# Compactly generated $\mathbb Z_p$-submodules of $\mathbb Z_p[[x]]$

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.

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 $$(1+x)^\alpha$$.
• 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? Commented Nov 24, 2021 at 4:38
• 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.
• @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 Commented Nov 24, 2021 at 6:22
• @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 $m \geq 0$? Commented Nov 25, 2021 at 5:05
• @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 $(1+x)^{\alpha+l p^k}$) Commented Nov 25, 2021 at 5:16