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I can show that $(x^m-1, x^n-1) \subseteq (x^{(m,n)} - 1)$, but I am stuck with the other inclusion. i.e. showing there exist polynomials $p, q \in \mathbb{Z}[X]$ such that $p(x)(x^m-1) + q(x)(x^n-1) = x^{(m,n)}-1$.

I can't use Bezout's identity because $\mathbb{Z}[X]$ is not a PID.

I computed some examples but I can't see a general patttern.

e.g.

\begin{align} (x^5-1) - (x^3 + x)(x^2-1) &= x-1 \\ (x^2-1)(x^5-1) -(x^4+x^2+x)(x^3-1) &= x-1 \\ (x^4-1) -x^2(x^2-1) &= x^2-1 \end{align}

Thanks in advance.

Sonja
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    I reckon that if $m>n$ then $(x^m-1,x^n-1)=(x^{m-n}-1,x^n-1)$ over $\Bbb Z[x]$ (or over $R[x]$ for any commutative ring $R$). – Angina Seng Aug 01 '20 at 22:21
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    Hint: consider the order of the element $X\in (\mathbb{Z}[X]/(X^m-1,X^n-1))^{\times}$. – Aphelli Aug 01 '20 at 22:22
  • @AnginaSeng Thanks! I see where to go from here :) – Sonja Aug 01 '20 at 23:51
  • Most links/searches here about this and related questions eventually lead to this. Nominally that is about gcd's of integers instead of polynomials, but some of the arguments work the same way. Mind you, more typically there are differences between gcd's of polynomials and their values at a selected point, but this time it plays out nicely. For the reason in Angina Seng's comment, fleshed out for example in this answer. – Jyrki Lahtonen Aug 02 '20 at 08:07
  • Approach0 doesn't always work optimally, but I still recomment using it. – Jyrki Lahtonen Aug 02 '20 at 08:09
  • @JyrkiLahtonen I read those posts but I was confused how they related to ideals. But Angina Seng's comment helped me to make the connection. – Sonja Aug 02 '20 at 17:19

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