$\newcommand{\ZZ}{\mathbb{Z}}$ $\newcommand{\QQ}{\mathbb{Q}}$ $\newcommand{\Hom}{\mathrm{Hom}}$ Some time ago I tried to construct for a given abelian group $M$ functorially a group $I(M)$ which should be an injective object (that is a divisible abelian group) and provide an injection $0 \to M \hookrightarrow I(M)$.

I formed $$ I(M) = \Hom_{\ZZ}(F_\ZZ \Hom_\ZZ(M, \QQ/\ZZ), \QQ/\ZZ) $$ where $F_\ZZ \Hom_\ZZ(M,\QQ/\ZZ)$ is the free abelian group over the abelian group $\Hom_\ZZ(M,\QQ/\ZZ)$.

The group $I(M)$ is an injective (divisible) group because of the following equation $$ \Hom_\ZZ(N^\bullet, \Hom_\ZZ (F_\ZZ\Hom_\ZZ(M,\QQ/\ZZ), \QQ/\ZZ)) = \Hom_\ZZ(N^\bullet \otimes_\ZZ F_\ZZ(\Hom_\ZZ(M, \QQ/\ZZ)), \QQ/\ZZ) $$ where $N^\bullet$ is the exact sequence $0 \to N' \to N \to N'' \to 0$. Now tensoring with a free group is exact and $\QQ/\ZZ$ is an injective group. So $\Hom_\ZZ(N^\bullet, I(M))$ preserves exactness, especially on $0 \to N' \to N$ and so $I(M)$ is an injective group.

Now the map $M \to I(M)$ is given as $$ (*) \quad m \mapsto ((\sum_i n_i \phi_i) \mapsto \sum_i n_i \phi_i(m)) $$ where $\sum_i n_i \phi_i$ is an element of $F_\ZZ \Hom_\ZZ(M, \QQ/\ZZ)$.

Now for every $m \in M$ with $m \neq 0$ one can construct a map $\phi_m:M \to \QQ/\ZZ$ with $\phi_m(m) \neq 0$ (because $\QQ/\ZZ$ is an injective group).

So the map $(*)$ is injective by choosing for given $m \in M$ the map $\phi_m$ as $\sum_i n_i \phi_i$ in $(*)$.

Two questions

1) Is this proof valid? It seems so simple that it is somewhat embarassing to ask such a question, but:

going through all my commutative algebra books I could not find this proof, although it seems obvious and simple and has the advantage of being functorial in $M$. So

2) Is there a place in the literature where a proof of enough injective objects in the abelian groups (or for modules over a commutative ring) is done with this idea (especially forming a free group with the intent of moving it to the left and tensor with it)?



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