Is $\mathbb{Q}_p/\mathbb{Z}_p$ an injective $\mathbb{Z}_p$-module

Question

I'm curious as to whether or not $\mathbb{Q}_p/\mathbb{Z}_p$ is an injective $\mathbb{Z}_p$-module. It's certainly an injective abelian group ($\mathbb{Z}$-module) since $\mathbb{Q}_p/\mathbb{Z}_p$ is divisible (see here). But because $\mathbb{Z}$ is dense in $\mathbb{Z}_p$, I think it must follow that $\mathbb{Q}_p/\mathbb{Z}_p$ is also an injective $\mathbb{Z}_p$-module, but for some reason I can't get the algebra to work, or I'm just being silly… Here are my thoughts. Let $A\subset B$, as both abelian groups and $\mathbb{Z}_p$-modules. We can extend any $\mathbb{Z}$-module map $A\rightarrow \mathbb{Q}_p/\mathbb{Z}_p$ to a corresponding map $B\rightarrow \mathbb{Q}_p/\mathbb{Z}_p$, again, as $\mathbb{Z}$-modules. If $f:A\rightarrow \mathbb{Q}_p/\mathbb{Z}_p$ is a map of $\mathbb{Z}_p$-modules, let $\bar f:B\rightarrow \mathbb{Q}_p/\mathbb{Z}_p$ be the $\mathbb{Z}$-extension of $f$ to $B$. We'd like to define a $\mathbb{Z}_p$-module homomorphism $\hat f:B\rightarrow \mathbb{Q}_p/\mathbb{Z}_p.$ The obvious option might be to define $\hat f:=\bar f$, for then $\hat f(n.b)=n.\hat f(b)$ for all $n\in \mathbb{Z}$ and $b\in B$. But if $c=\sum_{i=0}^\infty c_ip^i\in \mathbb{Z}_p$, is it necessarily true that $\hat f(c.b)=c.\hat f(b)$?

Answer 1

The proof that a divisible abelian group is injective as a $\mathbb{Z}$-module works with $\mathbb{Z}$ replaced by any PID: a module over a PID is injective iff it is divisible (as a module). Now $\mathbb{Z}_p$ is a PID, and $\mathbb{Q}_p$ is clearly a divisible $\mathbb{Z}_p$-module, so the quotient $\mathbb{Q}_p/\mathbb{Z}_p$ is also divisible and hence injective as a $\mathbb{Z}_p$-module.