Let $\omega$ be a root of the polynomial: $f(x)=x^2+x+1\in \mathbb{Q}[x]$.
Prove:
$\mathbb{Q}(\omega\sqrt[3]{2})$ is isomorphic to $\mathbb{Q}(\frac{\sqrt[3]{2}}{\omega})$
my attempt:
I'd like to prove that there exists $\phi:\mathbb{Q}(\omega\sqrt[3]{2}) \to \mathbb{Q}(\frac{\sqrt[3]{2}}{\omega})$.
I'm wondering if it's enough (or even true), that since: $\mathbb{Q}(\omega)=\{a+ib\sqrt{3}|a,b\in \mathbb{Q}\}$,
then
$\mathbb{Q}(\omega)= \mathbb{Q}(i\sqrt{3})$
it's enough to define $\phi$ as: $\phi(i\sqrt{3}\sqrt[3]{2})= \frac{\sqrt[3]{2}}{i\sqrt{3}}$.
what am i missing in order to determine that $\phi$ is an isomorphism?
is this true that that both fields mentioned above equals? this will complete the proof, but I don't think it's true and also not sure how to prove that:$\mathbb{Q}(\omega\sqrt[3]{2}) \neq \mathbb{Q}(\frac{\sqrt[3]{2}}{\omega})$.
Best Answer
The slightly easier way is to see that they are both solutions of the irreducible polynomial $x^3 -2$ so that
$$ \mathbb{Q} (\omega \sqrt[3]{2}) \cong \mathbb{Q} (\frac{\sqrt[3]{2}}{\omega} ) \cong \mathbb{Q}[x] / (x^3 -2)$$
From that construction you can easily see that the isomorphism is $\omega \sqrt[3]{2} \mapsto \frac{\sqrt[3]{2}}{\omega}$