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

Question

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

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

Method 2:

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

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

Thus the degree of $\Bbb Q( \sqrt[3] 2,ζ_3\sqrt[ 3] {2})$ divides $6$. We can see that $\sqrt[3]{2},(\sqrt[3]{2})^2, ζ_3\sqrt[ 3] {2},(ζ_3\sqrt[ 3] {2})^2\in \Bbb Q( \sqrt[3] 2,ζ_3\sqrt[ 3] {2})$ are linearly independent over $\Bbb Q$. Thus $\Bbb Q( \sqrt[3] 2,ζ_3\sqrt[ 3] {2})$ 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!

Answer 1

In the first approach, $x^2+\sqrt[3]2x+\sqrt[3]{2}^2$ has $\sqrt[3]2ζ_3$ as root, so $x^3-2$ is not its minimal polynomial.

You should've seen this because $\sqrt[3]2$ is also a root of $x^3-2$, so in $\Bbb Q(\sqrt[3]2)$, you can perform the polynomial division $(x^3-2)/(x-\sqrt[3]2)$, and get the above polynomial.

Answer 2

The minimal polynomial of $\zeta_3$ $\sqrt[3]{2}$ over $\mathbb{Q}(\sqrt[3]{2})$ is $$\left(\frac{X}{\sqrt[3]{2}}\right)^2 + \left(\frac{X}{\sqrt[3]{2}}\right) +1.$$