Points and DVR's - MathOverflow

Points and DVR's

2010-01-23T03:08:03Z

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?

Answer by Pete L. Clark for Points and DVR's

2010-01-23T03:32:05Z

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?

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

http://math.uga.edu/~pete/8410Chapter1.pdf

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?

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.

Answer by Emerton for Points and DVR's

2010-01-23T06:13:58Z

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?

Answer by Qing Liu for Points and DVR's

2010-01-23T23:28:25Z

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".