Is it true that every finitely generated submodule of a nonfinitely 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.
Is it true that every finitely generated submodule of a nonfinitely 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. 


I think I have a counterexample to the assertion. The following example of a nonfinitely generated projective module is Example (2.12D) in Lam's Lectures on Modules and Rings (attributed to Kaplansky): Let $R$ be the ring of continous, realvalued functions on $[0,1]$ and $P$ the ideal
$$ P = \{f\in R \mid f \text{ vanishes on } [0,\epsilon] \text{ for some } \epsilon > 0
\}. $$
As an illustration of the Dual Basis Lemma, Lam shows that $P$ is projective as $R$module. 


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 $\kappaQ = \text{id}_Q$. 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). 

