# Question of finding the degree of extension $\Bbb Q( \sqrt2,ζ_3\sqrt[ 3] {2})$

I am trying to find the degree of extension $\Bbb Q( \sqrt 2,ζ_3\sqrt[ 3] {2})$.

I tried by using the following methods:

Method 1:

Consider the tower of extension $\Bbb Q\subset \Bbb Q( \sqrt 2) \subset\Bbb Q( \sqrt 2,ζ_3\sqrt[ 3] {2})$.

The minimal polynomial of $\sqrt 2$ over $\Bbb Q$ is $x^3-2$, and the minimal polynomial of $ζ_3\sqrt[ 3] {2}$ over $\Bbb Q( \sqrt 2)$ is $x^3-2$,so $[\Bbb Q( \sqrt 2):\Bbb Q ]=[\Bbb Q( \sqrt 2,,ζ_3\sqrt[ 3]{2} ) :\Bbb Q( \sqrt 2)]=3$. Thus $[\Bbb Q( \sqrt2,ζ_3\sqrt[ 3] {2}) :\Bbb Q]=3\cdot 3=9$.

Method 2:

Consider the tower of extension $\Bbb Q\subset \Bbb Q( \sqrt 2) \subset\Bbb Q( \sqrt 2,ζ_3\sqrt[ 3] {2}) \subset \Bbb Q( \sqrt 2,ζ_3)$.

The minimal polynomial of $\sqrt 2$ over $\Bbb Q$ is $x^3-2$, and the minimal polynomial of $ζ_3$ over $\Bbb Q( \sqrt 2)$ is $x^2+x+1$. Thus $\Bbb Q( \sqrt 2,ζ_3)$ has degree $6$ over $\Bbb Q$.

Thus the degree of $\Bbb Q( \sqrt 2,ζ_3\sqrt[ 3] {2})$ divides $6$. We can see that $\sqrt{2},(\sqrt{2})^2, ζ_3\sqrt[ 3] {2},(ζ_3\sqrt[ 3] {2})^2\in \Bbb Q( \sqrt 2,ζ_3\sqrt[ 3] {2})$ are linearly independent over $\Bbb Q$. Thus $\Bbb Q( \sqrt 2,ζ_3\sqrt[ 3] {2})$ cannot have degree $2$ or $3$. Thus it has degree $6$.

Now by different methods, I have conclude $2$ different degrees of the same field. So I am now confused, there must be something wrong. Could someone please point it out? Thanks so much!

In the first approach, $x^2+\sqrt2x+\sqrt{2}^2$ has $\sqrt2ζ_3$ as root, so $x^3-2$ is not its minimal polynomial.
You should've seen this because $\sqrt2$ is also a root of $x^3-2$, so in $\Bbb Q(\sqrt2)$, you can perform the polynomial division $(x^3-2)/(x-\sqrt2)$, and get the above polynomial.
The minimal polynomial of $\zeta_3$ $\sqrt{2}$ over $\mathbb{Q}(\sqrt{2})$ is $$\left(\frac{X}{\sqrt{2}}\right)^2 + \left(\frac{X}{\sqrt{2}}\right) +1.$$