MHB 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 $$f = ( x - \alpha)^k g$$, where $$k$$ is an integer greater than $$1$$ and $$g$$ is a polynomial over $$K$$ ... ... Since $$f$$ is in $$F[x]$$ ... that is $$f$$ is over $$F$$ ... shouldn't g be over $$F$$ not $$K$$?

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

Question 2In the above text from Anderson and Feil we read the following:"... ... We then have that $$x - \alpha$$ is a factor of both $$f$$ and $$f'$$. But if we use term-by-term differentiation instead, it is clear that $$f'\in F[x]$$. ... ... "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 $$f'\in F[x]$$ ... ... ?
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 $$f = ( x - \alpha)^k g$$, where $$k$$ is an integer greater than $$1$$ and $$g$$ is a polynomial over $$K$$ ... ... Since $$f$$ is in $$F[x]$$ ... that is $$f$$ is over $$F$$ ... shouldn't g be over $$F$$ not $$K$$?

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 $$x - \alpha$$ is a factor of both $$f$$ and $$f'$$. But if we use term-by-term differentiation instead, it is clear that $$f'\in F[x]$$. ... ... "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 $$f'\in F[x]$$ ... ... ?

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 $$g$$ have it's coefficients in $$F$$ ... but possesses no root in $$F$$ ... and still be irreducible in $$F[x]$$ ... 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