Seeing if a polynomial is irreducible over a field extension of $\Bbb Q$. Exercise. Determine $m_{i\sqrt[3]{2}}$ over $\Bbb Q(i)$, where $m_{i\sqrt[3]{2}}$ represents the minimal polynomial of ${i\sqrt[3]{2}}$.
My attempt. We know that $m_i = x^2 +1$ over $\Bbb Q$ (this is obvious) and thus $\deg(i) = \deg(m_i) = 2$ over $\Bbb Q$. Given this, we can say that
\begin{equation*}
\Bbb Q(i) = <1,i>_\Bbb Q = \{a + b i : a,b \in \Bbb Q\}
\end{equation*}
Now, obviously $i\sqrt[3]{2} \notin \Bbb Q(i)$ since $\sqrt[3]{2} \notin \Bbb Q$ (the only chance would be taking $a=0$ and $b = \sqrt[3]{2}$). So, $\deg(i\sqrt[3]{2}) > 1$ over $\Bbb Q(i)$.Further more,
\begin{equation*}
(i\sqrt[3]{2})^3 = -i*2 = (-2) i
\end{equation*}
And so $i\sqrt[3]{2}$ is a root of the polynomial $f = x^3 + 2i$ over $\Bbb Q(i)$. Since $f$ is monic, all we have to do now is to prove that $f$ is irreducible over $\Bbb Q(i)$. Since $f$ has $\deg = 3$ and $\Bbb Q(i)$ is a field extension (and thus a field) all we have to is check if $f$ has roots in $\Bbb Q(i)$. Roots of $f$ are the following:
\begin{equation*}
x=\frac{\sqrt[3]{2}\sqrt{3}}{2}-i\frac{\sqrt[3]{2}}{2},\:x=\sqrt[3]{2}i,\:x=-\frac{\sqrt[3]{2}\sqrt{3}}{2}-i\frac{\sqrt[3]{2}}{2}
\end{equation*}
None of this is in $\Bbb Q(i)$ and so $f$ is irreducible, proving that $ m_{i\sqrt[3]{2}} = x^3 + 2i$ over $\Bbb Q(i)$.
My doubts. First, I would obviously like to know if what I did is right. Other than this, I would like to know how to compute the complex roots of $f$ manually (I used symbolab software to find the roots).
Thanks for any help in advance!
 A: What you've done is correct. There are a few ways to do it by hand, without using a computer.
The most elementary way is to take a general element $a + bi \in \mathbb{Q}(i)$ and show that it can't be a root of $f(x)$. That is, let $a, b \in \mathbb{Q}$ and suppose that $f(a + bi) = 0$. Then
$$
(a + bi)^3 = -2i,
$$
which implies
$$
(a^3 - 3ab^2) + (3a^2b - b^3)i = -2i.
$$
Comparing real parts, we get $a^3 - 3ab^2 = 0$, so $a=0$ or $a^2 = 3b^2$. If $a \neq 0$, then $a^2 = 3b^2$, so $b\neq 0$ and $(a/b)^2 = 3$, which famously is impossible, since $a/b \in \mathbb{Q}$. Therefore, $a = 0$, and comparing imaginary parts gives $b^3 = 2$, which is also impossible.
If you want to actually find the complex roots, just solve the equations $a^3 = 3ab^2$ and $3a^2b - b^3 = -2$ over $\mathbb{R}$. Assuming that $a$ is nonzero (the $a=0$ case is easier), we have $a^2 = 3b^2$, implying that $9b^3 - b^3 = -2$, so $8b^3 = -2$ and therefore $b^3 = - 1/4$. We get $b = - \frac{\sqrt[3]{2}}{2}$, and then you can get $a$ from $a^2 = 3b^2$.
A slightly less elementary way would be to put $z = re^{i\theta}$ for $r>0$ and $\theta \in [0, 2\pi)$, and suppose that $f(z) = 0$. Then $r^3e^{i3\theta} = -2i$, which gives $r^3 = 2$ and $3\theta \equiv \frac{3\pi}{2} \pmod{2\pi}$. This implies $r = \sqrt[3]{2}$ and $\theta = \frac{\pi}{2}, \frac{7\pi}{6}, \frac{11\pi}{6}$, from which it is easy to compute the real and imaginary parts of the roots.
