Timeline for Constructive treatment of Jacobson rings
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 12 at 3:00 | answer | added | Ryota Kuroki | timeline score: 2 | |
| Jun 10, 2018 at 13:58 | comment | added | darij grinberg | @JakobWerner: True; maybe "every quotient ring by a finitely generated ideal". | |
| Jun 10, 2018 at 13:49 | comment | added | Jakob Werner | @darijgrinberg You are right. However »every quotient ring« seems to be very strong, constructively. For example consider the statement that $ \mathbb{Z} $ is Jacobson. If $ I = \langle n \rangle$ is principal, then it is rather easy to prove that $ \operatorname{rad}(I) = \operatorname{Jac}(I) $. (Reduce to the case that $ n $ is prime and use that $\mathbb{Z}/n$ is a field. I think this should work intuitionistically.) If however $I$ is not assumed to be principal (or equivalently, finitely generated), I don't even know how to start. | |
| Jun 8, 2018 at 20:27 | comment | added | darij grinberg | "one should find a suitable constructive version of the definition of a Jacobson ring first": Actually, there is one on the Wikipedia page: "In every quotient ring, the nilradical is equal to the Jacobson radical". The Jacobson radical has several equivalent constructive definitions (see mathoverflow.net/questions/57877/… ). | |
| Jul 18, 2017 at 16:13 | comment | added | YCor | I'd be curious to know the meaning of "being constructively true at the same time". In particular, do you want something uniform in the ground Jacobson ring? | |
| Jul 18, 2017 at 16:05 | comment | added | Jakob Werner | Yes, they are. dummy characters | |
| Jul 18, 2017 at 16:03 | comment | added | darij grinberg | Are these rings commutative? | |
| Jul 18, 2017 at 15:36 | review | First posts | |||
| Jul 18, 2017 at 15:39 | |||||
| Jul 18, 2017 at 15:35 | history | asked | Jakob Werner | CC BY-SA 3.0 |