Timeline for Is there a choice-free proof that a Euclidean domain is a UFD?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Sep 16, 2018 at 21:38 | comment | added | Mark Meckes | @Vincent: The usual proof that a PID is a UFD uses Zorn's lemma, and as the question linked by Wojowu indicates, some form of choice is necessary (in the sense that ZF alone doesn't suffice). The point of KConrad's answer is that it's possible to prove that a Euclidean domain is a UFD without involving PIDs at all. | |
| Sep 16, 2018 at 21:18 | comment | added | Vincent | A bit late to the party here, but what proof using choice were you thinking of? I am pretty sure that the proof KConrad describes in his answer is the one I learned in college ~15 years ago. I am actually quite curious how you WOULD get choice into the mix | |
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Nov 21, 2016 at 19:41 | vote | accept | Mark Meckes | ||
| Nov 21, 2016 at 19:02 | answer | added | KConrad | timeline score: 17 | |
| Nov 21, 2016 at 17:30 | answer | added | Dave Witte Morris | timeline score: 5 | |
| Nov 21, 2016 at 15:02 | comment | added | Wojowu | Also, apparently, the book "Consequences of the Axiom of Choice" which was used to answer the previously linked question, doesn't have an answer to this one. There still might be hope :) | |
| Nov 21, 2016 at 14:59 | comment | added | Mark Meckes | @Wojowu: Thanks, I suspected that, but it's good to know for sure, and it makes me all the more interested in the answer to my question. | |
| Nov 21, 2016 at 14:57 | comment | added | Wojowu | This doesn't exactly answer your question, but shows that a detour via PIDs is not going to work: mathoverflow.net/questions/31507/… | |
| Nov 21, 2016 at 14:46 | history | asked | Mark Meckes | CC BY-SA 3.0 |