Let $R$ be a ring, and let $1 =e_1 +\cdots+ e_r = e'_1 + ...+ e'_r$ be two decompositions of $1$ into sums of orthogonal idempotents. If $e_1\cong e_i'$ for all $i$, show that there exists $u \in U(R)$ such that $e_i' = u^{-1}e_i u$ for all $i$.
(Exercise 21.15 from A First Course in Noncommutative Rings - Lam)
Can someone help me with this?
Obs 1: given two idempotents $e,f\in R$, we define $e\cong f$ iff $eR\cong fR$ as right $R$-modules. This is equivalent to the existence of elements $a,b\in R$ such that $e=ab$ and $f=ba$.
Obs 2: the author says that this is easy when $R$ is semisimple, but even in this case I didn't get anything.
Note that every morphism $eR \to fR$ of right $R$-modules is given by left multiplication with an element in $fRe$. Thus, if $e_i \cong e_i'$, we find $a_i \in e_i' R e_i$ and $b_i \in e_i R e_i'$ such that left multiplication with $a_i$ and $b_i$ gives isomorphisms $e_i R \to e_i' R$ and $e_i' R \to e_i R$ and as you already noted, we can choose $a_i$ and $b_i$ such that $a_i b_i = e_i'$ and $b_i a_i = e_i$.
Consider $a = \sum_i a_i$ and $b = \sum_{i} b_i$. Since we have $a_i b_j = 0 = b_j a_i$ unless $i = j$, we have $ab = \sum_i a_ib_i = \sum_i e_i' = 1$ and similarly $ba = 1$. It follows that $a$ is invertible with $b$ its inverse.
Finally note that $a e_i a^{-1} = a e_i b = a_i b_i = e_i'$ so you can take $u = a$.