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 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 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...

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    $\begingroup$ If I believe your characterization in terms of presentation, it seems clear to me that any such group, if torsion-free, is a direct sum (coproduct) of locally cyclic groups. Indeed, first we can erase all generators that represent zero. Then we have an equivalence relation between generators "to occur in the same relator". Each equivalence class corresponds to a presentation of a locally cyclic group, and the resulting group is coproduct of those. The group of $p$-adics, but also many subgroups of $\mathbf{Q}^2$, fail to be such a direct sum. $\endgroup$ – YCor May 5 '15 at 17:28
  • $\begingroup$ My first guess for something that might be a counterexample is: the direct product of an infinite number of copies of Z $\endgroup$ – Yemon Choi May 5 '15 at 17:29
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    $\begingroup$ @YemonChoi: indeed, such a group is not a direct sum of locally cyclic groups: if so, they would be necessarily cyclic, but indeed an infinite product of $\mathbf{Z}$'s is not free abelian. $\endgroup$ – YCor May 5 '15 at 17:32
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    $\begingroup$ @AsafKaragila the question comes down to this: which abelian groups admit a presentation by generators $e_i$ and relations of the form $n_{ij}e_i=e_j$? $\endgroup$ – Tim Campion May 5 '15 at 18:43
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    $\begingroup$ @Asaf: what is unclear about the highlighted description I gave? I was explicitly trying to cut out all of the category theory there. $\endgroup$ – Qiaochu Yuan May 5 '15 at 18:53

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).

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 $f(e_i) = e_i$

If $e_i$ is not in $A$, I define $f(e_i) = 0$

Then we have to check that this respects the equivalence relation, so we need $n f(e_i) = f(e_j)$

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 $n$).

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.

  • $\begingroup$ Thanks for writing this up, Tim! I admit I didn't follow the discussion about $p$-adic units before but it's clear now. $\endgroup$ – Qiaochu Yuan May 6 '15 at 5:35
  • $\begingroup$ I have a stupid question: why can you assume $ne_i=e_j$ instead of $ne_i=me_j$? In this $p$-adic case they are probably equivalent since only the $p$-adic valuation matters but in general isn't it like assuming that $J$ be ordered or filtered? $\endgroup$ – Filippo Alberto Edoardo May 6 '15 at 6:19
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    $\begingroup$ The answer is equally stupid: we have a functor $F: I \to \mathsf{Ab}$ such that $F(i) = \mathbb Z<e_i>$, and if $u: i \to j$, then $F(u): \mathbb Z <e_i> \to \mathbb Z<e_j>$ must be multiplication by some number $n$. The construction of the colimit is to take the coproduct over all the objects of their images, modulo all the relations imposed by the morphisms. One subtlety is that there can be multiple $n$'s for a given ordered pair $(e_i,e_j)$, if there were multiple morphisms $i \to j$ in the category. $\endgroup$ – Tim Campion May 6 '15 at 13:02
  • $\begingroup$ @Qiaochu No problem! I was confused by this for quite awhile, too. I was thinking that we needed to iterate the process of finding generators or something... $\endgroup$ – Tim Campion May 6 '15 at 13:21

Abelian groups of this type are known under the name "simply presented abelian group" and there are various investigations on them.

Among others, it is known that simply presented abelian $p$-groups that are reduced (that is the divisible supgroup is trvial) are characterized by their Ulm sequence.

As a consequence, two distinct abelian $p$-groups that are reduced and have the same Ulm sequence cannot be both simply presented.

An explcit example is the torsion subgroup of $\prod_n\mathbb{Z}/p^n\mathbb{Z}$ that has the same Ulm sequence as $\oplus_n\mathbb{Z}/p^n\mathbb{Z}$, and the latter is simply presented.

There is also a homological characterization of such groups; indeed, another name is totally projective $p$-groups.

I believe some information on this is in Fuchs' classic books on abelian groups. A more recent source would be Loth "Classifications of Abelian Groups and Pontrjagin Duality" (1998)

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Let $G$ be an abelian group with a presentation of abelian group given only by relators of the form $g^n=h$, $g,h$ generators and $n$ integer. Then $G$ is a direct sum of locally cyclic groups. We can assume we allow relators $g^n=1$ (because it can be encoded in adding a generator $k$ and adding the relators $h=k$ and $h^2=k$). Now given such a presentation (say "satisfying ($*$)") we can remove all generators representing the identity (replacing them with 1 in relators) and the resulting presentation still satisfies ($*$) and has the additional property that no generator maps to the identity. Now consider the equivalence relation on the set of generators generated by $g\simeq h$ if the (nontrivial) generators $g,h$ occur in a single relator. If $C$ is an equivalence class, consider the group $G_C$ with the presentation with generators in $C$ and relators involving generators of $C$. Then $G$ is the coproduct of the $G_C$. Note that so far this does not make use of the torsion-freeness of $G$, and the argument works for modules over an arbitrary ring.

Now we are reduced to understand the case when the equivalence relation defined above has a single equivalence class (i.e., is the undiscrete one). Then assuming $G$ torsion-free results in the fact that $G$ is locally cyclic (i.e., for a torsion-free group, isomorphic to a subgroup of $\mathbf{Q}$). To see this, I use the fact that if $G$ is a torsion-free group generated by elements $g_i$ such that any two $g_i$ have a common power, then $G$ is locally cyclic. Unlike the original question, this reduces (if necessary) to finitely generated groups: if $G$ is a torsion-free abelian group generated by a finite subset $S$ such that any two elements of $S$ have a common power, then $G$ is cyclic. This in turn is proved by a simple argument showing that in a torsion-free abelian group, if two elements have a common power, then they are powers of a common element (indeed the subgroup they generate is torsion-free abelian, generated by 2 elements and not isomorphic to $\mathbf{Z}^2$, hence is cyclic).

Thus in general, assuming that the abelian group $G$ is torsion-free implies that $G$ is a direct sum of locally cyclic groups (and more generally, in the category of $A$-modules when $A$ is a domain, assuming that $M$ is torsion-free implies that $M$ is a direct sum of torsion-free modules of rank 1; when $A$ is a PID, a torsion-free module of rank 1 is the same as a torsion-free locally cyclic module).

Among torsion-free abelian groups, those that are direct sums of locally cyclic groups are pretty rare. For instance $\mathbf{Z}_p$ does not satisfy this property, as any torsion-free abelian group $A$ of $\mathbf{Q}$-rank $\ge 2$, not containing any copy of $\mathbf{Z}[1/p]$ and such that $A/pA$ is cyclic); many subgroups of $\mathbf{Z}[1/p]^2$ are ruled out by this criterion. Also any non-free subgroup of $ \mathbf{Z}^X$ fails to be such a direct sum; this includes $\mathbf{Z}^{X}$ itself when $X$ is infinite.

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    $\begingroup$ Yes, every locally cyclic group is either isomorphic to a subgroup of $\mathbf{Q}$, or to a subgroup of $\mathbf{Q}/\mathbf{Z}$. (An arbitrary torsion-free locally cyclic module over a domain $A$ is isomorphic to an $A$-submodule of the field of fractions $K$ of $A$. If $A$ is a PID, the only other locally cyclic modules are the submodules of $K/A$.) $\endgroup$ – YCor May 5 '15 at 22:38
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    $\begingroup$ Can you give just a few more details? Aside from "cyclic" probably being "locally cyclic" in the first paragraph, I'd like to confirm that I know how you're using the hypothesis that $G$ is torsion-free. $\endgroup$ – Qiaochu Yuan May 6 '15 at 5:36
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    $\begingroup$ @QiaochuYuan: Perhaps an easier way to think about what happens in the torsion-free case once you've reduced to a single equivalence class is to tensor everything with $\mathbb{Q}$. Doing this turns your diagram into a commuting diagram where every object is $\mathbb{Q}$ and every map is an isomorphism, so the colimit is clearly $\mathbb{Q}$. Since your group was torsion-free, it injects into its tensor product with $\mathbb{Q}$ and is hence a subgroup of $\mathbb{Q}$. $\endgroup$ – Eric Wofsey May 6 '15 at 7:03
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    $\begingroup$ @TimCampion: not right. Try with the abelian group presentation $\langle x,y,z,v,w\mid 2x=z,2y=z,2z=w,v=w,v=2w\rangle$ (in which $v,w$ are trivial, reducing to the presentation $\langle x,y,z,v,w\mid 2x=z,2y=z,2z=0\rangle$ of $C_2\times C_4$, where $C_i$ is cyclic of order $i$) $\endgroup$ – YCor May 6 '15 at 14:03
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    $\begingroup$ @TimCampion: That group is not torsion-free because $2(x-y)=0$. $\endgroup$ – Eric Wofsey May 6 '15 at 15:14

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