I have the following question on ring theory, I would like somebody to help me:
Let $R =\mathbb Z_2[x]$, and consider the ideal $I$ of $R$ generated by the irreducible polynomial $f(x)=x^2+x+1\in R$. Show that the factor ring $R/I$ is a field.
Here is the solution:
$R/I =\{f(x)+I\mid f(x)\in R\}$ so by division algorithm $f(x)=q(x)( x^2+x+1)+r(x)$, then $f(x)= q(x)( x^2+x+1)+ax+b$. Hence in $R/I$: $$f(x)+I=q(x)(x^2+x+1)+ax+b+I=ax+b+I$$ $$R/I=\{ax+b+I\mid a,b\in\mathbb Z_2\}=\{0+I,1+I,x+I,1+x+I\}$$ Every element in $R/I$ has a multiplicative inverse, hence is a field.
I'm not clear with this - somebody help me, thanks.