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

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The discussion centers on Example 11.1.8 from Stephen Lovett's "Abstract Algebra: Structures and Applications," specifically regarding the minimal polynomial $$m_{ \alpha , \mathbb{Q} } (x) = x^4 - 10x^2 + 1$$ and its roots. The roots identified are $$\alpha_1 = \sqrt{2} + \sqrt{3}, \ \alpha_2 = \sqrt{2} - \sqrt{3}, \ \alpha_3 = -\sqrt{2} + \sqrt{3}, \ \alpha_4 = -\sqrt{2} - \sqrt{3}$$. The polynomial can be solved using the quadratic formula by substituting $$u = x^2$$. Additionally, the discussion explains that automorphisms $$\sigma \in \text{Aut}(F/ \mathbb{Q})$$ permute the roots of the polynomial due to their bijective nature, as established in Proposition 11.6.

PREREQUISITES
  • Understanding of Galois Theory and field extensions
  • Familiarity with minimal polynomials and their properties
  • Knowledge of polynomial equations and the quadratic formula
  • Concept of automorphisms in abstract algebra
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  • Study the properties of minimal polynomials in field theory
  • Learn about Galois groups and their relationship to field extensions
  • Explore Proposition 11.6 in detail to understand automorphism behavior
  • Investigate examples of polynomial root determination using the quadratic formula
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Students of abstract algebra, particularly those focusing on Galois Theory, mathematicians interested in field extensions, and educators teaching polynomial equations and their roots.

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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 $$\alpha = \sqrt{2} + \sqrt{3}$$ is $$m_{ \alpha , \mathbb{Q} } (x) = x^4 - 10x^2 + 1$$ and the four roots of this polynomial are $$\alpha_1 = \sqrt{2} + \sqrt{3}, \ \ \alpha_2 = \sqrt{2} - \sqrt{3}, \ \ \alpha_3 = - \sqrt{2} + \sqrt{3}, \ \ \alpha_4 = - \sqrt{2} - \sqrt{3} $$

... ... ... ... "
Can someone please explain why, exactly, these are roots of the minimum polynomial $$m_{ \alpha , \mathbb{Q} } (x) = x^4 - 10x^2 + 1$$ ... ... 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 $$\sigma \in \text{Aut}(F/ \mathbb{Q} )$$. Then according to Proposition 11.1.4, $$\sigma$$ must permute the roots of $$m_{ \alpha , \mathbb{Q} } (x)$$ ... ... "Can someone explain what this means ... how exactly does $$\sigma $$ permute the roots of $$m_{ \alpha , \mathbb{Q} } (x)$$ ... ... 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 $$\sqrt{2}, \sqrt{3} \in F$$ so all the roots of $$m_{ \alpha , \mathbb{Q} } (x)$$ are in $$F$$ ... ... "Can someone please explain in simple terms exactly why and how we know that $$\sqrt{2}, \sqrt{3} \in F$$ ... ... ?
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
 
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Hi Peter,

Peter said:
Can someone please explain why, exactly, these are roots of the minimum polynomial $$m_{ \alpha , \mathbb{Q} } (x) = x^4 - 10x^2 + 1$$ ... ... 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.
Peter said:
Question 2In the above text from Lovett we read the following:" ... ... Let $$\sigma \in \text{Aut}(F/ \mathbb{Q} )$$. Then according to Proposition 11.1.4, $$\sigma$$ must permute the roots of $$m_{ \alpha , \mathbb{Q} } (x)$$ ... ... "Can someone explain what this means ... how exactly does $$\sigma $$ permute the roots of $$m_{ \alpha , \mathbb{Q} } (x)$$ ... ... 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 $m_{\alpha, \Bbb Q}$.
Peter said:
Question 3In the above text from Lovett we read the following:" ... ... In Example 7.2.7 we observed that $$\sqrt{2}, \sqrt{3} \in F$$ so all the roots of $$m_{ \alpha , \mathbb{Q} } (x)$$ are in $$F$$ ... ... "Can someone please explain in simple terms exactly why and how we know that $$\sqrt{2}, \sqrt{3} \in F$$ ... ... ?
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., $\sqrt{2}\in F$.
 
Euge said:
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
 

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