Are the heredity ideals in an heredity chain always finitely generated?

Question

An ideal $J$ of a ring $A$ is a heredity ideal of $A$ if:

• $J^2=J$;

• $J$ is a projective $A$-module;

• $J (\operatorname{Rad}A)J=0$.

A (unitary) semiprimary ring $A$ is said to be quasihereditary if there exists a chain of ideals of $A$, $$0=J_0 \subseteq \cdots \subseteq J_i \subseteq \cdots J_n =A,$$ such that $J_i/J_{i-1}$ is a hereditary ideal of $A/J_{i-1}$ for all $i$.

Question:

Let $A$ be a (semiprimary) quasihereditary ring which is not noetherian. Is it possible that $A$ has a heredity chain $$0=J_0 \subseteq \cdots \subseteq J_i \subseteq \cdots J_n =A$$ such that $J_i/J_{i-1}$ is not finitely generated over $A$ for some $i$?

Answer 1

Yes, it is possible. (So the answer to the question in the title is no.)

Consider the ring $$A = \begin{bmatrix} \mathbb Q & \mathbb R \\ 0 & \mathbb Q \end{bmatrix}.$$ It has primitive idempotents $e=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ and $f=\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}.$

Now $J=AeA$ is a heredity ideal and there is a heredity chain $0 \subseteq J \subseteq A$. As an $A$-module $J$ has a direct sum decomposition $$J=AeAe \oplus AeAf.$$ The first summand is isomorphic to $Ae$. The second summand decomposes further into infinitely many direct summands, each isomorphic to $Ae$, the sum indexed by a $\mathbb Q$-basis for $\mathbb R$.