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?

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

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 $ab\neq 0$.


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