Semisimplicity is equivalent to each simple left module is projective? As it is well-known, a ring with unity $R$ is semisimple if and only if each left $R$-module is projective. My question: 

Is semisimplicity of $R$ equivalent to each simple left $R$-module being projective? 

I think it is true for a finite dimensional algebra $R$. But, in general, I could not reach any conclusion. Any help would be thanked!
 A: Yes.
A ring is semisimple iff it does not have an essential maximal right ideals, and making simple right modules projective causes all maximal right ideals to be summands of $R$.
The connection with essential ideals is this: given any right ideal $T$, you can find another right ideal $T'$ such that $T\oplus T'$ is an essential right ideal in $R$. So in particular a maximal right ideal must have such a complement, and that complement could be zero (if it is already essential) or it could complement it to the whole of $R$ (meaning it is a summand.)  Projectivity of simple right modules rules out the first case since the exact sequence $0\to T\to R\to R/T\to 0$ splits.
It is interesting, though, that "all simple right modules are injective" does not characterize semisimple rings. These are called "right $V$-rings," and there are examples of these which are even domains.
Simples being injective does not even characterize semisimplicity for commutative rings: the commutative $V$-rings are the von Neumann regular rings.
A nice theorem of Barbara Osofsky's is: if every cyclic right module is injective, then the ring is semisimple.
