Let $L$ the subfield of the complex with $\mathbb{Q}\leq L$ a normal extension. If $a=\sqrt{5}-\sqrt[3]{2}\in L$, show that $\omega=e^{2\pi i/3}\in L$.
I have done the following:
$Irr(\sqrt{5},\mathbb{Q})=x^2-5$ the roots are $\pm \sqrt{5}\in L$
We have that $a\in L, \sqrt{5}\in L$, $L$ is a field. So $a-\sqrt{5}\in L\Rightarrow \sqrt[3]{2}\in L$
$Irr(\sqrt[3]{2},\mathbb{Q})=x^3-2$ the roots are $\sqrt[3]{2}, \omega\sqrt[3]{2} , \omega^2 \sqrt[3]{2}\in L$
We have that $\sqrt[3]{2}\in L, \omega \sqrt[3]{2}\in L$, $L$ is a field. So $\omega \in L$.
Is this correct??
Once you've shown that $\sqrt5, \sqrt[3]2 \in L$, your proof is correct. To show this fact, you need to show that $$\mathbb Q(\sqrt 5,\sqrt[3]2)=\mathbb Q(\sqrt 5+\sqrt[3]2)$$and therefore that $a \in L \implies \sqrt5, \sqrt[3]2\in L$. Note that this is not immediate!
Can you show that both these extensions have degree $6$ over $\mathbb Q$? Can you then deduce that they are the same?