Is it true that $i \in \mathbb{Q}(\sqrt{2},\xi)$, where $\xi = 1_{\frac{2\pi}{3}} = -\frac{1}{2}+i\frac{\sqrt{3}}{2}$?

Question

I am trying to do the following exercise:

Give the Galois group of $\mathbb{Q}(\sqrt{3},\xi)$ over $\mathbb{Q}$, where $\xi = 1_{\frac{2\pi}{3}}$. Prove that $i \in \mathbb{Q}(\sqrt{2},\xi)$ and group together the elements of $\text{Gal}_{\mathbb{Q}}(\mathbb{Q}(\sqrt{2},\xi))$ that have the same values in $\mathbb{Q}(i)$.

It is very easy to prove that $i \in \mathbb{Q}(\sqrt{3},\xi)$, but I'm having more difficulties trying to prove that $i \in \mathbb{Q}(\sqrt{2},\xi)$. I've already found some errors in the headings of the other exercises on the same list, so I'm starting to suspect that in this case the exercise intended to say $\mathbb{Q}(\sqrt{3},\xi)$ instead of $\mathbb{Q}(\sqrt{2},\xi)$, as in the second part it suddenly changes from $\mathbb{Q}(\sqrt{3},\xi)$ to $\mathbb{Q}(\sqrt{2},\xi)$.

Even worse, it might a typo in the sense that we look at the splitting field of $X^3-2$, so at $L=\Bbb Q(\sqrt[3]{2},\xi)$. And this field also doesn't contain $\Bbb Q(i)$, see here. — Dietrich Burde, Mar 22 '24 at 17:10

score 1 · Accepted Answer · answered Mar 22 '24 at 17:24

Let $w=\exp(\tfrac{2\pi i}3)$. If $i\in\Bbb Q(\sqrt2,w)$, then for some rational numbers $a,b,c,d,e,f,$ $$a+bw+cw^2+d\sqrt2+e\sqrt2 w+f\sqrt2 w^2=i$$ Taking imaginary parts we have $$A\sqrt3+B\sqrt6=1$$ for some $A,B\in\Bbb Q.$ If, $A,B\neq0$, by squaring, this implies that $\sqrt{18}=3\sqrt2$ is rational which is not. If $A$ or $B$ is zero, then $\sqrt3$ or $\sqrt6$ is rational. Contradiction again.

Is it true that $i \in \mathbb{Q}(\sqrt{2},\xi)$, where $\xi = 1_{\frac{2\pi}{3}} = -\frac{1}{2}+i\frac{\sqrt{3}}{2}$?

1 Answers1