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.


  • $\begingroup$ At least $(x^2 + x + 1)(-x+1) = 1$. I guess you can play around a bit with signs and get others as well. $\endgroup$ – 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 $\mathbb{Z}[X]/(X^n)$.


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.

  • 1
    $\begingroup$ 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. $\endgroup$ – 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 $a,b$?

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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