Timeline for Can this way of comparing numbers of the form a+b sqrt(K) be generalized?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Oct 30, 2014 at 19:45 | vote | accept | tailcalled | ||
| Oct 29, 2014 at 11:45 | comment | added | tailcalled | Johannes Hahn: I know, but the numbers in $R[X]/(X^2-K)$ can be given an order (the order I described) which 'causes' the square root to be the positive one. Gerry Myerson: No need to deal with negative $K$. | |
| Oct 29, 2014 at 2:21 | answer | added | Robert Israel | timeline score: 0 | |
| Oct 29, 2014 at 0:13 | comment | added | Gerry Myerson | Do you need to deal with ${\bf R}[X]/(X^2+1)$? and if so, how do you do it? | |
| Oct 28, 2014 at 23:44 | comment | added | Johannes Hahn | Sidenote: You don't want $R[X]/(X^2-K)$, because in this ring there is no preferred choice of a square root of $K$. You want a ring that has such a preferred square root, i.e. the positive one. Thus this isn't really about rings and ring extensions, it is about totally ordered rings and extensions of totally ordered rings. | |
| Oct 28, 2014 at 23:40 | history | edited | Johannes Hahn |
added computer-algebra tag
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| Oct 28, 2014 at 22:39 | comment | added | Anthony Quas | I asked essentially the same question some time ago: mathoverflow.net/questions/69805/… You may find some useful information in Igor Rivin's answer to that also. | |
| Oct 28, 2014 at 20:48 | answer | added | Emil Jeřábek | timeline score: 2 | |
| Oct 28, 2014 at 20:01 | comment | added | tailcalled | In the second paragraph, I explained a way to compare elements in $R[X]/(X^2-K)$ (when you already have a method of comparing elements in $R$). That is the method I was talking about when I wrote 'this method'. | |
| Oct 28, 2014 at 19:59 | comment | added | Włodzimierz Holsztyński | ...this method...--which method? | |
| Oct 28, 2014 at 19:59 | comment | added | tailcalled | Also, in case what 'sufficiently nice' means becomes relevant, I can guarantee that any ordered ring that is a subring of an archimedean field is sufficiently nice, and I do not for now need anything more general than that. | |
| Oct 28, 2014 at 19:47 | history | asked | tailcalled | CC BY-SA 3.0 |