Are we able to say, how do all the divisors of $\displaystyle x^m - 1 $ like???

Thanks, nothing on my mind seems good now..(Doh)

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- Apr 14th 2009, 12:04 PMsididivisor
Are we able to say, how do all the divisors of $\displaystyle x^m - 1 $ like???

Thanks, nothing on my mind seems good now..(Doh) - Apr 14th 2009, 12:22 PMNonCommAlg
yes. see cyclotomic polynomials.

- Apr 14th 2009, 12:35 PMsidi
Thanks, but

I havenīt heard about these polynomials before... I need it to prove that thing about d=gcd(m,n) $\displaystyle x^d - 1$ as the $\displaystyle gcd(x^m - 1,x^n - 1) $from yesterday. But I think I needn`t use these polynomials...there might be easier way to show it, but it can`t come on my mind... - Apr 14th 2009, 03:33 PMNonCommAlg
i see! the one that i gave you: $\displaystyle \gcd(x^n - 1, x^m -1)=x^{\gcd(n,m)} - 1.$ here's a proof: let $\displaystyle \gcd(n,m)=d.$ since $\displaystyle d \mid n$ and $\displaystyle d \mid m,$ we have: $\displaystyle x^d -1 \mid x^n - 1$ and $\displaystyle x^d - 1 \mid x^m - 1.$

now suppose $\displaystyle f(x) \mid x^n - 1$ and $\displaystyle f(x) \mid x^m - 1.$ if we prove that $\displaystyle f(x) \mid x^d - 1,$ then we're done. we know that there exist $\displaystyle r,s \in \mathbb{N}$ such that $\displaystyle rn - sm=d.$ thus:

$\displaystyle x^{rn} - 1 = x^d(x^m)^s - 1 \equiv x^d - 1 \mod f(x),$ because $\displaystyle x^m \equiv 1 \mod f(x).$ on the other hand $\displaystyle x^{rn}-1 \equiv 0 \mod f(x),$ because $\displaystyle f(x) \mid x^n - 1.$ hence $\displaystyle x^d - 1 \equiv 0 \mod f(x).$ Q.E.D. - Apr 14th 2009, 09:31 PMThePerfectHacker
Here is a similar problem and an alternate solution.

- Apr 17th 2009, 04:43 AMsidi
Thank you very much!