$\DeclareMathOperator\gr{gr}$Let $A = \cup\_{i=0}^\infty F\_i A$ be a filtered commutative ring, $I \subseteq A$ an ideal. Then we have a canonical surjection
$$ \gr(A)/\gr(I) \to \gr(A/I).$$
Under what conditions is this surjection an isomorphism?


I most wish to know about the following special case: $A = \mathbb C[z\_1,\ldots, z\_n]$ equipped with the total degree filtration, $I$ an ideal of finite codimension. The ring $\gr(A/I)$ in this case should be functions on the scheme "$\lim\_{t \to 0} t\cdot V(I)$", and I'd like to know when the defining ideal is $gr(I)$.



Quotients and associated graded
I follow user @Z. M's comment.


If $M = \cup\_{i=0}^\infty F\_i M$ is a filtered module and
$$0 \to M' \to M \to M'' \to 0$$
is a short exact sequence such that $M', M''$ have the *induced filtrations*, then we get a short exact sequence of Rees modules
$$0 \to R\_hM' \to R\_hM \to R\_hM'' \to 0,$$
which yields the short exact sequence $0 \to \mathrm{gr}(M') \to \mathrm{gr}(M) \to \mathrm{gr}(M'') \to 0$ after taking the Snake Lemma (since our filtration is by submodules).


Now in our situation, $A/I$ and $I$ have the induced filtrations, so we get a short exact sequence
$ 0 \to \mathrm{gr}(I) \to \mathrm{gr}(A) \to \mathrm{gr}(A/I) \to 0$.


