Does the element $x^2$ have an inverse in the ring $\mathbb{Z}_2[x]/(x^3-x^2-x+1)$? In the ring $R = \mathbb{Z}_2[x]/(x^3-x^2-x+1)$, does the element $x^2$ have an inverse? How do I answer such a question? I know how to compute the inverse if it exists, and using the same method in this case gives me $[x^2]^{-1} = x^2$, but I don't know how to tell if that is an inverse or not. 
I know that an element multiplied by its inverse should yield the identity element, but I'm not sure what the identity element in these situations is. What I know is that an identity element leaves the other elements unchanged, so in $\mathbb{Z}$, that would be $1$. My intuition tells me $x^2$ does not leave every element in $R$ unchanged, but I don't know how to show that. 
 A: HINT:
$x^3-x^2 -x =-1$ so $x(x^2 -x-1) =-1$, $x^{-1} = -(x^2 -x-1)$ and $(x^2)^{-1}= (x^{-1})^2 =( x^2 -x-1)^2$. Divide this by $x^3 - x^2 - x +1$ with remainder  and get $-x^2 + 2$ being the inverse of $x^2$.  If the characteristic is $2$ this also equals $x^2$.
Obs: If $x$ satisfies a polynomial equation with invertible constant term then $x$ ( and so all its powers)  is invertible. 
A: Mind over matter
Over the field of two elements, $x^3-x^2-x+1=(x+1)^3$. Thus $R$ is a local ring with a unique maximal ideal $(x+1)$, and everything in that ideal is clearly nilpotent. 
In such a ring, each element is either a unit or a nilpotent element.
Now if $x^2$ were nilpotent, $x$ would be nilpotent, just as $x+1$ is. But then $x+1-x=1$ would be in the nilradical, and that is an absurdity.
Therefore $x$ and $x^2$ are units.
Brute force
If necessary, manual computation yields the answer very quickly as well.
Solving $(ax^2+bx+c)x^2=1$ gets you $a=1$ and $b=c=0$. Double checking, sure enough:
$x^2x^2=x(x^3)=x(x^2+x-1)=x^3+x^2-x=(x^2+x-1)+x^2-x=1$
All computations are done in the quotient, of course.
Brute finesse
If you get lucky, you can notice that $x^4-1=(x-1)^4$ by the "Freshman dream theorem".  Then $(x-1)^4=0$ since $(x-1)^3=0$, and both chains of equalities together yield $x^4-1=0$, so $x^4=1$ and $x^2$ is a unit.
Or
From $x^3-x^2-x+1=0$ compute $-1=x^3-x^2-x=x(x^2-x-1)$, so $x$ is a unit, so $x^2$ is a unit.
