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Conjugacy of trivial autohomeomorphisms of $\beta N\setminus N$.
Abstract
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Is Fp((Q)) NTP2?
Abstract
7 years ago, also in Oxford, Sylvy Anscombe and I asked this question, which is part of the general effort to try and understand the model theory of henselian valued fields through dividing lines. In 2024, Sylvy Anscombe and Franziska Jahnke completely classified NIP henselian valued fields. Their methods can be extended, with the help of works of Chernikov, Kaplan and Simon and of Kuhlmann and Rzepka, to NTP2 henselian valued fields, obtaining the following:
- if a henselian valued field is NTP2, then it is semitame and its residue field is NTP2;
- if a henselian valued field is separably algebraically maximal Kaplansky and its residue field is NTP2, then it is NTP2.
This covers a large class of fields, but there is still a gap. Notably, Fp((Q)) is in the middle: it is semitame but not Kaplansky.
To answer this question, we studied so called tame henselian fields with finite residue field, and derived quantifier elimination results, namely, we prove that any formula in the language of valued fields reduces to a formula of the form (∃y f(x,y)=0) ∧ φ(v(x)) ∧ ψ(res(x)), where φ and ψ are formulas in the language of ordered groups and of rings, respectively.
In Fp((Q)) specifically, the valuation ring itself is definable with a diophantine formula (ie of the form ∃y f(x,y)=0), reducing further our quantifier elimination result.
Finally, a large chunk of these formulas are known to be NTP2: when f(x,y) is additive in y, the formula ∃y f(x,y)=z is NTP2 (with respect to x and z). Unfortunately, that does not cover all formulas, so the answer to the titular question is still unknown.
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