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Let $S$ be a closed subset of $\mathbb Z_p[[x]]$, so $S$ is a compact set. Let $M$ be the $\mathbb Z_p$-submodule of $\mathbb Z_p[[x]]$ generated by $S$.

Is $M$ necessarily closed/compact?

It's true if $S$ is a finite set.

My guess: this need not be true in general. For example, the map $n \mapsto f_n := (1 + x)^n$ is $p$-adically continuous in $n$, so extends to a continuous map $\mathbb Z_p \to \mathbb Z_p[[x]]$, so we can define $f_\alpha := (1 + x)^\alpha$ for any $\alpha \in \mathbb Z_p$. Consider the set $S$ of all of the $f_\alpha$ for every $\alpha$ in $\mathbb Z_p$. Since $S$ is the continuous image of a compact set, $S$ is compact/closed. Moreover, $S$ includes a monic polynomial of every degree, so that if the $\mathbb Z_p$-span $M$ of $S$ as defined above is closed then $M$ would have to be all of $\mathbb Z_p[[x]]$. But could it really be that any power series in $\mathbb Z_p[[x]]$ is a FINITE $\mathbb Z_p$-linear combination of these $f_\alpha$?? It seems so unlikely! But there are also so so many, so uncountably many, of these $f_\alpha$ that I am not sure.

Thanks in advance for engaging!

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Why not just take $S=\{ x^n,n\ge 0\}\cup \{0\}$ ? The $\Bbb{Z}_p$-span is $\Bbb{Z}_p[x]$.


For your second question about the $\Bbb{Z}_p$ span of the $(1+x)^\alpha$, reduce $\bmod p$ to sit in $\Bbb{F}_p[[x]]$ where $(1+x)^{p^k}=1+x^{p^k}$, assume that $\sum_{n\ge 0} c_n x^n=\sum_{j= 1}^J b_j (1+x)^{\alpha_j}$, then all the vectors $v_m = (c_{p^{m+J}+1},\ldots,c_{p^{m+J}+2J})\in \Bbb{F}_p^{2J}$ will sit in a common $J$-dimensional $\Bbb{F}_p$ vector space. This is a strong constrain implying that most elements of $\Bbb{F}_p[[x]]$ are not in the $\Bbb{F}_p$ span of the $(1+x)^\alpha$.

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  • $\begingroup$ Oof. I see what you did there. I asked and you answered; I thank you and accept. But of course the question I should have asked, and may still, is a little different -- in particular, I want $S$ to be the injective image of a profinite group under a continuous map, not necessarily a group homomorphism. Any chance you might have an example where $S$ is uncountable up your sleeve? $\endgroup$
    – sibilant
    Commented Nov 24, 2021 at 4:38
  • $\begingroup$ If it is mere uncountability that you want to see up someone's sleeve, then behold: take $S = \{x^n : n \geq 0\} \cup \mathbf Z_p$. This is compact (the union of reuns's compact subset and the compact subset $\mathbf Z_p$) and its $\mathbf Z_p$-span is still $\mathbf Z_p[x]$. ¯_(ツ)_/¯ Admittedly this $S$ is not the injective continuous image of a profinite group. A better version of the question might be whether there are useful criteria for showing a $\mathbf Z_p$-submodule of $\mathbf Z_p[[x]]$ is compact. Or ask the question in the context of the 2nd to last sentence of your comment. $\endgroup$
    – KCd
    Commented Nov 24, 2021 at 5:50
  • $\begingroup$ @sibilant I tried an argument for your $span (1+x)^\alpha$ problem. The details are messy, tell me if you see the trick and agree $\endgroup$
    – reuns
    Commented Nov 24, 2021 at 6:22
  • $\begingroup$ @reuns Thanks! I'm afraid I haven't been able to make sense of your argument yet --- the bit with all the vectors $v_m = (c_{p^{m+J}+1},\ldots,c_{p^{m+J}+2J})\in \Bbb{F}_p^{2J}$ sitting in a common $J$-dimensional $\Bbb{F}_p$ vector space, of course. To start with, are you taking any $m \geq 0$? $\endgroup$
    – sibilant
    Commented Nov 25, 2021 at 5:05
  • $\begingroup$ @sibilant Yes, the set of $v_m,m\ge 0$ sits in a $J$-dimensional vector space. This is because $(1+x)^{l p^k}=1+l x^{p^k}+O(x^{2p^k})$ so multiplication by it is essentially shifting the $\sum_{n=0}^J c_n x^n$ polynomial to $\sum_{n=0}^J c_n x^{p^k+n}$ (the goal is to understand the difference between $(1+x)^\alpha$ and $(1+x)^{\alpha+l p^k}$) $\endgroup$
    – reuns
    Commented Nov 25, 2021 at 5:16

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