How to prove an isomorphism related to a filtration.

Question

Let $A$ be a filtered algebra with a filtration $F_0(A) \subset F_1(A) \subset \cdots \subset A$. Let $I$ be a two sided ideal of $A$. The algebra $A/I$ has a filtration $F_i(A/I)=F_i(A)/(F_i(A)\cap I)$. In this post it is said that \begin{align} F_i(A)/(F_{i-1}(A)+F_i(A) \cap I) \cong \frac{ F_i(A)/(F_i(A) \cap I) }{F_{i-1}(A)/(F_{i-1}(A) \cap I)}. \end{align} How to prove this identity? The third isomorphism theorem says that $(A/I)/(B/I)\cong A/B$. But here $F_{i}(A) \cap I \ne F_{i-1}(A) \cap I$. We cannot use the third isomorphism theorem directly. Thank you very much.

Answer 1

$\frac{ F_i(A)/(F_i(A) \cap I) }{F_{i-1}(A)/(F_{i-1}(A) \cap I)}$ is not well-defined since $F_{i-1}(A)/(F_{i-1}(A) \cap I)$ is not a subspace of $F_i(A)/(F_i(A) \cap I)$.

A proof of $\mathrm{gr}(A/I)= \oplus_{i \ge 0} F_i(A)/(F_{i-1}(A) + F_i(A) \cap I)$ is in the following.

\begin{align} & F_i(A/I) = F_i(A)/(F_{i}(A) \cap I) = (F_i(A) + I)/I, \end{align} \begin{align} & \mathrm{gr}(A/I) = \oplus_{i \ge 0} F_i(A/I)/F_{i-1}(A/I) \\ & = \oplus_{i \ge 0} ((F_i(A) + I)/I)/((F_{i-1}(A) + I)/I) \\ & = \oplus_{i \ge 0} (F_i(A) + I)/(F_{i-1}(A) + I) \\ & = \oplus_{i \ge 0} F_i(A)/(F_i(A) \cap ( F_{i-1}(A) + I )) \\ & = \oplus_{i \ge 0} F_i(A)/(F_{i-1}(A) + F_i(A) \cap I), \end{align} where the second and third isomorphism theorems are used.