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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.

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What is the standard example of a non-finitely generated projective module which is not free? –  Martin Brandenburg Aug 12 '11 at 21:05
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@Martin: big projective modules tend to be free (see H. BASS, 'Big projective modules are free', Illinois J. Math., 7 (1963), 24-31) but there do exist examples. All group rings for non soluble finite groups work, for example: Linnell, P. A. Nonfree projective modules for integral group rings. Bull. London Math. Soc. 14 (1982), no. 2, 124–126. –  Mariano Suárez-Alvarez Aug 12 '11 at 22:10
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@Martin: Let $R= \prod_i R_i$ be an infinite product of rings and let $P=\bigoplus_i R_i$. (This is Example 2.12C in Lam's book.) –  Frieder Ladisch Aug 12 '11 at 22:53

2 Answers 2

I think I have a counterexample to the assertion. The following example of a non-finitely 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, real-valued 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.
I claim that $P$ is indecomposable as $R$-module. Assume $P=M\oplus N$ for some ideals $M$, $N$. Then $MN=0$, so the support of any element of $M$ is contained in the zero set, $Z(g)$, of any function $g\in N$. Thus $$U:= \bigcup_{f\in M} \operatorname{Supp}(f) \subseteq \bigcap_{g\in N} Z(g) =: K. $$ Any element of $M\oplus N$ vanishes on $K\setminus U$. However, if $x\neq 0$, then there is $f\in P$ such that $f(x)\neq 0$. Thus $K\setminus U = \{0\}$. As $K$ is closed and $U$ open, it follows that either $K=[0,1]$ and $N=0$ or $U=\emptyset$ and $M=0$.

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I looked for examples of indecomposable big projectives but could not find them... I think you killed my lemma! –  Mariano Suárez-Alvarez Aug 12 '11 at 23:46
    
I just looked at the Bass paper from your comment above. The example from my answer is mentioned at the end of it, including the fact that $P$ is indecomposable (without proof). –  Frieder Ladisch Aug 14 '11 at 17:14

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).

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