One answer to this [Lemma on infinitely generated projective modules](https://mathoverflow.net/questions/72788/lemma-on-infinitely-generated-projective-modules) shows that every finitely generated module of a non-countably generated projective module is contained in a countably generated direct summand. Now I would like to ask : When can every countably generated submodule of a a non-countably generated projective module be contained in a countably generated direct summand ? 


My rings are commutative with unity. You my assume the ring to be Noetherian if the general case is not easy to handle. 



When can every countably generated submodule of a a non-countably generated projective module be contained in a countably generated direct summand ?
This is an elaboration of Ralph's answer to the question linked by the OP. I use the notation from there. Moreover, I assume the axiom of countable choice, i.e. countable unions of countable sets are countable. 


Let $P$ be a non-countably generated projective $R$-module with projective base $(x\_i, f\_i)\_{i\in I}$. A projective base has the two properties: 


For all $x \in P$: 
$$\text{supp}(x) := \{i \in I \mid f\_i(x) \neq 0\} \text{ is finite}\tag{1}$$
$$x = \sum\_{i \in\text{supp}(x)}f\_i(x)x\_i \tag{2}$$


Let $M$ be a submodule of $P$ with countably (finite or infinite) many generators $y\_k$. Define inductively 
$$I\_0 := \bigcup\_k \text{supp}(y\_k)$$
$$I\_{n+1} := \bigcup\_{i \in I\_n} \text{supp}(x\_i)$$
and let $J = \bigcup\_{n \ge 0}I\_n$. By the axiom of countable choice, the sets $I\_n, J$ are countable. Let $Q$ be the submodule of $P$ generated by (the countable many) $x\_i,\,i \in J$. 


By $(2)$, $y\_k$ is an $R$-linear combination of $x\_i\,(i \in \text{supp}(y\_k)\subseteq J)$ and hence $M \subseteq Q$. 


The essential step is to show that $Q$ is a direct summand of $P$. This is done by showing that the following $R$-linear map is a projection (i.e. fixes $Q$ pointwise):  

$$\kappa: P \to Q,\,\,\,x \mapsto \sum\_{i \in \text{supp}(x)\cap J}f\_i(x)x\_i$$


Let $x\_j$ be a generator of $Q$. That means $j \in J$, i.e. there is $n \ge 0$ such that $j \in I\_n$. Hence $\text{supp}(x\_j)\subseteq I\_{n+1}\subseteq J$ and  

$$\kappa(x\_j) \overset{\text{def}}{=}\sum\_{i \in \text{supp}(x\_j)\cap J}f\_i(x\_j)x\_i = \sum\_{i \in \text{supp}(x\_j)}f\_i(x\_j)x\_i\overset{(2)}{=}x\_j.$$
So $\kappa$ is the identity on $Q$ and the inclusion $Q \hookrightarrow P$ is a splitting of $\kappa$. 


Finally, $Q$ is a proper submodule of $P$, since $Q$ is countably generated and $P$ is not countably generated.$\blacksquare$


