Let $\widehat{\mathbb{Z}}[[\widehat{\mathbb{Z}}]] := \varprojlim\_{n,m}(\mathbb{Z}/n)[x]/(x^m-1)$ be the complete group algebra of the profinite free group of rank 1. In Corollary 5.9.2 of Ribes-Zalesski's *Profinite Groups*, they state that $\widehat{\mathbb{Z}}[[\widehat{\mathbb{Z}}]]\cong \widehat{\mathbb{Z}}[[t]]$, citing a paper of Lim which doesn't seem to prove exactly what they claim (though I'm sure it must follow, if one is sufficiently familiar with the theory).


Here, let's try giving $\widehat{\mathbb{Z}}[[t]]$ the topology corresponding to the product topology on $\prod\_{n\ge 0}\widehat{\mathbb{Z}}$, indexed by the coefficients of $t^n, n\ge 0$.


I would like to show directly that $\widehat{\mathbb{Z}}[[\widehat{\mathbb{Z}}]]\cong \widehat{\mathbb{Z}}[[t]]$. If this is true, then the map should be given by identifying a generator "$x$" of the group algebra with $1+t$.


Relative to the (product) topology described above on $\widehat{\mathbb{Z}}[[t]]$, a neighborhood basis of 0 is given by the ideals $(n,t^m)$ for $n,m\ge 1$. For every such ideal, one can find some $N,M\ge 1$ such that the map "$x\mapsto 1+t$" defines a quotient map
$$\widehat{\mathbb{Z}}[[\widehat{\mathbb{Z}}]]\rightarrow(\mathbb{Z}/N)[x]/(x^M-1) \rightarrow \widehat{\mathbb{Z}}[t]/(n,t^m)$$
(this follows from divisibility properties of binomial coefficients). However when I try to go in the other direction, it seems what I need is - for every $N,M\ge 1$, to find an $n,m$ such that $t\mapsto x-1$ induces a map
$$\widehat{\mathbb{Z}}[t]/(n,t^m)\rightarrow(\mathbb{Z}/N)[x]/(x^M-1) $$
However, this is clearly impossible, since $t$ is always nilpotent on the left, and yet $x-1$ is rarely nilpotent on the right.


Thus, either the "product" topology on $\widehat{\mathbb{Z}}[[t]]$ is not sufficient, or the map is more tricky than just sending "$x\mapsto 1+t$".


On the other hand, $\widehat{\mathbb{Z}}[[\widehat{\mathbb{Z}}]]$ is a commutative profinite ring, and hence a product of commutative profinite local rings. It certainly admits $\mathbb{Z}\_p[[\mathbb{Z}\_p]]$ as quotients for all primes $p$, which is known to be isomorphic to the local ring $\mathbb{Z}\_p[[t]]$, so it seems hard to imagine any other possibility for $\widehat{\mathbb{Z}}[[\widehat{\mathbb{Z}}]]$.


If it turns out the titular question has a negative answer, then naturally question is:


$$\text{What are the local direct factors of the profinite ring $\widehat{\mathbb{Z}}[[\widehat{\mathbb{Z}}]]$?}$$



Is $\widehat{\mathbb{Z}}[[t]]\cong\widehat{\mathbb{Z}}[[\widehat{\mathbb{Z}}]]$?
There may be some things to check here, but I think the following is correct and should answer your last question.


$\newcommand{\ZZ}{\mathbb{Z}}$
$\newcommand{\Zhat}{\widehat{\mathbb{Z}}}$


I believe the final result is
$$\Zhat[[\Zhat]] \cong \prod\_q\ZZ\_q[[t]]$$
as $q$ ranges over all prime powers $p^r$ with $r$ coprime to $p$, each appearing $N\_{p,r}$ times, where $N\_{p,r}$ is the number of Frobenius orbits of generators of $\mathbb{F}\_q^\times$


Firstly, $\Zhat[[\Zhat]]\cong\prod\_p\ZZ\_p[[\Zhat]]$. To see this, note that in every $(\ZZ/n)[x]/(x^m-1)$, the distinct prime powers dividing $n$ generate comaximal ideals, and hence if $n = \prod\_i p\_i^{r\_i}$ then 
$$(\ZZ/n)[x]/(x^m-1)\cong\prod\_i(\ZZ/p\_i^{r\_i})[x]/(x^m-1)$$


This decomposition at every finite stage should extend to the limit, and so it suffices to analyze $\ZZ\_p[[\Zhat]]$. Let
$$\ZZ' :=\prod\_{p'\text{ prime}\\p'\ne p}\ZZ\_{p'}$$
then since $\Zhat = \ZZ\_p\times\ZZ'$, and since $\ZZ\_p[A\times B] = \ZZ\_p[A]\hat{\otimes}\ZZ\_p[B]$ (completed tensor product over $\ZZ\_p$), we have:
$$\ZZ\_p[[\Zhat]] = \ZZ\_p[[\ZZ\_p]]\hat{\otimes}\ZZ\_p[[\ZZ']] = \ZZ\_p[[t]]\hat{\otimes}\ZZ\_p[[\ZZ']]$$
By definition,
$$\ZZ\_p[[\ZZ']] = \varprojlim\_m \ZZ\_p[x]/(x^m-1)$$
where $m$ is coprime to $p$. Since such $x^m-1$ factors into distinct irreducibles mod $p$, by Hensels lemma we find that $x^m-1 = \prod\_i f\_{m,i}$, where each $f\_{m,i}\mid x^m-1$ and is irreducible in both $\ZZ\_p[x]$ and $\mathbb{F}\_p[x]$. These $f\_{m,i}$ are actually pairwise comaximal (see [this answer](https://math.stackexchange.com/questions/2133889/do-any-two-coprime-factors-of-xn-1-over-the-p-adic-integers-mathbbz-p/2133983#2133983)), and hence we have a decomposition
$$\ZZ\_p[x]/(x^m-1) = \prod\_i \ZZ\_p[x]/(f\_{m,i})$$
where each term in the product is now isomorphic to $\ZZ\_q$, where $q = p^{\deg f\_{m,i}}$ (ie, the unique domain unramified over $\ZZ\_p$ of degree $\deg f\_{m,i}$) Taking the limit over all $m$ coprime to $p$, we get:
$$\ZZ\_p[[\ZZ']] = \prod\_f\ZZ\_p[x]/(f)$$
where $f$ ranges over the set:
$$\{f\in\ZZ\_p[x] \text{ irreducible} : \exists m\text{ coprime to $p$ such that } f\mid x^m-1\}$$
Thus, we get
$$\ZZ\_p[[\Zhat]] = \ZZ\_p[[t]]\hat{\otimes}\prod\_f\ZZ\_p[x]/(f)$$
By Proposition 7.7.5 in Wilson's book *Profinite Groups*, the completed tensor product commutes with *arbitrary* direct products, so we get
$$\ZZ\_p[[\Zhat]] = \prod\_f(\ZZ\_p[[t]]\hat{\otimes}\ZZ\_p[x]/(f))$$
Since each $\ZZ\_p[x]/(f)$ is a finite $\ZZ\_p$-algebra, the completed tensor product coincides with the usual tensor product (c.f. Ribes-Zalesski prop 5.5.3(d)), so we get
$$\ZZ\_p[[\Zhat]] = \prod\_f(\ZZ\_p[[t]]\otimes\ZZ\_p[x]/(f)) = \prod\_f (\ZZ\_p[x]/(f))[[t]] = \prod\ZZ\_q[[t]]$$
as $q$ ranges over all prime-to-$p$ powers of $p$, each appearing multiple times in the product.


Thus, I believe we have an isomorphism
$$\Zhat[[\Zhat]] \stackrel{\varphi}{\longrightarrow}\prod\_q\ZZ\_q[[t]] = \left(\prod\_q\ZZ\_q\right)[[t]]$$


For every $r$ coprime to $p$, the number of times each $\ZZ\_q = \ZZ\_{p^r}$ appears in the product is precisely the number $N\_r$ of irreducible degree $r$ factors of $x^{p^r-1}-1$ over $\mathbb{F}\_p$, or equivalently the number of $p$-power Frobenius orbits of generators of $\mathbb{F}\_q^\times$. Let $a\in\Zhat$ come from within the $[[\cdots]]$ of $\Zhat[[\Zhat]]$. Then, $\varphi(a) = a'(t+1)^{a\_p}$, where $a\_p$ is the image of $a$ in $\ZZ\_p$, and $a'\in\prod\_{p'\ne p}\ZZ\_{p'}$ is defined as follows. For every $q = p^r$ with $(r,p) = 1$, let $\overline{a}$ be the residue of $a$ mod $p^r-1$. Then, the images of $a'$ in the $N\_r$ copies of $\ZZ\_q$ are in bijection with a complete set of representatives of the Frobenius orbits of primitive $(q-1)$th roots of unity in $\ZZ\_q$, each raised to the $\overline{a}$-th power.


In particular, when $r = 1$, we find that the number of copies of $\ZZ\_p[[t]]$ in the product is precisely the number of linear factors of $x^{p-1}-1$ over $\mathbb{F}\_p$, which is precisely $p-1$. Thus, the projections onto each of these factors, followed by quotienting by $(t)$, yields the $p-1$ homomorphisms onto $\ZZ/p\ZZ$ referred to by Tom Goodwillie, and it seems like there can't be any others.


