Prove or disprove that the ring $\mathbb{Z}[x]/(x^2-1)$ is an integral domain 
Prove or disprove  that the ring $\mathbb{Z}[x]/(x^2-1)$ is an
  integral domain.

It is easy to see that $\mathbb{Z}[x]/(f)=\{P+(f): \deg(P)<\deg(f)\}$. If $\deg(P) \geq \deg(f)$, there exists $q,r \in \mathbb{Z}[x]$ such that $P=qf+r$ with $r=0$ or $\deg(r)< \deg(f) \implies P+(f)=qf+r+(f)=r+(f)$. So $\mathbb{Z}[x]/(f) = \{0,1, x, (x-1), (x+1)\}$. But $(x-1)(x+1)= x^2-1 \equiv 0 \mod (x^2-1)$.
Conclusion : Our ring is not an integral domain.
Question : Am I right? Otherwise, is anyone could help me on the ambiguity of this problem?
 A: You are making it hard on yourself. To show that the quotient ring is not an integral domain, you need only find any zero divisor. You are not correct in your description of what $\mathbb{Z}[x]/(x^2-1)$ is as a ring but you have correctly pulled out the necessary fact: $\overline{x+1},\overline{x-1} \in \mathbb{Z}[x]/(x^2-1)$ and $\overline{x-1} \cdot \overline{x+1}=\overline{x^2-1}$ which is $\overline{0}$ in $\mathbb{Z}[x]/(x^2-1)$ but neither $\overline{x-1},\overline{x+1}$ are zero. 
What is $\mathbb{Z}[x]/(x^2-1)$ really? As you said for any polynomial in $\mathbb{Z}[x]$, say $p(x)$, we can use division to express $p(x)$ as $p(x)=q(x)(x^2-1)+r(x)$, where $\deg r(x)<2$ (that is, $\deg r(x)$ is $0$ or $1$). Then in $\mathbb{Z}[x]/(x^2-1)$, $\overline{p(x)}=\overline{q(x)(x^2-1)+r(x)}=\overline{q(x)(x^2-1)}+\overline{r(x)}=\overline{r(x)}$. So all nonzero elements in $\mathbb{Z}[x]/(x^2-1)$ (those which aren't multiples of $x^2-1$, i.e. where $r(x)=0$) are any nonzero elements in $\mathbb{Z}[x]$ of degree $1$ or $0$. Then as a set, $\mathbb{Z}[x]/(x^2-1)=\{ax+b\;|\;a,b \in \mathbb{Z}\}$.
