Units in quotient ring of $\mathbb Z[X]$

An exercise from Dummit & Foote:

Determine the units of the ring $A = \mathbb{Z}[X]/(X^{3})$ and the structure of the unit group $A^{\times}$.

Help would be great.

Thanks!

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At least $(x^2 + x + 1)(-x+1) = 1$. I guess you can play around a bit with signs and get others as well. – Arthur Oct 22 '11 at 7:28

Actually nobody answered the second part of the problem about the structure of the unit group $A^{\times}.$ The answer is the following:

$A^{\times}$ is isomorphic to the abelian group $\mathbb{Z}_2\times\mathbb{Z}^2$.

Hint. The unit group can be written as a direct sum of the subgroups generated by $-1$, $1+x$, and $1+x^2$ respectively. The subgroup generated by $-1$ is the torsion subgroup of $A^{\times}$, whilst the subgroup generated by $1+x$ and $1+x^2$ is the free part of $A^{\times}$.

Of course, the problem (and its answer) can be easily generalized to the ring $\mathbb{Z}[X]/(X^n)$.

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The key trick is that there is a canonical ring-morphism $A=\mathbb Z[X]/(X^3)=\mathbb Z[x]\to \mathbb Z[X]/(X) \simeq\mathbb Z$ (why?) and that units are sent to units by ring morphisms.
So any unit of $A$ is of the form $u=a+bx+cx^2$ with $a$ a unit in $\mathbb Z$ .
I won't tell you that $x$ is nilpotent: my colleagues on this site would say that I'm making things too easy for you.

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Thanks! I actually proved in general that if $R$ is a ring (commutative wih unity of course) and $p$ a polynomial over $R$, then $p$ is a unit if and only if the free coefficient is a unit while the others are nilpotent. – Anna Oct 23 '11 at 1:58

Hints to get you started: every element of $A$ is represented by a unique polynomial $aX^2+bX+c$ with $a,b,c \in \mathbb{Z}$. Make sure you see why. If such an element is a unit, what can you say about $c$? Are there any requirements on $a,b$?

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Thanks. See the above comment of mine.\ – Anna Oct 23 '11 at 2:02