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Notation and motivation. Given an algebraic structure $\mathbb{M}$ of cardinality at most the continuum and with countably many operations, and a nonprincipal ultrafilter $\cal{U}$ on a countably infinite set, let $\mathbb{M}^{*}$ be the ultrapower of $\mathbb{M}$ modulo $\cal{U}$. It is well-known that if CH (continuum hypothesis) holds, then $\mathbb{M}^{*}$ is saturated (in the language of model theory) and therefore has many automorphisms.

Question. Let $\mathbb{Z}$ be the ring of integers, and $\mathbb{R}$ be the field of reals. Is it known if either $\mathbb{Z}^{*}$ or $\mathbb{R}^{*}$ has to have a nontrivial automorphism in the absence of CH?

Remark. The answer to the above question for the case of $\mathbb{Z}$ does not change if $\mathbb{Z}^{*}$ is replaced with $\mathbb{N}^{*}$ or with $\mathbb{Q}^{*}$, since automorphism groups are invariant under bi-interpretations, and:

(1) $\mathbb{Z}$ is bi-interpretable with $\mathbb{N}$ (thanks to Lagrange's four squares theorem), and

(2) $\mathbb{N}$ is bi-interpretable with $\mathbb{Q}$ (thanks to Julia Robinson's theorem).

In contrast, it is well-known that $\mathbb{C}^{*}$ has many automorphisms (independent of whether CH holds not) since $\mathbb{C}^{*}$ is isomorphic to $\mathbb{C}$ (by the Steinitz theorem that asserts that any two algebraically closed fields of the same uncountable cardinality are isomorphic), and the fact that the classical proof that $\mathbb{C}$ has $2^{2^{\aleph_0}}$ many automorphisms does not need CH in any of its steps.

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    $\begingroup$ This might be a silly question, but: can an ultrapower (of a structure in a countable language of size at most continuum with respect to a nonprincipal ultrafilter on $\omega$) ever be rigid? $\endgroup$ – Noah Schweber Mar 30 '18 at 4:30
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    $\begingroup$ @NoahSchweber You should add the adjective "infinite". $\endgroup$ – Joel David Hamkins Mar 30 '18 at 11:58
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    $\begingroup$ Generalizing that $\mathbb{C}^*$ has many automorphisms: for any structure $M$ and any nonprincipal ultrafiflter $\mathcal{U}$, every automorphism of $M$ lifts canonically to an automorphism of $M^\omega/\mathcal{U}$, i.e. we can view $\mbox{Aut}(M)$ as a subgroup of $\mbox{Aut}(M^\omega/\mathcal{U})$; this is because ultraproducts commute with expansions. So the interesting case is when $M$ is rigid (e.g. $\mathbb{Z}, \mathbb{N}$ as asked). $\endgroup$ – Douglas Ulrich Mar 30 '18 at 13:28
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    $\begingroup$ The "Viva/Vive la difference" series by Shelah (arxiv.org/abs/math/9201245, arxiv.org/pdf/math/9304207.pdf, arxiv.org/pdf/math/0112237.pdf) and "Automorphism Groups of Ultraproducts of Finite Symmetric Groups" (by Lucke and Thomas) seem relevant. In particular, in "Vive la difference III" Shelah seems to prove that consistently, there is an ultrafilter $\mathcal{U}$ on the set of primes such that $\prod_\omega \mathbb{F}_p/\mathcal{U}$ is rigid. Perhaps the proof concept can be modified... $\endgroup$ – Douglas Ulrich Mar 30 '18 at 13:50
  • $\begingroup$ @DouglasUlrich Thanks for your comments and for the pointers, I will take a look. $\endgroup$ – Ali Enayat Mar 30 '18 at 14:13

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