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:

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

 

[fresh_42]

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 $\sqrt{3}\notin L$.)

 

[Math Amateur]

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

 

[fresh_42]

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.

 

[Math Amateur]

Thanks fresh_42 ... the issues are getting clearer ...

I appreciate your considerable and significant help ..

Peter