TITLE: $P/P^2$ isomorphic to $R/P$ as $R$-modules
QUESTION [6 upvotes]: Let $P$ be an ideal of a ring $R$. When is it true that $P^n/P^{n+1}$ are isomorphic to $R/P$ as $R$-modules for any $n$? I was trying to show that for Dedekind domains the norm of ideals is a multiplicative function. I found a proof of this on page 26 of these lecture notes, which seems to imply that this is true for any ring and any prime ideal.
Is this true in general? I was considering this example: $R = k[x]/(x^2), P = (x)/(x^2)$ is a prime ideal of $R$. Then $P^2/P^3$ seems to be the zero $R$-module but $R/P$ is non-zero...

REPLY [3 votes]: No this is not going to be true in general. Consider $R = k[x,y]$ and $P =(x,y)$. Then $R/P \cong k$ that is a field but $P/P^2 \cong (x,y)/(x^2,xy, y^2)$ that is two dimensional over $k$.
More generally we see that $P^d/P^{d+1}$ is isomorphic to the space of homogeneous forms of degree $d$ in $x$ and $y$. Thus $\dim_k P/P^2 =d + 1$ and so $P/P^2$ is not going to be isomorphic to $k$ for any $d$.