TITLE: Example of non-isomorphic structures which are elementarily equivalent
QUESTION [21 upvotes]: I just started learning model theory on my own, and I was wondering if there are any interesting examples of two structures of a language L which are not isomorphic, but are elementarily equivalent (this means that any L-sentence satisfied by one of them is satisfied by the second).
I am using the notation of David Marker's book "Model theory: an introduction".

REPLY [47 votes]: First, I'm glad you are reading my book! :)
Let me make a couple of comments on Pete's answer--this is my first time here and I don't see how to leave comments.

Any two dense linear orders without endpoints are elementarily equivalent.
In particular $(Q,<)$ and $(R,<)$ are elementarily equivalent.  So there is no first order way of expressing the completeness of the reals.

Any two algebraically closed fields of the same characteristic are elementarily equivalent.  So the algebraic numbers
is elementarily equivalent to the complex numbers.  This means you can prove first order things about the algebraic numbers using complex analysis or the complex numbers using Galois Theory or countability.

Similarly the reals field is elementarily equivalent to the real algebraic numbers or to the field of real Puiseux series.
One can for example use the Puiseux series to prove asymptotic properties of semialgebraic functions.


Finally, Pete's comment 5) about infinite models of the theory of finite fields being elementarily equivalent isn't quite right.  This is only true if the relative algebraic closure of the prime fields are isomorphic.
For example,
a) take an  ultrapower of finite fields $F_p$ where the ultrafilter containing $\{2,4,8,\ldots\}$ then the resulting model has
characteristic $2,$ while if the ultrafilter contains the set of primes, then the ultraproduct has characteristic $0.$
b) if the ultrapower contains the set of primes congruent to $1 \bmod 4,$ in the ultraproduct $-1$ is a square, while if
the ultraproduct contains the set of primes congruent to $3 \mod 4$ then in the ultraproduct $-1$ is not a square..