Is there a clever way to find the units of the quotient ring $\mathbb{Z}_6[x]/\langle x^2+2x\rangle$? I've come to the problem of finding the units of the quotient ring $R:=\mathbb{Z}_6[x]/\langle x^2+2x\rangle$. I've no problem doing this using multiplication table -- i.e. working out the multiplication of the $6^2$ elements in $R$ with the extra rule that $x^2\equiv 4x$ over in $R$ (this comes from the fact that $x^2+2x$ is, obviously, in the principal ideal $\langle x^2+2x\rangle$). However, this is pretty tedious and not very instructive. Is there a prettiest way of doing this? 
Recall that $R=\{[a_0+a_1x]|a_0,a_1\in \mathbb{Z}_6\}$ is just the set of classes of the remainders of the Euclidean division by $x^2+2x$. 
Thanks a lot for any advice!
 A: [For convenience of notation, I will not distinguish between elements of $\mathbb{Z}_6[x]$ and their images in quotient rings.  For instance, I will write elements of $R$ as just $a_0+a_1x$ instead of your $[a_0+a_1x]$.]
A unit of $R$ is just an element which is not in any maximal ideal, so it is enough to identify the maximal ideals.  If $M\subset R$ is a maximal ideal, then $M$ contains either $2$ or $3$, since $2\cdot 3=0$ in $R$.  Similarly, either $x\in M$ or $x+2\in M$.
If $2\in M$, then in either case we get $x\in M$.  But $M=(2,x)$ is already a maximal ideal, with quotient $R/M\cong \mathbb{Z}_2$ (corresponding to the homomorphism $\mathbb{Z}_6[x]\to\mathbb{Z}_2$ sending $f(x)$ to $f(0)$ mod $2$), so $(2,x)$ is the only maximal ideal containing $2$.
If $3\in M$, then we get two different possibilities, either $M=(3,x)$ or $M=(3,x+2)$.  These are both maximal ideals with quotient $\mathbb{Z}_3$ (the first corresponds to the homomorphism $\mathbb{Z}_6[x]\to \mathbb{Z}_3$ sending $f(x)$ to $f(0)$ mod $3$ and the second corresponds to the homomorphism $\mathbb{Z}_6[x]\to \mathbb{Z}_3$ sending $f(x)$ to $f(-2)$ mod $3$).
So, the units are just the elements of $R$ that are not in $(2,x)$, $(3,x)$ or $(3,x+2)$.  For an element $a+bx\in R$ to not be in $(2,x)$ or $(3,x)$ means that $a$ is not divisible by $2$ or $3$ (so $a$ is $1$ or $5$).  For $a+bx$ to not be in $(3,x+2)$ means that $a-2b$ is not divisible by $3$.
