Timeline for An example where GCD depends on the domain
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Oct 29, 2014 at 16:28 | comment | added | YCor | A fantastic modern discovery is the notion of set. It could allow to use the notation $x\in gcd(a,b)_R$ instead of $x=gcd(a,b)_R$, which gives rise to the somewhat unpleasant $1= gcd(2,3)_{\mathbb{Z}}=-1$... | |
| Aug 24, 2014 at 10:22 | answer | added | deleted | timeline score: 1 | |
| Mar 3, 2010 at 4:51 | vote | accept | Alfonso Gracia-Saz | ||
| Mar 2, 2010 at 22:23 | answer | added | Arturo Magidin | timeline score: 2 | |
| Mar 2, 2010 at 20:17 | history | edited | Alfonso Gracia-Saz | CC BY-SA 2.5 |
added 67 characters in body
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| Mar 2, 2010 at 19:35 | answer | added | t3suji | timeline score: 14 | |
| Mar 2, 2010 at 19:31 | comment | added | Pete L. Clark | Maybe you want that $S$ is also a GCD-domain? | |
| Mar 2, 2010 at 19:30 | comment | added | Pete L. Clark | The way I understand your question, it would suffice to take any subdomain of a GCD-domain that is not integrally closed, for then it cannot be a GCD-domain. But this applies equally well to Bezout domains like Z[\sqrt{-1}] (take Z[2\sqrt{-1}]), so I don't understand that part of your question. Am I missing something? | |
| Mar 2, 2010 at 19:18 | history | asked | Alfonso Gracia-Saz | CC BY-SA 2.5 |