Timeline for Powers of elements in an Artinian Ring
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Jul 21, 2014 at 9:56 | history | suggested | user26857 | CC BY-SA 3.0 |
fixed a latex issue
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| Jul 21, 2014 at 9:53 | review | Suggested edits | |||
| S Jul 21, 2014 at 9:56 | |||||
| Jun 8, 2014 at 21:56 | comment | added | Neil Epstein | @user51197 Huh. I guess you're right; I was making things a bit too complicated. We have $e>2p-2$ since $0 \neq (xy)^{p-1} \in {\mathfrak m}^{2p-2}$, just as you say. | |
| Jun 7, 2014 at 16:01 | comment | added | user51197 | Doesn't your counterexample in the equicharacteristic $p$ case also work when $p=2$? In that example one has (for any prime $p$, including $p=2$) $e≥2p−1$, it seems. Thus, $e>p=t$. | |
| Jun 5, 2014 at 20:38 | history | edited | Neil Epstein | CC BY-SA 3.0 |
added 15 characters in body
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| Jun 5, 2014 at 20:27 | comment | added | Neil Epstein | @user51197 As you will see in the above edit, equality fails for any mixed characteristic pair $(p^c,p)$, which then completely resolves the question of equality. I hope this answer is now sufficient. | |
| Jun 5, 2014 at 20:26 | history | edited | Neil Epstein | CC BY-SA 3.0 |
resolved the mixed characteristic case completely
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| Jun 5, 2014 at 19:54 | comment | added | Neil Epstein | @user51197 See my edit above for an example where $e\neq t$ in the mixed characteristic case. | |
| Jun 5, 2014 at 19:47 | history | edited | Neil Epstein | CC BY-SA 3.0 |
provided a mixed characteristic counterexample
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| Jun 4, 2014 at 18:17 | comment | added | user51197 | Thank you. I am definitely interested in the mixed characteristic ($p^c,p)$ case. We also have $c \leq t$ to use/understand. | |
| Jun 4, 2014 at 17:35 | history | answered | Neil Epstein | CC BY-SA 3.0 |