A counterexample to $Ae\simeq Ae^{\prime}\to A(1-e)\simeq A(1-e^{\prime})$ Let  $e,e^{\prime}$ be two idempotents in a $k$- algebra  $A$ ($k$ is a field) . Then my guess  $Ae\simeq Ae^{\prime}$ (as a left $A$- module) does not imply $A(1-e)\simeq A(1-e^{\prime})$ in general, but I cannot find an example? can anyone please give me an example and tell me under what condition this implication is true?
According to this lecture notes (in Lemma2.1) the above implication is true in finite dimensional algebras by using Krull-Schmidt theorem.
 A: You're right to be skeptical. As you've already noted at your other question, Artinian rings are ruled out from consideration since the Krull-Schmidt theorem can be used to prove they have the aforementioned property.
Happily, I can demonstrate a ring without the property that is even von Neumann regular. I was inspired by a theorem in Goodearl's book von Neumann regular rings to this effect:

For a von Neumann regular ring $R$, $R$ is unit regular iff for all idempotents $e,e'$, $eR\cong e'R$ iff $(1-e)R\cong (1-e')R$.

With that in mind, let's go for my favorite von Neuman regular ring that isn't unit regular. Let $V$ be a countable dimensional left vector space over a field $F$, and let $R$ be its ring of linear transformations. The transformations will be written on the right of elements of $V$.
It's well known this ring $R$ is left self-injective and von Neumann reuglar and not unit regular. It has exactly one nontrivial ideal $I$: the set of transformations with finite dimensional image.
To define the following transformations, let $\{b_i\mid i\in \Bbb N\}$ be a basis for $V$. 
Define three transformations: 
$$e(b_i):=\begin{cases}0& \text{ if $i=0$}\\ b_i& \text{ if $i>0$}\end{cases}\\
v(b_i):=b_{2i}\\
u(b_i):=\begin{cases}0& \text{ if $i$ is odd }\\ b_{i/2}& \text{ if $i$ is even}\end{cases}
$$
It's easy to see that $e$ is idempotent and that $uv=1$. Furthermore $f=veu$ is idempotent.
We claim that $Re\cong Rf$ but $R(1-e)\ncong R(1-f)$.
Now, the map sending $re\mapsto reu$ from $Re$ to $Reu$ is injective since $u$ is right cancellable. Therefore $Re\cong Reu$. But $Reu=Rveu=Rf$.
In hopes of finding a contradiction, let's suppose that $R(1-e)$ and $R(1-f)$ are isomorphic. Since $R$ is left self-injective, the isomorphism out of $R(1-e)$ extends to an endomorphism of $R\to R$. As such, it is given by right multiplication by some element of $R$, call it $q$. So we have that $R(1-e)q=R(1-f)$. 
But this is a problem: the image of $(1-e)$ is finite dimensional (one dimensional, in fact). Then $R(1-e)q$ is trapped inside the proper ideal $I$ we mentioned before. But the image of $1-f$ is infinite dimensional, so it does not lie in the ideal. Thus $R(1-e)$ and $R(1-f)$ cannot be isomorphic.
So the heart of the trick here was to find idempotents $e,f$ such that $e,f,1-f$ all had infinite dimensional images, and yet $1-e$ had a finite dimensional image. We leveraged this asymmetry to produce a counterexample.
