Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer

up vote 4 down vote accepted

If $X=Spec(R)$ is an affine scheme (as it is in the example you refer to) and $Y$ is the subscheme defined by a prime ideal $P$, then the generic point of $Y$ is the point $[P]\in Spec(R)$. The local ring at that point is the localization $R_P$.

In the Hartshorne example, $R={\mathbb C}[x,y,z]/(xy-z^2)$, and $P$ is the ideal $(y,z)$. (You really should have mentioned this in your question). In the ring $R_P$, the maximal ideal is $(y,z)$, which is the same as $(z)$ (because $x$ is a unit). Because $y$ is a unit times $z^2$, the divisor of $y$ is twice the divisor of $z$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.