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:

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 to 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

 

[fresh_42]

We have $0=f(-d^{-1}t)=p(-d^{-1}t-a/p)(-d^{-1}t-b/p)$ in the field $F$. What does this mean for the factors?

 

[Math Amateur]

We have $0=f(-d^{-1}t)=p(-d^{-1}t-a/p)(-d^{-1}t-b/p)$ in the field $F$. What does this mean for the factors?

Hi fresh_42...

It seems that if we consider the polynomial $f(X) = fp (X - a/p) (X - b/p)$ in $F[X]$

then we must have that $d^{-1} t = a/p$ or $d^{-1} t = b/p$ in the field $F$ ...

But these cannot be solutions in $D$ ...

But how do I relate this to the factorization $f(X) = (cX + s) ( dX + t )$ in $D[X]$ ... ... ?

PeterEDIT ... oh ... the $d^{-1} t$ comes from $f(X) = (cX + s) ( dX + t )$ ! Maybe what I have said then means there is no factorization of the form $f(X) = (cX + s) ( dX + t )$ .

 

[fresh_42]

$f(x)=p(x-\frac{a}{p})(x-\frac{b}{p}) \in F[x]$ is simply set. It's a polynomial that is chosen on purpose. Since the condition (1.2.3) applies to all quadratic polynomials in $D[x]$, it also applies to $f(x)=px^2-(a+b)x+\frac{ab}{p} \in D[x]$. $f(x)$ is in $D[x]$ because $p,a,b,r=\frac{ab}{p}$ are all in $D$.

The factorization $f(x)=(cx+s)(dx+t)$ in $D[x]$ is the assumption we want to show cannot hold. It leads to $d^{-1}t \in D \;\wedge \; d^{-1}t \notin D$ which is obviously false. Thus $f(x)$ doesn't split over $D$. But according to condition (1.2.3) it has to split over $D$. Therefore $f(x)$ cannot be in $D[x]$. But the only way, that $f(x) \notin D[x]$ is by $r \notin D$. But then $p \nmid (ab)$, which means $p$ isn't irreducible and non-prime. Since $p$ is chosen irreducible, it has to conflict the other property, being non-prime, which was our first assumption. But non-non-prime means $p$ is prime, what had to be shown.

There's a lot of rollback in this proof. If it were a source code, I would insist on at least a three-level indent:
1. $p$ is not prime.

$\Longrightarrow r \in D$
2. $f(x)$ splits over $D$

$\Longrightarrow \textrm{ w.l.o.g. } d^{-1}t \in D$
3. w.l.o.g. $a/p \in D$

contradiction​

$\Longrightarrow d^{-1}t \notin D$
$\Longrightarrow f(x)$ doesn't split over $D$ // <-1 indent>
$\Longrightarrow r \notin D$ // <-2 indent>
$\Longrightarrow p$ not not prime // <-3 indent>
$\Longrightarrow p$ prime // <-3 indent>

​

 

[Math Amateur]

$f(x)=p(x-\frac{a}{p})(x-\frac{b}{p}) \in F[x]$ is simply set. It's a polynomial that is chosen on purpose. Since the condition (1.2.3) applies to all quadratic polynomials in $D[x]$, it also applies to $f(x)=px^2-(a+b)x+\frac{ab}{p} \in D[x]$. $f(x)$ is in $D[x]$ because $p,a,b,r=\frac{ab}{p}$ are all in $D$.

The factorization $f(x)=(cx+s)(dx+t)$ in $D[x]$ is the assumption we want to show cannot hold. It leads to $d^{-1}t \in D \;\wedge \; d^{-1}t \notin D$ which is obviously false. Thus $f(x)$ doesn't split over $D$. But according to condition (1.2.3) it has to split over $D$. Therefore $f(x)$ cannot be in $D[x]$. But the only way, that $f(x) \notin D[x]$ is by $r \notin D$. But then $p \nmid (ab)$, which means $p$ isn't irreducible and non-prime. Since $p$ is chosen irreducible, it has to conflict the other property, being non-prime, which was our first assumption. But non-non-prime means $p$ is prime, what had to be shown.

There's a lot of rollback in this proof. If it were a source code, I would insist on at least a three-level indent:
1. $p$ is not prime.

$\Longrightarrow r \in D$
2. $f(x)$ splits over $D$

$\Longrightarrow \textrm{ w.l.o.g. } d^{-1}t \in D$
3. w.l.o.g. $a/p \in D$

contradiction​

$\Longrightarrow d^{-1}t \notin D$
$\Longrightarrow f(x)$ doesn't split over $D$ // <-1 indent>
$\Longrightarrow r \notin D$ // <-2 indent>
$\Longrightarrow p$ not not prime // <-3 indent>
$\Longrightarrow p$ prime // <-3 indent>

​

Thanks fresh_42 ... really appreciate your help ...

... BUT ... just a clarification ...You write:"... ... It leads to $d^{-1}t \in D \;\wedge \; d^{-1}t \notin D$ ... ... "I can see that $d^{-1}t \in D$ ... ... since $d$ is a unit of $D$ we have $d^{-1} \in D$ and also we have $t \in D$ ... so $d^{-1}t \in D$ ...But in what way do we show that $d^{-1}t \notin D$ ... ... ? How do you argue that $a/p \in D$ ... ? Don't Alaca and Williams only argue that $a/p \in F$?

Can you help ...

Peter

 

[fresh_42]

We have $0=f(-d^{-1}t)=p(-d^{-1}t-a/p)(-d^{-1}t-b/p)$ in the field $F$.

So one factor has to be zero. Since $p\neq 0$, w.l.o.g. (the argument in the other case is the same) $d^{-1}t = -\frac{a}{p} \notin D$ because $p\nmid a$ in $D$.