Inverse element of $x^2 + x + 1$ in $\mathbb Z_2[x]/\langle x^3 + x + 1\rangle$ I know that $\mathbb Z_2[x]/\langle x^3 + x + 1\rangle$ is a field since $x^3 + x + 1$ is irreducible in $\mathbb Z_2$. But I still can't seem to find any inverse element for $x^2 + x + 1$. I want to find an element $g(x) \in \mathbb Z_2[x]/\langle x^3 + x + 1\rangle$ such that $(x^2 + x + 1)g(x) = 1$. I've tried multiplying every single element of $\mathbb Z_2[x]/\langle x^3 + x + 1\rangle$ by $x^2 + x + 1$, but I never seem to get any of them to be equal to 1 or equivalent. How is this possible if $\mathbb Z_2[x]/\langle x^3 + x + 1\rangle$ is a field?
Thanks in advance. 

Write $(x^2+x+1)(ax^2+bx+c)=1\implies ax^4+(a+b)x^3+(a+b+c)x^2+(b+c)x+c=a(-x^2-x)+(a+b)(-x-1)+(a+b+c)x^2+(b+c)x+c=(b+c)x^2+cx+c=(b+c)x^2+cx+c-a=1\implies a=1,b=0,c=0$, where I have used the fact that $x^3+x+1=0$ to substitute for $x^4$ and $x^3$.
So $x^2$ is your inverse. 
