# $\mathbb{Z}$-homomorphism and $\mathbb{Z}_p$-homomorphism

$$\newcommand{\cts}{\mathrm{cts}}$$Thanks for your reading. Let $$A,B$$ be two $$\mathbb{Z}_p$$-modules, where $$\mathbb{Z}_p$$ is the $$p$$-adic integer ring. I have two questions.

1. Is $$\mathrm{Hom}_{\mathbb{Z}}(A,B)=\mathrm{Hom}_{\mathbb{Z}_p}(A,B)$$? I think the answer may be no, but I can't find a counterexample. I don't have an idea about the difference between the two.

2. Is $$\mathrm{Hom}_{\mathbb{Z}_p}(A,B)=\mathrm{Hom}_{\cts}(A,B)$$, where $$A,B$$ have $$p$$-adic topology and $$\mathrm{Hom}_{\cts}$$ means continuous homomorphism between the two topological modules? I guess it is true, but I'm not sure how to prove it. And if $$A,B$$ have other linear topology, maybe it will not be true?

Let me explain my motivation for asking the second question, since I'm not sure the thought is right. I find that there are many cases in algebraic number theory using Pontryagin dual as an important tool. In fact, for a profinite group $$G$$, the dual is defined as $$\mathrm{Hom}_{\cts}(G,\mathbb{R}/\mathbb{Z})$$. Since its image should be finite, so $$\mathrm{Hom}_{\cts}(G,\mathbb{R}/\mathbb{Z})=\mathrm{Hom}_{\cts}(G,\mathbb{Q}/\mathbb{Z})$$. For $$p$$-adic Lie group $$G$$, since any integer $$n \neq p$$ is a unit in $$\mathbb{Z}_p$$, then for $$\mathbb{Z}_p$$-module $$G$$( I think there always exists a $$\mathbb{Z}_p$$-module structure on $$G$$? not 100% sure), so non-trivial maps in $$\mathrm{Hom}_{\cts}(G,\mathbb{Q}/\mathbb{Z})$$ should have image in $$\mathbb{Q}/\mathbb{Z}(p) \cong \mathbb{Q}_p/\mathbb{Z}_p$$. So the Pontryagin dual actually should be $$\mathrm{Hom}_{\cts}(G,\mathbb{Q}_p/\mathbb{Z}_p)$$. However I find in many cases it is also written as $$\mathrm{Hom}_{\mathbb{Z}_p}(G,\mathbb{Q}_p/\mathbb{Z}_p)$$. So I guess we have the general conclusion as the second one.

• How about $A=B=\mathbb{Z}_p$? Then $\operatorname{\rm Hom}_{\mathbb{Z}_p}(A,B)$ is $\mathbb{Z}_p$, but $\operatorname{\rm Hom}_{\mathbb{Z}}(A,B)$ is much bigger. Commented Mar 30 at 23:44
• @DaveBenson I believe that if $B$ is (derived) $p$-complete, the two homs are in fact the same. So for $B = \mathbb Z_p$ there should be no difference Commented Mar 30 at 23:52
• Any homomorphism between topologically finitely generated profinite groups is automatically continuous. For pro-p groups this is due to Serre. Commented Mar 31 at 0:42
• @Dave’s example works, for the reason he gave, if you replace $\mathbb{Z}_p$ by $\mathbb{Q}_p$. $\text{Hom}_{\mathbb{Z}_p}(\mathbb{Q}_p,\mathbb{Q}_p)\cong\mathbb{Q}_p$, whose cardinality is the continuum $\mathfrak{c}$. But since $\mathbb{Q}_p$ is a vector space over $\mathbb{Q}$ of dimension $\mathfrak{c}$, $\text{Hom}_\mathbb{Z}(\mathbb{Q}_p,\mathbb{Q}_p)$ has cardinality $2^\mathfrak{c}$. Commented Mar 31 at 1:30
• About @BenjaminSteinberg's comment: for finitely generated $\mathbb{Z}_p$- modules the reason is simple: any $\mathbb{Z}$-homomorphism maps $p^n A$ into $p^n B$ for all $n$, hence is continuous since $\{p^n A\}_{n\geq0}$ and $\{p^n B\}_{n\geq0}$ are bases of neighbourhoods of $0$. Commented Mar 31 at 6:35

As explained in the comments, if $$B$$ is $$p$$-complete (derived $$p$$-complete suffices), then the answer to 1- is yes.
This follows by adjunction from the fact that $$A\to A\otimes_\mathbb Z \mathbb Z_p$$ is an isomorphism upon $$p$$-completion.
This is true without $$p$$-completion if any element of $$A$$ is $$p$$-power torsion, and so gives another case where 1- has a positive answer.
However as pointed out in the comments, $$A=B=\mathbb Q_p$$ is a counterexample.
For 2-, note that any map of abelian groups is automatically continuous for the $$p$$-adic topology, thus 2- works with $$\hom_\mathbb Z$$ for any $$A,B$$, and thus (by the failure of 1-) not in general with $$\hom_{\mathbb Z_p}$$
• Brief summary is that for a finitely generated ideal I in a ring R, I-complete but not necessarily I-separated modules form nice, full abelian subcategory of R-modules with limit-preserving (but not sum-preserving) inclusion functor. It is a "covariant dual" of abelian category of I-torsion modules; precisely, in case of ideal (p) in integers, it is a heart of right semiorthogonal complement to $D(\Bbb Q)$ inside of $D(\Bbb Z_{(p)})$; while p-torsion modules form a heart of left semiorthogonal complement. Commented Mar 31 at 12:55