Cubic Polynomial $x^3-2$ is Irreducible Over the Field $\Q(i)$

Problem 399

Prove that the cubic polynomial $x^3-2$ is irreducible over the field $\Q(i)$.

Proof.

Note that the polynomial $x^3-2$ is irreducible over $\Q$ by Eisenstein’s criterion (with prime $p=2$).
This implies that if $\alpha$ is any root of $x^3-2$, then the degree of the field extension $\Q(\alpha)$ over $\Q$ is $3$:
$[\Q(\alpha) : \Q]=3. \tag{*}$

Seeking a contradiction, assume that $x^3-2$ is reducible over $\Q(i)$.
Then $x^3-2$ has a root in $\Q(i)$ as it is a reducible degree $3$ polynomial. So let us call the root $\alpha \in \Q(i)$.

Then $\Q(\alpha)$ is a subfield of $\Q(i)$ and thus we have
$2=[\Q(i) :\Q]=[\Q(i): \Q(\alpha)][\Q(\alpha):\Q]\geq 3$ by (*). Hence we have reached a contradiction.
As a result, $x^3-2$ is irreducible over $\Q(i)$.