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

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

• I think it's $\Bbb Q(\alpha)$, where $\alpha^2=2i$ – J. W. Tanner Jun 28 at 16:15
• @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? – Akash Gaur Jun 28 at 16:24
• @J.W.Tanner I'm only aware of the dimension of a vector space. How do you define the dimension of a field? – Akash Gaur Jun 28 at 16:33
• 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 – J. W. Tanner Jun 28 at 16:35

## 2 Answers

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

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