## Why are divisible abelian groups important?

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I just quote wikipedia:

"Divisible groups are important in understanding the structure of abelian groups, especially because they are the injective abelian groups."

I am asking for detail explanation of this application (i.e. concrete theorems, methods and so forth), or/and sources on that issue. Are there some other applications of DAG besides abelian groups theory or even besides algebra?

Thanks in advance.

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Over any PID, divisible and injective are the same thing. I think over $\mathbb{Z}$, things are nice because all divisible groups (and therefore all injective $\mathbb{Z}$-modules) are known up to isomorphism. There are lots of books which talk about this. Personally, I would recommend (if you can get your hands on it) Kaplansky's "Infinite Abelian Groups". Besides having a ton of info in about 100 pgs, I think the way he presents divisible groups is very slick and easy-to-follow. – Steve D Aug 8 2010 at 4:04

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An example where divisibility is used crucially that came to mind first for me is in constructing the invariant map $Br(K)\simeq\mathbb{Q}/\mathbb{Z}$, where $Br(K)=H^2(K,\mathbb{G}_m)$ is the Brauer group of the non-Archimedean local field $K$. One uses the short exact sequence $0\rightarrow\mathbb{Z}\rightarrow\mathbb{Q}\rightarrow\mathbb{Q}/\mathbb{Z}$ of trivial $G_K$-modules, and the fact that $\mathbb{Q}$ is uniquely divisible implies that its $\geq 1$-dimensional cohomology vanishes. – Keenan Kidwell Aug 8 2010 at 12:43

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<<Continued>> The unique divisibility of actually implies that $H^n(G,\mathbb{Q})=0$ for all finite groups $G$ and $n\geq 1$, since the latter group is killed by $\vert G\vert$, but the unique divisibility of $\mathbb{Q}$ implies that multiplication by $\vert G\vert$ is an isomorphism on cohomology. Taking direct limits one gets the same result for profinite $G$. This is crucial because upon taking the long exact cohomology sequence one gets $\mathbb{Q}/\mathbb{Z}\simeq Hom(\hat{\mathbb{Z}},\mathbb{Q}/\mathbb{Z})\simeq H^2(\hat{\mathbb{Z}},\mathbb{Z})$, which is a part of the invariant map. – Keenan Kidwell Aug 8 2010 at 12:51
As Steve has mentioned, divisible modules and injective modules are the same over a PID. More generally, this is true over any Dedekind domain. – Beren Sanders Aug 9 2010 at 4:17

## 3 Answers

In my work I come upon divisible abelian groups all the time, especially those of the form $\mathbb Q_p/\mathbb Z_p$, or direct sums of such groups. One frequently uses the fact that they are injective (to conclude that certain sequences of Hom groups, which a priori would only be left exact, are in fact exact on the right as well), and other properties as well; for example, groups of type mentioned above, related to $\mathbb Q_p/\mathbb Z_p$, have non-zero $p$-adic Tate modules, which can often be useful.

This last property is related to the role that divisible groups play in duality theory (in the sense of Pontrjagin duality), and this is another reason that divisible groups are important.

I think I probably speak for a lot of algebraic number theorists when I say that a lot of what we do is a kind of applied algebra. We don't care about algebraic properties of structures for their own sake, but in our work we encounter lots of different groups, rings, and modules, and we like to understand their properties as well as we can, so that we can use them to gain control of the number theoretic computations that we are trying to make. Divisibility of an abelian group is one such useful property, which we are trained to recognize and exploit.

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The meaning of the WP quote is that a divisible subgroup of an abelian group is always a direct summand. On the other hand, any divisible abelian group is a direct sum of copies of $\mathbb{Q}$ and $\mathbb{Q}_p/\mathbb{Z}_p$, thus unlike general infinite abelian groups, they can be completely classified. See the books of Kurosh and Kargapolov–Merzlyakov on group theory for the proofs. As Steve D mentioned in the comments, there is a straightforward generalization to modules over PIDs.

Here are two related "concrete applications outside of algebra".

1 Pontryagin duality for locally compact abelian groups, $G\mapsto\text{Hom}(G,T),$ where $T=\mathbb{R}/\mathbb{Z}$ is the circle group and Hom denotes continuous homomorphisms.

2 Tate duality for Galois modules in algebraic number theory, $A\mapsto\text{Hom}(A,\mu)$, where $\mu$ is the group of roots of unity ($\mu\simeq\mathbb{Q}/\mathbb{Z}$ as abelian groups, but the Galois module structure is different).

In both cases, divisibility of the target is needed to assure that the natural map into the double dual is injective.

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Per FAQ, this question is too general for MO. – Victor Protsak Aug 8 2010 at 4:38
Victor: Quote from FAQ: <<The site works best for well-defined questions: math questions that actually have a specific answer.>> This question is about concrete and, I think, well known for algebraists (I am not the one) relationship between two concrete algebraic structures expressible in precise mathematical theorems. On the other hand I agree that it is not a technical question, but it is community wiki. – Sergei Tropanets Aug 8 2010 at 5:08
Victor: <<A question should be made community wiki if you don't think that people should gain reputation for their answers. A typical case is requests for references where it is the reference that is being judged by the voting system rather than the person who supplied it.>> I think the question satisfy this. Note that according to FAQ the final characteristics of CW questions aren't defined yet. – Sergei Tropanets Aug 8 2010 at 5:11

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Sergei, I am not that worried about CW and all that. But you are, in essence, asking people to teach you a subject, without even providing a compelling motivation ("I just quote from Wikipedia"). While I am generally "on the left" in the matters which questions are suitable, I disagree with such an open-ended use of MO. Other people might even characterize it as a "fishing expedition". In any case, this is going to be my last answer to your questions: I've put some effort in answering your question, however imperfect it was, but you haven't reciprocated by reading and upvoting it. – Victor Protsak Aug 8 2010 at 5:40
I've read it and now I'm studying the details you provided. Thanks for the answer. – Sergei Tropanets Aug 8 2010 at 5:45
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Suppose we have a category (that is, a collection of objects and arrows ["morphisms"] between those objects; for example, the collection of abelian groups and all homomorphisms between those abelian groups). An object $I$ is injective if whenever we have an injection $i:A \to B$ and a homomorphism $f:A \to I$, then there is a morphism $g:B \to I$ such that $f = g \circ i$.

In the category of abelian groups, the divisible groups are precisely the injective objects in that category. To see a sketch of why, suppose $B$ is an infinite cyclic groups, and $A$ is the set of powers of $n$ resting within that cyclic group (hence also isomorphic to an infinite cyclic groups). Suppose $B$ has generator $b$, and $A$ has generated $a:=b^n$. Then a homomorphism $\varphi:A \to I$ amounts to choosing an element of $I$ where $a$ will be sent. Extending this to a homomorphism from $B$ to $I$ amounts to choosing an element of $I$ whose $n$th power is $\varphi(a)=\varphi(b)^n$; that is, dividing by $n$. This is why they are related.

Now, you might ask, why are injective objects important? Injective objects are some of the most important objects in a category which underlie many applications of homological algebra. Suppose we have an exact sequence $0 \to A \to B \to C \to 0$. It is well known that the sequence $0 \to Hom(C,I) \to Hom(B,I) \to Hom(A,I)$ is exact no matter what. However, we know that that last map is surjective if $I$ is injective. Thus, $Hom(-,I)$ is an exact (contravariant) functor iff $I$ is injective. It follows from this that $Ext(-,I)$ is trivial iff $I$ is injective, and in general, right derived functors are trivial on injective objects.

This latter fact is interesting because derived functors appear all over higher mathematics. To give an example, the group cohomology is always trivial on injective objects in the category of $G$-modules. However, all we talked about was injective objects in the category of $\mathbb{Z}$-modules (i.e. abelian groups). So why is this interesting? Well, the functor which sends an abelian group $M$ to the induced $G$-module $\mathrm{Ind}^G(M)$ (essentially $\mathbb{Z}[G] \otimes_{\mathbb{Z}} M$ by extension of scalars) is exact and sends injectives to injectives. This means that we get a whole host of injective $G$-modules by considering injective abelian groups, and what's more, it allows us to show that the category of $G$-modules has enough injectives (i.e. that every $G$-module can be embedded in an injective $G$-module).

Derived functors show up in many other areas, including sheaf cohomology (where injective objects are also very important), and divisible groups are the most basic example of the injective objects which underlie right derived functors.

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