What is $[\mathbb{Q}[\sqrt[4]{2},\sqrt{-1}]:\mathbb{Q}]$?

Question

What is $[\mathbb{Q}[\sqrt[4]{2},\sqrt{-1}]:\mathbb{Q}]$ ? Is it just $4\cdot2=8$? Is $\sqrt[4]{2} \in \mathbb{Q}[\sqrt{-1}]$?

Answer 1

The degree is $8$, and one can see that from using an intermediate field extension. But I like to save the complex extension for last, because it's most obviously non-trivial. It also has a nice added bonus of having prime degree, so the extension can't be "partial". For instance, it could conceivably be the case that $[\Bbb Q(\sqrt[4]2, i), \Bbb Q(i)] = 2$, as well as $1$ or $4$, so there are more things to check than just $\sqrt[4]2\notin \Bbb Q(i)$, since that only excludes $1$.

So clearly $[\Bbb Q(\sqrt[4]2): \Bbb Q] = 4$, and then $[\Bbb Q(\sqrt[4]2, i): \Bbb Q(\sqrt[4]2)]$ is clearly either $1$ or $2$. But it can't be $1$, since $\Bbb Q(\sqrt[4]2)\subseteq \Bbb R$, while $\Bbb Q(\sqrt[4]2, i)\not\subseteq \Bbb R$, so $\Bbb Q(\sqrt[4]2, i)\neq \Bbb Q(\sqrt[4]2)$. This yields $$[\Bbb Q(\sqrt[4]2, i):\Bbb Q] = [\Bbb Q(\sqrt[4]2, i):\Bbb Q(\sqrt[4]2)]\cdot [\Bbb Q(\sqrt[4]2):\Bbb Q] = 2\cdot 4 = 8$$