= Solution
For every $i\geq0$, the inclusion $P^{i+1}\subseteq P^i$ is strict: equality, followed by multiplication by the inverse of $P^i$, would give $P=R$. Hence one can choose
$$
\pi\in P^i\setminus P^{i+1}.
$$
Put $J=(\pi)+P^{i+1}$. Then
$$
P^{i+1}\subsetneq J\subseteq P^i.
$$
The classification just proved, applied modulo $P^{i+1}$, leaves no <ideal> strictly between consecutive powers. It follows that
$$
\boxed{P^i=(\pi)+P^{i+1}}.
$$
For $i=0$, the same argument says $(\pi)+P=R$, using $\pi\notin P$ and maximality.
The extension to every $r\geq1$ follows by <mathematical induction>. Suppose $P^i=(\pi)+P^{i+r}$. Multiplication of the initial equality by $P^r$ gives
$$
P^{i+r}=\pi P^r+P^{i+r+1}.
$$
Substitution yields
$$
P^i=(\pi)+\pi P^r+P^{i+r+1}=(\pi)+P^{i+r+1}.
$$
Thus
$$
\boxed{P^i=(\pi)+P^{i+r}\quad\text{for every }r\geq1}.
$$
Taking $r=n-i$ for $0\leq i<n$ gives
$$
\boxed{P^i/P^n=(\pi+P^n)}.
$$
This is a <principal ideal> of the <quotient ring> $R/P^n$. Its zero <ideal> is generated by zero, so every <ideal> of this <quotient ring> is a <principal ideal>.
Back to article page
