I'm working through a Galois Theory book and I'm stuck on the following:
$\left |\mathbb{Q}(\sqrt[3]{3},\sqrt{3},i:\mathbb{Q}) \right |$. I've (hopefully correctly) that this is a normal field extension but I can't see where to go from here to show what the dimension of this field is.
Any help appreciated.
$[\mathbb{Q}(\sqrt[3]{3}) : \mathbb{Q}] = 3$, whereas $[\mathbb{Q}(\sqrt{3}) : \mathbb{Q}] = 2$. Thus $[\mathbb{Q}(\sqrt[3]{3}, \sqrt{3}) : \mathbb{Q}] = 6$ because both 3 and 2 must divide this dimension, and the dimension can't be more than the product (the products of elements in a basis for each will span $\mathbb{Q}(\sqrt[3]{3},\sqrt{3}) / \mathbb{Q}$). But then when you adjoin $i$, you'll either multiply the dimension by $2$ or leave it alone, the latter happening only if $i \in \mathbb{Q}(\sqrt[3]{3},\sqrt{3})$, but $i \not\in \mathbb{Q}(\sqrt[3]{3},\sqrt{3})$ because $\mathbb{Q}(\sqrt[3]{3},\sqrt{3})$ is totally real. So the dimension of $\mathbb{Q}(\sqrt[3]{3},\sqrt{3}, i) / \mathbb{Q}$ is 12.