Quadratic Polynomials and Irreducibles and Primes ....

In summary, the conversation discusses the proof of Theorem 1.2.2 in "Introductory Algebraic Number Theory" by Saban Alaca and Kenneth S. Williams. The theorem states that if ##p## is an irreducible non-prime element in a Euclidean domain ##D##, then it is prime. The conversation focuses on understanding how the argument in the proof leads to the conclusion that ##p## is prime. Specifically, there is a question about how the statement "neither ##a/p## nor ##b/p## is in ##D##" relates to the factorization ##f(X) = (cX + s) (dX + t)## in ##D[X]##. In response
  • #1
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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:
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?temp_hash=bf44d3fe3806f5b5360d520962884fa0.png

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
 

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  • #2
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?
 
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  • #3
fresh_42 said:
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 )## .
 
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  • #4
##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>
 
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  • #5
fresh_42 said:
##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
 
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  • #6
fresh_42 said:
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##.
 
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1. What is a quadratic polynomial?

A quadratic polynomial is a polynomial of degree 2, meaning it has the form ax^2 + bx + c, where a, b, and c are constants and x is the variable. It is called quadratic because it contains a squared term (x^2).

2. What does it mean for a polynomial to be irreducible?

An irreducible polynomial is one that cannot be factored into simpler polynomials with integer coefficients. In other words, it cannot be broken down into smaller terms and constants. For example, x^2 + 1 is irreducible over the real numbers, but it can be factored into (x + i)(x - i) over the complex numbers.

3. How do you know if a polynomial is prime?

A polynomial is considered prime if it has only two factors: 1 and itself. In other words, it cannot be factored into smaller polynomials with integer coefficients. For example, x^2 + 2x + 1 is prime because it can only be factored into (x + 1)(x + 1).

4. Can a quadratic polynomial have more than one irreducible factor?

Yes, a quadratic polynomial can have multiple irreducible factors. For example, x^2 + 2 is irreducible over the rational numbers, but it can be factored into (x + √2)(x - √2) over the real numbers.

5. How are quadratic polynomials and irreducibles related to primes?

In general, a polynomial is considered prime if and only if it is irreducible. In the case of quadratic polynomials, this means that a polynomial is prime if and only if it is irreducible over the rational numbers. However, this is not always the case for higher degree polynomials, as some irreducible polynomials can still be factored into smaller terms and constants over certain fields.

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