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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 sometimes forget these are not, apparently, terms in common use in English (at least in the US). We used them all the time when I was growing up (whenever somebody wanted to go last and 'called it' by yelling "Last!", the next person would yell "Penultimate!", quickly followed by "Antepenultimate!", for those too slow to get the coveted last and next-to-last spots...)2011-11-15

1 Answers 1

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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.