Elements in $\hat{\mathbb{Z}}$, the profinite completion of the integers 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?
 A: 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
A: 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
$(0,1,0,1,0,3,0,1,0,5,0,9,\dots)$
A: 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
