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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?

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    Please don't double post: http://mathoverflow.net/questions/124151/a-question-on-generic-point-and-a-question-on-hartshorne Since questions about Hartshorne are not "research level" it is safest to ask here first and if you don't get a satisfactory answer after a significant time repost noting clearly that you also have it posted here. – Matt Mar 10 '13 at 18:04
  • Your advise is reasonable, I should be more careful about this. Thank you! – Yoshinobu Osawa Mar 11 '13 at 01:28

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$\bullet$ 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 $\rm Z$.

Damien L
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