Yes, it is possible. (So the answer to the question on 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 $\mathbb R$.

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 $\mathbb R$.

Yes, it is possible. (So the answer to the question on 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. As an $A$-module it 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 $\mathbb R$.

Yes, it is possible. (So the answer to the question on 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 $\mathbb R$.