prove: $\mathbb{Q}(\omega\sqrt[3]{2}) \cong \mathbb{Q}(\frac{\sqrt[3]{2}}{\omega})$

Question

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:$\mathbb{Q}(\omega\sqrt[3]{2}) \neq \mathbb{Q}(\frac{\sqrt[3]{2}}{\omega})$.

Answer 1

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 $\omega \sqrt[3]{2} \mapsto \frac{\sqrt[3]{2}}{\omega}$

Answer 2

The answer by @Sofia is excellent. I will like to add on your last query about how to show that they are only isomorphic but not equal.

If possible let $$\mathbb{Q}(\omega\sqrt[3]{2}) = \mathbb{Q}(\frac{1}{\omega}\sqrt[3]{2})$$

But notice that $\frac{1}{\omega}=\bar{\omega}$ .

So $$\omega\sqrt[3]{2}+\bar{\omega}\sqrt[3]{2}=(\sqrt[3]{2})\cdot2\text{Re}(\omega)=-\sqrt[3]{2}\in \mathbb{Q}(\omega\sqrt[3]{2}) $$.

So $\mathbb{Q}(\omega\sqrt[3]{2})=\mathbb{Q}(\omega\sqrt[3]{2},\sqrt[3]{2})$

And hence $\omega \in \mathbb{Q}(\omega\sqrt[3]{2})$ (As $\omega=\frac{\omega\sqrt[3]{2}}{\sqrt[3]{2}})$.

So $\mathbb{Q}(\omega\sqrt[3]{2})=\mathbb{Q}(\omega,\sqrt[3]{2})$ .

But you have $[\mathbb{Q}(\omega):\mathbb{Q}]=2$ and $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$ and hence they are relatively prime . So the degree of their compositum ($=\mathbb{Q}(\omega,\sqrt[3]{2})$) is the product of $[\mathbb{Q}(\omega):\mathbb{Q}]$ and $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]$ .

(The same argument proves that the splitting field of $x^{p}-2$ is $\mathbb{Q(\sqrt[p]{2},\zeta_{p}})$ of degree $p(p-1)$, where $p$ is a prime and $\zeta_{p}$ is a primitive $p-th$ root of unity.

So

$[\mathbb{Q}(\omega,\sqrt[3]{2})):\mathbb{Q}]=6$.

But $[\mathbb{Q}(\omega\sqrt[3]{2}):\mathbb{Q}]=3$.

A contradiction.

Hence $$\mathbb{Q}(\omega\sqrt[3]{2}) \neq \mathbb{Q}(\frac{1}{\omega}\sqrt[3]{2})$$

I know this seems long but once you get used to splitting fields these degree arguments are much shorter.

Alternatively you have to use the explicit basis $\{1,\omega\sqrt[3]{2},(\omega\sqrt[3]{2})^{2}\}$ and show that $\frac{\sqrt[3]{2}}{\omega}$ does not lie in $\text{span}(\{1,\omega\sqrt[3]{2},(\omega\sqrt[3]{2})^{2}\})$ which is easier said than done.