A characterization for idempotent lifting property Let $I$ be an ideal in a commutative ring $R$ with $1$ and let $g+I$ be an idempotent element of $R/I$. We say that this idempotent can be lifted  modulo $I$ in case there is an idempotent $e^2=e\in R$ such that $g + I = e + I$ ($g-e\in I$). 

Question: Is there any characterization for $I$ under which idempotents  lift modulo $I$, that is, every idempotent of $R/I$ lifts modulo $I$? And is there any comprehensive reference for this property?

 A: Yes! If $I\subseteq \sqrt{0}$, where $\sqrt{0}$ is the nilradical, then every idempotent in $R/I$ lifts modulo $I$. 
Instead of just stating this, I thought you'd like to see a proof. So, the following proof is taken directly from section 27 of Anderson and Fuller's book Rings and Categories of Modules, Second Edition.
Suppose $I\subseteq \sqrt{0}$, and $g\in R$ satisfies $g+I = g^2+I$. Then letting $n$ be the nilpotency index of $g-g^2$, we can use the binomial formula:
$$
0 = (g-g^2)^n  
= \sum_{k=0}^n{n \choose k} g^{n-k} (-g^2)^k
= \sum_{k=0}^n (-1)^k {n\choose k} g^{n+k} 
= g^n-g^{n+1}\left(\sum_{k=1}^n (-1)^{k-1}{n\choose k} g^{k-1} \right)
$$
Then $t:=\sum_{k=1}^n (-1)^{k-1}{n\choose k} g^{k-1}\in R$ is such that 
$$g^n = g^{n+1}t \hspace{10pt} \text{and} \hspace{10pt} gt =tg.$$
Now 
$$ e := g^nt^n = (g^{n+1}t)t^n = g^{n+1}t^{n+1} = g^{n+2}t^{n+2} = \cdots =g^{2n}t^{2n} = e^2,$$
so $e= g^nt^n$ is idempotent, and also
$$ g+I 
= g^n+I = g^{n+1}t+I = (g^{n+1} +I)(t+I)
= (g+I)(t+I) = gt+I$$
so that $g+I=(g+I)^n = (gt+I)^n = e+I$.
