Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In $\mathbb{Z}_4$, $1$, $2$, $3$ are all units and nilpotents in additive operation; but only 1, 3 are units and 2 is nilpotent in multiplicative operation. I did some experiments like polynomial combination that I might work out a way to prove this, but its complicated. Is there a better way to show this?

share|improve this question
    
Note: "nilpotent in additive operations" makes no sense in a group. You are in a group, every element is invertible. –  Arturo Magidin Apr 6 '12 at 21:33
    
$(1+2x^k)(1+2x^k)=1$. That's enough units. Nilpotents is simpler. –  André Nicolas Apr 6 '12 at 22:37
    
To complete AndreNicola's: $2x^k$ works for nilpotent :-P –  Daniel Montealegre Apr 6 '12 at 22:45
    
@Andre Why spoil Arturo's hints by doing all the work? –  Bill Dubuque Apr 6 '12 at 22:56
    
@Daniel That's implicit in the equality in Andre's hint, but please see my above comment. –  Bill Dubuque Apr 6 '12 at 22:57

1 Answer 1

up vote 4 down vote accepted

Establish the following hints.

Hint 1. If $a$ is nilpotent, then $1+a$ is a unit.

Hint 2. If $a$ is nilpotent, then $ab$ is nilpotent for any $b$ that commutes with $a$ and such that $ab\neq 0$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.