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Good evening,

In model theory there is a notion of Grothendieck ring defined here http://math.berkeley.edu/~scanlon/papers/greu12jun00.pdf. Do we know anything about the cardinality of these rings ? Can they be finite ?

If a structure M had a Grothendieck ring of cardinality say 2 this means that:

  1. there is a definable function on $M$ injective and whose image is $M$ minus two points.

  2. There is a definable function between $M$ and $M^2$.

(3. $M$ is infinite)

If we take a structure satisfying this, then its Grothendieck ring is either of cardinality 2 or trivial.

So in order to prove that there exists Grothendieck ring of cardinality say 2, we have to prove that there exists such a $M$ whose the Grothendieck ring is non trivial which is equivalent to the fact that there is no definable set in bijection with itself minus a point.

Could you have any suggestion of how to see that ?

Thank you in advance !

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The Grothendieck ring of $M$ has two elements iff on the one hand, for every definable subset $A\subseteq M^n$, there is a definable bijection from $A$ onto some $A'\subseteq M^m$, a definable set $B\subseteq M^m$ such that $|A'\cap B|\le1$, and a definable bijection from $B$ onto $B\cup A'$, and on the other hand, there is no definable bijection from some $A\subseteq M^n$ onto $A-\{a\}$, where $a\in A$. I don’t see how this implies conditions 1 and 2, or vice versa. –  Emil Jeřábek Nov 18 '13 at 16:20
    
Anyway, the question is interesting. –  Emil Jeřábek Nov 18 '13 at 16:24
    
Notice that if $M$ is a model of PA, $a\in M$, and $M_a$ is the structure with domain $[0,a]$ with all relations coded in $M$, then the Grothendieck ring of $M_a$ consists of the subsemiring $\{x\in M:\exists n\in\mathbb N\,x\le a^n\}\subseteq M$ together with the corresponding negative elements. This means that the Grothendieck ring can have an arbitrary infinite cardinality, so the question is really only about finite cardinalities. –  Emil Jeřábek Nov 18 '13 at 17:29

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