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Product of monic polynomials

  1. Mar 2, 2008 #1
    1. The problem statement, all variables and given/known data
    Show that [itex]x^{p^n}-x[/itex] is the product of all the monic irreducible polynomials in [itex]\mathbb{Z}_p[x][/itex] of a degree d dividing n.


    2. Relevant equations



    3. The attempt at a solution
    So, I want to prove that the zeros of all such monic polynomials are also zeros of [itex]x^{p^n}-x[/itex] and vice versa. I cannot do either, unfortunately. We know that the elements of GF([itex]p^n[/itex]) are precisely the zeros of [itex]x^{p^n}-x[/itex]. And we know that if f(x) is a monic polynomial of degree m in [itex]\mathbb{Z}_p[x][/itex], then when you adjoin any of its zeros to [itex]\mathbb{Z}_p[/itex], you get a field with p^m elements whose elements are precisely the zeros of [itex]x^{p^m}-x[/itex]. So, I guess if [itex]\alpha[/itex] is a zero of [itex]x^{p^n}-x[/itex], then does the irreducible monic polynomial that [itex]\alpha[/itex] is a zero of need to be a factor of [itex]x^{p^n}-x[/itex]?
     
    Last edited: Mar 2, 2008
  2. jcsd
  3. Mar 2, 2008 #2

    morphism

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    Yes: if m(x) is the minimal polynomial of [itex]\alpha[/itex] over Z_p (what you call the irreducible monic polynomial associated with [itex]\alpha[/itex]), and if f(x) is a another polynomial in Z_p[x] with [itex]\alpha[/itex] as a root, then m(x)|f(x).
     
  4. Mar 2, 2008 #3
    If you write out the unique factorization of m(x) and f(x) in [itex]\bar{\mathbb{Z}_p}[x][/itex], then both factorizations must contain [itex]x-\alpha[/itex], but why does m(x) need to contain factors with all of the other zeros of f(x)? Does it have something to with the fact that <m(x)> is a maximal ideal in [itex]\mathbb{Z}_p[x][/itex]?
     
  5. Mar 2, 2008 #4

    morphism

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    m(x) is the monic polynomial of least degree with [itex]\alpha[/itex] as a root (prove this if you don't already know it). Because Z_p is a field, Z_p[x] has a division algorithm. So let's write f(x)=a(x)m(x)+r(x), where r is either zero or its degree is less than that of m. Now note that f([itex]\alpha[/itex])=a([itex]\alpha[/itex])m([itex]\alpha[/itex])+r([itex]\alpha[/itex]), which implies that r([itex]\alpha[/itex])=0. The minimality of m now lets us conclude that r=0.
     
  6. Mar 3, 2008 #5
    So the next step is to show that if [itex]\alpha[/itex] is a zero of [itex]x^{p^n}-x[/itex], then the minimal polynomial of [itex]\alpha[/itex] has degree d dividing n. That would imply that the factorization of [itex]x^{p^n}-x[/itex] over [itex]\mathbb{Z}_p[/itex] consists only of polynomials of degree d dividing n. Let m be the degree of \alpha. Then \alpha is also a solution to [itex]x^{p^m}-x[/itex] but I am not really sure if that even helps.
     
  7. Mar 4, 2008 #6
    anyone?
     
  8. Mar 4, 2008 #7

    morphism

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    If you adjoin \alpha to Z_p, you'll get a subfield of GF(p^n). But now d=[Z_p(alpha) : Z_p] is a divisor of n=[GF(p^n) : Z_p].

    (If the minimal polynomial of \alpha over F has degree d, then alpha generates an extension of degree d of F.)
     
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