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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. |
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| Aug10-12, 11:35 PM | #1 |
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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.
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. |
| Aug11-12, 03:33 AM | #2 |
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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? |
| Aug11-12, 11:39 AM | #3 |
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What are m,n? Naturals I assume
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| Aug11-12, 11:49 AM | #4 |
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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.
Ok aassuming they are:
Perform division, paying attention to the form of the powers you get for the quotient |
| Aug11-12, 03:20 PM | #5 |
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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.
[IMG]http://www.physicsforums.com/attachment.php?attachmentid=49748&stc=1& d=1344716150[/IMG] 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? |
| Aug11-12, 06:10 PM | #6 |
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Why don't you do the exact same things as in the lemma?? Only some notation will be different.
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| Aug11-12, 06:34 PM | #7 |
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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?
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| Aug11-12, 06:37 PM | #8 |
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