.

Answer 1

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.

Answer 2

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 $\Bbb Q(i)$.