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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?

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    I see, so if I was working in $\mathbb Z_{3}$ would it be $0x+0, 0x+1, 0x+2, 1x+0, 1x+1, 1x+2$? and is this true for all $m(x)$? What if $m(x) = x^3 + 5$ instead?2012-11-08

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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.

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    Yes. ${}{}{}{}$2012-11-08