MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 without much luck.

share|cite|improve this question
What is the standard example of a non-finitely generated projective module which is not free? – Martin Brandenburg Aug 12 '11 at 21:05
@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
@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

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 $M=0$.

share|cite|improve this answer
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).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.