Timeline for When is $\ker AB = \ker A + \ker B$?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 6, 2011 at 3:13 | vote | accept | Manoj | ||
| Nov 6, 2011 at 3:13 | comment | added | Manoj | So you seem to have proved something stronger: If $AB=BA$ and either $\ker A = \ker A^2$ or $\ker B=\ker B^2$ then $\ker AB=\ker A+\ker B$. | |
| Nov 6, 2011 at 3:11 | comment | added | Manoj | Thanks, Konstantin. It is a very neat proof! I notice that you did not make use of the assumption that $\ker B^2 =\ker B$. | |
| Nov 5, 2011 at 18:00 | comment | added | Bill Johnson | +1. Why don't all books approach the JCF theorem the way suggested in this post? | |
| Nov 5, 2011 at 15:50 | comment | added | Todd Trimble | +1. Very neat proof. | |
| Nov 5, 2011 at 15:10 | comment | added | Martin Brandenburg | 1+. More generally, for an artinian object $V$ in an abelian category and two commuting endomorphisms $A,B$ of $V$ such that $\ker(A^2)=\ker(A)$ and $\ker(B^2)=\ker(B)$ (as subobjects of $V$), then $\ker(AB)=\ker(A) + \ker(B)$ (as subobjects of $V$). | |
| Nov 5, 2011 at 12:10 | history | answered | user91132 | CC BY-SA 3.0 |