lifting orthogonal idempotents (induction step) I'm trying to prove (by induction on $n$) that whenever $I$ is a nilpotent ideal of a ring $R$, and $r_1+I,\ldots,r_n+I$ form a complete set of orthogonal idempotents in the quotient $R/I$, there exists a complete set of orthogonal idempotents $e_1,\ldots, e_n\in R$ such that for each $1\leq i\leq n$, $e_i+I=r_i+I$. 
I'm fine with the base case, but am struggling with the inductive step. 
This result appears as Corollary 7.5 on page 107 of Etingof's representation theory book, available here: http://www-math.mit.edu/~etingof/replect.pdf. In the (rather terse) proof, they suggest applying the inductive assumption to a particular subring, but I don't understand exactly what they mean. I'm wondering if anyone could explain Etingof's argument, or perhaps suggest a different approach. 
Thanks very much. 
 A: The picture you should always keep in mind is a vector space $V$, $A$ is the algebra of matrices on $V$ and $\pi_1,...,\pi_n$ are projections onto subspaces $V_1,...,V_n$ with $V=V_1\oplus \cdots \oplus V_n$. Then $\pi_1,...,\pi_n$ are idempotents with $\pi_1+\cdots+\pi_n=1$, $V_i=\pi_iV$, and the algebra of matrices on $V_i$ is isomorphic to $\pi_iA\pi_i$. Note that $\pi_2+\cdots+\pi_n=1-\pi_1$ is an idempotent too - it's the projection onto $V_2\oplus \cdots \oplus V_n$ - and the algebra of matrices on $V_2\oplus \cdots \ \oplus V_n$ is thus $(1-\pi_1)A(1-\pi_1)$.
So now you have orthogonal idempotents $\overline{e}_1,...,\overline{e}_n$ in $A/I$ with $\sum \overline{e}_i=1$, and have a lift $e_1$ of $\overline{e}_1$. Note that $\overline{e}_2,...,\overline{e}_n$ are ortohogonal idempotents in 
$$(1-\overline{e}_1)A/I(1-\overline{e}_1) \ = \ (1-e_1)A(1-e_1)/I$$ with $\sum \overline{e}_i=1$ the identity in this algebra. Thus by induction there are idempotent lifts $e_2,...,e_n$ to $(1-e_1)A(1-e_1)$, which are orthogonal and $e_2+\cdots+e_n=1$ is the identity in $(1-e_1)A(1-e_1)\subseteq A$. Note
$$A\ = \ e_1Ae_1 \  \oplus \ (1-e_1)A(1-e_1)$$
and the identities in $A,e_1Ae_1,(1-e_1)A(1-e_1)$ are $1,e_1,(1-e_1)$ respectively. Thus
$$e_1\ + \ e_2 \ + \ \cdots \ + \ e_n\ = \ 1.$$
