Embedding $\mathbb{Z}$-modules into injective $\mathbb{Z}$-modules I want to show that any $\mathbb{Z}$-module $M$ can be embedded into an injective $\mathbb{Z}$-module (I'm in the process of showing this can be done for more general rings, but starting with this case).
My idea is to tensor with $\mathbb{Q}$, so define $N=\mathbb{Q}\otimes M$. This is a $\mathbb{Z}$-module via $$n(q\otimes m)=q\otimes nm=nq\otimes m$$ and extending linearly.
It is divisible since $$\sum q_im_i=n\sum \frac{q_i}{n}m_i$$ Now, $\mathbb{Z}$ is a PID and for a module over a PID injectivity is equivalent to divisibility, so $N$ is injective, and $M$ embeds via $m \rightarrow (0,m)$.
Is this correct?
 A: No, this is not correct. The result is indeed divisible, but the map is not injective in general. Consider the case of a finite cyclic group: $\mathbb{Z}/n\mathbb{Z} \otimes \mathbb{Q} = 0$.
For a proof, consider a presentation of your abelian group: $G = \mathbb{Z}\langle S \rangle / R$, where $R \subset \mathbb{Z}\langle S \rangle$ is an abelian group (necessarily normal). Then consider the quotient of abelian groups $I = \mathbb{Q}\langle S \rangle / R$ (where you view $R \subset \mathbb{Z}\langle S \rangle \subset \mathbb{Q}\langle S \rangle$ in the obvious way). The result is divisible, and the obvious map is injective.
It's best to work this out on example. If $G = \mathbb{Z} = \mathbb{Z}\langle x \rangle / 0$, then $I = \mathbb{Q}\langle x \rangle / (0) = \mathbb{Q}$. If $G = \mathbb{Z}/n\mathbb{Z} = \mathbb{Z}\langle x \rangle / (nx)$, then $I = \mathbb{Q} \langle x \rangle / \langle nx \rangle = \mathbb{Q}/n\mathbb{Z}$. The result is quite different from $G \otimes \mathbb{Q}$ (in fact $G \otimes \mathbb{Q}$ is a quotient of $I$: it's $I / (R \otimes \mathbb{Q})$).
