Suppose $n=p_1^{e_1}\,p_2^{e_2}\, \cdots \, p_t^{e_t}$ where $p_i$,$1\leq i \leq t$, are prime numbers and $P(X) \in \mathbb Z_d[X]$. Now, my question it is, If $X^{q_i} - 1$, is divisible by $P(X)$ mod $p_i^{e_i}$,$1\leq i \leq t$,where $q_i$ are positive integer numbers, and $Lcm(q_1\,q_2\, \cdots, \, q_t)$ be a number like $m$ where $Lcm$ is the Least common multiple, then can we conclude that $X^{m}-1$ is divisible by $P(X)$ mod $n$?
Thanks for any suggestion.
Hint $\ $ Apply CRT = Chinese Remainder Theorem. The ideals $\,I_i = (P,p_i^e)\,$ are pairwise coprime (i.e. comaximal) since the $\,(p^{\large e_i})\,$ are.
${\rm mod}\ I_i:\,\ \color{#c00}{X^{\large q_i}\equiv 1}\,\Rightarrow X^m \equiv (\color{#c00}{X^{\large q_i}})^{\large m/q_i}\equiv \color{#c00}1^{\large m/q_i}\equiv 1$
Thus $\,X^m\!-1\in I_i\,\Rightarrow\, X^m\!-1\in \bigcap I_i.\,$ But $\,\bigcap I_i = \prod I_i\, $ by $\,I_i\,$ pairwise coprime.
Note $\,\color{#0a0}{A+B=1}\,\Rightarrow\, (P,A)(P,B) = (P(P,\color{#0a0}{A,B}),AB) = (P,AB)\ $ thus we infer by induction that $\ \prod I_i = (P,\, p_1^{\large e_1}\cdots p_t^{\large e_t})$
We essentially employed the uniqueness of the CRT solution $\,Y = X^m\,$ of $\, Y\equiv 1 \pmod{I_i}.\,$ This is the ideal-theoretic analog of the integer constant-case CRT that I described here.
