$\mathbb Q [\sqrt{2} i]$ contains neither $\sqrt[4]{2}$ nor $\sqrt{2}$

Question

I want to prove that $x^4-2$ is irreducible over $\mathbb Q [\sqrt{2} i]$. In order to verify it has no linear factors and quadratic factors, I need to show $\mathbb Q [\sqrt{2} i]$ contains neither $\sqrt[4]{2}$ nor $\sqrt{2}$, can anybody show me a strict proof? Thanks in advance.

Answer 1

The field $Q[i\sqrt2]$ has degree two over $Q$ and so does $Q[\sqrt2]$. If the first contained $\sqrt2$ then the two would be equal. Yet the second one is contained in $R$ and the other is not.

On the other hand, the minimal polynomial $\sqrt[4]2$ over $Q$ has degree four, so that number cannot be in $Q[\sqrt2]$.

Answer 2

If $\mathbb{Q}[\sqrt{2}i]$ contains $\sqrt[4]{2}$, then $\mathbb{Q}[\sqrt{2}i]=\mathbb{Q}[\sqrt{2}i][\sqrt[4]{2}]$ and therefore

$$2=[\mathbb{Q}[\sqrt{2}i]:\mathbb{Q}]=[\mathbb{Q}[\sqrt{2}i][\sqrt[4]{2}]:\mathbb{Q}[\sqrt[4]{2}]]\cdot [\mathbb{Q}[\sqrt[4]{2}]:\mathbb{Q}]= k\cdot 4$$

where $k=[\mathbb{Q}[\sqrt{2}i][\sqrt[4]{2}]:\mathbb{Q}[\sqrt[4]{2}]]$, which is absurd.

(This is just Mariano's answer, I was writing mine when he posted).

Answer 3

Let $\alpha=\sqrt 2 i\notin \mathbb Q$. Using $\alpha^2=-2$, we note that $(a+b\alpha)^2=(a^2-2b^2)+2ab\alpha$ for $a,b\in\mathbb Q$. For the latter to equal $2$ we need $2ab=0$ and $a^2-2b^2=2$ (because $1,\alpha$ are $\mathbb Q$-linearly independant). But if $a=0$, we cannot have $-2b^2=2$, and if $b=0$ we cannot have $a^2=2$. Therefore $\sqrt 2\notin\mathbb Q[\alpha]$ and even more so $\sqrt[4]2\notin\mathbb Q[\alpha]$.