Timeline for Quotient of Cohen-Macaulay ring
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| S Jul 22, 2016 at 19:34 | history | edited | Pedro Lauridsen Ribeiro | CC BY-SA 3.0 |
format edited, added capitals, LaTeX math command delimiter repositioned
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| S Jul 22, 2016 at 19:34 | history | suggested | user 1 | CC BY-SA 3.0 |
format edited
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| Jul 22, 2016 at 19:17 | review | Suggested edits | |||
| S Jul 22, 2016 at 19:34 | |||||
| Jul 18, 2016 at 17:39 | comment | added | Paulo Rossi | Now I see, "considered as an $A-$module. Thank you! | |
| Jul 18, 2016 at 17:04 | comment | added | Jason Starr | "But I said $R$ is cohen-macaulay ring!" I do not know what you are trying to communicate. I wrote "R" once in my comment when I should have written "A" -- is that your concern? As demonstrated in my comment, for every local integral domain $A$, including for all Cohen-Macaulay local integral domains, the depth of the maximal ideal, considered as an $A$-module, is at most $1$. | |
| Jul 18, 2016 at 17:01 | comment | added | Paulo Rossi | But I said $R$ is cohen-macaulay ring! | |
| Jul 18, 2016 at 16:05 | comment | added | Jason Starr | @PauloRossi. You are wrong to write that $\text{depth}(I)$ equals $\text{depth}(A)$. If $R$ is a local integral domain, and if $(x,y)$ is any ordered pair of nonzero elements of the maximal ideal $I$, then $\overline{x} \in I/xI$ is nonzero, but $y\cdot \overline{x}$ equals $0$ in $I/xI$. Thus, the depth of $I$ is at most $1$. | |
| Jul 18, 2016 at 15:43 | comment | added | Paulo Rossi | A is cohen-macaulay, so, dim(A) = depth(A) = depth(I) (because I is a maximal ideal) | |
| Jul 18, 2016 at 15:35 | comment | added | Mohan | No. For example, consider the case of $I$ a maximal ideal. Then depth of $A/I=0$ and if $\dim A>0$ (locally near $I$), the depth of $I=1$. But, depth of $A$ can be as large as you want. | |
| Jul 18, 2016 at 15:29 | comment | added | Paulo Rossi | For example, depth(A) -depth(I) less than or equal depth(A/I)??? | |
| Jul 18, 2016 at 15:27 | comment | added | Mohan | What would you like to say? You can choose ideals so that the depth can be any number in the allowed range of $[0,\dim (A/I)]$. | |
| Jul 18, 2016 at 15:19 | history | asked | Paulo Rossi | CC BY-SA 3.0 |