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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:
View attachment 6591
View attachment 6592
In the above example from Lovett, we read the following:
" ... ... From our previous calculation, we see that in \(\displaystyle L[x]\),
\(\displaystyle m_{ \alpha , \mathbb{Q} } (x) = ( x^2 - 2 \sqrt{2} x - 1) ( x^2 + 2 \sqrt{2} x - 1) \)
... ... ... "
I cannot see how this formula for the minimum polynomial in \(\displaystyle L[x]\) is derived ...
Can someone please explain how Lovett derives the above expression for \(\displaystyle m_{ \alpha , \mathbb{Q} } (x) \) ... ... ?
Hope someone can help ... ...
Peter
*** EDIT ***
Just noted that earlier in the example we have that \(\displaystyle \alpha = \sqrt{2} + \sqrt{3}\) is a root of
\(\displaystyle p(x) = x^4 - 10 x^2 + 1 \)
which factors (in one case of two possibilities) as follows:
\(\displaystyle p(x) = (x^2 + cx - 1) (x^2 + dx - 1) \)
and this solves to \(\displaystyle (c,d) = \pm ( 2 \sqrt{2}, - 2 \sqrt{2} ) \)
... but how and why we can move from a field in which we have \(\displaystyle \sqrt{2}\) and \(\displaystyle \sqrt{3}\) to one in which we only have \(\displaystyle \sqrt{2}\) ... ... I am not sure ... seems a bit slick ... indeed I certainly don't follow Lovett's next step which is to say
" ... ... Hence, \(\displaystyle \alpha\) is a root of one of those two quadratics. By direct observation, we find that
\(\displaystyle m_{ \alpha , L } (x) = ( x^2 - 2 \sqrt{2} x - 1)\) ... "
Can someone please explain what is going on here ... in particular, why must \(\displaystyle \alpha\) be a root of one of those two quadratics?
Peter
I am currently focused on Chapter 7: Field Extensions ... ...
I need help with Example 7.2.7 ...
Example 7.2.7 reads as follows:
View attachment 6591
View attachment 6592
In the above example from Lovett, we read the following:
" ... ... From our previous calculation, we see that in \(\displaystyle L[x]\),
\(\displaystyle m_{ \alpha , \mathbb{Q} } (x) = ( x^2 - 2 \sqrt{2} x - 1) ( x^2 + 2 \sqrt{2} x - 1) \)
... ... ... "
I cannot see how this formula for the minimum polynomial in \(\displaystyle L[x]\) is derived ...
Can someone please explain how Lovett derives the above expression for \(\displaystyle m_{ \alpha , \mathbb{Q} } (x) \) ... ... ?
Hope someone can help ... ...
Peter
*** EDIT ***
Just noted that earlier in the example we have that \(\displaystyle \alpha = \sqrt{2} + \sqrt{3}\) is a root of
\(\displaystyle p(x) = x^4 - 10 x^2 + 1 \)
which factors (in one case of two possibilities) as follows:
\(\displaystyle p(x) = (x^2 + cx - 1) (x^2 + dx - 1) \)
and this solves to \(\displaystyle (c,d) = \pm ( 2 \sqrt{2}, - 2 \sqrt{2} ) \)
... but how and why we can move from a field in which we have \(\displaystyle \sqrt{2}\) and \(\displaystyle \sqrt{3}\) to one in which we only have \(\displaystyle \sqrt{2}\) ... ... I am not sure ... seems a bit slick ... indeed I certainly don't follow Lovett's next step which is to say
" ... ... Hence, \(\displaystyle \alpha\) is a root of one of those two quadratics. By direct observation, we find that
\(\displaystyle m_{ \alpha , L } (x) = ( x^2 - 2 \sqrt{2} x - 1)\) ... "
Can someone please explain what is going on here ... in particular, why must \(\displaystyle \alpha\) be a root of one of those two quadratics?
Peter
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