Multiplication and Addition tables the following: What would be the addition and multiplication tables of $Z_2[x]/\langle x^2 + x\rangle$?
I know how to do the addition and multiplication tables for normal modular arithmetic, butam not sure about this.
 A: The tables are 
$$
\begin{array}{r|c c c c}
+ & 0 & 1 & x & 1+x \\
\hline
0 & 0 & 1 & x & 1+x \\
1 & 1 & 0 & 1+x & x \\
x & x & 1+x & 0 & 1 \\
1 + x &  1+x & x & 1 & 0
\end{array}
$$
and
$$
\begin{array}{r|c c c c}
\times & 0 & 1 & x & 1+x \\
\hline
0 & 0 & 0 & 0 & 0 \\
1 & 0 & 1 & x & 1+x \\
x & 0 & x & x & 1 \\
1 + x &  0 & 1+x & 0 & 1+x
\end{array},
$$
where we have identified the elements $0,1,x,1+x$ with their cosets $0 + I$, etc, where $I$ is the ideal in question.
But how do we arrive at these tables? Well, the first step is to find manageable representatives for the cosets $p + I$ for any polynomial $p$. These representatives are $0,1,x,$ and $1+x$.
The way we find these representatives is by taking an arbitrary polynomial 
$$
p(z) = a_nx^n + ... + a_0 \in \mathbb{Z}/2\mathbb{Z}[x].
$$
Notice however, that $x + x^2 \in I$, so $x + I = x + (x^2+x) I = x^2 + I.$ Similarly, we also have that $x^n + I = x^{n+1} + I$ for any $n \geq 1.$ This relation shows that
$$
p(x)+I = p(z) + a_nx^n + a_nx^{n-1} + I = (a_{n}+a_{n+1})x^{n-1} + a_{n-2}x^{n-2} +... +I.
$$
Now, apply this trick again with the polynomial on the right hand side of the above. We gradually reduce the degree of our representative of $p(x) + I$. When we can no longer apply this trick, we must have that the highest degree monomial in our representative $q(x)$ must be less than $2$. Otherwise, we could simply apply the trick again. Thus $q$ must be one of $0,1, x,$ or $1+x$, as these are the polynomials of degree less than $2$ in $\mathbb{Z}/2\mathbb{Z}[x]$. To compute the addition and multiplication tables, we now just take representatives of any two cosets and reduce them in this fashion.
For example, if we are computing what $(x+I)(1+x+I)$ is, we know by definition that this is equal to $x+x^2 + I$. Since $x+x^2 \in I$, $x+x^2 + I = I = 0 +I$. Thus $x(1+x) = 0$ in the ring $\mathbb{Z}/2\mathbb{Z}[x]/I$. Similarly, if we are computing $(x+I)(x+I)$ in the ring, we see by definition that this is $x^2 + I$. Since $x^2 + x \in I$ $x^2 + I = x^2 + x^2 + x + I = x + I$. Thus $x^2 = x$ in $\mathbb{Z}/2\mathbb{Z}[x]/I.$
