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Hello, I'm looking for someone who can help me to understand Zariski's theory of valuations.

First I outline the theory: we take a field $K$ which is a finitely generated transcendent extension of another field $k$. We only consider the case $k=\mathbb{Q}$ or $\mathbb{C}$. By definition, A model of $K$ is a variety $V\subset \mathbb{CP}^n$ defined over $k$, such that the rational function field of $V$ over $k$ is isomorphic to $K$. We define the underlying topological space of $V$ to be a space, whose points are irreducible subvarieties of $V$, endowed with Zariski topology.

Now comes the interesting thing: Zariski gave an homeomorphism between the space of valuations on $K/k$ and the inverse limit of underlying topological spaces of all models of $K$.

Question: Plz give me some concret examples of the above correspondence.

The only example I know is that, given an irreducible hypersurface of a model $V$, one can count the order of rational functions on $V$ over the hypersurface. This gives a discret rank one valuation.

Is there some other easily-described points in the inverse limit, whose corresponding valuations are non-discret, or of higher rank?

Other comments are welcome!

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Easiest example: a complex curve. Every valuation is discrete of rank one, the Zariski-Riemann space is homeomorphic to the smooth model of C. This is the motivating example (no inverse limit needed, and well known before Zariski). First "interesting" case: surfaces. As an example, if p is a point of the curve C on S, a rank 2 valuation of K(S) counts order on C and intersection with C at p. The corresponding point in the Zariski-Riemann space is the limit of all points on C infinitely near to p. References: Zariski-Samuel "Commutative Algebra", Casas-Alvero "Singularities of plane curves". –  quim Sep 20 '11 at 9:23
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I'll put several remarks in separate comments. First, the definition of that valuation: assume C is defined near p by x=0 for some $x\in {\mathcal O}_{S,p}$. Every $f\in {\mathcal O}_{S,p}$ can be written uniquely as $x^v⋅\tilde f$ with $v$ a nonnegative integer and $\tilde f$ not divisible by $x$ ($v$ is the order of $f$ over $C$). Now the map $f\mapsto (v,I_p(C,\tilde f))$ is a rank 2 valuation. –  quim Sep 20 '11 at 14:01
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Second, the intersection index $I_p(C,\tilde f)$ can be defined in various ways. One often found is as $\dim {\mathcal O}_{S,p}/(f,x)$. –  quim Sep 20 '11 at 14:03
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But, third! $C$ can be parameterized locally (analytically) by Puiseux series. That is, if $y,z\in {\mathcal O}_{S,p}$ are local coordinates at $p$ (equivalently, $y,z$ generate the maximal ideal at $p$) then there exist an integer $n$ and a series $h(t)$ such that $t\mapsto (t^n, h(t))$ is a parameterization of $C$ (i.e., the kernel of the map ${\mathcal O}_{S,p}\rightarrow {\mathbb C}[[t]]$ given by $y\mapsto t^n$, $z\mapsto h(t)$ is the ideal generated by $x$. Then, $I_p(C,\tilde f)$ can also be defined as the order (in t) of the image of $\tilde f$ in ${\mathbb C}[[t]]$. –  quim Sep 20 '11 at 14:11
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Fourth, seen this way, the rank 2 valuation defined above is a particular case of Zariski-Samuel, "Commutative Algebra II", VI.15, example 2, second paragraph. –  quim Sep 20 '11 at 14:13

1 Answer 1

up vote 2 down vote accepted

Simply put (more or less already said above). Let $K=C(x,y)$ and let $xi(t)$ be a generalized power series (a power series with well-ordered exponents) where the exponents are non-negative real numbers. Assume (this is important) that $P(t,\xi(t))\neq 0$ for any $P\in K$. Then the map: $\nu(P(x,y))=ord_{t}(t,\xi(t))$ is a valuation (the "order of contact of P with $(t,\xi(t))$").

Notice that if $\xi(t)=t^{\pi}$, for example, the valuation has rational rank $2$. If $\xi(t)$ corresponds to an analytic branch of a curve (non-algebraic), the valuation has rational rank $1$, etc.

By the way, here you may find something useful. That book may be of help.

You should read something about point blowing-ups and then you understand the projective limit thing. But without it, it gets somewhat too algebraic.

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