For $m(x) = x^2+1$ in $\mathbb Z_{2}[x]$, we have
$\frac{\mathbb Z_{2}[x]}{ \langle x^2+1 \rangle} = \{0, 1, x, x+1\}$
How do we get that set? I think it's supposed to be a set containing all possible remainders?
For $m(x) = x^2+1$ in $\mathbb Z_{2}[x]$, we have
$\frac{\mathbb Z_{2}[x]}{ \langle x^2+1 \rangle} = \{0, 1, x, x+1\}$
How do we get that set? I think it's supposed to be a set containing all possible remainders?
Yes, it is a set of all possible remainders. Since you are dividing by a polynomial of degree two, the remainder must be of degree at most 1; since you are working over ${\bf Z}_2$, the coefficients must be 0 or 1; that's the list of all polynomials of degree at most 1, with coefficients 0 or 1.