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Claim: The category of torsion abelian groups has enough injectives.

I thought I had a proof of this, but discovered a mistake in my proof. I was trying to use the techniques of the usual proof that the category of abelian groups has enough injectives, such as the proof at this page: https://stacks.math.columbia.edu/tag/01D8. This proof constructs a canonical embedding $$ M \to F(M^\vee)^\vee $$ where $M^\vee = \operatorname{Hom}_{\mathbb{Z}}(M,\mathbb{Q}/\mathbb{Z})$. However, $F(M^\vee)^\vee$ is not generally torsion, and even if $M$ is torsion, I suspect $F(M^\vee)^\vee$ may not be torsion. Is there a simpler way to show that a torsion abelian group $M$ embeds into a torsion injective (equivalently, divisible) abelian group?

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  • $\begingroup$ Don't be greedy. Make with the injectives you have! $\endgroup$ – Asaf Karagila Apr 21 at 21:17
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Let $I$ be an injective abelian group and let $T(I)$ be its torsion subgroup. Then $T(I)$ is injective. Indeed, $T(I)$ is divisible since $I$ is divisible (and if $nx\in T(I)$ then $x\in T(I)$).

So, if $M$ is torsion, then it embeds into the torsion subgroup of $F(M^\vee)^\vee$, which is still injective.


Alternatively, it is pretty easy to directly prove that any (torsion) abelian group $M$ embeds into a divisible (torsion) abelian group. The key step is the following:

Lemma: Let $M$ be an abelian group, $x\in M$, and $n\in\mathbb{Z}\setminus\{0\}$. Then there is an abelian group $N$ which contains $M$ as a subgroup and an element $y\in N$ such that $ny=x$. If $M$ is torsion, so is $N$.

Proof: Let $N=M\oplus\mathbb{Z}/K$ where $K$ is the subgroup generated by $(x,-n)$. Then the natural homomorphism $M\to N$ is injective since $n\neq 0$, and if $y$ is the image of $(0,1)$ in $N$ then $ny=x$. Since $N$ is generated by $M$ and $y$ and $ny\in M$, it is clear that if $M$ is torsion, so is $N$. $\blacksquare$

Now given an abelian group $M$, you can embed it in a divisible abelian group by just using the Lemma repeatedly by transfinite recursion, adjoining new elements to correct every failure of divisibility until there are none left. If $M$ is torsion, then each step of the construction will also be torsion.


Both of these ideas generalize to modules over any ring, replacing "divisible" with "satisfying Baer's criterion for injectivity". In particular, over any integral domain, you can conclude that the category of torsion modules has enough injectives (either since the torsion submodule of an injective module is still injective, or alternatively because if you start with a torsion module and adjoin elements to it in order to satisfy Baer's criterion, the result is still torsion).

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