Quadratic Polynomials and Irreducibles and Primes

[Math Amateur]

I am reading "Introductory Algebraic Number Theory"by Saban Alaca and Kenneth S. Williams ... and am currently focused on Chapter 1: Integral Domains ...

I need some help with the proof of Theorem 1.2.2 ...

Theorem 1.2.2 reads as follows:
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In the above text from Alaca and Williams, we read the following:

"... ... Then the roots of $$f(X)$$ in $$F$$ are $$-ds/p$$ and $$-d^{-1} t $$. But $$d^{-1} t \in D$$ while neither $$a/p$$ nor $$b/p$$ is in $$D$$. Thus no such factorization exists. ... I am unsure of how this argument leads top the conclusion that $$f(X)$$ does not factor into linear factors in $$D[X]$$ ... in other words how does the argument that "" ... $$d^{-1} t \in D$$ while neither $$a/p$$ nor $$b/p$$ is in $$D$$ ... "lead to the conclusion that no such factorization exists. ...

Indeed ... in particular ... how does the statement "neither $$a/p$$ nor $$b/p$$ is in $$D$$" have meaning in the assumed factorization $$f(X) = (cX + s) ( dX + t )$$ ... ... ? ... What is the exact point being made about the assumed factorization ... ?I am also a little unsure of what is going on when Alaca and Williams change or swap between $$D[X]$$ and $$F[x]$$ ...Can someone help with an explanation ...

Help will be appreciated ...

Peter

 

[Euge]

Hi Peter,

The contradiction here is that $-d^{-1}t$ is in $D$ and also not in $D$. As $d, t\in D$, $-d^{-1}t\in D$. On the other hand, $-d^{-1}t$ is one of the roots of $f$ in $F[X]$, namely $a/p$ or $b/p$; since neither $a/p$ nor $b/p$ is in $D$, $-d^{-1}t$ is not in $D$.

 

[Math Amateur]

Hi Peter,

The contradiction here is that $-d^{-1}t$ is in $D$ and also not in $D$. As $d, t\in D$, $-d^{-1}t\in D$. On the other hand, $-d^{-1}t$ is one of the roots of $f$ in $F[X]$, namely $a/p$ or $b/p$; since neither $a/p$ nor $b/p$ is in $D$, $-d^{-1}t$ is not in $D$.

Thanks Euge ... that is clear now ...

Appreciate your help ...

Peter