Automorphisms of Field Extensions .... Lovett, Example 11.1.8 .... ....

[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 11.1.8 on page 559 ... ...Example 11.1.8 reads as follows:
View attachment 6657
My questions regarding the above example from Lovett are as follows:
Question 1In the above text from Lovett we read the following:" ... ... The minimal polynomial of \(\displaystyle \alpha = \sqrt{2} + \sqrt{3}\) is \(\displaystyle m_{ \alpha , \mathbb{Q} } (x) = x^4 - 10x^2 + 1\) and the four roots of this polynomial are \(\displaystyle \alpha_1 = \sqrt{2} + \sqrt{3}, \ \ \alpha_2 = \sqrt{2} - \sqrt{3}, \ \ \alpha_3 = - \sqrt{2} + \sqrt{3}, \ \ \alpha_4 = - \sqrt{2} - \sqrt{3} \)

... ... ... ... "
Can someone please explain why, exactly, these are roots of the minimum polynomial \(\displaystyle m_{ \alpha , \mathbb{Q} } (x) = x^4 - 10x^2 + 1\) ... ... and further, how we would go about methodically determining these roots to begin with ... ...

Question 2In the above text from Lovett we read the following:" ... ... Let \(\displaystyle \sigma \in \text{Aut}(F/ \mathbb{Q} )\). Then according to Proposition 11.1.4, \(\displaystyle \sigma\) must permute the roots of \(\displaystyle m_{ \alpha , \mathbb{Q} } (x)\) ... ... "Can someone explain what this means ... how exactly does \(\displaystyle \sigma \) permute the roots of \(\displaystyle m_{ \alpha , \mathbb{Q} } (x)\) ... ... and how does Proposition 11.1.4 assure this, exactly ... ...
NOTE: The above question refers to Proposition 11.1.4 so I am providing that proposition and its proof ... ... as follows:
View attachment 6658

Question 3In the above text from Lovett we read the following:" ... ... In Example 7.2.7 we observed that \(\displaystyle \sqrt{2}, \sqrt{3} \in F\) so all the roots of \(\displaystyle m_{ \alpha , \mathbb{Q} } (x)\) are in \(\displaystyle F\) ... ... "Can someone please explain in simple terms exactly why and how we know that \(\displaystyle \sqrt{2}, \sqrt{3} \in F\) ... ... ?
NOTE: Lovett mentions Example 7.2.7 so I am providing the text of this example ... as follows:
View attachment 6659
View attachment 6660

I hope that someone can help with the above three questions ...

Any help will be much appreciated ... ...

Peter

 

[Euge]

Hi Peter,

Can someone please explain why, exactly, these are roots of the minimum polynomial \(\displaystyle m_{ \alpha , \mathbb{Q} } (x) = x^4 - 10x^2 + 1\) ... ... and further, how we would go about methodically determining these roots to begin with ... ...

Fortunately, you can use high school algebra. The polynomial $x^4 - 10x^2 + 1$ is a quadratic in disguise, namely, if $u = x^2$ then the polynomial is $u^2 - 10u + 1$. So you can use the quadratic formula to find the roots.

Question 2In the above text from Lovett we read the following:" ... ... Let \(\displaystyle \sigma \in \text{Aut}(F/ \mathbb{Q} )\). Then according to Proposition 11.1.4, \(\displaystyle \sigma\) must permute the roots of \(\displaystyle m_{ \alpha , \mathbb{Q} } (x)\) ... ... "Can someone explain what this means ... how exactly does \(\displaystyle \sigma \) permute the roots of \(\displaystyle m_{ \alpha , \mathbb{Q} } (x)\) ... ... and how does Proposition 11.1.4 assure this, exactly ... ...

Let $R$ be the set of roots of $m_{\alpha, \Bbb Q}$ in $F$. Since $\sigma$ is an automorphism of $F$, it is a bijection. By Proposition 11.6, $\sigma$ restricted to $R$ is a map $R\to R$, which is a bijection since $\sigma$ is a bijection. Hence, $\sigma$ permutes the elements of $R$, i.e., the roots of $m_{\alpha, \Bbb Q}$.

Question 3In the above text from Lovett we read the following:" ... ... In Example 7.2.7 we observed that \(\displaystyle \sqrt{2}, \sqrt{3} \in F\) so all the roots of \(\displaystyle m_{ \alpha , \mathbb{Q} } (x)\) are in \(\displaystyle F\) ... ... "Can someone please explain in simple terms exactly why and how we know that \(\displaystyle \sqrt{2}, \sqrt{3} \in F\) ... ... ?
NOTE: Lovett mentions Example 7.2.7 so I am providing the text of this example ... as follows:

It follows from closure under addition and scalar multiplication in $F$. For example, since $\alpha_1, \alpha_2\in F$, then $(1/2)(\alpha_1 + \alpha_2)\in F$, i.e., $\sqrt{2}\in F$.

 

[Math Amateur]

Hi Peter,
Fortunately, you can use high school algebra. The polynomial $x^4 - 10x^2 + 1$ is a quadratic in disguise, namely, if $u = x^2$ then the polynomial is $u^2 - 10u + 1$. So you can use the quadratic formula to find the roots.
Let $R$ be the set of roots of $m_{\alpha, \Bbb Q}$ in $F$. Since $\sigma$ is an automorphism of $F$, it is a bijection. By Proposition 11.6, $\sigma$ restricted to $R$ is a map $R\to R$, which is a bijection since $\sigma$ is a bijection. Hence, $\sigma$ permutes the elements of $R$, i.e., the roots of $m_{\alpha, \Bbb Q}$.

It follows from closure under addition and scalar multiplication in $F$. For example, since $\alpha_1, \alpha_2\in F$, then $(1/2)(\alpha_1 + \alpha_2)\in F$, i.e., $\sqrt{2}\in F$.

Thanks Euge ... I appreciate your help ...

Peter