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Let $k$ be a field and let $A=k[x,y,z]/(xy-z^2)$. Let $X=\operatorname{Spec}A$. What exactly do we mean by a ruling of the cone $Y:y=z=0$? Why is $Y$ a prime divisor of $X$? Edited: What do we mean when we say that $Y$ can be cut-out set-theoretically by the function $y$? Why is the divisor of $y$ equal to $2Y$? I am trying to understand example 6.5.2, p. 133 from Hartshorne and i am interested in seeing the details that Hartshorne omits. Could somebody please explain this example?

Edited: By definition, $\operatorname{div}(y)=\sum_{B} v_B(y)B$, where the summation is over all prime divisors $B$ of $X$. How do we go from this definition to $\operatorname{div}(y)=2Y$?

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    I meant by $Z(y)$ the zero-set of $y$ (to distinguish with $V(y)$ the closed subscheme defined by the ideal $yA$). All points of $Y$ (except $x=y=z=0$) are zeros of order $2$ of $y$.2012-11-26

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As I said in the comment, the closed subset $Z(y)$ of zeros of $y$ is equal to $Y$ because $y=0$ implies that $z=0$. If you prefer, $\sqrt{yA}=(y,z)$.

So the support of $\mathrm{div}(y)$ is $Y$. To compute the multiplicity of this divisor, we consider the generic point $\eta$ of $Y$ and have to compute the valuation $v_{\eta}(y)$, where $v_{\eta}$ is the valuation on $K(X)$ (field of rational functions on $X$) defined by the discrete valuation ring $O_{X, \eta}$.

Let $\mathfrak p=(y, z)$ be the prime ideal of $A$ defining the point $\eta$. Then $\mathfrak p A_{\mathfrak p}$ is generated by $z$ because $y=z^2x^{-1}$. Therefore $v_{\eta}(y)=2v_{\eta}(z)=2$. So $\mathrm{div}(y)=2Y$.

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    @Manos, Outside from this point $Y$ is smooth and parametrized by $z$. As $y=z^2/x$, it vanishes with order two at any point $(a,b,c)\in Y$ with $a\ne 0$.2012-11-30
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Isn't $Y = \operatorname{Spec} k[x,y,z]/(xy-z^2,y,z) = \operatorname{Spec} k[x]$? So it is an integral closed subscheme of $X$ of codimension $1$.

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    Dear Manos, this follows from Chapter I, Proposition 1.13 in Hartshorne.2012-11-28