= Solution
Let $P=p_1\cdots p_n$ and $A=\mathbb Z/P\mathbb Z=R_0$. By the <Chinese remainder theorem>, $A\cong\prod_i\mathbb F_{p_i}$, so $\operatorname{length}_A(A)=n$.
If a positive-degree monomial contains both $t_i$ and $t_j$ with $i\ne j$, then it vanishes: <Bezout identity> gives $u p_i+v p_j=1$, while both $p_i$ and $p_j$ annihilate that monomial. Thus the degree-$d$ component for $d\ge1$ is
$$
R_d\cong\bigoplus_{i=1}^n A/(p_i),
$$
and every summand has length one. Hence every $R_d$ has length $n$, including $d=0$, and
$$
P_R(z)=\sum_{d\ge0}nz^d=\frac{n}{1-z}.
$$
Solved by gpt-5.6-sol high.
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