Let $k$ be a field, let $A=k[x,y,z]/\langle xy-z^2\rangle$ and let $X=\operatorname{Spec}A$. Let $Y:y=z=0$ I want to know the divisor of $y$
In Hartshorne book, because $y=0 \Rightarrow z^2=0$ and $z$ generates the maximal ideal of the local ring at the generic point of $Y$. So, the divisor of $y$ is $2Y$.
My question is
$Y$ means that $V(\langle y,z\rangle)=\{P \in X : \langle y,z\rangle \subseteq P\}$ in $X$?
What is the generic point of $Y$?
Source: Algebraic Geometry, Robin Hartshorne
yes. $Y=Spec\ A/(y,z)=k[x, y, z]/(y, z, xy-z^2)=k[x,y,z]/(y,z)$ since $xy-z^2$ is in the ideal $(y,z)$
The generic point of $Y$ is the prime ideal $(y, z)$.
To compute the divisor of $y$ along $Y$, we look at the local ring $A_{(y,z)}$ in which $x$ has become invertible, so $y=z^2/x$ in this local ring and hence this local ring is $k[x,z]_{(z)}$ and then as you said $z$ is a generator of the maximal ideal, and $y$ is $z^2$ times an invertible element, so the coefficient of div(y) along $Y$ is $ 2$
<and>mean "less than" and "greater than", and produce spacing correct for that meaning only. When you want angle brackets, you need to use\langleand\rangle. Also, to get proper spacing after $\operatorname{Spec}$, you should use\operatorname{Spec}. – Zev Chonoles Jul 01 '13 at 03:27