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?
Show that $\mathbb{Z}_4[x]$ has infinitely many units and infinitely many nilpotents.
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abstract-algebra
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0@D$a$niel That's im$p$licit in the equ$a$lity in Andre's hint, but please see my above comment. – 2012-04-06
1 Answers
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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 $ab\neq 0$.