Timeline for Units and Nilpotents
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 21, 2016 at 4:01 | comment | added | Derek Elkins left SE | This identity can be heuristically arrived at, and the whole problem approached directly, by using the infinite series expansion of $1/(1-x)$ with, to start, $x = -a$. Since $a$ is nilpotent, the series reduces to a perfectly acceptable polynomial. Rewriting $1/(u+a)$ into the above form is a simple exercise in high school algebra but leads to the the need to verify the (easy) questions Arturo raises. | |
| Mar 25, 2015 at 13:28 | history | edited | AMPerrine | CC BY-SA 3.0 | Corrected an exponent that's been wrong for over three years! |
| Mar 14, 2012 at 3:10 | vote | accept | Shannon | ||
| Mar 14, 2012 at 3:10 | comment | added | Shannon | oh, I see. I will work on it. Thank you so much. | |
| Mar 14, 2012 at 2:56 | comment | added | Arturo Magidin | @Shannon: That's what my last paragraph is about. Multiply $u+a$ by $v$; then you get $1+(va)$... is $va$ nilpotent? | |
| Mar 14, 2012 at 2:51 | comment | added | Shannon | I had tried your way before I asked. But I used u+a instead of 1+a. I was stuck cause I couldnt conclude 1. Here, 1+a is a special case of u+a, do you have to prove u+a? | |
| Mar 14, 2012 at 2:26 | history | answered | Arturo Magidin | CC BY-SA 3.0 |