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In the familiar case of (smooth projective) curves over an algebraically closed fields, (closed) points correspond to DVR's.

What if we have a non-singular projective curve over a non-algebraically closed field? The closed points will certainly induce DVR's, but would all DVR's come from closed points? Is there a characterization of the DVR's that aren't induced by closed points?

And how about for a general projective variety that is regular in codimension 1 (both for algebraically closed and non-algebraically closed)? Point of codimension 1 induce DVR's. Do they induce all of them? What is the characterization of the ones they do induce?

How about complete integral schemes that are regular in codimension 1?

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3 Answers

up vote 8 down vote accepted


Yes: let $L/k$ be a function field in one variable, so it can be given as a finite separable extension of $K = k(t)$. Then the discrete valuations on $L$ which are trivial on $k$ are in canonical bijection with the closed points on the unique regular projective model $C_{/l}$, where $l$ is the algebraic closure of $k$ in $L$ (since $l/k$ is finite, any valuation which is trivial on $k$ is also trivial on $l$).

By coincidence, this is almost exactly where I am in a course I am now teaching, although I won't insist on the geometric language: see Section 1.7, especially Exercise 1.22, of



No, they do not induce all of them (even if the variety is smooth, which I will assume for simplicity). The problem here is that, unlike in dimension one, there is no unique nonsingular projective model, so e.g. there will be discrete valuations on $k(t_1,t_2)$ which correspond not to codimension one points on $\mathbb{P}^2$ but to points on (at least) some arbitrary blowup of $\mathbb{P}^2$. Because of this, I am pretty sure that there is no simple valuation-theoretic characterization of those valuations which correspond to closed points on a particular projective model of the function field.

By the way, I do not understand valuation theory on function fields in more than one variable very well, so I especially welcome further responses which elaborate on this issue.

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Suppose that $X$ is a projective variety, and that $v$ is a discete valuation on $K(X)$ (trivial on $k$) whose corresponding valuation ring we will denote by $R$. The valuative criterion shows that the map Spec $K(X) \rightarrow X$ extends to a map Spec $R \rightarrow X$. If I have the terminology correct, the image of the closed point of Spec $R$ is called the centre of the valuation $v$ on $X$. It has codimension anywhere between $1$ and dim $X$. Note that if $x \in X$ is the centre, then $R$ dominates $\mathcal O\_x$ in $K(X)$ (i.e. we have a local inclusion of local rings $\mathcal O_x \subset R$).

Let's suppose for a moment that $X$ is a smooth surface. If the centre $x$ is codimension 1, then both $\mathcal O\_x$ and $R$ are (discrete) valuation rings. Since valuation rings are (characterized by being) maximal for the partial order of dominance, $R$ and $\mathcal O\_x$ coincide, and so the discrete valuation $v$ is just that given by the divisor of which $x$ is the generic point.

Suppose instead that $x$ is a closed point. Now we can blow up $x$ in $X$, to get a projective variety $X\_1$, and the centre of $v$ in $X\_1$ will now be contained in the exceptional divisor of $X\_1$ (i.e. the preimage of $x$). If it coincides with the exceptional divisor, then we have found a curve on $X\_1$ giving rise to $v$; otherwise it is a point $x\_1$, which we can blow up again.

Either we eventually obtain a divisor on some iterated blow-up of $X$, or we obtain a sequence of points $x \in X, x\_1 \in X\_1, \ldots,$ with each $X\_n$ a blow-up of the previous. In this case one sees that $R = \bigcup \mathcal O\_{x\_n}.$

There are a couple of exercises related to this issue in Hartshorne, namely II.4.5, II.4.12, and V.5.6. If I understand them correctly, any such sequence of $x\_n$ gives a valuation ring $R$ in this way, and $R$ is a discrete valuation ring unless one constructs the sequene $x_n$ in the following manner: choose an irreducible curve $C$ in $X$ and define $x\_n$ to be the intersection of the proper transform of $C$ in $X\_n$ with the exceptional divisor. For a sequence $x_n$ constructed in this latter manner, one obtains not a discrete valuation ring, but rather a rank 2 valuation ring: the valuation is determined by first taking the valuation at the generic point of $C$, and then (for those functions which are defined and non-zero at this generic point) restricting to $C$ and computing the order of zero or pole at $x$.

What is the geometric intuition for the discrete valuation rings that correspond to an infinite sequence $x_n$ rather than to some curve on $X$? One can think of them as a transcendental curve on $X$, passing through $x$.
Indeed, imagine you had such a curve. Then you could restrict a rational function to it; since it is transcendental, a non-zero rational function would not have a zero or pole along this curve, and so would restrict to give a non-zero meromorphic function on the curve. We could then compute the order of the zero or pole of this meromorphic function at $x$. In other words, because the curve is transcendental, we get a rank one valuation, in contrast to the rank two valuations that arise when we apply this process with an algebraic curve $C$ passing through $x$.

I'm not sure about the details of the higher dimensional case. (Among other things, I am worried about the possibility of the center being codim > 1, but singular, which seems like it could complicate the analysis.) Does anyone here know how it goes?

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I wish this site existed when I was a grad student! –  Hailong Dao Jan 23 '10 at 6:16
I understand things better after reading M. Emerton's answer, but still not completely. I think it goes to show that the typical modern (i.e., since Grothendieck) presentation of algebraic geometry leaves out some key knowledge of valuation theory that our forebears in the Zariski era knew very well. So probably the answer is to consult an older text on the subject. I know Zariski and Samuel contains a bit of material on this; I'm not sure if it has the whole story. Can anyone else suggest a reference? –  Pete L. Clark Jan 23 '10 at 6:27
That's a wonderful explanation! Is there a formalization of these "transcendental curves" you're talking about? Where can I read about them? –  H. Hasson Feb 26 '10 at 19:59
Just imagine something like the graph of $y = e^x$ passing through the point $(0,1)$ in the plane. To see these kinds of arguments used to prove theorems in algebraic geometry, one can look at some of the papers of Zariski, e.g. in volume 1 of his collected works. The technical flavour of these papers is quite different to the post-Grothendieckian, cohomologically-intensive modern view-point, but some of them are very readable even now, and many of them are masterpieces, estabilishing results (such as Zariski's Main Theorem --- the original version!) of fundamental importance. –  Emerton Feb 27 '10 at 1:25
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To complete partly the answer of Emerton, the picture for DVR is relatively clear. Let $X$ be an integral noetherian scheme and let $R$ be a DVR with field of fractions equal to the field of rational functions $k(X)$ on $X$. Suppose that $R$ has a center $x\in X$ (e.g. if $X$ is proper over a subring of $R$). Let $k_R$ be the residue field of $R$. Then $k_R$ has transcendental degree over $k(x)$ bounded by $\dim O_{X,x} -1$. Suppose further that $X$ is universally catenary and Nagata (e.g. $X$ is excellent), then the equality holds if and only if the center of $R$ in some $X'$ proper and birational over $X$ is a regular point of codimension 1. This is a theorem of Zariski. See M. Artin: ''Néron Models'', § 5, in Cornell & Silverman: "Arithmetic Geometry".

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To summarize this very complete answer: for any proper $k$-variety $X$, Zariski theorem says that every discrete valuation of $k(X)$ corresponds to a prime divisor on some blowup of $X$. For this reason, discrete valuations of $k(X)$ are called "geometric". –  VA. Jan 23 '10 at 23:55
VA: you mean one can omit the condition on the transcendental degree of the residue extension ? –  Qing Liu Jan 24 '10 at 0:19
Qing: my bad. I was trying to simplify your answer: if $X$ is a proper variety then obviously $R$ has a center on $X$, and $X$ is excellent. But you are right, the condition on the transcendental degree is not removable. –  VA. Jan 24 '10 at 0:41
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