More precisely, is every abelian group a colimit $\text{colim}\_{j \in J} F(j)$ over a diagram $F : J \to \text{Ab}$ where each $F(j)$ is isomorphic to $\mathbb{Z}$?


Note that this does not follow from the statement that every abelian group has a presentation, which is equivalent to the statement that every abelian group is a coequalizer of a pair of maps between free abelian groups, hence every abelian group is an *iterated* colimit of copies of $\mathbb{Z}$. A single colimit $A = \text{colim}\_{j \in J} F(j)$ of copies of $\mathbb{Z}$ is in particular the coequalizer of a pair of maps between free abelian groups, but the maps have a very special form, which works out explicitly to imposing the following constraint:



> 
> $A$ must have a presentation by generators and relations in which the only relations say that some generator is a multiple of some other generator. 
> 
> 
> 


Examples of abelian groups admitting a presentation of this form include cyclic groups and localizations of $\mathbb{Z}$, and the class of all such groups is closed under coproducts.


But I see no reason to believe that every abelian group admits a presentation of this form, and in particular I believe that the $p$-adic integers doesn't. Tyler Lawson [sketched a proof](http://chat.stackexchange.com/transcript/message/21413805#21413805) of this in the homotopy theory chat but it had a gap; the subsequent discussion may have filled the gap but I didn't follow it, and in any case I'd like someone to write up the details. 


Mike Shulman wrote a [lovely note](http://home.sandiego.edu/~shulman/papers/generators.pdf) about various different senses in which an object or objects of a category can generate it; in the terminology of that note, the question is whether $\mathbb{Z}$ is *colimit-dense* in $\text{Ab}$. Until a week or so ago, if you had asked me, I would have answered without hesitation that $R$ is colimit-dense in $\text{Mod}(R)$, and I doubt that I was alone in this...



Is every abelian group a colimit of copies of Z?
Here's my version of Tyler's argument that $\mathbb{Z}\_p$ is a counterexample. Maybe I'm still missing something, but I think it works with Tyler's suggested change. I'll make it community wiki, since it's not really my argument. As Tyler says,



> 
> Let's suppose that $\mathbb Z\_p$ were such a colimit. Then $\mathbb Z\_p$ could be written as having a presentation as follows:
> 
> 
> It would have a set of generators $e\_i$ (indexed by objects in the diagram), and it would have a set of relations all of the form $n e\_i = e\_j$ (indexed by morphisms in the diagram).
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Now, (and here's the change Tyler suggested later) one of the the $e\_i$'s must be a $p$-adic unit - otherwise the image of the $e\_i$'s would be contained in the proper subgroup $p\mathbb Z\_p \subset \mathbb Z\_p$. Pick such a generator $e$ and define $A = \mathbb Z\_{(p)} e \subset \mathbb Z\_p$ (I might be the only one not to realize this, but $\mathbb Z\_{(p)}$ is $\mathbb Z$ localized at $p$, i.e. elements of the form $a/b$ where $a,b \in \mathbb Z$ and $b$ *not* divisible by $p$). Now, multiplication by $e$ is an automorphism of $\mathbb Z\_p$, so we might as well assume $e = 1$ and $A = \mathbb Z\_{(p)}$. Then, as Tyler says,



> 
> Then I would be able to define a self-map $f$ of $\mathbb Z\_p$ as follows:
> 
> 
> If $e\_i$ is in $A$, I define $f(e\_i) = e\_i$
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> If $e\_i$ is not in $A$, I define $f(e\_i) = 0$
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> 
> Then we have to check that this respects the equivalence relation, so we need $n f(e\_i) = f(e\_j)$
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> 


To be honest, I don't quite follow Tyler's argument about $A$ and $\mathbb Z\_p / A$ both being torsion-free. But


* If $e\_i,e\_j \in A$ or $e\_i,e\_j \not \in A$, then the relation is trivial.
* If $n=0$, then the relation is respected.
* If $n \neq 0$, then either $e\_i,e\_j$ are both in $A$ or both not in $A$ becase $A$ is closed in $\mathbb Z\_p$ under both multiplication and division by $n \neq 0 \in \mathbb Z $.


Then as Tyler concludes,



> 
> Therefore this gives a well-defined such map $f$.
> 
> 
> However, any abelian group homomorphism $\mathbb Z\_p \to \mathbb Z\_p$ which is the identity on $\mathbb Z\_{(p)}$ must be the identity (because it must be the identity mod $p^n$ for all $n$).
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Actually, it suffices to note that $f$ is the identity on $\mathbb Z$. This is particularly clear, because $\mathbb Z$ is generated by $e$, which is definitely fixed by $f$. In fact, $\mathbb Z\_{(p)}$ is fixed because if $nx = 1$, then $nf(x)=1$, and $\mathbb Z\_p$ is a UFD.


