TITLE: $\operatorname{Hom}(Z_p, Z_p) = Z_p$?
QUESTION [6 upvotes]: Is $\operatorname{Hom}(\mathbf{Z}_p, \mathbf{Z}_p) = \mathbf{Z}_p$?
My proof: $\operatorname{Hom}(\mathbf{Z}_p, \mathbf{Z}_p) = \operatorname{Hom}(\mathbf{Z}_p, \varprojlim\mathbf{Z}/p^n) = \varprojlim \operatorname{Hom}(\mathbf{Z}_p, \mathbf{Z}/p^n) = \varprojlim \operatorname{Hom}(\mathbf{Z}_p, \mathbf{Z}/p^n) = \mathbf{Z}_p$.
Is this correct?

REPLY [2 votes]: You can just see this by hand in this case.  Given an element $a$ of ${\mathbb Z}_p$, you get a group homomorphism $\phi_a$ from $\mathbb Z_p$ to itself by translation: set $\phi_a(x) := ax$. This gives you a map from $\mathbb Z_p$ to ${\mathop{\rm Hom}}(\mathbb Z_p, \mathbb Z_p)$. It's easy to convince yourself that this map is injective. 
To see that this map is also surjective is only a little trickier. In fact, it turns out that any group homomorphism $f$ from $\mathbb Z_p$ to $\mathbb Z_p$ is $p$-adically continuous. That's because $f$ has to map $p^n \mathbb Z_p$ to itself because it's a group homomorphism, so that $f^{-1}(p^n \mathbb Z_p)$ is subgroup of $\mathbb Z_p$ containing the open set $p^n \mathbb Z_p$, hence open.
Since any $f \in {\mathop{\rm Hom}}(\mathbb Z_p, \mathbb Z_p)$ is $p$-adically continuous, it's determined by the image of $1$, so that $f = \phi_{f(1)}$, so that the map $a \to \phi_a$ described above is surjective.