I want to find $a$ in the algebraic closure of $\mathbb{Q}$ such that $\mathbb{Q}(a)= \mathbb{Q}( \sqrt{3}, \sqrt[3]{5}, \omega)$
Where $\omega$ is the 3rd primitive root of unity.
This is what i have so far:
I have shown that $\mathbb{Q}(\sqrt{3} + \omega) = \mathbb{Q}(\sqrt{3} , \omega)$ using Steinitz'.
So i am left to find the simple extension of $\mathbb{Q}(\sqrt{3} + \omega, \sqrt[3]{5})$.
I want to find the roots of irr$(\sqrt{3} + \omega, \mathbb{Q})$, but found it so tedious to do so! Do you know of any easy way to find its roots?
It shouldn't be very hard to find the automorphisms of $\mathbb{Q}(\sqrt{3},\omega)$, fixing $\mathbb{Q}$, as after all it's the splitting field of the polynomial $(x^2-3)(x^2+x+1)$. If you apply them to the primitive element you will get all the distinct conjugates of it, i.e. the roots of $\min(\sqrt{3}+\omega, \mathbb{Q})$.
Indeed you will end up with $\sqrt{3}+\omega, \sqrt{3} + \omega^2, -\sqrt{3} + \omega, -\sqrt{3} + \omega^2$.
From here you should be able to conclude that $\sqrt{3}+\sqrt[3]{5}+\omega$ is a primitive element of $\mathbb{Q}(\sqrt{3},\sqrt[3]{5},\omega)$