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.

Thanks in advance for engaging!

Solution 1:

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 $(1+x)^\alpha$.