Lemma on infinitely generated projective modules - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T16:46:55Z http://mathoverflow.net/feeds/question/72788 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/72788/lemma-on-infinitely-generated-projective-modules Lemma on infinitely generated projective modules Mariano Suárez-Alvarez 2011-08-12T19:11:52Z 2011-08-13T02:17:10Z <p>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?</p> <p><strong>N.B.:</strong> I asked this <a href="http://math.stackexchange.com/questions/56987/lemma-on-infinitely-generated-projective-modules" rel="nofollow">already</a> on math.stackexchange.com without much luck.</p> http://mathoverflow.net/questions/72788/lemma-on-infinitely-generated-projective-modules/72797#72797 Answer by F. Ladisch for Lemma on infinitely generated projective modules F. Ladisch 2011-08-12T22:40:46Z 2011-08-12T22:40:46Z <p>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 <em>Lectures on Modules and Rings</em> (attributed to Kaplansky): Let $R$ be the ring of continous, real-valued functions on $[0,1]$ and $P$ the ideal <code>$$P = \{f\in R \mid f \text{ vanishes on } [0,\epsilon] \text{ for some } \epsilon &gt; 0 \}.$$</code> As an illustration of the Dual Basis Lemma, Lam shows that $P$ is projective as $R$-module.<br> 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 <code>$K\setminus U = \{0\}$</code>. As $K$ is closed and $U$ open, it follows that either $K=[0,1]$ and $N=0$ or $U=\emptyset$ and $M=0$.</p> http://mathoverflow.net/questions/72788/lemma-on-infinitely-generated-projective-modules/72805#72805 Answer by Ralph for Lemma on infinitely generated projective modules Ralph 2011-08-13T02:17:10Z 2011-08-13T02:17:10Z <p>The lemma is at least true, if the projective module has an uncountable projective base (sometimes also called a dual base). </p> <p>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 $$I_0 = \lbrace i \in I \mid \exists 1 \le k \le n: f_i(y_k) \neq 0 \rbrace$$ $$I_{n+1} = I_n \cup \lbrace i \in I \mid \exists j \in I_n: f_i(x_j) \neq 0 \rbrace$$ $$J = \cup_{n\ge 0}I_n\hspace{140pt}$$</p> <p>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 $M \le Q$. </p> <p>Next I want to show $$x_j = \sum_{i \in J}f_i(x_j)x_i \quad\text{ for each } i \in J \hspace{80pt}(\ast)$$</p> <p>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 $P$. </p> <p>BTW: In the great example from F. Ladisch, $P$ has a countable projective base (see Lam's book). </p>