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

  • $\begingroup$ 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
  • 2
    $\begingroup$ 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

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 $[\mathbb Q(\alpha):\mathbb Q]=2.$

  • $\begingroup$ $x^4 + 4 = (x^2-2x+2)(x^2+2x+2)$ and is thus not irreducible, and therefore isn't the miminal polynomial of $\alpha$. $\endgroup$ – Daniel Schepler Jun 28 at 18:02
  • 1
    $\begingroup$ @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

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: $$\sqrt{2} \in F$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.