Injectivity of $\mathbb{Q}_{p}$ and $\mathbb{Z}(p^{\infty})$

Question

Please help me. In the category of $R$-modules, we know that the Prüfer group $\mathbb{Z}(p^{\infty})$ is an injective envelope of $\mathbb{Z}/p\mathbb{Z}$ the ring integers modulo $p$ where $p$ is a prime number. But, I found that $\mathbb{Q}_{p}/\mathbb{Z}_{p}\cong \mathbb{Z}(p^{\infty})$, where $\mathbb{Z}_{p}$ is a $p$-adic integers and $\mathbb{Q}_{p}$ is a $p$-adic field. So, we we can construct an exact sequence: $0\rightarrow \mathbb{Z}_{p}\rightarrow \mathbb{Q}_{p}\rightarrow\mathbb{Z}(p^{\infty})\rightarrow 0$ and we also have monomorphism from $\mathbb{Z}/p\mathbb{Z}\hookrightarrow \mathbb{Z}_{p}$. The $p$-adic field $\mathbb{Q}_{p}$ is an injective module as a $\mathbb{Z}$-modules, so there is a minimal injective module than $\mathbb{Z}(p^{\infty})$ containing $\mathbb{Z}/p\mathbb{Z}$, which is $\mathbb{Q}_{p}$. But, the injective envelope of $\mathbb{Z}/p\mathbb{Z}$ is $\mathbb{Z}(p^{\infty})$. I don't know where my mistake lies. Thank you so much for considering my questions.

Answer 1

In fact, $\mathbb{Z}/p\mathbb{Z}$ does not embed in $\mathbb{Q}_p$ at all. $\mathbb{Q}_p$ is torsion-free, but $\mathbb{Z}/p\mathbb{Z}$ is torsion, so the only homomorphism $\mathbb{Z}/p\mathbb{Z} \to \mathbb{Q}_p$ is the zero map.

$\mathbb{Z}/p\mathbb{Z}$ does embed in $\mathbb{Z}(p^\infty)$ however, so it makes sense to claim that $\mathbb{Z}(p^\infty)$ is the injective envelope of $\mathbb{Z}/p\mathbb{Z}$ (and this is true!)