If $\mathfrak a_i/J(A)\cong\mathfrak a_j/J(A)$, then $\mathfrak a_i\cong\mathfrak a_j$ Let $A$ be a ring, and $A_A=\bigoplus_i\mathfrak a_i$ be a decomposition into indecomposable right ideals. Let $J(A)$ be the Jacobson radical of $A$, and $\bar{\mathfrak a}_i$ the respective quotient w/r/t $J(A)$.  I want to show:
$$\text{If $\bar{\mathfrak a}_i\cong\bar{\mathfrak a}_j$, then $\mathfrak a_i\cong\mathfrak a_j$.}$$
Attempt. My attempt so far is the following. I can write $\mathfrak a_i=e_iA$ for primitive orthogonal idempotents $e_i$. I guess that if $\bar{\mathfrak a}_i\cong\bar{\mathfrak a}_j$, then there are $\bar\lambda$, $\bar\kappa$ such that $\bar{\mathfrak a}_i\xrightarrow{\cong}\bar{\mathfrak a}_j$ is given by $\bar f\mapsto\bar\kappa\bar f$, and conversely with $\bar\lambda$. Then $\lambda\kappa \equiv e_i \pmod{J(A)}$.
Why does this give me that $\mathfrak a_i\cong\mathfrak a_j$? How do I get rid of $J(A)$?
 A: The following is more general than you are asking.
Assume first that $J(A)Q$ is superfluous in $Q$. We can lift $\bar f\colon P/J(A)P\xrightarrow\sim Q/J(A)Q$ to a map $f\colon P\to Q$. Then $Q=f(P)+J(A)Q$, so $f$ must be onto, and hence a split epi. The kernel $K$ of $f$ is then projective and satisfies $K=J(A)K$, so $K=0$ (by a result of Bass). Hence $f$ is an isomorphism.
Note that if $Q$ is finitely generated, then $J(A)Q$ is superfluous by Nakayama's Lemma.
If $J(A)$ is nilpotent, then $J(A)Q$ is again superfluous. For, if $Q=U+J(A)Q$, then multiplying by $J(A)$ gives $J(A)Q=J(A)U+J(A)^2Q\subseteq U+J(A)^2Q$, so that $Q=U+J(A)^2Q$. Repeating gives $Q=U+J(A)^nQ$ for all $n$, so $J(A)$ nilpotent implies $Q=U$.

There is a much more general result, however, which holds for all projectives over any ring, contained in the following paper
https://doi.org/10.1016/j.jpaa.2006.12.004
They show that, given any two projectives $P$ and $Q$, any isomorphism $\bar f\colon P/J(A)P\xrightarrow\sim Q/J(A)Q$ can be lifted to an isomorphism $f\colon P\xrightarrow\sim Q$.
A: The proof runs like this:


*

*$J(R)$ is the intersection of all the maximal right ideals of $R$.  See Wikipedia for details why.

*For any group $G$, the set on non-generators of $G$ is a group, which coincides with the intersection of all maximal (proper) subgroups of $G$.

*If $R$ is commutative, then the two coincide.  This means that any lift $a$ of a generator $\bar{a}$ of $\bar{\frak a}$ will already generate the full ideal $\frak a$.  This immediately implies your statement.
Let me know what part(s) you are not sure how to prove, so I can try to explain it.
