Splitting Fields: Anderson and Feil, Theorem 45.4

[Math Amateur]

I am reading Anderson and Feil - A First Course in Abstract Algebra.

I am currently focused on Ch. 45: The Splitting Field ... ...

I need some help with some aspects of the proof of Theorem 45.4 ...

Theorem 45.4 and its proof read as follows:View attachment 6677
My questions on the above proof are as follows:Question 1In the above text from Anderson and Feil we read the following:"... ... This means that \(\displaystyle f = ( x - \alpha)^k g\), where \(\displaystyle k\) is an integer greater than \(\displaystyle 1\) and \(\displaystyle g\) is a polynomial over \(\displaystyle K\) ... ... Since \(\displaystyle f\) is in \(\displaystyle F[x]\) ... that is \(\displaystyle f\) is over \(\displaystyle F\) ... shouldn't g be over \(\displaystyle F\) not \(\displaystyle K\)?

(I am assuming that f being "over \(\displaystyle F\)" means the coefficients of \(\displaystyle f\) are in \(\displaystyle F\) ... )

Question 2In the above text from Anderson and Feil we read the following:"... ... We then have that \(\displaystyle x - \alpha\) is a factor of both \(\displaystyle f\) and \(\displaystyle f'\). But if we use term-by-term differentiation instead, it is clear that \(\displaystyle f'\in F[x]\). ... ... "What do Anderson and Feil mean by term-by-term differentiation in this context ... ... and if they do use term-by-term differentiation (what ever they mean) how does this show that \(\displaystyle f'\in F[x]\) ... ... ?
Hope someone can help ...

Help will be much appreciated ... ...

Peter

 

[Euge]

Question 1In the above text from Anderson and Feil we read the following:"... ... This means that \(\displaystyle f = ( x - \alpha)^k g\), where \(\displaystyle k\) is an integer greater than \(\displaystyle 1\) and \(\displaystyle g\) is a polynomial over \(\displaystyle K\) ... ... Since \(\displaystyle f\) is in \(\displaystyle F[x]\) ... that is \(\displaystyle f\) is over \(\displaystyle F\) ... shouldn't g be over \(\displaystyle F\) not \(\displaystyle K\)?

If $g$ were a polynomial over $F$, then $f$ would be reducible in $F[x]$, contrary to assumption.

Question 2In the above text from Anderson and Feil we read the following:"... ... We then have that \(\displaystyle x - \alpha\) is a factor of both \(\displaystyle f\) and \(\displaystyle f'\). But if we use term-by-term differentiation instead, it is clear that \(\displaystyle f'\in F[x]\). ... ... "What do Anderson and Feil mean by term-by-term differentiation in this context ... ... and if they do use term-by-term differentiation (what ever they mean) how does this show that \(\displaystyle f'\in F[x]\) ... ... ?

They mean to differentiate each summand in the standard form of $f$. So if $f(x) = c_0 + c_1x + \cdots + c_n x^n$, then you are to compute $Df$ by computing the derivative of each summand $c_i x^i$ ($0 \le i \le n$). This would give a polynomial in $F[x]$ of lesser degree than that of $f$.

 

[Math Amateur]

If $g$ were a polynomial over $F$, then $f$ would be reducible in $F[x]$, contrary to assumption. They mean to differentiate each summand in the standard form of $f$. So if $f(x) = c_0 + c_1x + \cdots + c_n x^n$, then you are to compute $Df$ by computing the derivative of each summand $c_i x^i$ ($0 \le i \le n$). This would give a polynomial in $F[x]$ of lesser degree than that of $f$.

Thanks for the help Euge ...You write:"... ... If $g$ were a polynomial over $F$, then $f$ would be reducible in $F[x]$, contrary to assumption. ... ... But ... ... ... couldn't \(\displaystyle g\) have it's coefficients in \(\displaystyle F\) ... but possesses no root in \(\displaystyle F\) ... and still be irreducible in \(\displaystyle F[x]\) ... why is this not possible ... ?

Can you help further ...?

Peter

 

[Euge]

Since $k >1$, $x - \alpha$ would be a proper factor of $f$.

 

[Math Amateur]

Since $k >1$, $x - \alpha$ would be a proper factor of $f$.

Thanks Euge ... but I still do not follow ... sorry ... probably being slow to catch on ...

I am having difficulties seeing the link between k > 1 and f being reducible in F[x] and g being over K (or F) ...

Are you able to clarify ...

Again ... apologies for not following your points ...

Peter

 

[Euge]

Sorry, I made a slight error, assuming $\alpha\in F$. It's by definition of $\alpha$ being a repeated root in $K$ that $f(x)$ can be expressed as $(x - \alpha)^k g(x)$ for some $g(x)\in K[x]$ and integer $k > 1$. So while it's possible that $g(x)\in F[x]$ (as $F[x]\subset K[x]$), it is not necessarily true that $g(x)\in F[x]$.

 

[Math Amateur]

Sorry, I made a slight error, assuming $\alpha\in F$. It's by definition of $\alpha$ being a repeated root in $K$ that $f(x)$ can be expressed as $(x - \alpha)^k g(x)$ for some $g(x)\in K[x]$ and integer $k > 1$. So while it's possible that $g(x)\in F[x]$ (as $F[x]\subset K[x]$), it is not necessarily true that $g(x)\in F[x]$.

Thanks Euge ...

Appreciate all your help ...

Peter