I'm trying to find an injective resolution for the $\mathbb Z$-module $\mathbb Z/p$ where $p$ is a prime. I've come across the result that an injective resoltion for a PID $R$ is of the form
$$ 0 \rightarrow R \rightarrow K \rightarrow K/R \rightarrow 0 $$
where $K$ is the field of fractions of $R$. Since $\mathbb Z /p$ is a field, in our case $K = \mathbb Z /p$. So the injective resolution here is
$$ 0 \rightarrow \mathbb Z /p \rightarrow \mathbb Z /p \rightarrow 0 \rightarrow 0 $$
where the non-trivial map is the identity (because of the canonical isomorphism between $R$ and $K$ when $R$ is a field?).
But we also have the result that for an integral domain $R$, an $R$-module $M$ is injective if and only if it is divisible. In our case, take $p=2$. Then $\mathbb Z / 2 $ is surely not divisible because because for $4 \in \mathbb Z$ and $1 \in \mathbb Z /2$, there is no $ n \in \mathbb Z /2$ such that $4n=1$. But surely it should be divisible because it should be injective since we've shown it to be part of an injective resolution? I'd appreciate it if anybody could shed some light on where I've gone wrong here. Many thanks.
Edit: with regards to the comments, I suppose I have another question which would be: when saying $$0 \rightarrow R \rightarrow K \rightarrow K/R \rightarrow 0$$ is an injective resolution, what should the objects in the sequence be injective over? I thought that since we're treating things as $\mathbb Z$-modules originally (i.e. working in the category $\mathbb{Z} \text{-Mod}$), our objects should be injective in this category?
Best Answer
The group $\mathbb{Z}(p)$ (this is a more common notation in this context) is not injective as it is not divisible.
Its injective envelope is the Prüfer $p$-group $\mathbb{Z}(p^{\infty})$. Since every nonzero quotient of $\mathbb{Z}(p^{\infty})$ is isomorphic to $\mathbb{Z}(p^{\infty})$, an injective resolution is $$ 0\to\mathbb{Z}(p)\to\mathbb{Z}(p^{\infty})\to\mathbb{Z}(p^{\infty})\to0 $$
Using the field of quotients $\mathbb{Q}$ is impossible, because $\mathbb{Z}(p)$ is torsion, so it cannot be embedded in the torsionfree group $\mathbb{Q}$.
The fact that $\mathbb{Z}/p\mathbb{Z}$ (as a field) is injective as a module over itself is of no concern. If $A$ is any commutative ring and $I$ is a maximal ideal thereof, then $A/I$ is a field (hence an injective module over itself), but it's generally not true that every simple $A$-module is injective (as a module over $A$).
If $R$ is a Dedekind domain, then $0\to R\to K\to K/R\to 0$ is an injective resolution of $R$ as an $R$-module, with the quotient field $K$ considered as an $R$-module, of course.