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Homework Help: If F is ANY field then (x^m-1)|(x^n-1) in F[x] if and only if m|n.

  1. Aug 10, 2012 #1
    Hello. The question that I am having trouble with is the one found in the title. I will repeat it again here in the post.

    "If F is ANY field then (x^m-1)|(x^n-1) in F[x] if and only if m|n."

    I have determined that if m|n, then (x^m-1)|(x^n-1) as long as n >= 2 and s >= r >= 1. However, the part that I am now having trouble proving is that this statement holds for ANY field. I am not looking for an answer, but maybe a more precise definition of what I am trying to prove, and maybe the steps I need to take to prove it. Thank you very much in advance.
  2. jcsd
  3. Aug 11, 2012 #2
    Here's an idea which doesn't work:

    If [itex]x^n-1[/itex] divides [itex]x^m-1[/itex] then any linear factor of [itex]x^n-1[/itex] is a linear factor of [itex]x^m-1[/itex]. The linear factors correspond to the roots. So if [itex]x^n-1[/itex] divides [itex]x^m-1[/itex] then all roots of [itex]x^n-1[/itex] must be roots of [itex]x^m-1[/itex].

    The is useless since a polynomial doesn't even need to have roots. Can you do something so that our approach will work?
  4. Aug 11, 2012 #3
    What are m,n? Naturals I assume
  5. Aug 11, 2012 #4
    Ok aassuming they are:

    Perform division, paying attention to the form of the powers you get for the quotient
  6. Aug 11, 2012 #5
    I worked through the proof myself for the most part, but to make it easier to view, I attached the proof that was provided to us.


    What I must prove now is that this holds for ANY field. This is where I get stuck. Basically, how would I go about proving that this statement holds for ANY field?

    Attached Files:

  7. Aug 11, 2012 #6
    Why don't you do the exact same things as in the lemma?? Only some notation will be different.
  8. Aug 11, 2012 #7
    It seems that what I need to do is prove that this works for ANY field. So would I need to do something extra with the already existing proof in order to solve the problem?
  9. Aug 11, 2012 #8
    Maybe. Go over the proof and see whether every step is justified in [itex]F(x)[/itex] (and why it is justified).
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