TITLE: Lemma on infinitely generated projective modules
QUESTION [6 upvotes]: Is it true that every finitely generated submodule of a non-finitely generated projective over a (not necessarily commutative!) ring is contained in a proper summand?
N.B.: I asked this already on math.stackexchange.com without much luck.

REPLY [7 votes]: The lemma is at least true, if the projective module has an uncountable projective base (sometimes also called a dual base). 
Proof: Let $P$ be a projective $R$-module with uncountable projective base $(x_i, f_i)$, $(i\in I)$ and $M = \sum_{k=1}^nRy_k \subseteq P$. Define inductively 
$$I_0 = \lbrace i \in I \mid \exists 1 \le k \le n: f_i(y_k) \neq 0 \rbrace$$
$$I_{n+1} = I_n \cup \lbrace i \in I \mid \exists j \in I_n: f_i(x_j) \neq 0 \rbrace$$
$$J = \cup_{n\ge 0}I_n\hspace{140pt}$$
Set $Q = \sum_{j \in J}Rx_j \le P$. Since $y_k = \sum_{i \in I}f_i(y_k)x_i$ it follows from $I_0 \subseteq J$ that $M \le Q$. 
Next I want to show 
$$x_j = \sum_{i \in J}f_i(x_j)x_i \quad\text{   for each } i \in J \hspace{80pt}(\ast)$$
Let $j \in I_n$. Write $x_j = \sum_{i \in I}f_i(x_j)x_i$. If $f_i(x_j) \neq 0$ it follows $j \in I_{n+1} \subseteq J$. Thus $(\ast)$ is shown. Define 
$$\kappa: P \to Q, x \mapsto \sum_{i \in J}f_i(x)x_i.$$
$\kappa$ is $R$-linear and from $(\ast)$ one concludes $\kappa|Q = \text{id}_Q$. Thus $Q$ is a direct summand of $P$ and since $Q$ is countably generated, $Q$ is a proper subset of $P$. 
BTW: In the great example from F. Ladisch, $P$ has a countable projective base (see Lam's book).