Timeline for Free ordered field?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 29, 2019 at 14:20 | vote | accept | Zemyla | ||
| Oct 29, 2019 at 9:10 | comment | added | Emil Jeřábek | If you care only about the field structure of the ordered fields: fields that can be ordered are exactly the formally real fields (which implies characteristic $0$), and if I’m not missing something, the field of rational functions $\mathbb Q(X)$ over a set of indeterminates $X$ is a free formally real field over $X$. | |
| Oct 29, 2019 at 4:07 | history | became hot network question | |||
| Oct 28, 2019 at 21:05 | answer | added | nombre | timeline score: 10 | |
| Oct 28, 2019 at 21:02 | answer | added | Johannes Hahn | timeline score: 15 | |
| Oct 28, 2019 at 20:57 | comment | added | Zemyla | So if the functor forgets the + and * functions and just keeps the < relation, there may still be a free object construction? | |
| Oct 28, 2019 at 20:52 | comment | added | HenrikRüping | I think you are right with your last sentence. If F is the free ordered field on two elements x,y, the map of sets which swaps x and y should induce a morphism of ordered fields $F\rightarrow F$ extending that map of sets. Thus it cannot be order preserving. However it might make sense to talk about the free ordered field on an ordered set, i.e. to consider the left adjoint to the forgetful functor from ordered fields to ordered sets (that assigns to an ordered field F the set $F\setminus \{0\}$). | |
| Oct 28, 2019 at 20:07 | history | asked | Zemyla | CC BY-SA 4.0 |