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

Consider the following property of an abelian group $G$:

S: $G$ 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 summand of $F_j$.

Clearly, every free abelian group satisfies S. Does also the converse hold? If not, is S satisfied by $A^\mathbb{N} / A^{(\mathbb{N})}$, where $A=\mathbb{Q}_+=\mathbb{Z}^{(\mathbb{P})}$?

share|cite|improve this question

Edit: The answer below has been modified to reflect the comments.

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

$0 \to A \to B \to C \to 0$

of abelian groups, we get an induced sequence

$0 \to \lim_i \mathrm{Hom}(F_i,A) \to \lim_i \mathrm{Hom}(F_i,B) \to \lim_i \mathrm{Hom}(F_i,C) \to R^1 \lim_i \mathrm{Hom}(F_i,A) \to ...$

So it suffices to show that $R^1 \lim_i \mathrm{Hom}(F_i,A)$ vanishes for any abelian group $A$. Is this true?

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 $\mathrm{Hom}(F_j,A) \to \lim_{i < j} \mathrm{Hom}(F_i,A)$, for j sufficiently big. After making the harmless assumption that the system $(F_i)$ consists of all 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.

share|cite|improve this answer
Is the surjectivity condition sufficient to guarantee vanishing of $\lim^1$ for a directed limit which is not necessarily a sequential limit? – Reid Barton Jan 30 '10 at 21:51
Probably not. If I is a directed poset with at most $\aleph_d$ morphisms, then $lim^n$ over $I$ vanishes for $n \geq d+2$ (Mitchell, Rings with several objects). I don't think that bhargav's proofs works this easily. – Martin Brandenburg Jan 31 '10 at 2:19
You're both right. I think I implicitly assumed the system was indexed by the natural numbers, which is definitely not the case for this question. – Bhargav Jan 31 '10 at 7:55
Taking the direct limit of modules over a directed index set is an exact functor, non? – Joel Dodge Jan 31 '10 at 9:13
In any case, I think direct limits are pretty clearly right exact, no matter what index set you look at. The left exactness is what is less clear and one can prove exactness just using the fact that your index set is directed. – Joel Dodge Jan 31 '10 at 9:16

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.