# Module complements to rings embedded in a projective module

Let $$R$$ be noncommutative unital ring and $$M$$ a projective (right) $$M$$-module. Assume that $$R$$ embedds into $$M$$ as a right -module.

A) If $$R$$ is a semisimple ring, then of course $$R$$ admits an $$R$$-module complement. But this is a very strong assumption. What are weaker but sufficient criteria for an $$R$$-module complement to exist.

B) What is a non-free example where $$M$$ does not admit an $$R$$-module complement in $M. C) What is a non-free module example where $$M$$ does admit a complement? • The simplest example is the embedding$2:\mathbb{Z}\to \mathbb{Z}$. Sep 26 at 9:52 • Why is it clear that it does not admit a complement? Sep 26 at 9:59 • The complement should be a$\mathbb{Z}/2$and$\mathbb{Z}$has no 2-torsion. More generally this works for any commutative ring$R$and$x\in R$non invertible nonzerodivisor. Sep 26 at 10:04 • @Denis: thanks for the example! However, I was looking for a noncommutative example. Sep 26 at 12:00 • @DickJohnson Just view$\mathbf{Z}$as quotient of a noncommutative ring$\mathbf{Z}\times R\$ to make this obvious counterexample "noncommutative".
– YCor
Sep 26 at 15:37

For (A) a ring $$R$$ is selfinjective if it is injective as an $$R$$-module (on e should say left or right). By definition of injective this means $$R$$ has a complement in any module. Examples include Frobenius and quasi-frobenius rings.
If $$R$$ is von Neumann regular, then any finitely generated submodule of a projective module has a complement. So again if $$R$$ embeds in a projective module there is a complement.
B) A standard situation is the injective envelope $$I(R)$$ of $$R$$. There is always an embedding $$R \rightarrow I(R)$$ and rings where $$I(R)$$ is additionally projective are called QF-3 rings (a generalisation of Frobenius rings).
C) Cant we just take $$M=R \oplus N$$ for a projective module $$N$$ so that $$M$$ is not free? That is very easy to obtain for finte dimensional algebras.