Timeline for A question about Euclidean domains
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Aug 29, 2023 at 17:30 | history | edited | Mohammad Safdari | CC BY-SA 4.0 |
edited body
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| Aug 29, 2023 at 17:17 | history | edited | Mohammad Safdari | CC BY-SA 4.0 |
added 564 characters in body
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| Aug 29, 2023 at 16:45 | comment | added | David E Speyer | @PaceNielsen Thanks! | |
| Aug 29, 2023 at 15:07 | comment | added | Pace Nielsen | @DavidESpeyer Note that Mohammad only requires a nonstrict inequality $\deg(a-r)\leq \deg(a)$. | |
| Aug 29, 2023 at 7:27 | comment | added | darij grinberg | @PaceNielsen: Alas I don't have a good approach to understanding all Euclidean norms on a ring... | |
| Aug 29, 2023 at 0:59 | comment | added | David E Speyer | I don't understand why you say this is true for $F[x]$. If $a(x) = a_n x^n + \cdots + a_1 x + a_0$ and $b(x) = x$, then there is only one choice for $q$ and $r$: We must take $q(x) = a_n x^{n-1} + \cdots + a_2 x + a_1$ and $r(x) = a_0$. Then $\deg(a-r) = \deg(a)=n$. What am I misunderstanding? | |
| Aug 28, 2023 at 21:53 | comment | added | Pace Nielsen | @darijgrinberg Can you make your example work with any Euclidean norm on $\mathbb{Z}[\sqrt{-2}]$? | |
| Aug 28, 2023 at 18:05 | comment | added | darij grinberg | I'd prefer to keep this question open, as I'm curious which quadratic rings $\mathbb Z[\alpha]$ satisfy this stronger version of Euclideanness. | |
| Aug 28, 2023 at 15:41 | comment | added | Mohammad Safdari | @PaceNielsen $q$ is zero in your argument, so it does not work. $N(\cdot )$ is just $|\cdot |$ for integers; this should clarify what goes wrong in the argument. | |
| Aug 28, 2023 at 15:40 | comment | added | Mohammad Safdari | @GerryMyerson The degree of an integer is its absolute value. | |
| Aug 28, 2023 at 15:39 | comment | added | Mohammad Safdari | @MatemáticosChibchas Thanks, it is interesting. | |
| Aug 27, 2023 at 23:52 | comment | added | Matemáticos Chibchas | This paper may be useful. | |
| Aug 27, 2023 at 22:17 | comment | added | Gerry Myerson | I don't understand what is meant by the degree of an integer. | |
| Aug 27, 2023 at 17:50 | comment | added | Mohammad Safdari | @darijgrinberg Thank you. If you can turn your comment into an answer I will accept it. Including a concrete example would be nice too; I think $a=1+\sqrt 2 i$ and $b=1-\sqrt 2 i$ will work (the remainders are $r=-1$ and $r=-\sqrt2 i$). | |
| Aug 27, 2023 at 17:47 | comment | added | Mohammad Safdari | @PaceNielsen I should have emphasized that in $\deg a \le \deg (ab)$ both $a,b$ are nonzero. | |
| Aug 27, 2023 at 15:15 | comment | added | darij grinberg | I don't think it's true for $\mathbb{Z}\left[\sqrt{-2}\right]$ then, with the usual Euclidean norm (= square of the absolute value). If you look at the four radius-$1$ circles covering the fundamental domain (a rectangle with sidelengths $1$ and $\sqrt 2$), you will see that all four circles are needed (no three cover the whole domain), so that $a-r$ might be further away from the origin than $a$. | |
| Aug 27, 2023 at 15:06 | comment | added | Mohammad Safdari | @darijgrinberg That in the division $\deg r < \deg b$ or $r=0$. And $\deg(a)\le \deg(ab)$. | |
| Aug 27, 2023 at 14:35 | comment | added | darij grinberg | What do you require of $\deg$? (Different authors impose rather different axioms.) | |
| Aug 27, 2023 at 14:02 | history | asked | Mohammad Safdari | CC BY-SA 4.0 |