A non-projective module Let a ring with identity $R$  be decomposed as $S_1⊕\cdots⊕S_n$ (as a right $R$-module),  where $S_i=e_iR$ with $e_i$  nonzero idempotents of $R$ adding up to $1$. If $J$ is the Jacobson radical of $R$, then $e_iJ≠0$ for some $i$, and Nakayama's lemma implies $N=S_i/e_iJ≠0$. Is it true that $N$ is not a projective $R$-module? Thanks in advance!
 A: Yes. If $N$ were projective, then the short exact sequence
$$0\to e_iJ \to S_i \to N \to 0$$
of right $R$-modules would split, so $e_iJ$ would be a right $R$-module direct summand of $R$, and therefore of the form $fR$ for some idempotent $f$.
But the Jacobson radical contains no nonzero idempotents.
A: Here is an alternative way that runs along the lines of Jeremy's answer for a while:
If $S_i/e_iJ$ were projective so that $0\to e_iJ \to S_i \to S_i/e_iJ \to 0$ splits in Mod-$R$, $e_iJ$ would be an $R$-summand of $S_i$, hence an $R$-summand of $R$.
But $e_iJ \subseteq J$, so this is saying a submodule of $J$ is a summand of $R_R$. This is a problem since $J$ (and all of its submodules) are superfluous in $R$. In other words, there does not exist a submodule $N\subseteq R$ such that $e_iJ+N=R$ except for $N=R$. This would imply that $e_iJ=\{0\}$.
So if you're working under assumptions that make $e_iJ\neq\{0\}$, $S_i/e_iJ$ can't be $R$ projective. Of course, if $e_iJ=\{0\}$, then $S_i/e_iJ$ is projective. 
