Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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

1 Answer 1

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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.