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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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Set-theoretically, $y=0$ on $X$ implies that $z=0$, so $Y=Z(y)$ as set. The quotien $A/yA=k[x,z]/z^2$. So $y$ vanishes at order $2$ on $Y$ and this is the meaning of $\mathrm{div}(y)=2Y$. – user18119 Nov 24 '12 at 9:39
@QiL: What exactly do you mean by $Z(y)$? What is $Z$? What does "$y$ vanishes at order 2 on $Y$" mean? – Manos Nov 26 '12 at 19:15
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 $y$. – user18119 Nov 26 '12 at 20:49
up vote 4 down vote accepted

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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Thanks for the answer. I have a few questions. I can now see that $Z(y)=Y$. 1) How can we show irreducibility of the set $Y$? What is the ideal that corresponds to $Y$? Is it $(y,z) mod (xy-z^2)$? What is its relation to $yA$? 2) I still can not see why $y$ is vanishing at order 2 on $Y$. Could you please elaborate a little bit further? – Manos Nov 28 '12 at 16:22
In particular i can not understand your statement "all points of $Y$ (except $x=y=z=0$) are zeros of order 2 of $y$". Could you please explain this? – Manos Nov 28 '12 at 16:28
Here is an important point that i am missing: by definition, $\operatorname{div}(y)=\sum_{B} v_B(y) \cdot B$, where the summation is over all prime divisors $B$ of $X$. How do we go from this definition to $\operatorname{div}(y)=2 \cdot Y$? – Manos Nov 28 '12 at 23:30
@Manos: As $y$ is a regular function, $v_B(y)\ge 0$ and it is non-zero if $y$ vanishes at $B$. Now the zero locus of $y$ is $Y$, so $v_B(y)>0$ only for $B=Y$. The multiplicity $v_B(y)$ is computed at the end of my answer. – user18119 Nov 29 '12 at 21:19
That's great, i understand. One more thing remains and i am covered: what do you mean by "all points of $Y$ except $x=y=z=0$ are zeros of order 2 of $Y$"? – Manos Nov 30 '12 at 2:50

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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Why is the codimension 1? – Manos Nov 23 '12 at 18:00
Because it has dimension $1$ ! – Georges Elencwajg Nov 26 '12 at 21:40
@GeorgesElencwajg: I don't understand :) – Manos Nov 28 '12 at 16:49
Dear Manos, the codimension of $Y$ is $codim(Y)= dim (X)-dim (Y)=2-1=1$. This follows from the (non-trivial) commutative algebra result Theorem I.8A (b) page 6 in Hartshorne's book . – Georges Elencwajg Nov 28 '12 at 17:12
Dear Manos, this follows from Chapter I, Proposition 1.13 in Hartshorne. – Georges Elencwajg Nov 28 '12 at 19:07

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