Is it true that simple projective modules are injective? It is known that simple modules are either projective or singular. Is it true that simple projective modules over (commutative) rings are injective ?
 A: If we assume that $\text{Spec}(R)$ is connected, then $R$ is always a field (note the spectrum of a local ring is always connected).  This is equivalent to the nonexistence of nontrivial idempotents.  Indeed, over a commutative ring any simple $R$-module is also of the form $R/\mathfrak m$ for a maximal ideal $\mathfrak m$ of $R$.  Write $R = R/\mathfrak m \oplus I$, for some finitely generated ideal $I$ of $R$.  By the determinant trick, there is an $x\in R$ with $x-1\in I$ such that $xI = I$, hence $I = I^2$.  This implies $I$ is generated by an idempotent, hence $I = 0$, since $R/\mathfrak m$ is nonzero.  
A: If $P$ a projective $R$-module is simple, then it is a direct summand of $R$ itself (indeed, any nonzero map $R\to P$ is surjective and therefore splits) and is then isomorphic to a minimal ideal $I$ which is projective. $I$ is generated by an idempotent, which is central if the ring is commutative. It follows that $I$ is a direct factor of $R$ as a ring, and therefore $I$ is also injective.
A: No, the cyclic group of order $p$ is not a direct summand in the cyclic group of order $p^2$ ($p$ a prime).
