When do projective modules give generators? I am reading arXiv:math/0111066 [math.RA] and am stuck understanding the proof of Lemma 1.4.  The only part I don't follow is the claim:

If $B$ is a projective (right) module over a ring $R$ which is simple, then $B$ is a generator for $\operatorname{Mod}-R$.

I don't see how to leverage the hypothesis that $R$ is simple.  $B$ is the direct summand of a free module (and free modules generate) but why does that show that $B$ generates (and not the complement of $B$?)
Update: One possible reference is Lang's Algebra section XVII, 7.  Note that $B$ becomes a left $R'(B)=\operatorname{End}_R(B)$ module in a natural way.  Morita's Theorem says that $B$ is a generator if and only if the natural map $R\rightarrow \operatorname{End}_{R'}(B)$ is onto ($B$ is "balanced") and $B$ is finitely generated projective over $R'(B)$.  However, I have no idea how to verify this...
 A: I only know how to prove this if $B$ is finitely generated; fortunately in Lemma 1.4 this follows because $B$ is a direct summand (hence a quotient) of $P$ which is assumed to be finitely generated.
Given that, $B$ is a direct summand of some finite free module $R^n$, hence can be presented as the image of an idempotent $e \in M_n(R)$. In this blog post (skip down to "More explicit Morita equivalences") you can see a proof of the following (this should be standard but I don't know a reference and it was faster to just prove it):

Lemma: $\text{im}(e)$ is a generator iff $e \not \equiv 0 \bmod I$ for any proper two-sided ideal $I$ of $R$.

If $R$ is simple the only such ideal is $I = (0)$ so this condition is automatically satisfied for any nonzero idempotent $e$ and hence for any nonzero finitely generated projective module $B$.
(The blog post is mostly a discussion about different definitions of "generator," which are inequivalent in general but which basically all agree in $\text{Mod}(R)$ and more generally in any cocomplete abelian category. These equivalences allow me to use the weakest definition of generator, which is an object $B$ such that if $C$ is any nonzero object then there's a nonzero map $B \to C$.)
A: *

*It is easy to see that $R$ itsef is a generator.

*I will provide you with an epimorphism $P \twoheadrightarrow R$, notice that this is the same of providing you with a non-zero map, because $R$ is simple. Such a "cover" of a generator shows that also $P$ is a generator.

*Being projective, $P$ is a split submodule of $R^{(\Gamma)}$, the free module generated by some set $\Gamma$. Now, consider all the projections $\pi_i: R^{(\Gamma)} \to R$, composing with the inclusion $P \hookrightarrow R^{(\Gamma)} \to R$, there must be at least one $i$ such that the projection is non-zero, because $P$ is a nontrivial module.

