# 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.

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Here's an idea which doesn't work:

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

The is useless since a polynomial doesn't even need to have roots. Can you do something so that our approach will work?

What are m,n? Naturals I assume

Ok aassuming they are:

Perform division, paying attention to the form of the powers you get for the quotient

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 stuck. Basically, how would I go about proving that this statement holds for ANY field?

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Why don't you do the exact same things as in the lemma?? Only some notation will be different.

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?

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?
Maybe. Go over the proof and see whether every step is justified in $F(x)$ (and why it is justified).