Intuitively, this makes sense since in $\Bbb{Z}[x]/(x-2) $ we have $x=2$, and we clearly have that $\{a_0 +a_1(2)+a_2(2)^2+...\: a_i \in \Bbb{Z}\} \cong \Bbb{Z}$, as there is a bijection $\Bbb{Z} \to \{a_0 +a_1(2)+a_2(2)^2+...\: a_i \in \Bbb{Z}\}$ given by $n \rightsquigarrow n$ (i.e., just send each integer $n$ to the 'polynomial' where $a_i = 0$ for all $i>0$, and $a_0 =n$).
However, Artin states: "the principle ideal $(g)$ of $\Bbb{Z}[x]$ is the kernel of the homomorphism $\Bbb{Z}[x] \to \Bbb{Z}$ that sends $x \rightsquigarrow 2$, so when we kill $x-2$ in $\Bbb{Z}[x]$, we obtain a ring ismorphic to $\Bbb{Z}$."
I would like some help understanding what this actually means, and if my 'intuitive understanding' is correct. Also, is it always possible to simplify a quotient ring such as $\Bbb{Z}[x]/(ax-b)$ by saying $\Bbb{Z}[x]/(ax-b) \cong \Bbb{Z}[\frac{b}{a}]$? Thanks!