Is $\widehat{\mathbb{Z}}[[t]]\cong\widehat{\mathbb{Z}}[[\widehat{\mathbb{Z}}]]$?

Question

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:

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

Answer 1

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), 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.

Answer 2

It seems to me that for $p$ prime $\hat{\mathbb Z}[[t]]$ has exactly one continuous ring homomorphism to $\mathbb Z/p\mathbb Z$, while $\hat{\mathbb Z}[[\hat{\mathbb Z}]]$ has exactly $p-1$ such homomorphisms.

Answer 3

There is no continuous surjection from $\hat Z[[t]]$ to $(\mathbb Z/3\mathbb Z)[\mathbb Z/2\mathbb Z] = \mathbb Z/3\mathbb Z \oplus \mathbb Z/3\mathbb Z$.