Fixed field of model of ACFA is not stable

Question

If $(K,\sigma)$ is a model of ACFA, then $\operatorname{Fix}(\sigma)$ is apparently not stable, we can find $E=acl_\sigma(E)\subseteq K$ and $a,b,c\in \operatorname{Fix}(\sigma)\backslash E$ such that $a$ and $b$ are independent over $E$ and $b$,$c$ are independent over $E$ but $\sqrt{a-c}\in \operatorname{Fix}(\sigma)$ while $\sqrt{b-c}\not\in \operatorname{Fix}(\sigma)$. How do we find $E$ and $a,b,c$?

Answer 1

Let $E$ be any $\text{acl}_\sigma$-closed subset of $K$. Let's assume $K$ is a monster model, or at least $|E|^+$-saturated.

Now thinking of $E$ as an abstract field (given together with an automorphism $\sigma$), we can adjoin to it a transcendental element $z$, forming the field $E_1 = E(z)$. Further, we can extend $\sigma$ to an automorphism $\sigma_1$ of $E_1$ by setting $\sigma_1(z) = z$ (explicitly, thinking of elements of $E_1$ as rational functions in $z$ with coefficients from $E$, $\sigma_1$ acts as $\sigma$ does on the coefficients). It's a theorem that any automorphism of a field can be extended to an automorphism of its algebraic closure, so extend $\sigma_1$ to an automorphism $\overline{\sigma_1}$ of $\overline{E_1}$.

Now adjoin another transcendental $y$, forming the field $E_2 = \overline{E_1}(y)$, and extend $\sigma_1$ to $\sigma_2$ on $E_2$ by $\sigma_2(y) = y$ as before, and then to $\overline{\sigma_2}$ on $\overline{E_2}$. We can further ensure that $\overline{\sigma_2}$ swaps $\sqrt{y-z}$ with $-\sqrt{y-z}$: if it doesn't, compose it with an extention to $\overline{E_2}$ of this Galois automorphism of $E_2(\sqrt{y-z})$ over $E_2$.

Finally, adjoin another transcendental $x$, forming the field $E_3 = \overline{E_2}(x)$ and extending $\overline{\sigma_2}$ first to $E_3$ and then to its algebraic closure. This time, ensure that $\overline{\sigma_3}$ fixes both $x$ and $\sqrt{x-z}$.

All in all, we've built a difference field extension of $E$ (with the same cardinality as $E$), and by saturation it can be embedded in $K$ over $E$. Let $a$, $b$, and $c$ be the images of $x$, $y$, and $z$ respectively. We established the behavior of $\sigma$ on the relevant elements by hand, and independence comes from the fact that each new element we adjoined was transcendental over the previous field (which was $\text{acl}_\sigma$-closed).