$X^3 - 2$ is irreducible over $\Bbb{Q}(i)$ without cubing $(a + bi)$?

Question

I want to easily prove $X^3 - 2$ is irreducible over $\Bbb{Q}(i)$ without cubing a general element of $\Bbb{Q}(i)$ in the proof.

I know that if $X^3 - 2$ is reducible over $\Bbb{Q}(i)$, then it must have a root in $\Bbb{Q}(i)$. Also, I know that $X^3 - 2$ is irreducible over $\Bbb{Q}$ by Schonemann-Eisenstein criterion.

I think its: $3 = \deg_{\Bbb{Q}}(\sqrt[3]{2}) \leq \deg_{\Bbb{Q}}(i) = 2$, contradiction. Am I right?

Answer 1

Your idea is good, but your argument isn't.

We know that $X^3-2$ is irreducible over $\mathbb{Q}$, by Eisenstein. Suppose it's reducible over $\mathbb{Q}(i)$, so it has a root $r\in\mathbb{Q}(i)$, because it has degree $3$. Then $\mathbb{Q}(r)\subseteq\mathbb{Q}(i)$ and we have $$ 2=[\mathbb{Q}(i):\mathbb{Q}]= [\mathbb{Q}(i):\mathbb{Q}(r)][\mathbb{Q}(r):\mathbb{Q}] =3[\mathbb{Q}(i):\mathbb{Q}(r)] $$ by the irreducibility of $X^3-2$ over $\mathbb{Q}$, a contradiction.

Answer 2

You don't need to explicitly cube anything. The cube roots of $2$ are $2^{1/3}\omega^j,j=0,1,2$. The real part of these is never rational (it is actually always a rational multiple of $2^{1/3}$, thus they are not contained in $\Bbb Q[i]$. This is an easy consequence of the fact that $\Bbb R[i]=\Bbb C$ is a real vector space with basis $1,i$; so if $r+is=2^{1/3}\omega^j$, we must have $r=2^{1/3}\Re(\omega^j)$.