Find the minimum polynomial of $\left(\sqrt[3]{2}\right)$ over $\mathbb{Q}(i)$

Question

Find the minimum polynomial of $\left(\sqrt[3]{2}\right)$ over $\mathbb{Q}(i)$

I know that over $\mathbb{Q}$ this is simply $X^3 - 2$ but fail to see how this can be reduced over $\mathbb{Q}(i)$

Answer 1

The idea is to show that $\mathbb{Q}(\sqrt[3]{2},i)$ has degree $6$ over $\mathbb{Q}$. You can realise that $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$ and $[\mathbb{Q}(i):\mathbb{Q}]=2$. Now since $2$ and $3$ are relatively prime, then $[\mathbb{Q}(\sqrt[3]{2},i):\mathbb{Q}]=6$, or you can say that $\mathbb{Q}(\sqrt[3]{2})$ is a real field, and $i$ does not belong to it, and hence $[\mathbb{Q}(\sqrt[3]{2},i):\mathbb{Q}(\sqrt[3]{2})]=2$, leading to $(\mathbb{Q}(\sqrt[3]{2},i):\mathbb{Q}]=6$, and hence $[\mathbb{Q}(\sqrt[3]{2},i):\mathbb{Q}(i)]=3$, so $\sqrt[3]{2}$ satisfies an irreducible polynomial of degree $3$ in $\mathbb{Q}[i]$. Since $x^{3}-2$ has degree $2$ and is statisfied by $\sqrt[3]{2}$ it must be irreducible in $\mathbb{Q}(i)[X]$

Answer 2

It is $X^3-2$, if not, it is a polynomial $P$ of degree $2$ which has coefficients with imaginary parts, $P=(X-\sqrt[3]{2})(X-a)$, this implies that $a$ is a root of $P$ and $X^3-2$, $a=j\sqrt[3]{2}$ or $a=j^2\sqrt[3]{2}$ where $j$ is the third root of unity, you can verify in these cases, the coefficients of $P$ are not in $\mathbb{Q}(i)$.

Answer 3

In this very particular case, perhaps the most direct proof would be as follows. If the polynomial $X^3 - 2$ were reducible over $\mathbf Q(i)$, it would decompose over $\mathbf Q(i)$ into the product of 3 linear factors, or of a linear factor by a quadratic factor. In any case, it would have a root $\alpha \in \mathbf Q(i)$. But $[\mathbf Q(i):\mathbf Q]=2$, whereas $[\mathbf Q(\alpha):\mathbf Q]=3$ by Eisenstein criterion : impossible.