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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 one can ask:

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

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:

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

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 $\widehat{\mathbb{Z}}[[\widehat{\mathbb{Z}}]]$.

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 one can ask:

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

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, I think we want $\widehat{\mathbb{Z}}[[t]]$ to be given the topology corresponding to the product topology on $\prod_{n\ge 0}\widehat{\mathbb{Z}}$, indexed by the coefficients of $t^n, n\ge 0$, though I could be mistaken.

Anyhow, 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 $1+t$.

However, one difficulty is that I don't know of a neighborhood basis of open ideals of $\widehat{\mathbb{Z}}[[t]]$. At first I thought the ideals $(n,t^m)$ for $n,m\ge 1$ should suffice, and indeed for every such ideal, one can find some $N,M\ge 1$ such that the map "$x\mapsto 1+t$" defines a quotient map $$\widehat{\mathbb{Z}}[[\widehat{\mathbb{Z}}]]\rightarrow(\mathbb{Z}/N)[x]/(x^M-1) \rightarrow \widehat{\mathbb{Z}}[t]/(n,t^m)$$ 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, I guess the ideals $\{(n,t^m)\}_{n,m\ge 1}$ do not form a neighborhood basis of 0 in $\widehat{\mathbb{Z}}[[t]]$, which leaves me stuck...

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 $\widehat{\mathbb{Z}}[[\widehat{\mathbb{Z}}]]$.