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

[Math Amateur]

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 6592In 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

 

[jvicens]

Dear Peter,

Thank you for reaching out for help with Example 7.2.7 in "Abstract Algebra: Structures and Applications." I will do my best to explain the steps leading up to the formula for the minimum polynomial m_{ \alpha , \mathbb{Q} } (x) in L[x].

In the example, we are given that \alpha = \sqrt{2} + \sqrt{3} is a root of the polynomial p(x) = x^4 - 10x^2 + 1. This means that p(\alpha) = 0, and \alpha is an element of the field extension L = \mathbb{Q}(\sqrt{2},\sqrt{3}). From this, we can deduce that \alpha is also a root of the polynomial q(x) = (x - \sqrt{2} - \sqrt{3})(x - \sqrt{2} + \sqrt{3})(x + \sqrt{2} - \sqrt{3})(x + \sqrt{2} + \sqrt{3}). This is because if we substitute \sqrt{2} + \sqrt{3} into q(x), we get q(\sqrt{2} + \sqrt{3}) = 0.

Next, we can use the fact that q(x) can be factored into two quadratics (as you noted in your edit). This is because q(x) is a polynomial of degree four, and every polynomial of degree n has n roots (counting multiplicity). Therefore, we can write q(x) = (x^2 + cx - 1)(x^2 + dx - 1). We can solve for c and d by comparing coefficients with the original polynomial p(x) = x^4 - 10x^2 + 1. This gives us the solutions (c,d) = \pm ( 2 \sqrt{2}, -2 \sqrt{2} ).

Finally, we can use this information to write q(x) as a product of two quadratics that have \alpha as a root. By substituting the values for c and d, we get q(x) = (x^2 - 2 \sqrt{2} x - 1)(x^2 + 2 \sqrt{2} x - 1). This is the same expression that we see in the formula for m_{ \alpha , \mathbb