What is $\mathbb{Q}(\sqrt[4]{3}i)$?

Question

Is it $\mathbb{Q}(\sqrt[4]{3}i) = \{ a+b\sqrt[4]{3}i+c\sqrt[4]{9}i + d\sqrt[4]{27} : a,b,c,d \in \mathbb{Q} \}$? Is it isomorphic to $\mathbb{Q}(\sqrt[4]{3})$?

Answer 1

Abstractly, $\mathbb{Q}(\sqrt[4]{3}i)$ is obtained by adjoining a root of $x^4-3$ to $\mathbb{Q}$.

The same description applies to $\mathbb{Q}(\sqrt[4]{3})$.

Hence, $\mathbb{Q}(\sqrt[4]{3}i)$ is isomorphic to $\mathbb{Q}(\sqrt[4]{3})$.

A concrete isomorphism sends $\sqrt[4]{3}i$ to $\sqrt[4]{3}$.

If $\alpha$ is algebraic of degree $n$, then $\mathbb{Q}(\alpha)$ is the set of rational linear combinations of $\alpha^k$ for $k=0,\dots,n$.

Therefore, $ \mathbb{Q}(\sqrt[4]{3}i)= \{ a+b\sqrt[4]{3}i+c\sqrt[4]{9} + d\sqrt[4]{27}i : a,b,c,d \in \mathbb{Q} \} $.