Timeline for A question arising in the distribution theory of L. Schwartz
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 14, 2018 at 11:17 | comment | added | Jochen Wengenroth | Okay, convinced. | |
| Jun 14, 2018 at 10:36 | comment | added | jarauh | @JochenWengenroth I disagree. The full ring is an ideal. Otherwise, the sum of ideals would not be an ideal, etc., and the definition of "the ideal generated by..." would be difficult. On the other hand, the definition of a maximal ideal explicitly excludes the full ring. That is, "maximal ideal" is short hand for "maximal proper ideal". | |
| Jun 14, 2018 at 10:32 | comment | added | Jochen Wengenroth | @jarauh By definition, the full ring isn't an ideal. Otherwise it would always and trivially be the maximal ideal. | |
| Jun 14, 2018 at 10:26 | comment | added | jarauh | Well, $J$ is an ideal, by definition. But maybe it is not a strict ideal. Ok, I was wrong. | |
| Jun 14, 2018 at 10:22 | comment | added | Jochen Wengenroth | I doubt that $J$ is an ideal! The function $f(x)=x$ for $x\ge 0$ and $f(x)=0$ for $x\le 0$ belongs to $J$ and $\delta_0'' \ast f = f'' = \delta_0$. | |
| Jun 14, 2018 at 10:09 | vote | accept | Lucia | ||
| Jun 14, 2018 at 10:37 | |||||
| Jun 14, 2018 at 10:09 | vote | accept | Lucia | ||
| Jun 14, 2018 at 10:09 | |||||
| Jun 14, 2018 at 9:51 | history | answered | jarauh | CC BY-SA 4.0 |