Show, by negating, that $x^2-3$ is irreducible over $\mathbb{Q}(\sqrt[3]{2})$

Question

Show, by negating, that $x^2-3$ is irreducible over $\mathbb{Q}(\sqrt[3]{2})$

I don't think this question has any sense for $\mathbb{Q}(\sqrt[3]{2})$. Shouldn't it be $\mathbb{Q}(\sqrt[2]{3})$, or there is a way to think in the other case?

Because if I suppose that $x^2-3$ is reducible over $\mathbb{Q}(\sqrt[3]{2})$ we have that $x^2-3 = (x-a)(x-b)$ for $a,b\in \mathbb{Q}(\sqrt[3]{2})$. I know that $x=3$ must be a solution for either $x-a=0$ or $x-b=0$, so let's suppose it's for $a$, so $a=3$. I don't see what's happening...

For $\mathbb{Q}(\sqrt[2]{3})$, we'd have that $x=3$ must be a solution for either $x-a=0$ or $x-b=0$, so let's suppose it's for $a$, so $a=3$. Now I must find a way to see that $b\notin \mathbb{Q}(\sqrt[2]{3})$, am I right? But how?

Answer 1

A simple naїve proof may be as follows.

Because if I suppose that $x^2-3$ is reducible over $\mathbb{Q}(\sqrt[3]{2})$ we have that $x^2-3 = (x-a)(x-b)$ for $a,b\in \mathbb{Q}(\sqrt[3]{2})$

Thus $a$ is a root of the polynomial $x^2-3$. Since $a\in \mathbb{Q}(\sqrt[3]{2})$, $a=p+q\sqrt[3]{2}+r\sqrt[3]{4}$ for some rational $p$, $q$, and $r$. Then

$$3=(p+q\sqrt[3]{2}+r\sqrt[3]{4})^2=(p^2+4qr)+(2pq+2r^2)\sqrt[3]{2}+(2pr+q^2)\sqrt[3]{4}.$$

Since $1$, $\sqrt[3]{2}$, and $\sqrt[3]{4}$ are linearly independent over $\Bbb Q$, we have that $p^2+4qr=3$, $pq+r^2=0$, and $2pr+q^2=0$. If $q=0$ then $p^2=3$, that is impossible. So $q\ne 0$. Then $p=-r^2/q$, $-2r^3/q+q^2=0$, and $\sqrt[3]{2}=q/r$ is a rational number, a contradiction.