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I read a set of notes on projective curves, where it introduced discrete valuations on a field $K$ as a surjective homomorphism $v:K^\times \to \mathbb Z$ where $v(x+y) \geq \min(v(x), v(y))$ for all $x,y$. Then proceeds to give $\mathcal O_P$, the local ring at $P$, the structure of a discrete valuation ring by counting the order of vanishing of rational functions at $P$.

The way that I see it, I don't think the structure of the discrete valuation or the DVR had any significance in what's to follow (counting degrees of functions, constructing the divisor class group and Jacobian). One could have said to consider $v_P(f)$ as the order of vanishing of $f$ at $P$, without mentioning the structure of discrete valuation, and proceed without any problems.

So I am wondering what am I missing here. Is there a deeper significance of discrete valuation that I'm not seeing? And are there examples in other settings of mathematics where DVRs are used (where the valuation is not order of vanishing)?

pooper
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    What's the order of vanishing of a function on the curve $y^2 = x^3$ at the point $(0,0)$? (Here $K = \frac{k(x)[y]}{(y^2 - x^3)}$, where $k$ is a ground field.) Should the order of vanishing of the function $x^3$ be $3$? Or since $x^3 = y^2$, should it be $2$? Being a DVR ensures the existence of a uniformizing parameter, which allows you to define the order of vanishing sensibly. (It also is an algebraic manifestation of nonsingularity.) Without one it's not possible to define the order of vanishing and have it possess the nice properties we expect. – Viktor Vaughn Apr 10 '20 at 06:04
  • @RichardD.James: I see, smoothness ensures that this order of vanishing makes sense, and DVR is the construction to make this concrete. Do you know if there are ways we can measure how far the local ring is from being a DVR, and maybe give us a sense of how singular the curve is at a point? Perhaps (me thinking naively) $y^3 = x^4$ is "more singular" than $y^2 = x^3$ at $0$, how can we measure that? – pooper Apr 10 '20 at 07:28
  • Upon inspection it seems that all PID local rings can be given a DVR structure. The problem with $y^2 = x^3$ is that while the localization at $0$ is a local ring (unique maximal ideal), it is not a PID (need both $x$ and $y$ to generate the ideal of functions vanishing at $0$). Surely this is directly related to all partial derivatives vanishing. For instance $y = x^2$ doesn't have this problem, the maximal ideal of the corresponding local ring at $0$ is generated by $x$. – pooper Apr 10 '20 at 12:40
  • Yes, a local PID is the same thing as a DVR. Yes, your comment about the local ring and the partials vanishing is correct. A point $P$ of a variety $V$ is smooth iff the local ring $\mathcal{O}_{V,P}$ is a regular local ring. Atiyah--MacDonald cover this nicely, and I mention some of the relevant results here. I'll try to turn these comments into an answer later. – Viktor Vaughn Apr 11 '20 at 02:20

2 Answers2

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Historically one/the(?) reason for the prominent appearance of discrete valuations in the case of algebraic curves is that one gets a theory that possesses strong similarities to algebraic number theory. Here are some points of the story:

  • In algebraic number theory the $p$-adic valuations on $\mathbb{Q}$ and their extensions to extension fields $K|\mathbb{Q}$ of finite degree are a tool of central importance. All these valuations are discrete.
  • Along with every algebraic curve $C$ comes its field of rational functions $K(C)$, which is an extension of finite degree of a rational function field $K(x)$ in one variable $x$.
  • Vice versa: Given an extension field $F$ of finite degree over a rational function field $K(x)$ there exists a unique projective algebraic curve $C$ without singularities such that $F=K(C)$. The points of this curve are in bijection with the discrete valuations on $F$ that are trivial on $K$. (In fact more is true here.)
  • As a consequence of the vague points just stated one could say that doing geometry with projective algebraic curves without singularities is the same as doing arithmetic in function fields $F|K$ of one variable.
  • In the case of a finite field $K$ there are strong similarities doing arithmetic in finite extensions of $\mathbb{Q}$ or in finite extensions of $K(x)$.
Hagen Knaf
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Let $X$ be a smooth projective curve over a field $k$, let $P$ be a point on $X$, let $\DeclareMathOperator{\O}{\mathcal{O}} \O_P$ be the local ring at $P$ with maximal ideal $\DeclareMathOperator{\m}{\mathfrak{m}} \m_P$, and let $K = k(X)$ be the function field of $X$. Let's try to define order of vanishing without assumptions and see what goes wrong. We have a filtration of $\O_P$ by powers of $\m_P$: $$ \O_P \supsetneq \m_P \supsetneq \m_P^2 \supsetneq \cdots \, . $$ So for $f \in \O_P$, let $v(f) = \max\{r \in \mathbb{Z}_{\geq 0} : f \in \m_P^r\}$. However, without further assumptions on $X$, we already run into trouble. As I pointed out in the comments, let $X: y^2 = x^3$ be the cuspidal cubic plane curve and let $P = (0,0)$. Then $\m_P = (x,y)$, so $v(x) = v(y) = 1$, but $v(y^2) = v(x^3) = 3$, so $v$ is not a homomorphism: $1+1 \neq 3$. One way to remedy this is to insist that the local ring $\O_P$ be a DVR for all $P$, which is equivalent to insisting that $\m_P$ be principal for all $P$. (See this answer for more equivalent conditions.) Then one can easily show that $v$ is a homomorphism by writing an element of $\O_P$ as $f = u t^r$ where $t$ is the uniformizer and $u$ is a unit. In geometric terms, this is the same as insisting that $X$ be nonsingular.

We similarly run into trouble defining the class group. Let $X_0$ be a nonempty affine open subset of $X$, and let $A = k[X_0]$ be its coordinate ring. In order to get a group, we need every fractional ideal of $A$ to be invertible, but this is equivalent to insisting that $A$ be a Dedekind domain. And tying back in with the first paragraph, $A$ is a Dedekind domain iff $A_P$ is a DVR for every nonzero prime $P$.

The definition of a DVR may seem a bit unnatural, at first. But the various alternative characterizations---a local PID, a local Dedekind domain, or a noetherian normal local domain of Krull dimension $1$---arise very naturally.

Even in scheme theory, valuations are still very useful. For noetherian schemes, they show up in the valuative criteria for properness and separatedness, which allow one to determine whether a map is proper or separated (the scheme-y versions of compact and Hausdorff, respectively) purely in terms of DVRs.

Viktor Vaughn
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