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Recall that for a set $x$ its rank $\alpha$ is the least ordinal such that $x \in V_{\alpha+1}$. Or in other words: $x$ is built up out of $\alpha$ levels of braces and the empty set.

I think with the usual constructions of numbers (cartesian products, sets of equivalence classes, Dedekind cuts, etc.), we have

$rank(\mathbb{N})=\omega, rank(\mathbb{Z})=\omega+4, rank(\mathbb{Q})=\omega+8, rank(\mathbb{R})=\omega+10$

Now it is possible to find a bijection $\mathbb{Z} \cong \mathbb{N}$, so that there is a copy of $\mathbb{Z}$ of smaller rank, namely $\omega$. But this is, of course, nonsense. We should also consider the ring structure on $\mathbb{Z}$, which is given by two maps $\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$. Therefore we might ask the following:

Is there a ring $(R,+,\*)$, which is isomorphic to $(\mathbb{Z},+,\*)$, but $rank(R,+,\*) < rank(\mathbb{Z},+,\*)$? What about the other rings above?

Of course, this question is just out of curiosity. I doubt that anybody cares about these bounds of ranks (if not, please let me know).

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When computing the "rank of a structure" such as N, do you mean the rank of the domain N, or the rank of the ordered tuple (N, +, times) of the domain plus the operations? The rank of the latter will be greater than omega. – John Goodrick Sep 3 '10 at 15:27
Yes. Only in the first paragraphs I meant the sets, but then the structures (which should be clear from the notation). – Martin Brandenburg Sep 3 '10 at 16:20

If you want to have a copy of $(\mathbb Z,+,\cdot)$ of low rank the problem is not the domain. Here you can choose any infinite subset of $V_\omega$, which would then be in $V_{\omega+1}$. You cannot get below that. $V_\omega$ is closed under pairs, so with this incarnation of $\mathbb Z$, $\mathbb Z\times\mathbb Z$ and $(\mathbb Z\times\mathbb Z)\times\mathbb Z$ are still in $V_{\omega+1}$ (i.e., of rank $\omega$), and so are $+$ and $\cdot$ (as sets of triples of integers).
What drives up the rank of $(\mathbb Z,+,\cdot)$ now is our definition of pair (Kuratowski pair). There are alternative pairs ("flat" pairs) that play a role in type theory which don't drive up the rank as much (Randall Holmes told me about flat pairs). But if you insist on the usual notion of pair, and you really want to have the object $(\mathbb Z,+,\cdot)$ with some countable set for the integers, then you cannot get below the rank of my construction. Note that implementing the triple $(\mathbb Z,+,\cdot)$ is done of lowest rank by a map from $3$ into the universe, rather than by iterating the pair construction.

So, the main trick that I see is not to use the infinite equivalence classes in the case of countable structures but rather any countably infinite set of rank $\omega$. Similarly, for uncountable structures, use a set of the appropriate size of lowest possible rank for the domain. Then everything else basically falls into place, except for some playing around with how to implement pairs, triples, and so on. In particular, all countable rings have isomorphic copies of the same minimal rank (with domains of rank $\omega$).
The reals have an isomorphic copy with a domain of rank $\omega+1$, for instance the set $V_{\omega+1}$. Unfortunately, $V_{\omega+1}$ is not closed under pairs, so you get a larger rank for the operations on the reals.

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