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I am interested to know if there is a standard name for the inverse limit, $\hat{\mathbb{Z}}_!$, say, of the inverse system of rings $$\ldots \rightarrow \mathbb{Z}/((n+1)!)\mathbb{Z} \rightarrow \mathbb{Z}/(n!)\mathbb{Z} \rightarrow \ldots$$

$\hat{\mathbb{Z}}_!$ stands in the same relationship to the factorial system of representations for the integers as the ring $\hat{\mathbb{Z}}_p$ of $p$-adic integers has to the base $p$ system. It has each $\hat{\mathbb{Z}}_p$ as a quotient. Am I correct in thinking that $\hat{\mathbb{Z}}_!$ is not the same as the profinite completion of the integers (i.e., the inverse limit of the inverse system comprising all quotient rings of $\mathbb{Z}$), since the latter is not an integral domain?

Any pointers to references would be appreciated.

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2 Answers 2

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I've seen this inverse limit called the $\mathbb Z$-adic completion of $\mathbb Z$. Unless I'm very confused, this is the same as (i.e., canonically isomorphic to) the inverse limit of all the finite quotients of $\mathbb Z$, because the sequence of quotients in your question is cofinal (or coinitial, depending on which direction you're facing) in the inverse system of all finite quotients. Also, unless I'm even more confused, this ring is (canonically isomorphic to) the direct product, over all primes $p$, of the $p$-adic completions $\hat{\mathbb Z}_p$.

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  • $\begingroup$ The profinite completion is isomorphic to the product of the $\hat{\mathbb{Z}}_p$ and hence is not an integral domain. Maybe I am wrong in thinking that $\hat{\mathbb{Z}}_!$ is an integral domain. $\endgroup$
    – Rob Arthan
    Jul 29, 2013 at 15:51
  • $\begingroup$ No, it is not an integral domain. $\endgroup$ Jul 29, 2013 at 15:53
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Let $\hat{\mathbb Z}$ be the inverse limit of $\{\mathbb Z/n\}$ with the natural maps $\mathbb Z/n\to \mathbb Z/d$ when $d|n$.

Then give any sequence of numbers $n_1,n_2,\dots$, such that $n_i|n_{i+1}$ and for every $n$, there exists $i$ such that $n|n_i$, we have that the inverse limit of the chain:

$$\mathbb Z/n_1\leftarrow \mathbb Z/n_2\leftarrow \dots$$

is isomorphic to $\hat{\mathbb Z}$.

This is a general property of inverse limits.

More generally, if $n_1\mid n_2\mid n_3\dots$ then you can define for each prime $p$ the value $k_p=\sup\{k|\exists i:p^k|n_i\}$. Then the inverse limit of $\mathbb Z/n_i$ is $\prod_p\mathbb Z/p^{k_p}$ where, by $\mathbb Z/p^{\infty}$ we mean the $p$-adic integers.

Another way to see it in this case is in the isomorphism $\mathbb Z/n \cong \prod_p \mathbb Z/p^{\nu_p(n)}$. Clearly, this factorization agrees when $d|n$ - the map $\mathbb Z/n\to \mathbb Z/d$ corresponds to the product of the maps $\mathbb Z/p^{\nu_p(n)}\to\mathbb Z/p^{\nu_p(d)}$. So the inverse limit is the product of the limits of the $\mathbb Z/p^{\nu_p(n_i)}$.

You can find zero divisors in your inverse limit easily enough. Fix $p$, and for each $n$, use Chinese remainder theorem to find $a_n\in\mathbb Z/n!$ so that $a_n\equiv 1\pmod{p^{\nu_p(n!)}}$ and $a_n\equiv 0\pmod {n!/p^{\nu_p(n!)}}$. $a_n$ is unique, and $a_{n+1}\to a_n$ when you map $\mathbb Z/(n+1)!\to \mathbb Z/n!$. Since each $a_n$ is an idempotent in $\mathbb Z/n!$, the sequence $(a_n)$ is an idemptotent in the inverse limit, and it is non-zero. So $(a_n)(1-a_n)=0$ in the inverse limit.

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