I am struggling with the following claim: There is a group isomorphism $R/P \cong P^n/P^{n+1}$ for any natural number $n$, where R is a ring (commutative with unit) and $P$ is a prime ideal.
My attempt: Let $\alpha \in P^n\setminus P^{n+1}$. There is an isomorphism $R/P \rightarrow \alpha R/\alpha P$ (multiplication with $\alpha$ is surjective and has $P$ equal to the kernel). Because $\alpha R \subset P^n$ the inclusion induces a homomorphism $\psi:\alpha R \rightarrow P^n/P^{n+1}$ with $ker(\psi) = \alpha R \cap P^{n+1}$ and $im(\psi) = \alpha R + P^{n+1}$. I thus want to show that $\alpha R \cap P^{n+1} = \alpha P$ and $\alpha R + P^{n+1} = P^n$ to conclude that $\psi$ factors to give an isomorphism $\alpha R/\alpha P \rightarrow P^n/P^{n+1}$.
If $x\in \alpha P$, $x=\alpha p, p\in P$ then $x\in \alpha R$ and $x \in P^{n+1}$. If $x \in \alpha R \cap P^{n+1}$, $x=\alpha r, r\in R$, then because $x\in P^{n+1}$ and $\alpha \not\in P^{n+1}$ we find that $r\in P$ and we have shown the first equality.
As both $\alpha R$ and $P^{n+1}$ are contained in $P^{n}$ also their sum is, which gives one inclusion of the second inequality.
However I don't know how to show the last inclusion $P^n \subset \alpha R + P^{n+1}$. Any help is appreciated.
Edit: Seems like this is not true in this generality (see comments), I need it for the case where $R$ is the ring of integers $D \subset \mathbb{Q}(\sqrt{d})$ (e.g. a Dedekind domain)
The following proof supposes that your familiar with ideal arithmetic in Dedekind domain (It seems impossible to avoid it, since you have counter examples for general $R$).
Since $\alpha\in P^n$, we have $\alpha R\subset P^n$, and since $R$ is a Dedekind domain, it is equivalent to $P^n\mid \alpha R$. Hence $\alpha R= I P^n$ for some ideal $I$. Note now that by choice of $\alpha$, the $P$-adic valuation of $\alpha R$ is exactly $n$ (since $\alpha\notin P^{n+1}$). In particular, $I$ and $P$ are coprime. Therefore, $I+P=R$.
Now $\alpha R+P^{n+1}=P^n ( I+ P)=P^n R=P^n$.