direct limit of free complemented subgroups - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T21:40:29Z http://mathoverflow.net/feeds/question/13500 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/13500/direct-limit-of-free-complemented-subgroups direct limit of free complemented subgroups Martin Brandenburg 2010-01-30T19:48:08Z 2010-02-01T01:41:12Z <p>Consider the following property of an abelian group $G$:</p> <p><b>S</b>: $G$ is torsionfree and a directed limit of finitely generated (hence free) subgroups $\{F_i\}_i$, such that for all $i \leq j$, $F_i$ is a direct <i>s</i>ummand of $F_j$.</p> <p>Clearly, every free abelian group satisfies <b>S</b>. Does also the converse hold? If not, is S satisfied by $A^\mathbb{N} / A^{(\mathbb{N})}$, where $A=\mathbb{Q}_+=\mathbb{Z}^{(\mathbb{P})}$?</p> http://mathoverflow.net/questions/13500/direct-limit-of-free-complemented-subgroups/13504#13504 Answer by Bhargav for direct limit of free complemented subgroups Bhargav 2010-01-30T20:59:22Z 2010-01-31T08:17:25Z <p>Edit: The answer below has been modified to reflect the comments.</p> <p>My guess is that $G$ is forced to be projective, hence free, in this situation. To show this, we need to verify that $\mathrm{Hom}(G,-)$ is an exact functor. As we have an identification of functors $\mathrm{Hom}(G,-) \simeq \lim_i \mathrm{Hom}(F_i,-)$, applying $\mathrm{Hom}(G,-)$ to an exact sequence </p> <p>$0 \to A \to B \to C \to 0$ </p> <p>of abelian groups, we get an induced sequence </p> <p>$0 \to \lim_i \mathrm{Hom}(F_i,A) \to \lim_i \mathrm{Hom}(F_i,B) \to \lim_i \mathrm{Hom}(F_i,C) \to R^1 \lim_i \mathrm{Hom}(F_i,A) \to ...$</p> <p>So it suffices to show that $R^1 \lim_i \mathrm{Hom}(F_i,A)$ vanishes for any abelian group $A$. Is this true?</p> <p>Here's a not-so-well-thought-out idea: if I chased elements correctly, an affirmative answer to the question above follows from the bijectivity of the natural map $\mathrm{Hom}(F_j,A) \to \lim_{i &lt; j} \mathrm{Hom}(F_i,A)$, for j sufficiently big. After making the harmless assumption that the system $(F_i)$ consists of <em>all</em> finitely generated saturated subgroups of $G$, the preceding bijectivity question translates to: given a free abelian group F of finite rank, when is the natural map $\mathrm{colim}_i H_i \to F$ an isomorphism, where the indexing set I is the poset of all proper saturated subgroups $H_i \subset F$. I think the answer to this question is yes when the rank of $F$ is at least $3$ (which is enough for the application at hand), but I'm not sure.</p>