Two part question: Are the fields $\mathbb{Q} (\sqrt[3]{2}, i \sqrt{3})$ and $\mathbb{Q} (\sqrt[3]{2}, i, \sqrt{3})$ identical in algebraic structure? I have in notes that they both have degree of 6 over $\mathbb{Q}$.
How do I show explicitly that $\mathbb{Q} ( i \sqrt{3})$ is only degree 2 over $\mathbb{Q}$. The usual trick is to adjoin the real roots and then adjoin the complex root, but it's a different story when it's not just $i$ by itself.
Edit: I'm starting to mistrust that the degree of the two extensions are identical.
For starters, neither $i$ nor $\sqrt{3}$ is in $\mathbb{Q}(\sqrt[3]{2},i\sqrt{3})$. And I think you're right to mistrust that the degrees are equal since $|\mathbb{Q}(\sqrt[3]{2},i,\sqrt{3})/\mathbb{Q}|=12$.