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Let $G=1+t\mathbb{F}_q[[t]]$ be the unipotent unit group of the ring $R=\mathbb{F}_q[[t]]$ of formal power series.

This question is a continuation of a past question, and for more context see another question. Since this group is a profinite group, (see the second question cited) it is a topological group, so asking if the group is finitely generated as a topological group is better suited.

Recall that a topological group $G$ is finitely generated if "there exists a dense finitely generated subgroup [of $G$]." (see groupprops)

Note: $G=\varprojlim (1+\mathbb{F}_q[t]/(t^n)),$ under natural projection, so showing $G$ is finitely generated is equivalent to showing there exists $d$, such that for any $n$, $1+\mathbb{F}_q[t]/(t^n)$ is generated by $d$ elements. (see wikipedia) The other formulation may be easier by employing a counting argument, but such method has eluded me.

If finitely generated is still to weak of a condition, by way of $G$ being uncountable, then the natural follow up question is whether $G$ is countably generated, since $\mathbb{R}$ is generated by $\mathbb{Q}$, topologically, so a simple cardinality argument won't suffice.

Torsten Schoeneberg
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Mattan Feldman
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    Technically, that is not "the unit group" of the power series ring. It is a subgroup. Any element not in $t\mathbb F_q[[t]]$'is a unit. – Thomas Andrews Jun 23 '25 at 20:02
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    This should be covered in any decent source on local fields, because $\mathbb F_q((t))$ (formal Laurent series) basically are the local fields of positive characteristic, with $\mathbb F_q[[t]]$ as discrete valuation ring. And the structure of the multiplicative group, which up to a finite quotient is the group you are looking at, is of utmost importance in that theory. – Torsten Schoeneberg Jun 23 '25 at 20:54
  • @ThomasAndrews thanks for catching the mistake, changed the wording to unipotetnt – Mattan Feldman Jun 24 '25 at 16:45
  • @TorstenSchoeneberg I've skimmed through Local Fields by Serre, and the topology defined seems significantly different from the product topology we use. Also, I didn't see anything about the resulting topology groups being finitely generated. Maybe this is of note: $G$ is $I$-addically complete. – Mattan Feldman Jun 24 '25 at 16:55
  • I would be very surprised if the topologies do not match. In my answer I certainly assume the standard topology from the discrete ($T$-adic) valuation, but as said I would be surprised if you convince me that is not equivalent to the I-adic / profinite one you are talking about. – Torsten Schoeneberg Jun 24 '25 at 18:21

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The abelian group you are looking at has a natural structure of $\mathbb Z_p$-module ($\mathbb Z_p$ denoting the $p$-adic integers). As such, it is topologically countably (but not finitely) generated, with a set of topological generators given by $$(1+\theta_j T^i)_{1\le j\le r, \;i>0, \; (i,p)=1}$$ where $\theta_1, ..., \theta_r$ are an $\mathbb F_p$-basis of $\mathbb F_q$ ($q=p^r$). See (slighly more general statement in) Fesenko / Vostokov, Local Fields 2nd ed., prop. 6.2 and corollary.

Torsten Schoeneberg
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