Galois Theory - Structure Within Aut(K/Q) .... Lovett, Example 11.1.10 .... ....

[Math Amateur]

I am reading "Abstract Algebra: Structures and Applications" by Stephen Lovett ...

I am currently focused on Chapter 8: Galois Theory, Section 1: Automorphisms of Field Extensions ... ...

I need help with Example 11.1.10 on page 560 ... ...Example 11.1.10 reads as follows:
View attachment 6662
My questions regarding the above example are as follows:

Question 1In the above text from Lovett we read the following:

" ... ... It is easy to show that \(\displaystyle \text{Aut} ( K / \mathbb{Q} ( \sqrt[3]{13} ) \ \cong \mathbb{Z}_2\) ... ... "Can someone please help me to show that \(\displaystyle \text{Aut} ( K / \mathbb{Q} ( \sqrt[3]{13} ) \ \cong \mathbb{Z}_2\) ... ... ?

Question 2In the above text from Lovett we read the following:

" ... ... By Proposition 11.1.4, there are at most three options where to map \(\displaystyle \sqrt[3]{13}\) and at most two options where to map \(\displaystyle \sqrt{ -3 }\) ... ... "My question is ... what are the three options where to map \(\displaystyle \sqrt[3]{13}\) ... surely there is only one option to map \(\displaystyle \sqrt[3]{13}\) as the field extension is over \(\displaystyle \mathbb{Q}\) and two of the "options" are imaginary or complex numbers ... or is Lovett just relying on "at most" ... why not rule out the complex roots straight away ... similarly I am puzzled about the two options for \(\displaystyle \sqrt{ -3 }\) ... I am also puzzled about the role of Proposition 11.1.4 in this matter ... How does Proposition 11.1.4 ensure that there are at most three options where to map \(\displaystyle \sqrt[3]{13}\) and at most two options where to map \(\displaystyle \sqrt{ -3 }\) ... ... ... can someone please clarify the situation ...The above question refers to Proposition 11.1.4 ... ... so I am providing the text of that proposition ... ... as follows ... ...
https://www.physicsforums.com/attachments/6663

Help with the above questions will be much appreciated ... ...

Peter
***EDIT***

Just thinking ... since \(\displaystyle K = \mathbb{Q} ( \sqrt[3]{13}, \sqrt{ -3 } )\) it already contains a complex number, namely \(\displaystyle \sqrt{ -3 } = \sqrt{ 3 }i\) ... maybe that partly explains my questions ... and I think it is \(\displaystyle K\) containing the relevant roots, not \(\displaystyle \mathbb{Q}\) as I was implying above ...

 

[jvicens]

Dear Peter,

Thank you for reaching out for help with Example 11.1.10 in Lovett's "Abstract Algebra: Structures and Applications." I am happy to assist you in understanding this example better.

Question 1:

To show that \text{Aut} ( K / \mathbb{Q} ( \sqrt[3]{13} ) \ \cong \mathbb{Z}_2, we need to show that there exists an isomorphism between the two groups. An isomorphism is a bijective homomorphism, which means it is a function that preserves the group structure and is one-to-one and onto. In this case, we need to show that there exists a bijective homomorphism between the automorphism group of K over \mathbb{Q} (\sqrt[3]{13}) and the group \mathbb{Z}_2.

To do this, we can first define a map between the two groups. Let \phi : \text{Aut} ( K / \mathbb{Q} ( \sqrt[3]{13} ) \ \rightarrow \mathbb{Z}_2 be defined as follows:

\phi(\sigma) = \begin{cases} 0, & \text{if } \sigma(\sqrt[3]{13}) = \sqrt[3]{13} \\ 1, & \text{if } \sigma(\sqrt[3]{13}) = -\sqrt[3]{13} \end{cases}

In words, this map takes an automorphism \sigma and assigns it the value 0 if it fixes \sqrt[3]{13} and 1 if it sends \sqrt[3]{13} to its negative. We can then show that this map is a bijective homomorphism, which will prove that \text{Aut} ( K / \mathbb{Q} ( \sqrt[3]{13} ) \ \cong \mathbb{Z}_2.

Question 2:

Proposition 11.1.4 states that for any automorphism \sigma \in \text{Aut}(K/\mathbb{Q}), \sigma(\sqrt[3]{13}) is either \sqrt[3]{13}, \omega \sqrt[3]{13}, or \omega^2 \sqrt[3]{13}, where \omega is a primitive cube root of unity. Similarly, \sigma(\sqrt{-3