Timeline for In commutativity theorems in ring theory
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 29, 2016 at 18:43 | comment | added | MH.Fakharan | An ideal $I$ is null, means every element of $I$ is nilpotent. Of course every nilpotent ideal is null. But the converse is not true. | |
| Jun 29, 2016 at 3:28 | comment | added | LSpice | My question is what 'null' means of an ideal (since it doesn't seem to mean "equals the 0 ideal"). EDIT: Maybe it means 'nilpotent'? | |
| Jun 28, 2016 at 23:00 | comment | added | MH.Fakharan | Suppose that $A=\{ ab-ba : a,b\in R\}$. $I=<x :x\in A>$, the ideal generated by the elements of $A$ or $I= \cap J$ where $A\subseteq J$ and $J$ is an ideal of $R$. We should show that $I$ is null. Yes, $R$ is commutative if and only if $I=0$ and in this case $I$ is null. Of course in general we should prove that $I$ is null. | |
| Jun 28, 2016 at 20:45 | comment | added | LSpice | Of course, the JH theorem shows that $R/Z(R)$ is commutative. What do you mean by "the ideal generated by all additive commutators is null", if not that it is the 0 ideal (and hence that the ring is commutative)? | |
| Jun 28, 2016 at 20:38 | history | edited | MH.Fakharan | CC BY-SA 3.0 |
deleted 9 characters in body
|
| Jun 28, 2016 at 15:22 | history | asked | MH.Fakharan | CC BY-SA 3.0 |