Cuts in recursively saturated models

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Free cuts

A cut $I$ in a model $M\models PA$ is free if for all $a, b\in I$ if $(M,a)\equiv (M,b)$, then $(M,I,a)\equiv (M,I,b)$. There are free elementary cuts in every countable recursively saturated model of $PA$, and generic cuts of Kaye and Tin Lok Wong are free.

Problem: Let $M\models PA$ be countable and recursively saturated. Does $M$ have a free elementary cut $I$ such that the pair $(M,I)$ is recursively saturated?

The problem was posed in , freeness of elementary generic cuts is discussed in .

Omitting theories of subsets

Suppose $M$ is countable recursively saturated and $X$ is an undefinable subset of $M$. Is there a countable recursively saturated $N$ such that $N$ is an elementary end extension of $M$, and if $Y \subseteq M$ is coded in $N$, then $(M,Y) \not\equiv (M,X)$?

The answer if `yes' is either $(M,X)\not\models PA^*$ or ${\rm Th}(M,X)\notin {\rm SSy}(M)$.

This problem is listed in , but, unfortunately, with many typos.

Elementarily equivalent nonisomorphic pairs?

Let $M\models PA$ be countable and recursively saturated and let $K$ and $K'$ be short elementary cuts of $M$ such that $(M,K)\equiv (M,K')$. Are $(M,K)$ and $(M,K')$ isomorphic?

The problem was posed in .