Issues with ideals

Solution 1:

This is because $x^3 + x^2 + 1 \in I$ by definition and therefore $(x^3 + x^2 + 1) + I = I$ meaning that, $$ (x^3 + x^2) + I = (x^3 + x^2) + (x^3 + x^2 + 1) + I = 1 + I $$ because the characteristic is $2$.

Solution 2:

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 $q(x)(x^{3}+x^{2}+1)\in\langle x^{3}+x^{2}+1\rangle$ .