Are different constructions of an algebraic structure always isomorphic?

Question

Any two complete ordered fields are isomorphic (as proved, e.g., in Spivak's Calculus; see also this question).

While I understand this proof, I cannot yet appreciate why it is necessary. Given any set of axioms for some algebraic structure, is it not always the case that any two constructions satisfying those axioms are isomorphic? If not, can someone supply a counterexample?

Answer 1

Usually we can't assume that a set of axioms only has one model up to isomorphism. In fact, if a set of first-order axioms has just one infinite model, then it is a consequence of the Löwenheim–Skolem theorem that it has models of all infinite cardinalities. Since all isomorphisms are bijections, it follows that there must be two non-isomorphic models. The reason why this doesn't break down for complete ordered fields is that completeness is a second-order axiom: it quantifies over infinite sets of variables ('any bounded increasing sequence has a least upper bound').

However, there are sets of axioms whose models of a given size are all isomorphic. This phenomenon is known as categoricity. An example of this is unbounded dense linear orders, whose theory is countably categorical (i.e. all its countably infinite models are isomorphic). In other words, if $M$ is a countable set endowed with a dense linear order $<_M$, such that $M$ has no $<_M$-least or $<_M$-greatest element, then $(M,<_M) \cong (\mathbb{Q}, <)$.

Categoricity is a special property held by certain sets of axioms, but certainly not all sets of axioms. Most of the time, you cannot expect categoricity. (E.g. two given groups of a given cardinality, especially when infinite, are typically non-isomorphic.)

Answer 2

Consider, for example, the group axioms. It is not the case that any two structures that satisfy the axioms are isomorphic -- in other words, not all groups are isomorphic.

It is a quite unusual property of the axioms for complete ordered fields that they allow exactly one realization up to isomorphism.

Answer 3

The complete ordered field example "cheats", because it implicitly invokes set theory in addition to ordered field theory.

You may find the theory of real closed fields to be interesting: it is the theory of complete ordered fields without "cheating", and there are many non-isomorphic examples. The smallest example is the field of all numbers that are both real and algebraic.

Answer 4

Nope.

Suppose your only axiom for an algebraic structure with a binary operation * is

1. $\forall$x $\forall$y (x*y)=(y*x).

Two constructions (if I've understood your term correctly) which satisfy this axiom are:

1  a  b
a  a  b
b  b  a

and

2  a  b
a  b  b
b  b  a

But clearly these structures are not isomorphic.