Timeline for Is there a choice-free proof that a Euclidean domain is a UFD?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 21, 2016 at 19:52 | comment | added | Wojowu | BTW, to explain the downvote (which is mine), it is not there because I think the answer is wrong, but rather because I think the answer is not complete. | |
| Nov 21, 2016 at 19:45 | comment | added | Wojowu | @MarkMeckes That's indeed the case, but IMO they should be addressed, or at the very least mentioned, in the answer. For example, for the more general definition of Euclidean function, $p$ is a nonunit divisor of minimal norm doesn't mean $p$ is irreducible, so some clarification would be desirable. | |
| Nov 21, 2016 at 19:42 | comment | added | Mark Meckes | @Wojowu: It looks to me like the proof sketched here is basically the same one in Theorem 4.2 of the note KConrad linked to. Your objections are explicitly answered in section 3 of that note. | |
| Nov 21, 2016 at 19:30 | comment | added | Wojowu | I don't see why you can do it like this. The Euclidean function needn't be multiplicative, so 1. you can't deduce $p$ is irreducible and 2. you don't necessarily know what $r/p$ has smaller norm. | |
| Nov 21, 2016 at 17:30 | history | answered | Dave Witte Morris | CC BY-SA 3.0 |