Let $R$ be a ring, $J \subset I$ ideals. Let $\mathfrak{I} = R\supset I \supset I^2 \supset \cdots$ be the $I$-adic filtration of $R$. I want to compare the associated graded module $\operatorname{gr}_\mathfrak{I}(R)$ to the associated graded module $\operatorname{gr}_\mathfrak{I}(R/J)$ where $R/J$ is viewed as an $R$-module.
According to an exercise in Eisenbud, we have $\operatorname{gr}_\mathfrak{I}(R/J) \simeq \operatorname{gr}_\mathfrak{I}(R)/\operatorname{in}(J)$ where $\operatorname{in}(J) = \left\langle \operatorname{in}(f) \,\mid\, f \in J\right\rangle$. I think I'm close, but I'm not quite there yet. Here's my reasoning so far.
Suppose that $J \subset I^n \,\forall n \in \mathbb{N}$. Then it is not difficult to show that $\operatorname{in}(J) = 0$, since by definition, $f \in I^n \,\forall n \in \mathbb{N}$, we have $\operatorname{in}(f) = 0$ so $\operatorname{in}(J) = 0$. So then we have for each index $k$, that $I^{k-1}(R/J)/(I^k(R/J)) = I^{k-1}/I^k$ and thus $\operatorname{gr}_\mathfrak{I}(R/J) = \operatorname{gr}_\mathfrak{I}(R)$.
Otherwise, for some $n \in\mathbb{N}$ we have $J \subset I^n$ but $J\not\subset I^{n+k}$ for all $k \ge 1$. So then the $n+k$ graded piece of $\operatorname{gr}_\mathfrak{I}(R/J)$ is $I^{n+k}(R/J) / I^{n+k+1}(R/J) = (I^{n+k} \cap J)/(I^{n+k+1}\cap J)$ and in the special case of $k=0$, we have $I^n/(I^{n+1}\cap J)$. I can't exactly see how to relate this to the quotient by the initial form of $J$.
For instance, for $f \in \operatorname{gr}_\mathfrak{I}(R)/\operatorname{in}(J)$, we have $f = \sum_{i=1}^k f_i + I^{n_i+1}$ where $f_i \in I^{n_i}$ but $f_i \not \in I^{n_i+1}$. I want to be able to consider this as something in the intersection with $J$ to relate it with the remarks in the previous paragraph, but I can't really see how to do it.
Can anyone provide a hint as to continue?