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

    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?
  2. T

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

    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. https://www.physicsforums.com/attachment.php?attachmentid=49748&stc=1& d=1344716150 What I must prove now is that this holds for ANY field. This is where I get...
  3. T

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

    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...
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