Identity of a ring as two different sums of idempotents Let $R$ be any ring with identity $1_R$.
Prove that if there exist idempotents $e_1,..., e_n, e'_1,...,e'_n \in R$ such that
$$ 1_R = e_1 +...+ e_n = e'_1+... e'_n$$ then the following conditions are equivalent:
(i) $Re_i \simeq_R Re'_i$ for all $1 \leq i \leq n$
(ii) There exists an invertible element $a \in R^\times$ such that $e'_i = ae_ia^{-1}$ for all $1 \leq i \leq n$
My attempt:
$\bullet$ $(ii) \Rightarrow (i)$
Define a ring homomoprhism $$ \psi : Re'_i \longrightarrow Re_i$$
such that $\psi(x) = xa$.
It is well defined and we check that its image lies in $Re'_i$ :
$$\psi(x) = \psi(xe'_i) = xe'_ia=x(aa^{-1})e'_ia=xae_i \in Re_i$$
To prove surjectivity just note that for any $r \in R$, we have $$re_i= \psi(ra^{-1}e'_i)$$
Injectivity follows because $a$ is a unit, and therefore $xa=0 \Longleftrightarrow x=0$, thus Ker $\psi = \{0\}$
$\bullet$ $(i) \Rightarrow (ii)$
For each $1 \leq i \leq n$ let $\phi_i$ be an isomoprhism between $Re_i$ and $Re'_i$.
Define $\phi \in $ End($R$) such that $\phi(x) = \phi_i(x)$ whenever $x \in Re_i$.
Now is the part I get stuck, I feel I am heading in the right direction and I have tried to play with the images of $e_i$ but I can't seem to get close to proving what I want.
Any suggestions please?
 A: A simple example to show that this is not in general true without the orthogonality condition:
Let $R$ be a ring of characteristic $2$ with two conjugate but non-equal idempotents $e$ and $f$, so that $Re$ and $Rf$ are isomorphic as left $R$-modules. For example, take $R=M_2(\mathbb{F}_2)$, $e=\begin{pmatrix}1&0\\0&0\end{pmatrix}$ and $f=\begin{pmatrix}0&0\\0&1\end{pmatrix}$.
Then 
$$1=1+e+e+f+f=1+e+f+e+f.$$
A: As it is already commented by Mathematician 42 and san, this statement is usually stated with an orthogonality hypothesis (you can find, for example, Exercise 7 in here). So it is likely that your professor forget to mention about it.
You (and perhaps your professor) should be careful that the definition of an idempotent decomposition $e = e_1 + \dotsb + e_n$ requires not only $e_i^2 = e_i$ but also $e_ie_j = \delta_{ij}e_j$ (op. cit.), which I also had some trouble.
A: If you know that $e_ie_j=0$ for $i\ne j$ and $e′_ie′_j=0$ for $i\ne j$ (a condition forgotten to mention by the OP), then 
$a=\sum_i \phi_i^{-1}(e′_i)$ and $a^{−1}=\sum_i \phi_i(e_i)$ will do the job, where $\phi_i:R e_i\to Re′_i$ are the given left $R$-module isomorphisms. 
