Timeline for Automorphisms of the hyperreals over the rationals and nontrivial automorphism groups
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Dec 18, 2023 at 14:14 | history | edited | Joel David Hamkins |
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| Feb 13, 2018 at 14:54 | history | edited | Mikhail Katz |
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| Jan 1, 2017 at 2:30 | answer | added | Ali Enayat | timeline score: 11 | |
| Dec 31, 2016 at 17:39 | comment | added | YCor | @nfdc23 this is why "that are not subfields of the real field" was specified in my second comment on Shelah's paper. | |
| Dec 31, 2016 at 16:56 | comment | added | nfdc23 | Whoops, indeed I forgot about the condition to respect the order structure. Mea culpa. Thanks to Emil and Todd for pointing out the error. | |
| Dec 31, 2016 at 14:50 | comment | added | Todd Trimble | @nfdc23 I don't think your argument works. The point is that a real closure of an ordered field has a unique ordering whose positive elements are precisely the sums of squares, and since an automorphism will then preserve the ordering, its restriction to the set of roots $r_1 < r_2 < \ldots < r_n$ of an $f$ must take $r_i$ to $r_i$. See Lang's Algebra, chapter XI section 2. | |
| Dec 31, 2016 at 14:50 | comment | added | Emil Jeřábek | @nfdc23 No, the field of real algebraic numbers has no nontrivial automorphism, by the same argument as for R. The uniqueness of real closures only gives that K has an automorphism extending an order-preserving one of any subfield, but those are all trivial. Only nonarchimedean real-closed fields may admit nontrivial automorphisms. | |
| Dec 31, 2016 at 13:32 | comment | added | nfdc23 | The field $K$ of algebraic real numbers is a real closure of $\mathbf{Q}$, so by the uniqueness of real closures up to isomorphism it admits an automorphism extending one of any subfield. For example, any non-trivial finite Galois extension of number fields $F'/F$ with $F'$ having a real embedding (e.g., $F=\mathbf{Q}$ and $F'$ real quadratic field or $\mathbf{Q}(\zeta + 1/\zeta)$ for a primitive $n$th root of unity $\zeta$ with $n\ge 7$) admits $[F':F]$ automorphisms over $F$, all extending to $K$ upon choosing an embedding of $F'$ into $K$. Hence, ${\rm{Aut}}(K)$ is huge (even uncountable). | |
| Dec 31, 2016 at 13:09 | answer | added | Joel David Hamkins | timeline score: 10 | |
| S Dec 31, 2016 at 7:57 | history | suggested | Amir Sagiv | CC BY-SA 3.0 |
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| Dec 31, 2016 at 7:47 | comment | added | YCor | There also exist real-closed fields with no nontrivial field automorphism and that are not subfields of the real field (S. Shelah, Models with second order properties IV. A general method and eliminating diamonds, 1983 sciencedirect.com/science/article/pii/0168007283900131) | |
| Dec 31, 2016 at 7:45 | comment | added | YCor | The field of real Puiseux series is real-closed, and has an obvious automorphism $\sum a_rt^r\mapsto \sum a_r t^{2r}$. | |
| Dec 31, 2016 at 7:20 | review | Suggested edits | |||
| S Dec 31, 2016 at 7:57 | |||||
| Dec 31, 2016 at 4:57 | review | First posts | |||
| Dec 31, 2016 at 7:20 | |||||
| Dec 31, 2016 at 4:54 | history | asked | Brian Pinsky | CC BY-SA 3.0 |