I've recently read in a book a proof that used the following fact:
Since $(X^m-1, X^n+1)=1$ in $\mathbb{Q}[X],$ there exist $u,v \in \mathbb{Z}[X]$ such that $$(X^m-1)u + (X^n+1)v = 2$$ hence $(b^m-1)u(b) + (b^n+1)v(b) = 2 \quad (*)$
For context, $b$ is a random element of a finite ring $A$ with unity.
My question is the following: how were they able to get to $(*)?$ Are the rings $A[X]$ and $\mathbb{Q}[X]$ somehow connected? How is $u(b)$ even defined, as $u \in \mathbb{Z}[X]?$