Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X$ be the curve $xy-z^2 \subset \mathbb{P}^2$, and let $f$ be the rational function $x/y$ (Edit: I'm trying to simplify this as much as possible, but of course $x$ itself isn't a rational function; my question is only about the divisor of zeros, but the same answer will tell me how to get the divisor of poles).

I am fine with the intuitive (and hand-wavy) method of computing the divisor of zeros (of $x$) by saying something like, "the line $x=0$ intersects $X$ at a single point with multiplicity 2, and so the coefficient of the point defined by their intersection (an irreducible codimension 1 subvariety) is 2."

I'm having trouble proving this in complete rigor using the definitions provided. In particular, if I have an irreducible codimension 1 subvariety $C$ in mind, I want to choose an affine open set $U \subset X$ intersecting $C$ and look at the local equation $\pi$ of $C$ in $U$. Here the ideal of $C$ in $U$ would hence be $(\pi)$, and necessarily for any $f \in k[U]$ there would be a maximal $m$ for which $f \in (\pi^m)$. The valuation of $f$ (and the coefficient of $C$ in the divisor) is defined to be $m$.

So I'm trying this with $C$ simply defined by $x=0$ (as a subvariety/point of $X$). The obvious affine chart is $y=1$, and then $k[U] = k[x,z]/(x-z^2)$. So now $x \in k[U]$, and it is its own local equation. However, it appears that $x \in (x)$, but $x \not \in (x^2) = (z^4)$, so the coefficient of $C$ would be 1.

This obviously doesn't fit with my intuitive understanding of how to compute the divisor of a rational function. I think the problem may be my misunderstanding of local equations, but I haven't yet been able to pinpoint the problem. What am I doing wrong?

share|improve this question
I'm still learning this stuff, so it's entirely possible my "intuitive reasoning" is just plain wrong. If that's the case, please let me know. –  JeremyKun Nov 13 '12 at 1:04
What do you mean by "the rational function $x$"? $x$ is not a rational function on the curve $X$ nor on $\mathbb{P^2}$. –  Makoto Kato Nov 13 '12 at 3:02
good point. Let's say it's $x/y$. I still have to compute the valuation of the $x$ part in the affine slice $y=1$, where there everything is regular. –  JeremyKun Nov 13 '12 at 5:24

2 Answers 2

up vote 1 down vote accepted

In your example, it looks like you want to compute the intersection multiplicity of $V(x)$ with $V(x-z^2)$ at $(x,z)=(0,0)$. To do this, compute the length of the local ring $\left(k[x,z]/(x,x-z^2)\right)_{(x,z)}$. This ring equals $\left(k[x,z]/(x,z^2)\right)_{(x,z)}\cong \left(k[z]/(z^2)\right)_{(z)}\cong k[z]/(z^2)$, which is a length $2$ ring.

Or, you may simply want to notice that the ring $k[x,z]/(x-z^2)\cong k[z]$ and the function $x$ is equivalent to the function $z^2\in k[z]$ under this isomorphism, so in this chart you are really looking at the function $z^2.$

share|improve this answer
I don't see how this fits into the "local equation" picture. Are you saying that $z$ is actually the local equation of $x = z^2$ in $k[U]$? That would fit with my expected answer, but why not $z^2$? –  JeremyKun Nov 13 '12 at 15:26
@Bean, I'm saying that locally in the chart $y=1$ the variety $V(x-z^2)$ has coordinate ring $k[z]$, and the local equation of $V(x)$ is $z^2,$ which we want to use to get the order of vanishing. –  Andrew Nov 13 '12 at 18:45
Ah, I see what you're asking, $z$ is not a local equation for $V(x-z^2).$ Since the coordinate ring is $k[z]$, the local equation is $0\in k[z].$ But $z$ is a regular system of parameters for the local ring $k[z]_{(z)}$, which implies that this local ring is a DVR, and we can compute the order of $V(x)$ by using the fact that $x=z^2$, which must have order $2.$ –  Andrew Nov 13 '12 at 18:55
I think this just boils down to an issue with Shafarevich's (the book I'm using) notation. So we're not actually working with any local equations. We're saying that $pi$ is not a local equation of $x$, but instead we have to take it to be the generator of the defining ideal of $C$ in $U$. Then everything falls into place. –  JeremyKun Nov 13 '12 at 21:36
@Bean, yes I think that's right. I think Shafarevich calls $\pi$ a "local parameter," which is different from a "local equation" in my mind, the latter referring to a locally regular function that cuts out a subspace. (Although, the local parameter can also be viewed as a local equation on $V(x-z^2)$ for the intersection point $(0,0),$ but not for the entire curve!) –  Andrew Nov 13 '12 at 21:42

Let $L$ be the line $y = 0$. Let $V = X - X\cap L$. We can identify $V$ with the affine curve $x = z^2$. Let $p = (0, 0) \in V$. Let $O_p$ be the local ring of $V$ at $p$. Let $m_p$ be the maximal ideal of $O_p$. Then $m_p = (x, z)$. Since $x = z^2, x \in m_p$. Hence $m_p = (z)$. Hence $O_p$ is a discrete valuation ring. Since $x = z^2 \in m_p^2$, the order of $x$ at $p$ is $2$.

share|improve this answer

Your Answer


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.