MHB Quadratic Polynomials and Irreducibles and Primes

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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:
View attachment 6514
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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
 
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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 $D$.
 
Euge said:
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
 
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This is the question, I understand the concept, in ##\mathbb{Z_n}## an element is a is a unit if and only if gcd( a,n) =1. My understanding of backwards substitution, ... i have using Euclidean algorithm, ##471 = 3⋅121 + 108## ##121 = 1⋅108 + 13## ##108 =8⋅13+4## ##13=3⋅4+1## ##4=4⋅1+0## using back-substitution, ##1=13-3⋅4## ##=(121-1⋅108)-3(108-8⋅13)## ... ##= 121-(471-3⋅121)-3⋅471+9⋅121+24⋅121-24(471-3⋅121## ##=121-471+3⋅121-3⋅471+9⋅121+24⋅121-24⋅471+72⋅121##...
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