Consider ${\bf Z}$ as a first-order structure in the language $\{0,+,-\}$ of abelian groups.
Is ${\bf Z} \times {\bf Z}$ elementarily equivalent to ${\bf Z}$ and why?
I think they are not elementarily equivalent.
Is ${\bf Z} \times {\bf Z}$ elementarily equivalent to ${\bf Z}$?
-1
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logic
model-theory
1 Answers
6
They are not elementarily equivalent.
In ${\bf Z}$ we have $\exists x \forall y (\exists z (z + z = y \vee z + z + x = y))$
Namely, $x = 1$.
In ${\bf Z} \times {\bf Z}$, no such $x$ is possible. Specifically, there are four classes (odd, odd), (even, even), (even, odd), and (odd, even), and no x covers the classes that aren't (even, even).