Show that $R$ is self-injective iff every finitely generated projective right $R$-module is injective While studying for an upcoming exam, I've crossed the following problem:

A ring $R$ is self-injective if, and only if, every finitely generated projective right $R$-module is an injective right $R$-module.

By definition, $R$ is self-injective if $R_R$ is an injective $R$-module.
I know some similar results, but all of them use additional hypotheses on $R$. I really wish to understand what is going on, so I prefer hints. Of course, any help is appreciated.
 A: The nontrivial direction is that if $R$ is right self injective, then f.g. projective modules are injective.
Now in general, a projective module is just a summand of a free module.  If only this free module were guaranteed to be injective because it is a direct sum of injective modules!  You may know in general arbitrary direct sums of injective need not be injective.  IF it were injective, then all its summands would be injective too.
But..... what if the projective module is f.g.?  What’s that say about the free module you can make it a summand of?  Are you in the clear?
A: Following @rschwieb, I have tried to put together a full answer to my question.
Since one direction is trivial, assume that $R$ is right self injective. If $P$ is a finitely generated projective, then $P$ is a direct summand of a free finitely generated right $R$-module. Let $L$ be such a module. If $\{x_1, \ldots, x_n\}$ is a generating set, we may write $$L = I_1 \oplus \cdots \oplus I_n$$
in which $I_j = \langle x_j \rangle$.
Now, each $I_j$ is isomorphic to $R_R$. To see this, define $\varphi: I_j \rightarrow R$ as $\varphi(x_jr) := r, \ \forall r \in R$, which is clearly an isomorphism. It follows that each $I_j$ is an injective (since $R_R$ is, by hypothesis), and since the finite direct sum of injectives is again injective, $L$ is injective.
Since $P$ is a direct summand of an injective $R$-module, it follows that $P$ is also injective.
