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Aug 25, 2021 at 14:03 comment added Emil Jeřábek @ErikWalsberg I didn’t know about this paper, thanks for mentioning it.
Aug 24, 2021 at 4:58 comment added Erik Walsberg There is also a description of definable sets in arbitrary oag's, see the paper "Quantifier elimination in ordered abelian groups". You can eliminate quantifiers down to very special kinds of quantifiers that take some work to define.
Aug 23, 2021 at 22:23 comment added Erik Walsberg A lot more oag's have qe if you add predicates defines the set of nth multiples for all n, for example any archimedean ordered abelian group has QE in this language.
Aug 23, 2021 at 22:19 comment added Erik Walsberg @EmilJeřábek Any ordered abelian group with quantifier elimination in the language of ordered abelian groups is divisible. Proof: It's easy to see that QE implies o-minimality, and it is well known and easy to see that o-minimality implies divisibility.
Aug 20, 2021 at 8:18 vote accept Sh.M1972
Aug 20, 2021 at 8:01 comment added Emil Jeřábek @Sh.M1972 I don’t think that’s true in general. There is quantifier elimination of this form for divisible ordered abelian groups.
Aug 20, 2021 at 5:00 history became hot network question
Aug 20, 2021 at 3:19 comment added Sh.M1972 @Ycor yes this is correct, but at least we know that in such a group, every formula $\varphi(\overline{x})$ (in the language of groups)is equivalent to a quantifier-free formula $\psi(\overline{x})$ (possibly containing $<$) and this is also good.
Aug 19, 2021 at 21:45 comment added YCor "ordered abelian groups" is not a class of abelian groups, it's an additional structure. If one only assumes the existence of such a structure without fixing it, one gets orderable abelian groups, which are precisely torsion-free abelian groups.
Aug 19, 2021 at 21:20 answer added Emil Jeřábek timeline score: 14
Aug 19, 2021 at 20:58 history asked Sh.M1972 CC BY-SA 4.0