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

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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 ...

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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