$\mathbb{Q}(\sqrt{a+bi})=\mathbb{Q}(\sqrt{a-bi})$?

Question

For $a,b,n\in\mathbb{Z}$:

• Do we have $\mathbb{Q}(\sqrt{a+bi})=\mathbb{Q}(\sqrt{a-bi})$?
• More generally, do we have $\mathbb{Q}(\sqrt[n]{a+bi})=\mathbb{Q}(\sqrt[n]{a-bi})$?
• Given $\mathbb{Q}(\sqrt{a+bi})$, does there always exist a Gaussian Prime $z$ so that $\mathbb{Q}(\sqrt{a+bi})=\mathbb{Q}(\sqrt{z})$?

Answer 1

Over a field $K$ of characteristic zero, $K(\sqrt\alpha)=K(\sqrt\beta)$ for non-squares $\alpha$ and $\beta$ iff $\alpha\beta$ is a square in $K$.

Here, take $K=\Bbb Q(i)$, $\alpha=a+bi$ and $\beta=a-bi$. Then $\alpha\beta=a^2+b^2$ and that is a square in $\Bbb Q(i)$ iff it's a square in $\Bbb Q$. So, for instance, $\Bbb Q(\sqrt{2+i})\ne\Bbb Q(\sqrt{2-i})$.