Points and DVR's - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T04:26:33Z http://mathoverflow.net/feeds/question/12717 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/12717/points-and-dvrs Points and DVR's Randy Brown 2010-01-23T03:08:03Z 2010-01-23T23:28:25Z <p>In the familiar case of (smooth projective) curves over an algebraically closed fields, (closed) points correspond to DVR's.</p> <p>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?</p> <p>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?</p> <p>How about complete integral schemes that are regular in codimension 1?</p> http://mathoverflow.net/questions/12717/points-and-dvrs/12723#12723 Answer by Pete L. Clark for Points and DVR's Pete L. Clark 2010-01-23T03:32:05Z 2010-01-23T06:17:41Z <blockquote> <p>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?</p> </blockquote> <p>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 $l$). </p> <p>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</p> <p><a href="http://math.uga.edu/~pete/8410Chapter1.pdf" rel="nofollow">http://math.uga.edu/~pete/8410Chapter1.pdf</a></p> <blockquote> <p>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?</p> </blockquote> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/12717/points-and-dvrs/12730#12730 Answer by Emerton for Points and DVR's Emerton 2010-01-23T06:13:58Z 2010-01-23T06:13:58Z <p>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 $\mathcal O_x \subset R$).</p> <p>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.</p> <p>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.</p> <p>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 $R = \bigcup \mathcal O_{x_n}.$</p> <p>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 <I>unless</I> 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 $x$. </p> <p>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 $x$.<br /> 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 $x$.</p> <p>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?</p> http://mathoverflow.net/questions/12717/points-and-dvrs/12792#12792 Answer by Qing Liu for Points and DVR's Qing Liu 2010-01-23T23:28:25Z 2010-01-23T23:28:25Z <p>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 &amp; Silverman: "Arithmetic Geometry". </p>