$\operatorname{Hom}(Z_p, Z_p) = Z_p$?

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?

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I take it from the working $Z_p$ is the p-adic integers? – Juan S Sep 27 '11 at 12:04
Yes, $\mathbf{Z}_p = \varprojlim \mathbf{Z}/p^n$. – user5262 Sep 27 '11 at 12:18
Did you mean to write $\varprojlim \mathbf{Z}/p^n$ before your last equals sign? – Zhen Lin Sep 27 '11 at 14:38
What category is this Hom in? – Matt Sep 27 '11 at 15:53
@user5262: You need to show/argue that $\mathrm{Hom}(\mathbf{Z}_p,\mathbf{Z}/p^n\mathbf{Z})\cong\mathbf{Z}/p^n\mathbf{Z‌​}$; otherwise fine. But why do you have the same expression in the penultimate and antepenultimate terms of your equality chain? – Arturo Magidin Sep 27 '11 at 16:21

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.