TITLE: A question on generic point and a question on Hartshorne
QUESTION [6 upvotes]: 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?

REPLY [3 votes]: $\bullet$ The coordinate ring of the cone is $k[\rm{ X, Y ,Z}] / (XY - Z^2)$ ;
$\bullet$ The prime ideal representing $\rm Y$ is $\mathfrak p =  \rm (Y, Z)$ ;
so the local ring you want is $(k[\mathrm{X, Y, Z}]/\rm (XY-Z^2))_{(Y,Z)}$. 
Now in this local ring $\rm X$ is invertible and $\rm XY - Z^2 = 0$ which implies $\rm Y = X^{-1} Z^2$;
so the maximal ideal $\mathfrak p (k[\mathrm{X, Y, Z}]/(\rm XY - Z^2))_{\mathfrak p}$ which is by definition generated by $\rm Y$ and $\rm Z$, is  only generated by $\rm Z$.