We're trying to understand the field $\mathbb{Z}_2[x]/\langle x^3+x^2+1 \rangle$, and in particular why it has eight elements. I know that $x^3+x^2+1$ is irreducible in $\mathbb{Z}_2$ and so factor is going to be a field.
The issue I'm having is that the notes claim that $(x+I)$ has an inverse in the field, where $I = \langle x^3+x^2+1 \rangle$. They write $$ (x+I)(x^2+x+I)=x^3+x^2+I=1+I $$ I don't understand the last line, I know that multiplication of cosets is defined that way, I just don't see why we get 1 at the end. I think it's possible I didn't fully understand cosets and ideals but I can't fully articulate what the issue is. Any help towards filling in the gaps of my understanding would be greatly appreciated.
You might know that $a+I=b+I\,\iff a-b\in I$.
So $(x^{3}+x^{2})-1=(x^{3}+x^{2}+1)\in I$
So $(x^{3}+x^{2})+I=1+I$
also any element in $\mathbb{Z}_2[x]/(x^{3}+x^{2}+1)$ is of the form $(ax^{2}+bx+c)+I$. There are $2$ choices for each $a,b,c$. So there are $8$ distinct elements in $\mathbb{Z}_2[x]$ . This is because if you divide a polynomial in $\mathbb{Z}_{2}[x]$ by a cubic polynomial, you get a remainder which is a quadratic polynomial of the form $ax^{2}+bx+c$.
So if $p(x)\in \mathbb{Z}_{2}[x]$ then $p(x)=q(x)(x^{3}+x^{2}+1)+r(x)$ where $r(x)=ax^{2}+bx+c$. for some $a,b,c\in\mathbb{Z}_{2}$.
So $p(x)+I = ax^{2}+bx+c + I$ , as $q(x)(x^{3}+x^{2}+1)\in\langle x^{3}+x^{2}+1\rangle$ .