Symmetric powers of ideal quotients in a local ring.

Question

Let $R$ be a local ring and $I \subset R$ any ideal. When is it the case that $(I \: \backslash I^2)^n = I^n \: \backslash I^{n+1}$?

Put another way, when is the natural map $\text{Sym}^n(I/I^2) \to I^n/I^{n+1}$ an isomorphism (of $R$-modules)?

This is easily seen to be true when $R$ is a discrete valuation ring. Does it hold in a more general setting?

Answer 1

Found an answer, after quite a bit of digging. This is Theorem 8.21A(e) in Hartshorne:

Let $A$ be Cohen-Macaulay and let $I$ be generated by a regular sequence. Then $I/I^2$ is a free $A/I$-module (of rank equal to the length of the sequence) and $\text{Sym}^n(I/I^2) \to I^n/I^{n+1}$ is an isomorphism.

Any ideal with a generating set not containing zero-divisors is generated by a regular sequence (namely a minimal generating subset of that generating set), so this is at least true for those ideals (and at least true in general when $A$ is also an integral domain).