Galois Groups .... A&F Example 47.7 .... ....

[Math Amateur]

I am reading Anderson and Feil - A First Course in Abstract Algebra.

I am currently focused on Ch. 47: Galois Groups... ...

I need some help with an aspect of the Example 47.7 ...

Example 47.7 and its proof read as follows:
View attachment 6869
View attachment 6870In the above example, Anderson and Feil write the following:

"... ... We note that \(\displaystyle [ \mathbb{Q} ( \sqrt[3]{2} ) : \mathbb{Q} ] = 3\) and \(\displaystyle [ \mathbb{Q} ( \zeta ) : \mathbb{Q} ] = 2\). ... ... "
Can someone please explain to me how/why \(\displaystyle [ \mathbb{Q} ( \zeta ) : \mathbb{Q} ] = 2\) ... ... ?

Anderson and Feil give the definition of \(\displaystyle \zeta\) in Chapter 9 in Exercise 25 ... as follows ... :
https://www.physicsforums.com/attachments/6871
Hope someone can help ...

Peter

 

[Euge]

Since $\zeta^3 = 1$, $\zeta$ is a root of $x^3 - 1$. Note $x^3 - 1 = (x - 1)(x^2+x+1)$. So as $\zeta \neq 1$, $\zeta$ is a root of $x^2+x+1$, which is irreducible over $\Bbb Q$. Hence, $|\Bbb Q(\zeta): \Bbb Q| = 2$.

 

[Math Amateur]

Since $\zeta^3 = 1$, $\zeta$ is a root of $x^3 - 1$. Note $x^3 - 1 = (x - 1)(x^2+x+1)$. So as $\zeta \neq 1$, $\zeta$ is a root of $x^2+x+1$, which is irreducible over $\Bbb Q$. Hence, $|\Bbb Q(\zeta): \Bbb Q| = 2$.

Thanks Euge ... appreciate your help ...

Peter