Assume that $S$ and $R$ are finite commutative local rings with maximal ideals $M$ and $m$ respectively.
I am working through the proof of Theorem XIV.3 in McDonald's Finite Rings with Identity
The local ring $S$ is a separable extension of $R$ if and only if $S/mS$ is a separable $R/m$-algebra.
For background, $S$ is a separable extension of $R$ if $R$ is a unital subring of $S$ and $S$ is projective over the enveloping algebra $S^e = S \otimes_R S$. (Recall that the enveloping algebra has natural surjective homomorphism $\phi: S^e \to S$ defined by $a \otimes b \mapsto ab$). This is equivalent to the existence of a "separability idempotent" $e \in S^e$ such that $\ker(\phi) e = 0$ and $\phi(e) = 1$, the identity of $S$.
My question has to do with proof of the necessity of the Theorem. Assuming that $S/mS$ is a separable $R/m$ algebra, there exists a separability idempotent $\bar e \in (S/mS)^e$, McDonald then claims that we may lift $\bar e$ to an idempotent $e \in S^e$. I am wondering the justification.
Certainly if $mS = M$, idempotents can be lifted modulo the radical of S (in this case $M$ since $S$ has a unique maximal ideal), but this theorem eventually is used to show that $mS = M$ (i.e. S is an unramified extension of R). Furthermore, we are dealing with the enveloping algebra and I am not sure how lifting idempotents works in that context.