Ring theory - cosets problem; prove specific result Let $F=\mathbb{Z}_2[x]/I$ where $f(x)=x^3+x^2+[1]\in\mathbb{Z}_2[x]$ and $I=(x^3+x^2+[1])\mathbb{Z}_2[x]$. With $X=x+I\in F$ and writing $a$ for $a+I$ with $a\in\mathbb{Z}_2$ show that the following hold:
(i) $X^3=X^2+[1]$
(ii) $X^4=X^2+X+[1]$
The way I've been taught is to say $X^3+X^2+1=0$ from the initial polynomial $f$ (not sure why we assume this to be zero). From this, $X^3=-X^2-[1]$, but its negative is $X^2+[1]$ and parity does not change under negation. Thus $X^3=X^2+[1]$
For (ii) we have $X^4=X+X^3$ (since the 'multiplication' is in fact addition) and so $X^4=X^2+x+[1]$. The idea that multiplication was intended to be addition comes from the fact that this hold immediately in this case and if we simply wrote $X^4=XX^3$ we would get the wrong answer.
How can I better understand this topic and write better solutions?

You could use the congruence notation, in the same way as you would write $13 \equiv 3 \bmod 5$. In your case, your question amounts to show that
(i) $X^3 \equiv X^2+1 \bmod X^3 + X^2 + 1$
(ii) $X^4 \equiv X^2+X+1 \bmod X^3 + X^2 + 1$
You already observed that, in your ring of characteristic $2$, the conditions $g = h$, $g-h= 0$ and $g+h= 0$ are equivalent. In particular, (i) is equivalent to
(i') $X^3 + X^2 + 1 \equiv 0 \bmod X^3 + X^2 + 1$
which is trivially true. Similarly, (ii) is equivalent to
(ii)' $X^4 + X^2 + X + 1 \equiv 0 \bmod X^3 + X^2 + 1$
which is true since $(X^3 + X^2 +1)(X + 1) = X^4 + X^2 + X + 1$.
