Cosets of $\mathbb{Z}_2[x]/<x²+1>$

Question

This PDF says that the cosets of $$\frac{\mathbb{Z}_2[x]}{<x²+1>}$$

looks like the following:

$$0+<x²+1>, x^{57}+x³+1+<x²+1>, (x²+1)+<x²+1>$$

Then it says that we can add and multiply cosets like the following:

$$[(x²+x+1)+<x²+1>]+[(x²+1)+<x²+1>] = x+<x²+1>$$

When you divide $2x²+x+2$ by $x²+1$ you get $x$, this is the reduction mod $x²+1$, rigth?

and

$$[(x+1)+<x²+1>][(x²+1)+<x²+1>] = (x³+x²+x+1)+<x²+1>$$

why there's no reduction mod $x²+1$ here?

Also, it says that $[(x+1)+<x²+1>][(x+1)+<x²+1>] = (x^2+1)+<x^2+1>$

what happened here? $(x+1)(x+1) = x²+2x+1$, how did it turn in $x²+1$ again?

Also, it says that since multiples of $x²+1$ are congruent to 0, we have:

$$x^2+<x²+1> = [x²+(x²+1)]+<x²+1> = 1+<x²+1>$$

what's happening???

Answer 1

The pdf does not say that those are the cosets. It says "here are some cosets."

One set of representatives of cosets is $\{0,1, x , x+1\}$. Any other representatives can be reduced to these by dividing by $x^2+1$.

Answer 2

You have polinomyals with coefficents in $\mathbb{Z}_2$, and the relation $X^2 + 1 = 0$, in other words $X^2 = -1$. So you can reduce every polinomyal by replacing the above equivalence several times. You only get elements of the form $aX + b$ with $a,b \in \mathbb{Z}_2$ thus there are only $4$ representatives, namely $0, 1, X, X+1$. Note that in this ring with $4$ elements you get $(X+1)^2 = X^2 + 2X + 1 = -1 + 0 + 1 = 0$, thus this is not an integral domain (nor a field).