# Elementary embedding and elementary equivalence [duplicate]

Elementary embedding implies elementary equivalence, but elementary equivalence between structures does not implies, necessarily, the existence of an elementary embedding between these structures. I`m looking for an example of two structures that are elementary equivalent but there is no elementary embedding between them.

any ideas?

Thanks!

## marked as duplicate by tomasz, Amzoti, rschwieb, Dan Rust, Lord_FarinJun 14 '13 at 17:27

Consider the theory of equivalence relations with exactly two infinite equivalence classes, which is a complete theory, so every pair of models is elementarily equivalent. Let $M$ be a model with classes of size $\aleph_0$ and $\aleph_2$ and let $N$ be a model with classes of size $\aleph_1$ and $\aleph_1$. Then $M$ cannot embed in $N$ because a class of size $\aleph_2$ cannot embed in a class of size $\aleph_1$, and $N$ cannot embed in $M$ for a similar reason. So there is no embedding at all, much less an elementary embedding.
Consider the following two fields (as structures in the language $(0,1,+,\cdot)$): the real numbers $\Bbb{R}$ on the one hand, and on the other some countable nonarchimidean real-closed field $F$. They're both real-closed, so they're elementarily equivalent. But $\Bbb{R}$ doesn't embed in $F$ (due to cardinality) and $F$ doesn't embed in $\Bbb{R}$ (due to not being archimedean).