Show that $\mathbb{Z}_4[x]$ has infinitely many units and infinitely many nilpotents.

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?

• 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

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 $ab\neq 0$.