Lifting idempotents in semiperfect rings $R$ modulo ideals other than $\mathrm{rad}(R)$ A ring $R$ is semiperfect by definition, if $R/\mathrm{rad}(R)$ is semisimple and all idempotents of $R/\mathrm{rad}(R)$ can be lifted to idempotents in $R$. I wonder whether this lifting is also possible with respect to other ideals than $\mathrm{rad}(R)$ in a semiperfect ring.
There is a homological characterization of semiperfect rings as those rings for which all finitely generated modules have projective covers. I am quite sure that the proof of that characterization also shows, that we can lift idempotents modulo any ideal of $R$ contained in $\mathrm{rad}(R)$.
Can anyone confirm this observation? If it is true, is there another proof which is closer to the definition of semiperfect rings? I would also be happy to see a reference.
 A: I found a simple argument why we can lift idempotents modulo any ideal $I \subseteq \mathrm{rad}(R)$ provided that we can lift idempotents modulo $\mathrm{rad}(R)$:
Let $e' \in R/I$ be any idempotent. Then we can lift the idempotent $e' + \mathrm{rad}(R)$ to an idempotent $e \in R$. Let $\overline{x}$ denote reduction modulo $I$ for any $x \in R$. Of course $e'$ and $\overline{e}$ are not necessarily identical, but they are isomorphic since they have the same image in $R/\mathrm{rad}(R)$. For the same reason, $\overline{1-e}$ and $1-e'$ are isomorphic, hence $e'$ and $\overline{e}$ are even conjugated in $R/I$. Let $\overline{a} \in (R/I)^\times$ such that $(\overline{a})^{-1} \cdot \overline{e} \cdot \overline{a} = e'$. Since $I \subseteq \mathrm{rad}(R)$, we have $a \in R^\times$. Now $a^{-1} e a \in R$ is an idempotent lift of $e'$.
Actually, in a semiperfect ring it is even possible to lift idempotents modulo any left ideal [1]. 
[1] Nicholson, W. Keith. "Lifting idempotents and exchange rings." Transactions of the American Mathematical Society 229 (1977): 269-278.
