Is the internal sum of projective submodules projective.

Question

Let $R$ be a unital ring, and $M$ a (right) $R$-module. Assume that $M$ admits a (possibly infinite) family of projective submodules $N_i$, for $i \in I$, such that $$ M = \sum_{i \in I} N_i. $$ Does it then follow that $M$ is projective?

Answer 1

Question: "Does it then follow that M is projective?"

Answer: if $A:=k[x,y], M:=(x,y) \cong Ax + Ay$ since $A$ is an integral domain it follows $Ax,Ay$ are rank one free $A$-modules, hence projective. But $M$ is not projective. $M$ cannot be generated by one element and any set of generators cannot be linearly independent: Given any two elements $f,g\in (x,y)$ it follows $(-g)f+(f)g=0$. A projective module over $k[x,y]$ is (this is a theorem) free, hence $M$ cannot be projective.

Answer 2

No. For instance, if $R$ is an integral domain and $M$ is torsion-free, then $M$ is a sum of projective submodules, namely the cyclic submodules generated by each of its elements (which are all free of rank 1 except when the element is 0). But a torsion-free module does not have to be projective (for instance, if $R=\mathbb{Z}$ and $M=\mathbb{Q}$, or $R=\mathbb{Z}[x]$ and $M=(2,x)\subset R$).