About the injection $M \hookrightarrow \mathbb Q \otimes_{\mathbb Z} M$. I want to prove that every abelian group can be embedded in a divisible abelian group.
So I tried $M \rightarrow \mathbb Q \otimes_{\mathbb Z} M, m \mapsto 1 \otimes m$. It is obvious that $\mathbb Q \otimes_{\mathbb Z} M$ is divisible, so I just need to show that it is an injection indeed, that is $1 \otimes m = 0 \implies m = 0$. So I search for a $\mathbb Z$-bilinear map $\mathbb Q \times M \to N$ with $N$ an abelian group such that the image of $(1,m)$ is zero only for $m=0$.
The only $N$ I can think of is the following one : take the $\mathbb Q$-free module (actually free vector space) with basis $M$ and quotient by the kernel of the $\mathbb Z$-module morphism $$\mathbb Z[M] \twoheadrightarrow M, \sum k_m \cdot m \mapsto \sum k_mm,$$ and consider the quotient as a $\mathbb Z$-module. Then, the bilinear map $(r,m) \mapsto \overline{r \cdot m}$ is as we want : $\overline{1 \cdot m} = 0$ implies $m = 0$ in $M$.
But it seems to me that the constructed $N$ is actually $\mathbb Q \otimes_{\mathbb Z} M$ ... It tells me that I probably could see directly that $1 \otimes m$ isn't zero for $m \neq 0$. Just how ?
 A: Let $R$ be a commutative ring, $S \subset R$ a multiplicatively closed subset, and $M$ an $R$-module.  It is a standard fact -- which has a short, straightforward proof directly from the definition -- that the kernel of the localization map $M \rightarrow S^{-1} M$, $x \mapsto \frac{x}{1}$ is the set of $m \in M$ such that $sm = 0$ for some $s \in S$.  (For instance this is Exercise 7.8 in my commutative algebra notes.)
Applying this with $R = \mathbb{Z}$ and $S = \mathbb{Z}^{\bullet} = \mathbb{Z} \setminus \{0\}$ shows that for a $\mathbb{Z}$-module $M$,
$\operatorname{Ker}(M \rightarrow M \otimes \mathbb{Q}) = M[\operatorname{tors}]$.
[Note that the same holds any integral domain $R$ with fraction field $K = (R^{\bullet})^{-1} R$: $\operatorname{Ker}(M \rightarrow M \otimes_R K) = M[\operatorname{tors}]$.]
In particular, this map is an injection iff $M$ is torsionfree, so in general this will not work to embed an arbitrary $\mathbb{Z}$-module into a divisible one.  To do this, one can make either of the following two arguments:

*

*Define the Pontrjagin dual $\mathbb{Z}$-module $M^* = \operatorname{Hom}(M,\mathbb{Q}/\mathbb{Z})$ and show that the natural map $M \hookrightarrow M^{* *}$ is an injection, and then proceed as in Quimey's answer.


*Argue a bit more directly: write $M = F/K$ with $F$ a free abelian group.  Since $F$ is torsionfree, the map $F \rightarrow F \otimes \mathbb{Q}$ is an injection, hence $M$ embeds in $(F \otimes \mathbb{Q})/K$.  The latter is the quotient of a divisible module, hence divisible (hence injective, since $\mathbb{Z}$ is a PID).  This seems to be the correct solution which is closest to the type of argument you're trying to make.
A: If $M$ is a torsion module then $\mathbb{Q}\otimes_{\mathbb{Z}}M=0$:
Assume there exists $n\in\mathbb{Z}$ such that $nm=0$ for all $m\in M$. We have $1\otimes m= \frac{n}{n}\otimes m = \frac{1}{n}\otimes nm=0$.
The usual trick is to embed $M$ in $\hom_\mathbb{Z}(N,\mathbb{Q}/\mathbb{Z})$, for suitable $N$. This is done, for instance, in Lang's Algebra (Third edition), Thm 4.1, Chapter XX.
Let $M^*=\hom_\mathbb{Z}(M,\mathbb{Q}/\mathbb{Z})$.
The idea is show first that the natural map $M\to M^{**}$ is an injection. Now, take a projection $F\to M^*$ with $F$ free abelian. Applying $(-)^*$, we get an injection $M^{**}\to F^*$ (it is easy to see that $F^*$ is injective). If we compose this injection with $M\to M^{**}$ we get the desired injection.
