7
$\begingroup$

$\newcommand{\cts}{\mathrm{cts}}$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 $n \neq p^n$ 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.

Thanks for any comments and answers.

Improve this question
$\endgroup$
10
  • 5
    $\begingroup$ 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. $\endgroup$
    – Dave Benson
    Commented Mar 30 at 23:44
  • 3
    $\begingroup$ @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 $\endgroup$
    – Maxime Ramzi
    Commented Mar 30 at 23:52
  • 2
    $\begingroup$ Any homomorphism between topologically finitely generated profinite groups is automatically continuous. For pro-p groups this is due to Serre. $\endgroup$
    – Benjamin Steinberg
    Commented Mar 31 at 0:42
  • 9
    $\begingroup$ @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 $2^\mathfrak{c}$. $\endgroup$
    – Jeremy Rickard
    Commented Mar 31 at 1:30
  • 2
    $\begingroup$ 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 $0$. $\endgroup$
    – Laurent Moret-Bailly
    Commented Mar 31 at 6:35

1 Answer 1

Reset to default
5
$\begingroup$

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 $\hom_{\mathbb Z_p}$

Improve this answer
$\endgroup$
2
  • $\begingroup$ There's a very nice and conceptually clear way to think about "derived complete modules": contramodules. It is explained very well (in elementary cases relevant to the question) in Positselski's papers arxiv.org/abs/1503.00991 arxiv.org/abs/1605.08018 arxiv.org/abs/1605.03934 $\endgroup$
    – Denis T
    Commented Mar 31 at 12:41
  • 1
    $\begingroup$ 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. $\endgroup$
    – Denis T
    Commented Mar 31 at 12:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .