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Does there exist a pair of non-isomorphic structures $\mathfrak{A}$ and $\mathfrak{B}$ as well as sets $I$ and $J$ and ultrafilters $\mathcal{U}$ on $I$ and $\mathcal{F}$ on $J$ such that $\mathfrak{A}^I/\mathcal{U}\cong\mathfrak{B}$ and $\mathfrak{B}^J/\mathcal{F}\cong\mathfrak{A}$?

This question is inspired by the Keisler-Shelah theorem, which says that you can arrange $\mathfrak{A}^I/\mathcal{U} \cong \mathfrak{B}^J/\mathcal{F}$ (moreover you can arrange that $\mathfrak{A}^I/\mathcal{U}\cong \mathfrak{B}^I/\mathcal{U}$). The answer to this question very likely depends on set theoretic assumptions, so a reasonable weakening would be mere consistency relative to some strong set theoretic assumptions or in some forcing extension. Another reasonable weakening would be to ask about iterated ultrapowers rather than just ultrapowers.

On the other hand assuming a positive answer an obvious follow-up would be the question of the existence of a triple of pairwise non-isomorphic structures, each of which is isomorphic to ultrapowers of the other two. Another obvious follow-up is whether or not it can be arranged that $\mathfrak{A}^I/\mathcal{U}\cong\mathfrak{B}$ and $\mathfrak{B}^I/\mathcal{U\cong{\mathfrak{A}}}$, i.e. whether or not there exists a structure and an ultrafilter such that iterating ultrapowers by that ultrafilter alternates between two isomorphism classes.

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    $\begingroup$ We can assume $I=J$ by taking their union if necessary. Also, since both are $\frak A$ and $\frak B$ satisfy this being isomorphic to an ultrapower of the other, they have the same cardinality. $\endgroup$
    – Asaf Karagila ♦
    Commented Nov 21, 2018 at 7:11
  • $\begingroup$ One goal post would be to look at the characteristics of ultrapowers which follow from properties of the ultrafliter, then try to take two ultrafilters which produce non-isomorphic ultrapowers (say of $\Bbb N$) and then iterate that with some complementing ultrafilters to get the wanted result. $\endgroup$
    – Asaf Karagila ♦
    Commented Nov 21, 2018 at 7:13
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    $\begingroup$ I guess I don't know. But i would imagine it is something about solving the equation $U+x=F+y$ in $\beta\omega$, or something. $\endgroup$
    – Asaf Karagila ♦
    Commented Nov 21, 2018 at 7:16
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    $\begingroup$ A (trivial) restatement of the question: is there a structure $\mathfrak{A}$ and ultrafilters $\mathcal{U},\mathcal{F}$ such that $\mathfrak{A}$ is isomorphic to $\mathfrak{A}^{\mathcal{U}\cdot\mathcal{F}}$ but not $\mathfrak{A}^{\mathcal{U}}$? $\endgroup$
    – Miha Habič
    Commented Nov 21, 2018 at 15:15
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    $\begingroup$ Since we can assume that $I=J$, this gives us a reasonable notion of iterating the ultrapowers as an ultrapower itself, moreover the iteration is just taking the product of the ultrafilters in the Stone space (modulo some reasonable bijection of $I$ with $I^2$, that is). Which goes back to my remarks about solving some equations in the Stone space (although with multiplication rather than with addition), and with discerning properties of ultrapowers from properties of filters, etc. $\endgroup$
    – Asaf Karagila ♦
    Commented Nov 21, 2018 at 15:57

1 Answer 1

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An example assuming large cardinals: suppose $U$ and $W$ are normal ultrafilters on a measurable cardinal $\kappa$ such that $(V_{\kappa+2})^{M_U}\neq (V_{\kappa+2})^{M_W}$, where $M_U$ denotes the transitive collapse of the ultrapower of $V$. Let $Z = U\times W$. Let $M$ be the iterated ultrapower of length $\omega$ hitting $Z$ and its images. Then $M = j_Z(M) = j_U(j_W(M))$. But $j_W(M)$ is not isomorphic to $M$ since $(V_{\kappa+2})^M = (V_{\kappa+2})^{M_U}\neq (V_{\kappa+2})^{M_W} = (V_{\kappa+2})^{j_W(M)}$. For any structure $N$ in the language of set theory and any ultrafilter $D$, $j_D(N)$ is isomorphic to the (external) ultrapower of $N$ by $D$. So $M^\kappa / W$ is not isomorphic to $M$ but $(M^{\kappa}/W)^\kappa/U$ is isomorphic to $M$, as desired. If you want to replace $M$ with a set sized structure, consider instead $(V_\lambda)^M$ where $\lambda$ is a fixed point of $j_Z$.

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  • $\begingroup$ If you don't require $S$ to be a well-ordering, how do you iterate along it? (I vaguely recall hearing that one could do this, but I don't recall the details.) $\endgroup$
    – Noah Schweber
    Commented Nov 21, 2018 at 16:40
  • $\begingroup$ @Noah I suppose you would need an appropriate directed system of embeddings and then you would use directed limits. $\endgroup$
    – Andrés E. Caicedo
    Commented Nov 21, 2018 at 17:19
  • $\begingroup$ Why is $(V_{\kappa+2})^M = (V_{\kappa+2})^{M_U}$ and why is $(V_{\kappa+2})^{M_W} = (V_{\kappa+2})^{j_W(M)}$? $\endgroup$
    – Danielle Ulrich
    Commented Nov 22, 2018 at 15:03
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    $\begingroup$ $V_{\kappa+2}^M = V_{\kappa+2}^{M_U}$ since the natural map $M_U \to M_Z \to M$ has critical point $j_U(\kappa)$. Since $V_\kappa^M = V_\kappa$, $V_{j_W(\kappa)}^{j_W(M)} = j_W(V^M_\kappa) = j_W(V_\kappa) = V_{j_W(\kappa)}^{M_W},$ which implies the second equality. $\endgroup$
    – Gabe Goldberg
    Commented Nov 22, 2018 at 17:25

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