The problem is this:
Suppose $I \subseteq R$ is a nilpotent ideal and there is $r \in R$ with $r \equiv r^2 \pmod I$. Show $r \equiv e \pmod I$ for some $e \in R$ idempotent.
I have spent a few hours rolling around in abstracta with no destination. I believe that if I could write down a concrete example of this, the example could guide me through the abstract definitions and show me where to look for an idempotent in $R$. I have looked at 2x2 matrices and could not find any such examples. Might anybody have one?
The proof I've seen is a sort of surgery that is not hard, but also not obvious. I will adjust notation a bit:
Suppose $\bar r$ is idempotent in $R/I$, and consider its preimage $r\in R$. To make things easier to look at, define $s=1-r$. Notice that $\bar s$ is also an idempotent in $R/I$, and $\bar r + \bar s \equiv \bar 1$.
We have that $rs=r-r^2\in I$, and obviously $rs=sr$, so there's some $k$ such that $r^ks^k=0$. Since $\bar r^k+\bar s^k\equiv \bar r+\bar s\equiv \bar 1$, we have $x=1-r^k-s^k\in I$ is nilpotent, and then $1-x=r^k+s^k$ is invertible in $R$, with inverse $u=1+x+x^2+\ldots +x^{l-1}$ where $x^l=0$. Notice that $\bar u\equiv\bar1$, and $u$ commutes with $r$ and $s$ (since $x$ does).
Finally, we have $ur^k+us^k=1$. Multiplying both sides by $ur^k$ we see $(ur^k)^2+u^2r^ks^k=ur^k$, but $u^2r^ks^k=0$ by our efforts. Furthermore $\overline{ur^k}\equiv\bar u\bar r^k\equiv\bar 1\bar r\equiv \bar r$.