(Zelmanowitz) Regular Module but not Projective

Question

An $R-module$ $M$ is called regular (Zelmanowitz) if given any $m \in M$, there exists R-module homomorphism $f:M \longrightarrow R$ such that $m=f(m)m$. I have a problem to find a regular module but not projective. I had proof that any submodule of regular module is regular and if $R-module $ $M$ is countably (finitely) generated then $M$ is projective. I also have a hint from the paper (Regular Modules, J. Zelmanowitz, page 343) : the ring of linear transformation of a countable dimensional vector space contains nonprojective left ideal. Thank you.

Answer 1

Take a nonprojective left ideal $M$. For each $m\in M$, there exists $x$ such that $m=mxm$ (since the ring is von Neumann regular.)

Now define $f:R\to R$ by $f(r)=rx$, then restrict the map to have domain $M$.

By design, $f(m)m=mxm=m$. This is possible for any $m$, so $M$ is regular.