Timeline for Height of ideal in graded ring
Current License: CC BY-SA 3.0
7 events
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| S Jul 20, 2014 at 22:40 | history | suggested | user26857 | CC BY-SA 3.0 |
updated an username and fixed a small latex mistake
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| Jul 20, 2014 at 22:30 | review | Suggested edits | |||
| S Jul 20, 2014 at 22:40 | |||||
| Jul 20, 2014 at 22:29 | comment | added | user26857 | @Youngsu If the height of an arbitrary ideal behaves so badly when localize and take factor rings, how can you conclude in your answer to the original problem that $\text{height }I\le\text{height }I'+ 1$? In fact you got the following: $\text{height }(I+q)/q\le 1$ in $R/q$. Is this enough to prove that $\text{height }I\le\text{height }I'+ 1$? Or maybe I didn't get your reasoning? | |
| Jun 3, 2012 at 7:01 | history | edited | Youngsu | CC BY-SA 3.0 |
to answer http://mathoverflow.net/questions/97730/height-of-ideal-in-graded-ring/98695#98695 .; added 2 characters in body
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| May 29, 2012 at 9:32 | comment | added | Thomas Kahle | A useful inequality for one is: In any Noetherian ring $R$ with proper ideal $I$ you have $\text{ht}(I) + \text{dim}(R/I) \leq \text{dim}(R)$. | |
| May 27, 2012 at 13:06 | history | answered | Youngsu | CC BY-SA 3.0 |