I'm trying to prove that $\mathbb{Q}(\sqrt[3]{2})$ is isomorphic to $\mathbb{Q}(e^{2\pi i/3}\sqrt[3]{2})$. I want to use the first isomorphism theorem that say that if $\phi: A\to B$ is a homomorphism, then $\phi(A)$ is isomorphic to $A/\!\ker{\phi}$. I'm trying to build a homomorphism such that $\phi(\mathbb{Q}(\sqrt[3]{2}))=\mathbb{Q}(e^{2\pi i/3}\sqrt[3]{2})$ and $\ker{\phi}=\{0\}$.
Does such an homorphism exist or is there another way to prove this?
Thanks for your help.
Another way to approach it: $$\Bbb Q(\sqrt[3]2)\cong\Bbb Q[x]/(x^3-2)$$ This follows from the evaluation homomorphism: $$\phi:\Bbb Q[x]\rightarrow\Bbb Q(\sqrt[3]2)\\f(x)\mapsto f(\sqrt[3]2)$$ And then the 1st isomorphism theorem.
What about $\Bbb Q(e^{2\pi i/3}\sqrt[3]2)$?