Timeline for On Krull dimension \dim M and \dim Supp(M)
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 11, 2019 at 8:07 | vote | accept | Tri Nguyen | ||
| Jul 5, 2017 at 12:20 | answer | added | Fred Rohrer | timeline score: 2 | |
| Jul 1, 2017 at 14:16 | review | Suggested edits | |||
| Jul 1, 2017 at 14:53 | |||||
| Jun 29, 2017 at 14:43 | comment | added | Tri Nguyen | Yes, $\dim M=\dim R/0:_R M$ is the definition of dimension in my question. Thanks. | |
| Jun 28, 2017 at 18:02 | comment | added | Fred Rohrer | I guess you define the dimension of an $R$-module $M$ to be the dimension of the ring $R/(0:_RM)$. Then, the answer is no. For examples, see Bourbaki, Algèbre commutative, II.4 Exercice 22. | |
| Jun 28, 2017 at 15:42 | review | First posts | |||
| Jun 28, 2017 at 15:44 | |||||
| Jun 28, 2017 at 15:35 | history | asked | Tri Nguyen | CC BY-SA 3.0 |