Structure of $F=\mathbb Q(\sqrt {2i})$?

Question

Let $F=\mathbb Q(\sqrt {2i})$. My guess would be a field with elements of the form $a+b\sqrt 2i$ with $a,b \in \mathbb Q$. But the last option suggests that it is a vector space. Do not need hints about the solution yet.

Let $$F = \Bbb{Q}(\sqrt{2i}).$$

Which of the following is not true?

(A) $\sqrt{2} \in F$

(B) $i\in F$

(C) $x^8-16=0$ has a solution in $F$

(D) $\dim_{\Bbb{Q}}(F)=2$

Answer 1

Consider $\Bbb Q(\alpha)$, where $\alpha^2=2i$. Then $\alpha^4=(2i)^2=-4,$ so

$\alpha$ is a root of $x^4+4=(x^2+2)^2-4x^2=(x^2-2x+2)(x^2+2x+2).$

Elements of $\Bbb Q(\alpha)$ can be expressed as $q_1+q_2\alpha$ with $q_1,q_2,\in\mathbb Q$

$(\alpha^2 $ can be expressed as a linear combination of $\alpha$ and $1$), so $[\mathbb Q(\alpha):\mathbb Q]=2.$

Answer 2

Notice that $$(1+i)^2 = 1 + 2i -1 = 2i,$$ hence $$1+i = \sqrt{2i}.$$ Therefore $\Bbb{Q}(\sqrt{2i})=\Bbb{Q}(1+i)$. Let $\alpha := 1+i$. The minimal polynomial of $\alpha$ is $$p(x)=x^2-2x +2,$$ which has degree $2$, hence $$\dim_{\Bbb{Q}}(F)=2.$$ Also notice that $$i=-1 + \alpha \in \Bbb{Q}(\alpha).$$ Furthermore $$x^{8}-16 = (x^4-4)(x^4+4)$$ and $\alpha$ is a root of $x^4+4$, since $$\alpha^4 + 4 = (1+i)^4 + 4 = -4+4=0$$ Therefore, by exclusion the wrong statement is: $$\sqrt{2} \in F$$