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If $\mathbb{F}_{q^n}$ is a finite field with $q^n$ elements ($q$ being a power of a prime $p$) we have the trace map $tr^n_m:\mathbb{F}_{q^n}\rightarrow \mathbb{F}_{q^m}$ such that $x\mapsto x+F^m(x)+..+F^{n-m}(x)$, if $m\mid n$. So we can form the inverse limit of these maps, that will be $$ \left\{(x_n)_{n\in\mathbb{N}}\in \prod_{n\in\mathbb{N}} \mathbb{F}_{q^n}\;\middle|\;\text{$tr^n_m(x_n)=x_m$ (if $m|n$)}\right\} $$ (an abelian group and also an $\mathbb{F}_q$-module).

My questions are

  • "what" abelian group it is,

  • what is the relation with the algebraic closure of $\mathbb{F}_q$, and

  • can we put a ring structure on the inverse limit?

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One answer to the question of "what abelian group" is "the inverse limit of a countable sequence of surjective maps of $\mathbb F_p$-vector spaces, of increasing dimension". (We can use the factorial function to reduce the inverse system to a countable sequence.) It is easy to see that we can choose a basis such that each map is projection onto a subset of the basis vectors, and the vector space is therefore the vector space of $\mathbb F_p$-valued functions on a countable infinite sets.

Alternately, we can use duality. Every finite field is self-dual with the map $(a,b) \to tr_n^1 (ab)$. Your inverse system is then dual to the directed system of maps $\mathbb F_{q^m} \to \mathbb F_{q^m}$, so its inverse limit is the dual of the directed limit of that system, which is $\overline{\mathbb F}_q$. So $\operatorname{Hom}(\overline{\mathbb F}_q,\mathbb F_q)$ is another possible answer, reducing to the previous one by choosing a basis of $\mathbb F_q$.

One can see by the description as a space of functions that there is a ring structure, coming from the ring structure on functions, but I don't think you will find a natural ring structure on this (topological) group.

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