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

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MHB
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 ... ...

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My questions regarding the above example from Lovett are as follows:

Question 1

In 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 2

In 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:

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Question 3

In 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:

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

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

Peter

Last edited:

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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 2

In 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 3

In 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., $\sqrt{2}\in F$.

Gold Member
MHB
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