On page 134, Weil divisors, example 6.5.2, he said:
The divisor of $y$ is $2Y$, because $y=0$ implies $z^2=0$, and $z$ generate the maximal ideal of the local ring at the generic point of $Y$.
I was stupid and can not figure this out. Can someone give a down to earth computation what is the generic point of $Y$ (Depict it using prime ideals), and what is the local ring at the generic point of $Y$? Further, you are give a closed subset of $X$, cut out by several polynomials, how can you compute the generic point of this subset at once?