Algebraic Extensions - Lovett, Example 7.2.7 .... ....

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In summary, Peter is reading "Abstract Algebra: Structures and Applications" by Stephen Lovett and is currently focused on Chapter 7: Field Extensions. He needs help understanding Example 7.2.7, which involves finding the minimal polynomial of a complex number over a field. The example shows that the minimal polynomial can be written as a product of two quadratics, and Lovett then explains how to determine which one is the minimal polynomial for the given number. Peter is still trying to fully understand the concept of splitting fields and the term "decomposition" in this context. He thanks fresh_42 for their help and is continuing to study and reflect on the topic.
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I am reading "Abstract Algebra: Structures and Applications" by Stephen Lovett ...

I am currently focused on Chapter 7: Field Extensions ... ...

I need help with Example 7.2.7 ...Example 7.2.7 reads as follows:
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?temp_hash=d094c6dc6c31a758354a9e0817969ade.png

In the above example from Lovett, we read the following:

" ... ... From our previous calculation, we see that in ##L[x]##,

##m_{ \alpha , \mathbb{Q} } (x) = ( x^2 - 2 \sqrt{2} x - 1) ( x^2 + 2 \sqrt{2} x - 1)##

... ... ... "
I cannot see how this formula for the minimum polynomial in ##L[x]## is derived ...

Can someone please explain how Lovett derives the above expression for ##m_{ \alpha , \mathbb{Q} } (x) ## ... ... ?Hope someone can help ... ...

Peter

*** EDIT ***Just noted that earlier in the example we have that ##\alpha = \sqrt{2} + \sqrt{3}## is a root of

##p(x) = x^4 - 10 x^2 + 1##

which factors (in one case of two possibilities) as follows:

##p(x) = (x^2 + cx - 1) (x^2 + dx - 1)##

and this solves to ##(c,d) = \pm ( 2 \sqrt{2}, - 2 \sqrt{2} )##

... but how and why we can move from a field in which we have ##\sqrt{2}## and ##\sqrt{3}## to one in which we only have ##\sqrt{2}## ... ... I am not sure ... seems a bit slick ... ...

... indeed I certainly don't follow Lovett's next step which is to say" ... ... Hence, ##\alpha## is a root of one of those two quadratics. By direct observation, we find that

##m_{ \alpha , L } (x) = ( x^2 - 2 \sqrt{2} x - 1)## ... "Can someone please explain what is going on here ... ... in particular, why must ##\alpha## be a root of one of those two quadratics?

Peter
 

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  • #2
For the minimal polynomial of ##\alpha## over ##\mathbb{Q}## we have
$$
m_{\alpha,\mathbb{Q}}(x)=x^4-10x^2+1=p(x)=(x^2+ax+1)(x^2+bx+1) \textrm{ or } = (x^2+cx-1)(x^2+dx-1)
$$
in a potential splitting field. From this, the only solutions are ##(a,b) \in \{\pm 2\sqrt{3}\, , \,\mp 2\sqrt{3}\}## and ##(c,d) \in \{\pm 2\sqrt{2}\, , \,\mp 2\sqrt{2}\}##.

This has been shown by the calculation Lovett refers to.
Now ##\sqrt{2} \in \mathbb{Q}(\sqrt{2}) =: L## which leaves only the decomposition with ##(c,d)## as a possibility in ##L[x]##, because Lovett has shown in the lines before, that ##\sqrt{3}\notin L##, which would have allowed the ##(a,b)## solution also to be possible. He did this by showing that ##\sqrt{3}\in L = \mathbb{Q}(\sqrt{2}) \Longleftrightarrow \sqrt{3}=r+s\sqrt{2}\, , \,r,s\in \mathbb{Q}## leads to a contradiction.
(## \{1,\sqrt{2}\}## is a ##\mathbb{Q}-##basis of ##L##)

Thus ##m_{\alpha,\mathbb{Q}}(x)=x^4-10x^2+1=(x^2-2\sqrt{2}-1)(x^2+2\sqrt{2}-1)## is a valid decomposition in ##L[x]=\mathbb{Q}(\sqrt{2})[x]## and ##m_{\alpha,\mathbb{Q}}(\alpha)=0##, i.e. one of the factors has to be zero. Since ##\alpha =\sqrt{2}+\sqrt{3}## we can calculate which one and find the minimal polynomial of ##\alpha## over ##L##. (It has to be irreducible, since we already know, that ##\sqrt{3}\notin L##.)
 
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  • #3
fresh_42 said:
For the minimal polynomial of ##\alpha## over ##\mathbb{Q}## we have
$$
m_{\alpha,\mathbb{Q}}(x)=x^4-10x^2+1=p(x)=(x^2+ax+1)(x^2+bx+1) \textrm{ or } = (x^2+cx-1)(x^2+dx-1)
$$
in a potential splitting field. From this, the only solutions are ##(a,b) \in \{\pm 2\sqrt{3}\, , \,\mp 2\sqrt{3}\}## and ##(c,d) \in \{\pm 2\sqrt{2}\, , \,\mp 2\sqrt{2}\}##.

This has been shown by the calculation Lovett refers to.
Now ##\sqrt{2} \in \mathbb{Q}(\sqrt{2}) =: L## which leaves only the decomposition with ##(c,d)## as a possibility in ##L[x]##, because Lovett has shown in the lines before, that ##\sqrt{3}\notin L##, which would have allowed the ##(a,b)## solution also to be possible. He did this by showing that ##\sqrt{3}\in L = \mathbb{Q}(\sqrt{2}) \Longleftrightarrow \sqrt{3}=r+s\sqrt{2}\, , \,r,s\in \mathbb{Q}## leads to a contradiction.
(## \{1,\sqrt{2}\}## is a ##\mathbb{Q}-##basis of ##L##)

Thus ##m_{\alpha,\mathbb{Q}}(x)=x^4-10x^2+1=(x^2-2\sqrt{2}-1)(x^2+2\sqrt{2}-1)## is a valid decomposition in ##L[x]=\mathbb{Q}(\sqrt{2})[x]## and ##m_{\alpha,\mathbb{Q}}(\alpha)=0##, i.e. one of the factors has to be zero. Since ##\alpha =\sqrt{2}+\sqrt{3}## we can calculate which one and find the minimal polynomial of ##\alpha## over ##L##. (It has to be irreducible, since we already know, that ##\sqrt{3}\notin L##.)
Thanks for the help, fresh_42 ... appreciate it ...

Still reflecting on your post ...

BUT ... Lovett does not get to "splitting fields" for another 30 pages or so ... so may have to wait a bit before I fully appreciate what you have written ...

I am also not sure of what you mean when you mention "decomposition" either ... but hopefully soon either Lovett or Dummit and Foote will cover the term in the context of field extensions ... ...

Thanks again,

Peter
 
  • #4
Well, ##m_{\alpha , \mathbb{Q}}## is irreducible. By splitting field I meant a field extension, where it is possible to factor it. But it's sufficient here to assume a factorization. A "split" into linear factors isn't needed.

In any case it's important to keep in mind, that irreducibility depends on "where in". E.g. ##x^2+1## is irreducible over the reals, but as ##x^2+1=(x+i)(x-i)## it is not over ##\mathbb{C}##, i.e. it splits.

Decomposition simply means a factorization, usually into linear factors, but as it's not a mathematical term, I used it as synonym for factorization (here into quadratic factors). Splitting field on the other hand is a defined term. It means the field, that provides the elements (numbers) needed to split a polynomial into linear factors, to decompose it.
 
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Thanks fresh_42 ... the issues are getting clearer ...

I appreciate your considerable and significant help ..

Peter
 

Related to Algebraic Extensions - Lovett, Example 7.2.7 .... ....

1. What is an algebraic extension?

An algebraic extension is a field extension where all elements of the extension are algebraic over the base field. This means that they are all solutions to polynomial equations with coefficients in the base field.

2. How is the degree of an algebraic extension determined?

The degree of an algebraic extension is the degree of the minimal polynomial of any element in the extension. This is the smallest degree polynomial with coefficients in the base field that has the element as a root.

3. What is the difference between an algebraic extension and a transcendental extension?

An algebraic extension consists of elements that are all algebraic over the base field, while a transcendental extension contains elements that are not algebraic over the base field. This means that they cannot be expressed as solutions to any polynomial equations with coefficients in the base field.

4. Can an algebraic extension have a finite degree?

Yes, an algebraic extension can have a finite degree. For example, if the base field is the rational numbers and the extension is the square root of 2, the degree of the extension is 2 since it is a solution to the polynomial equation x^2 - 2 = 0.

5. How are algebraic extensions used in mathematics?

Algebraic extensions are used in various areas of mathematics, including algebraic geometry, number theory, and algebraic topology. They are also essential in Galois theory, which studies the symmetries of polynomial equations and their solutions.

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