# 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

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. Aug 11, 2012

### micromass

Staff Emeritus
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?

3. Aug 11, 2012

### Mandlebra

What are m,n? Naturals I assume

4. Aug 11, 2012

### Mandlebra

Ok aassuming they are:

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

5. Aug 11, 2012

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?

#### Attached Files:

• ###### Problem.png
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6. Aug 11, 2012

### micromass

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

7. Aug 11, 2012

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