Timeline for The Ax-Kochen isomorphism theorem and the continuum hypothesis
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 18, 2019 at 16:47 | comment | added | David Fernandez-Breton | @MohammadGolshani right, I interpreted too literally the "$\mathsf{CH}$ can't be removed" part (although in a strict reading, "$\mathsf{CH}$ can't be removed" could read as "$\mathsf{CH}$ is necessary" which would literally mean "the Ax--Cochen isomorphism theorem implies $\mathsf{CH}$"). Anyway, the question is really interesting. | |
| Jul 18, 2019 at 14:17 | comment | added | Todd Eisworth | Drat! I was hoping to be able to cite Rudy Rucker's mathematics! | |
| Jul 18, 2019 at 4:34 | comment | added | Mohammad Golshani | @PaulLarson Yes, I've seen this question in Shelah's paper as well, but have no ideas for it. | |
| Jul 18, 2019 at 4:29 | comment | added | Mohammad Golshani | @ToddEisworth Thanks for introducing this nice reference. However, it does not answer the question, as for the isomorphism theorem, we require it holds for all non-principal ultrafilter over prime numbers | |
| Jul 17, 2019 at 23:19 | comment | added | Paul Larson | Also interesting : is the absoluteness of the result of Moloney mentioned in Shelah's paper still open? | |
| Jul 17, 2019 at 14:33 | comment | added | Todd Eisworth | It's been a long time since I've thought about model theory, but can this be addressed by techniques of Ellentuck and Rucker "Martin's Axiom and Saturated Models"? They show that under MA there is an ultrafilter $F$ on $\omega$ such that the $F$-ultrapower of any countable structure is saturated. | |
| Jul 15, 2019 at 14:04 | comment | added | Mohammad Golshani | It is a consistency result, there is a model of $ZFC$ in which CH fails and the isomorphism theorem does not hold. It does not say that the failure of CH implies the negation of the theorem. | |
| Jul 15, 2019 at 13:28 | comment | added | David Fernandez-Breton | I'm confused: if Shelah showed that $\mathsf{CH}$ can't be removed from the theorem, then that means the theorem isn't consistent with $\neg\mathsf{CH}$? Am I missing something? | |
| Jul 15, 2019 at 7:30 | history | asked | Mohammad Golshani | CC BY-SA 4.0 |