# Connecting the intuitive way to compute divisors of rational functions to the rigorous definition.

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?

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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

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 $z^2.$

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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 $2$.