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

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  • $\begingroup$ I think it's $\Bbb Q(\alpha)$, where $\alpha^2=2i$ $\endgroup$ – J. W. Tanner Jun 28 at 16:15
  • $\begingroup$ @J.W.Tanner So a field with elements of the form $a+b\sqrt {2i}$ with $a,b \in \mathbb Q$. Then what is its dimension of a field, see last option? $\endgroup$ – Akash Gaur Jun 28 at 16:24
  • $\begingroup$ @J.W.Tanner I'm only aware of the dimension of a vector space. How do you define the dimension of a field? $\endgroup$ – Akash Gaur Jun 28 at 16:33
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    $\begingroup$ a field extension is a vector space over the subfield; the dimension is how many elements in a basis, just like any other vector space $\endgroup$ – J. W. Tanner Jun 28 at 16:35
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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 $[\mathbb Q(\alpha):\mathbb Q]=2.$

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  • $\begingroup$ $x^4 + 4 = (x^2-2x+2)(x^2+2x+2)$ and is thus not irreducible, and therefore isn't the miminal polynomial of $\alpha$. $\endgroup$ – Daniel Schepler Jun 28 at 18:02
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    $\begingroup$ @DanielSchepler: thanks so much for pointing that out; I have tried to correct my answer accordingly $\endgroup$ – J. W. Tanner Jun 28 at 18:22
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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: $$\sqrt{2} \in F$$

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