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I Quadratic Polynomials and Irreducibles and Primes ...

  1. Apr 2, 2017 #1
    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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    Last edited: Apr 2, 2017
  2. jcsd
  3. Apr 3, 2017 #2

    fresh_42

    Staff: Mentor

    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?
     
  4. Apr 3, 2017 #3

    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]## ... ... ?

    Peter


    EDIT ... 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 )## .
     
    Last edited: Apr 3, 2017
  5. Apr 3, 2017 #4

    fresh_42

    Staff: Mentor

    ##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>

     
    Last edited: Apr 3, 2017
  6. Apr 3, 2017 #5


    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
     
    Last edited: Apr 3, 2017
  7. Apr 3, 2017 #6

    fresh_42

    Staff: Mentor

    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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