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Let $\hat{\mathbb{Z}}$ be the profinite completion of $\mathbb{Z}$. Since $\hat{\mathbb{Z}}$ is the inverse limit of the rings $\mathbb{Z}/n\mathbb{Z}$, it's a subgroup of $\prod_n \mathbb{Z}/n\mathbb{Z}$. So we can represent elements in $\hat{\mathbb{Z}}$ as a subset of all possible tuples $(k_1,k_2,k_3,k_4,k_5,...)$, where each $k_n$ is an element in $\mathbb{Z}/n\mathbb{Z}$. The precise subset of such tuples which corresponds to $\hat{\mathbb{Z}}$ is given by the usual definition of the inverse limit.

There is a canonical injective homomorphism $\eta: \mathbb{Z} \to \hat{\mathbb{Z}}$, such that to each $z \in \mathbb{Z}$ corresponds the tuple $\text{(z mod 1, z mod 2, z mod 3, ...)}$. However, it is well known that this homomorphism is not surjective, meaning there exist elements in $\hat{\mathbb{Z}}$ which do not correspond to anything in $\mathbb{Z}$.

Does anyone know how to explicitly construct an example of such an element in $\hat{\mathbb{Z}}$, which isn't in the image of the homomorphism $\eta$, and to represent it as a tuple as outlined above?

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    $\begingroup$ You can play with something $p$-adic. e.g. take your favorite $p$-adic integer $a$, and consider the tuple where the $n$-th coordinate is 0 if $n$ is not prime power, and $a$ mod $p^k$ if $n = p^k$. $\endgroup$
    – user27126
    Dec 12, 2012 at 1:26
  • $\begingroup$ I had considered this exact thing earlier today, but I don't think it's going to work. For instance, say we construct some 2-adic number by setting all nth coefficients, where n is a power of 2, to 1, and all other coefficients to 0. So the sixth coefficient would be 0, but the second coefficient would be 1. However, via the homomorphism $\mathbb{Z}/6\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}$ that's part of the inverse system, the sixth coefficient being 0 should imply the second coefficient is also 0. Since it's not, this tuple isn't actually in the inverse limit. $\endgroup$ Dec 12, 2012 at 7:30
  • $\begingroup$ Sorry, I was careless. This would probably work - again take your favorite $p$-adic $a$, and consider the tuple where the $n = \prod p_i^{k_i}$-th coordinate, considered as an element of $\prod \mathbb{Z}/p_i^{k_i}\mathbb{Z}$, corresponds to $(a,0,\cdots,0)$ where $a$ is the part that corresponds to powers of $p$, and $0$ for other parts. $\endgroup$
    – user27126
    Dec 12, 2012 at 9:09
  • $\begingroup$ I don't understand your comment. In $\prod {p_i}^{k_i}$, what is the product being taken over? $\endgroup$ Dec 13, 2012 at 1:12
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    $\begingroup$ You could also use something like $\sum n!$ that converges $p$-adically regardless of the value of $p$. It seems like this would make it easier to write down all the details... $\endgroup$
    – Micah
    Dec 13, 2012 at 9:44

3 Answers 3

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The Chinese Remainder Theorem tells you that if $n=\prod_pp^{e(p)}$, where the product is taken over only finitely many primes, and each $p$ appears to the power $e(p)$ in $n$, Then $\mathbb Z/n\mathbb Z$ is isomorphic to $\bigoplus(\mathbb Z/p^{e(p)}\mathbb Z)$. This direct sum is also direct product, and when you take the projective limit, everything in sight lines up correctly, and you get this wonderful result: $$ \projlim_n\>\mathbb Z/n\mathbb Z\cong\prod_p\left(\projlim_m\mathbb Z/p^m\mathbb Z\right)\cong\prod_p\mathbb Z_p\>. $$ Thus to hold and admire a non-$\mathbb Z$ element of $\hat{\mathbb Z}$, all you need is any old collection of $p$-adic integers.

Given the isomorphism $\hat{\mathbb Z}\cong\prod_p\mathbb Z_p$, because the operations addition and multiplication go componentwise, $\hat{\mathbb Z}$ must have lots of zero divisors. E.g. if $e_2 := (1,0,0,...)\ne 0$ with the $1$ in $\mathbb Z_2$ and $e_3 := (0,1,0,...) \ne 0$ with the $1$ in $\mathbb Z_3$ then $e_2\cdot e_3 = (1\cdot 0,0\cdot 1,0\cdot 0,...)=0\in \mathbb Z$. Of course, $e_2$ and $e_3$ cannot be members of $\mathbb Z$, because the latter does not contain zero divisors (and both cannot be "finite").

@Mike Battaglia: To your question as of Dec 12 '12 at 7:30, it seems to me that two isomorhisms are mixed up: first the isomorhism $\hat{\mathbb Z}\cong\prod_{p\in\mathbb P}\mathbb Z_p$, where you can freely chose 2-adic, 3-adic etc numbers and build a profinite integer being congruent to all these freely chosen components, and second the inclusion $\hat{\mathbb Z}\subset\prod_{n\in\mathbb N}\mathbb Z/n\mathbb Z$, where you have in addition to observe the compatibility conditions as you mention them: $\mathbb Z/6\mathbb Z\to \mathbb Z/2\mathbb Z$. – Herbert Eberle

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  • $\begingroup$ Hrm. I like this representation much better than factorial numerals. I wonder why I usually see $\hat{\mathbb{Z}}$ described in terms of quotients by factorials? Does it have to do with the topology? Anyways, +1 $\endgroup$
    – user14972
    Dec 25, 2012 at 18:32
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    $\begingroup$ I can’t answer your question, but I first saw $\hat{\mathbb Z}$ as the Galois group associated to a finite field and its algebraic closure. Then the $\mathbb Z_p$-factors are the $p$-Sylow subgroups. In that context, the representation I pointed out is very natural. $\endgroup$
    – Lubin
    Dec 26, 2012 at 2:27
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    $\begingroup$ Coming back to this, is it possible to give something like a "prime factorization" to a profinite integer, but perhaps allowing infinite prime exponents or infinitely many nonzero exponents, etc? $\endgroup$ Sep 23, 2015 at 2:09
  • $\begingroup$ I’m not sure what “infinite prime exponents” means. On the other hand, the nature of a product is that infinitely many of the components may fail to be the identity element. $\endgroup$
    – Lubin
    Sep 23, 2015 at 14:05
  • $\begingroup$ Lublin, for instance, see something like this - en.m.wikipedia.org/wiki/Supernatural_number $\endgroup$ Sep 24, 2015 at 15:44
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You can think of a presentation of an element of $\hat{\mathbb{Z}}$ by a tuple $(a_1,a_2,a_2,\dots)$ as a description of an "ideal" integer's residues mod $1, 2, 3, \dots$

If you're looking at $\prod_n \mathbb{Z}/n\mathbb{Z}$, which is the limit of the diagram consisting of the rings $\mathbb{Z}/n\mathbb{Z}$ with no connecting maps, you're allowed to choose $a_1, a_2, a_3,\dots$ totally arbitrarily.

But the diagram of which $\hat{\mathbb{Z}}$ is the limit enforce restrictions, and these restrictions are exactly the finite implications between residues which exist in $\mathbb{Z}$, i.e. if $x\equiv 4$ (mod 6), then $x\equiv 1$ (mod 3).

By the Chinese Remainder Theorem, the residue of an integer mod $a$ is entirely determined (according to these restrictions) by its residues mod $p_1^{r_1}, \dots, p_k^{r_k}$, where these are the prime powers appearing in $a$.

So all you need to do to explicitly determine an element of $\hat{\mathbb{Z}}$ is to give a consistent choice of residues mod all prime powers. Then for each other integer, compute what the residue should be. It is easy to do this in a way that is not satisfied by any element of $\mathbb{Z}$.

Example: Let's make our element divisible by all powers of odd primes, but give it residue $1$ modulo all powers of $2$. Then it starts

$(0,1,0,1,0,3,0,1,0,5,0,9,\dots)$

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  • $\begingroup$ I'd also like to point out Lubin's answer explains precisely what I mean when I say "all you need to do... is to give a consistent choice of residues mod all prime powers." This is just a choice of a p-adic integer for each prime p. $\endgroup$ Dec 25, 2012 at 4:41
  • $\begingroup$ So in the representation of this group as the direct product of p-adics, this corresponds to $(1_2,0_3,0_5,0_7,0_11,...),$ right? $\endgroup$ Sep 23, 2015 at 2:11
  • $\begingroup$ @MikeBattaglia That's right. $\endgroup$ Sep 23, 2015 at 3:30
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It may be more convenient to simplify the limit to be a linear chain of quotient maps, such as

$$ \cdots \to \mathbb{Z} / 5! \mathbb{Z} \to \mathbb{Z} / 4! \mathbb{Z} \to \mathbb{Z} / 3! \mathbb{Z} \to \mathbb{Z} / 2! \mathbb{Z} \to \mathbb{Z} / 1! \mathbb{Z} \to \mathbb{Z} / 0! \mathbb{Z}$$

and so it suffices to represent an element of $\hat{\mathbb{Z}}$ by a sequence of residues modulo $n!$ such that $$s_{n+1} \equiv s_{n} \pmod{n!}$$ In this representation, an easy-to-construct element not contained in $\mathbb{Z}$ is the sequence $$s_n = \sum_{i=0}^{n-1} i! $$ It may be interesting to think of this as the infinite sum $$s = \sum_{i=0}^{+\infty} i!$$ which makes sense in the representation you use too, since it's a finite sum in every place.

I suppose the elements of $\hat{\mathbb{Z}}$ should be in one to one correspondence with the left-infinite numerals in the factorial number system

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  • $\begingroup$ Of course, $\:\: \mathbb{Z}/1!\mathbb{Z} = \mathbb{Z}/0!\mathbb{Z} \;\;$. $\;\;\;\;$ $\endgroup$
    – user57159
    Sep 23, 2013 at 1:16
  • $\begingroup$ Wouldn't this just be the tuple corresponding to -1? Unless I've misunderstood your construction, adding (...1,1,1,1,1) to your number yields zero. $\endgroup$ Sep 24, 2015 at 15:42

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