Timeline for Smallest ring whose field of fractions includes all the reals (subring of omnific integers?)
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 18, 2022 at 18:33 | vote | accept | Mike Battaglia | ||
| Sep 18, 2022 at 18:31 | vote | accept | Mike Battaglia | ||
| Sep 18, 2022 at 18:33 | |||||
| Jun 18, 2020 at 7:55 | answer | added | nombre | timeline score: 2 | |
| Jun 18, 2020 at 5:39 | history | edited | Mike Battaglia | CC BY-SA 4.0 |
added detail
|
| May 24, 2020 at 0:31 | comment | added | Kevin Casto | Related: math.stackexchange.com/questions/1396872/… | |
| May 23, 2020 at 20:38 | history | edited | Mike Battaglia | CC BY-SA 4.0 |
clearer title and some grammar
|
| May 23, 2020 at 19:39 | comment | added | Kevin Casto | Algebraically, if we let $K = \overline{\mathbb{Q}} \cap \mathbb{R}$, then $\mathbb{R}$ is isomorphic to the real closure of $K$ adjoin uncountably many variables. So I imagine a "smallest" such ring would be some version of taking "integers" here -- e.g. taking the ring of real algebraic integers, adjoining infinitely many variables, and then adding a root of all monic odd-degree polynomials successively. Of course, this is totally non-canonical and you get many distinct subrings of $\mathbb{R}$ depending on which transcendence basis you pick. | |
| May 23, 2020 at 19:09 | comment | added | Goldstern | As I understand it, Oz is a proper class, and you certainly do not need all of Oz to find the real numbers as a quotient field. If you take a sufficiently large and sufficiently closed ordinal $\delta$, then the set $Oz_\delta$ of all surreal integers born before day $\delta$ will be a ring, and its quotient field will include all the reals. I think that $\delta:=(2^{\aleph_0})^+$ will do, but perhaps much smaller smaller ordinals are enough, such as $\omega^ \omega$? (Clearly you need the ring to have at least continuum many elements.) | |
| May 23, 2020 at 15:22 | comment | added | LSpice | If your $R$ has field of fractions $K$ containing $\mathbb R$, then it makes sense to embed both $R$ and $\mathbb R$ into $K$, and so to speak of $R \cap \mathbb R$. Could it happen that the field of fractions of $R \cap \mathbb R$ is smaller than $\mathbb R$? (Also, in your (1), do you want some kind of universality—e.g., a unique isomorphism (satisfying some extra conditions, I guess—or really just abstract isomorphism?) | |
| May 23, 2020 at 15:17 | history | asked | Mike Battaglia | CC BY-SA 4.0 |