# Two examples: discrete valuation rings on curves

Let’s look at the curves ${\mathcal C : y = x^2}$ and ${\mathcal D : y^2 = x^3}$. They’re pretty simple, so naturally, we’re going to use some big tools to analyze him. Today the plan is to look at these curves by studying how functions behave on them locally.

1. Vanishing at a point

Every point on ${\mathcal C}$ is of the form ${(x,x^2)}$. Let’s pick a point ${p = (a,a^2)}$. Here’s some of the usual objects:

• The ring of polynomial functions ${{\mathbb C}[\mathcal C] = {\mathbb C}[x,y]/(x-y^2)}$
• The ideal corresponding to ${p}$ is ${\mathfrak m_p = (x-a,y-a^2)}$

As is the common practice, we’ll let ${\mathcal O_p}$ denote the ring

$\displaystyle \mathcal O_p = {\mathbb C}[\mathcal C]_{\mathfrak m_p} = \left\{ \left. \frac f g \right| f,g \in {\mathbb C}[\mathcal C] \text{ and } g(p) \neq 0 \right\} = \left\{ \left. \frac f g \right| f,g \in {\mathbb C}[\mathcal C] \text{ and } g \not\in \mathfrak m_p \right\} \mathrm.$

This ring (and its properties) are going to be our main tool for studying the local properties of functions.

This ring contains a special ideal:

$\displaystyle \mathfrak v_p = \left\{ \left. \frac{f}{g} \right| f(p) = 0 \right\} \mathrm.$

This ideal is the unique maximal ideal in ${\mathcal O_p}$, simply because it contains every non-unit in ${\mathcal O_p}$ (the only way ${\frac{f}{g}}$ can fail to be a unit is if ${f \in \mathfrak m_p}$, which is equivalent to ${f(p) = 0}$).

We now consider the following question: what can we say about the generators of this ideal? Perhaps we can convince ourselves that

$\displaystyle \mathfrak v_p = \left(\frac{x-a}{1}, \frac{y-a^2}{1}\right) \mathrm,$

but, in fact, ${\mathfrak v_p}$ is actually a principal ideal. Indeed, we have the relation

$\displaystyle \frac{y-a^2}{1} = \frac{x^2-a^2}{1} = \frac{(x-a)(x+a)}{1} \mathrm,$

which implies (since ${\frac{x+a}{1}}$ is a unit in ${\mathcal O_p}$) that

$\displaystyle \mathfrak v_p = \left(\frac{y-a^2}{1}\right) = \left(\frac{x-a}{1}\right) \mathrm.$

Since every element of ${\mathcal O_p}$ is either a unit or is in ${\mathfrak v_p}$, and ${\mathfrak v_p}$ is now known to be a principal ideal, we know that every element of ${f \in \mathcal O_p}$ can be written in the form

$\displaystyle f = u\pi^d \text{ where } \mathfrak v_p = (\pi),\ d \in {\mathbb Z}_{\geq 0}$

(for instance, we may take ${\pi = \frac{y-a^2}{1}}$ or ${\pi = \frac{x-a}{1}}$).

We define a map ${v_p : \mathcal O_p^\times \rightarrow {\mathbb Z}_{\geq 0}}$ which sends ${v_p(f) = d}$. This map is a homomorphism of the (multiplicative) monoid ${\mathcal O_p^\times}$ to the (additive) monoid ${{\mathbb Z}_{\geq 0}}$. This map extends to the field of fractions of ${\mathcal O_p}$, whose elements are of the form ${\frac{f}{g}}$ without any restrictions on ${g}$. Here we have

$\displaystyle \bar v_p : FF(\mathcal O_p)^\times \rightarrow {\mathbb Z}$

acting by

$\displaystyle \bar v_p\left(\frac f g\right) = v_p(f) - v_p(g) \mathrm.$

Here’s a simple example, still on our curve ${\mathcal C}$. Consider the function

$\displaystyle F(x,y) = \frac{(x-2)(x+3)}{(y-4)(x+2)} \mathrm.$

At the point ${p = (2,4)}$ we have

$\displaystyle \bar v_{(2,4)}(F) = v_{(2,4)}\big( (x-2)(x+3) \big) - v_{(2,4)}\big( (y-4)(x+2) \big) = 1 - 1 = 0 \mathrm,$

indicating no vanishing or poles at this point. At the point ${q = (-3,9)}$ we have

$\displaystyle \bar v_{(-3,9)}(F) = v_{(-3,9)}\big( (x-2)(x+3) \big) - v_{(-3,9)}\big( (y-4)(x+2) \big) = 1 - 0 = 1 \mathrm,$

indicating vanishing to order ${1}$ at this point. At the point ${r = (4,16)}$ we have

$\displaystyle \bar v_{(4,16)}(F) = v_{(4,16)}\big( (x-2)(x+3) \big) - v_{(4,16)}\big( (y-4)(x+2) \big) = 0 - 1 = -1 \mathrm,$

indicating a pole of order ${1}$ at this point.

2. Discrete Valuation Rings

A ring ${R}$ is said to be a discrete valuation ring (or DVR) if there is a surjective group homomorphism ${v : R^\times \rightarrow {\mathbb Z}}$. In the example above, we saw that ${FF(\mathcal O_p)}$ is a discrete valuation ring with the map ${\bar v_p}$.

Does this always happen for curves? As it happens, no; a necessary and sufficient condition for ${FF(\mathcal O_p)}$ to be a DVR is that ${p}$ must be a smooth point of the curve (this is a consequence of Nakayama’s lemma, which allows us to take a basis for the ${1}$-dimensional vector space ${\mathfrak v_p/\mathfrak v_p^2}$ and lift it to a generator for the ideal ${\mathfrak v_p}$).

We can see an example of why smoothness is necessary. Consider the curve ${\mathfrak D : y^2 = x^3}$. Here we know that the curve is smooth everywhere besides ${o = (0,0)}$. We’ll now see that ${FF(\mathcal O_p)}$ is a DVR everywhere except at ${p=o}$.

Our basic trick will be the following: in order for ${FF(\mathcal O_p)}$ to be a DVR, we need the ideal ${\mathfrak v_p}$ to be principal. This will only happen when ${p \neq o}$.

Here the situation is pretty simple. Let ${p = (a^2,a^3)}$ be a point on ${\mathfrak D}$, so that ${\mathfrak v_p = (x-a^2, y-a^3)}$. Then

$\displaystyle \frac{y-a^3}{1} = \frac{(y-a^3)(y+a^3)}{y+a^3} = \frac{y^2-a^6}{y+a^3} = \frac{x^3 - a^6}{y+a^3} = \frac{(x-a^2)(x^2 + a^2x + a^4)}{y+a^3} \mathrm,$

which gives us a relationship between the two generators of ${\mathfrak v_p}$. We now note that

$\displaystyle p \neq 0 \iff a \neq 0 \iff \frac{x^2+a^2x+a^4}{y+a^3} \text{is a unit in} \mathcal O_p \mathrm,$

so that ${\mathfrak v_p}$ is principal precisely when ${p \neq o}$.