What are examples illustrating the usefulness of Krull (i.e., rank > 1) valuations? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T11:44:38Z http://mathoverflow.net/feeds/question/12279 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/12279/what-are-examples-illustrating-the-usefulness-of-krull-i-e-rank-1-valuation What are examples illustrating the usefulness of Krull (i.e., rank > 1) valuations? Pete L. Clark 2010-01-19T04:51:48Z 2010-02-23T22:14:16Z <p>In modern valuation theory, one studies not just absolute values on a field, but also <b>Krull valuations</b>. The motivation is easy enough:</p> <p>If $k$ is a field, a <b>valuation ring</b> of $k$ is a subring $R$ such that for every $x \in k^{\times}$, at least one of $x, x^{-1}$ is an element of $R$. (It follows of course that $k$ is the fraction field of $R$.) If $| \ |$ is a non-Archimedean norm on a field, then the set <code>$\{x \in k \ | \ |x| \leq 1 \}$</code> is a valuation ring. However, the converse does not hold, since if $R$ is a valuation ring, then $k^{\times}/R^{\times}$ need not inject into $\mathbb{R}$: rather it is (under a straightforward extension of the divisibility relation on $R$) a totally ordered abelian group. Moreover, a certain formal power series construction shows that for any totally ordered abelian group $\Gamma$, there exists $k$ and $R$ with $k^{\times}/R^{\times} \cong \Gamma$.</p> <p>My question is this: what are some instances where having the generality of Krull valuations is useful for solving some problem (which is not <i>a priori</i> concerned with valuation theory)? How do Krull valuations arise in algebraic geometry?</p> <p>I can almost remember one example of this. I believe it is possible to give a quick proof of the Lang-Nishimura Theorem -- that having a smooth $k$-rational point is a birational invariant among complete [hmm, valuative criterion!] $k$-varieties. I think I saw this in some of Bjorn Poonen's lecture notes, but I forget where. [Last year at this time, I would have emailed Bjorn. I am trying out this new approach on the theory that Bjorn can reply if he wishes, and if not someone else will surely be eager to tell me the answer.] </p> <p>Are there other nice examples? Maybe something to do with resolution of singularities?</p> http://mathoverflow.net/questions/12279/what-are-examples-illustrating-the-usefulness-of-krull-i-e-rank-1-valuation/12280#12280 Answer by Emerton for What are examples illustrating the usefulness of Krull (i.e., rank > 1) valuations? Emerton 2010-01-19T05:05:43Z 2010-01-23T06:52:44Z <p>Dear Pete,</p> <p>This is a fairly general comment, rather than a precise answer:</p> <p>General valuations were studied very deeply by Zariski, who used them as one of the cornerstones of his investigations of birational geoemtry (including various forms of his Main Theorem, resolution of singularities, and so on). His picture was that different valuation rings (dominating the given local ring $\mathcal O_P$ at a point $P$ in a surface $X$, say) correspond to different germs of curves passing through the point $P$. The rank one valuations are just the germs of algebraic curves passing through $P$, but the other valuations are like transcental curves that give additional information.</p> <p>I believe that Krull himself didn't see that higher rank valuations would be related to geometry, but that it was Zariski (who studied Krull very carefully) who saw the applicability, and introduced them (and the closely related concept of normality) into algebraic geometry. (I believe that Zariski writes about this somewhere, although I can't remember where.) </p> <p>This point of view persisted in algebraic geometry up until the Grothendieck revolution. Maybe one of the last results proved using this view-point was the Nagata embedding theorem.</p> <p>Apparently Grothendieck was very unhappy with valuation theory (and indeed tried to keep it out of Bourbaki's commutative algebra texts, without success), and the only vestige that survived in his view-point was the valuative criteria for separatedness and properness.</p> <p>Let me close by using this soap-box to encourage people to read Zariski's papers. They are quite wonderful.</p> <p>EDIT: I just remembered that Lang's book on algebraic geometry, which is a kind of an abridged version of Weil's Foundations, uses valuation theory at various points. (I remember that it is treated at the beginning of the book in the section dealing with foundational concepts in algebra, but I don't remember now exactly what he proves with it later on. But it is a fairly short book, so one could flip through it and see. The whole book doesn't go all that far, and so it's possible that it appears just because it was so endemic to the commutative algebra and algebraic geometry of that time period, rather than because he does anything particularly special with it.)</p> <p>EDIT: After thinking about <a href="http://mathoverflow.net/questions/12717" rel="nofollow">this question</a>, I realized that the above statement about rank one valuations corresponding to germs of algebraic curves passing through a point is misleading. See my answer <a href="http://mathoverflow.net/questions/12717/points-and-dvrs/12730#12730" rel="nofollow"> here</a> for a (hopefully) more correct discussion of the geometric intution behind various valuations.</p> http://mathoverflow.net/questions/12279/what-are-examples-illustrating-the-usefulness-of-krull-i-e-rank-1-valuation/12287#12287 Answer by Bjorn Poonen for What are examples illustrating the usefulness of Krull (i.e., rank > 1) valuations? Bjorn Poonen 2010-01-19T06:15:47Z 2010-01-19T06:15:47Z <p>Since you asked for it, here is a little bit about the role of valuations in the Lang-Nishimura theorem, one version of which is as follows (my version implies yours):</p> <p><strong>Theorem (Lang-Nishimura):</strong> Let $X \to \to Y$ be a rational map between $k$-varieties, where $X$ is integral and $Y$ is proper. If $X$ has a smooth $k$-point $x$, then $Y$ has a $k$-point.</p> <p><strong>Sketch of proof:</strong> Let $K$ be the function field of $X$. The rational map gives a $K$-point on $Y$. If $\dim X=1$, then <code>$\mathcal{O}_{X,x}$</code> is a valuation ring, so the valuative criterion for properness gives an <code>$\mathcal{O}_{X,x}$</code>-point of $Y$, which reduces to a $k$-point of $Y$. For $\dim X=n>1$, modify the argument by <em>embedding</em> $K$ into a valued field with value group $\mathbb{Z}^n$ and residue field $k$, namely the iterated Laurent series field $F:=k((t_1))((t_2))\cdots((t_n))$, where the $t_i$ are local parameters at $x$. $\square$</p> <p>1) If one prefers, one can replace this use of the valuative criterion for a rank $n$ discrete valuation with $n$ uses of the valuative criterion for rank $1$ discrete valuations: prove the lemma that if a proper variety has an $L((t))$-point, then it has an $L$-point, and apply the lemma $n$ times. So for this particular application, you don't really need the fancy valuations.</p> <p>2) Geometrically, the Lang-Nishimura theorem can be understood as follows: Find a smooth irreducible curve $C$ in $X$ through $x$ such that $C$ is not entirely contained in the locus of indeterminacy of $\phi \colon X \to\to Y$. Then <code>$\phi|_C$</code> extends to a morphism, and the image of $x$ is a $k$-point of $Y$. (The existence of $C$ is not completely obvious, though, so the valuation-theoretic proof is cleaner.)</p> http://mathoverflow.net/questions/12279/what-are-examples-illustrating-the-usefulness-of-krull-i-e-rank-1-valuation/12292#12292 Answer by Georges Elencwajg for What are examples illustrating the usefulness of Krull (i.e., rank > 1) valuations? Georges Elencwajg 2010-01-19T09:35:51Z 2010-01-19T09:35:51Z <p>A strange and difficult question is whether there exists a scheme without any closed point. It is very tempting to think that since an affine scheme does have closed points ( they correspond to maximal ideals of its ring), one could reduce to the affine case and show that an arbitrary scheme necessarily has a closed point. But this is false: there exist schemes without closed points. And all constructions I know, with one exception, use a valuation ring $V$ with huge valuation group: exactly what you require . They take the affine scheme $Spec(V)$ and the required scheme is $Spec(V)\setminus {M}$, where $M$ corresponds to the maximal ideal of $V$. The simplest example is in Qing Liu's book, exercises 3.26, 3.27 page 113,114 (he attributes this construction to Florian Pop).</p> <p>The exception I mentioned before is based on Hochster's characterization of topological spaces arising as spectra of affine schemes. You take such a space with just one closed point and (yes, you guessed it) remove it. Details are in Ravi Vakil's publication list, at the bottom of the page, Miscellaneous 3.</p> <p><a href="http://math.stanford.edu/~vakil/preprints.html" rel="nofollow">http://math.stanford.edu/~vakil/preprints.html</a></p> <p>The drawback of this approach without valuation rings is that you have to read Hochster's article, which (although quite interesting) is very technical, as Hochster himself acknowledges.</p> http://mathoverflow.net/questions/12279/what-are-examples-illustrating-the-usefulness-of-krull-i-e-rank-1-valuation/12297#12297 Answer by Hagen Knaf for What are examples illustrating the usefulness of Krull (i.e., rank > 1) valuations? Hagen Knaf 2010-01-19T10:56:14Z 2010-01-19T10:56:14Z <p>Let $F/K$ be an algebraic function field. The set $Z$ of all valuation rings of $F$ that contain $K$ can be identified with the projective limit of all projective (proper) $K$-varieties $X$ having $F$ as their function field. In general $Z$ is a ringed space but not a scheme. By construction it is clear that $Z$ encodes geometric information.</p> http://mathoverflow.net/questions/12279/what-are-examples-illustrating-the-usefulness-of-krull-i-e-rank-1-valuation/12309#12309 Answer by Hagen Knaf for What are examples illustrating the usefulness of Krull (i.e., rank > 1) valuations? Hagen Knaf 2010-01-19T13:55:41Z 2010-01-19T13:55:41Z <p>The following is about an application of Krull valuations in non-noetherian ring theory:</p> <p>A noetherian normal domain $R$ can be written as an intersection of discrete valuation domains. Moreover these valuation domains are localizations of $R$ - at the prime ideals of height $1$.</p> <p>Of course every normal domain $R$ can be written as an intersection of valuation domains of its field of fractions $K$. Motivated by the situation in the noetherian case one can however try to find families $F$ of valuation domains of $K$ such that $R$ is the intersection of the members of $F$ and $F$ has one or more of the following properties:</p> <ol> <li>All members of $F$ are localizations of $R$ at certain primes (or weaker: for every member of $F$ the values of the elements $r\in R$ are precisely all non-negative values in the value group).</li> <li>All members of $F$ satisfy certain requirements concerning the value groups or the Krull dimension.</li> <li>Every non-zero $r\in R$ lies in the maximal ideal of at most finitely many elements of $F$.</li> </ol> <p>These properties can be used to classify normal domains to a certain extent. In the period roughly between 1950 and 1970 a bunch of articles were published that seemed to follow this idea. Some author names: Paolo Ribenboim, Jim Brewer, Malcolm Griffin.</p> http://mathoverflow.net/questions/12279/what-are-examples-illustrating-the-usefulness-of-krull-i-e-rank-1-valuation/12735#12735 Answer by Hailong Dao for What are examples illustrating the usefulness of Krull (i.e., rank > 1) valuations? Hailong Dao 2010-01-23T07:34:48Z 2010-01-23T08:02:26Z <p>I would like to point out Teissier's Oberwolfach <a href="http://people.math.jussieu.fr/~teissier/documents/Oberwolfach0909.pdf" rel="nofollow">report</a> about valuations and resolutions of singularity in positive characteristics. </p> <p>Also, more related to your comment <a href="http://mathoverflow.net/questions/12717/points-and-dvrs/12730#12730" rel="nofollow">here</a> ,Cutkosky-Tessier has a recent <a href="http://math.missouri.edu/~dale/semigroupex6.pdf" rel="nofollow">paper</a> which has some references with Zariski's viewpoint results on valuation theory. The websites of <a href="http://www.math.missouri.edu/~dale/papers.html" rel="nofollow">Cutkosky</a> and <a href="http://people.math.jussieu.fr/~teissier/" rel="nofollow">Teissier</a> have a lot more. </p> http://mathoverflow.net/questions/12279/what-are-examples-illustrating-the-usefulness-of-krull-i-e-rank-1-valuation/15280#15280 Answer by BCnrd for What are examples illustrating the usefulness of Krull (i.e., rank > 1) valuations? BCnrd 2010-02-14T19:32:12Z 2010-02-14T19:32:12Z <p>M. Temkin has some recent beautiful work on a new approach to semistable reduction for relative curves (without even assuming properness, and not using input about properties of moduli spaces, instead yielding a very new proof of the semistable reduction theorem and much more), and in this he makes very creative use of Zariski-Riemann spaces of valuations.</p> <p>His method rests on the following very nifty strategy. By valuation-theoretic techniques of Zariski, and with enough uniqueness at one's disposal to do gluing, for certain kinds of problems it suffices to work over valuation rings. Now if $R$ is a valuation ring, we can exhaust its fraction field $K$ by fraction fields $K_i$ of finitely generated $\mathbf{Z}$-subalgebras, and let $R_i$ be the induced valuation ring of $K_i$. Many "finitely presented" problems over $R$ can be reduced to the case of the $R_i$, so we may suppose (for many purposes) that $R$ contains a finitely generated $\mathbf{Z}$-subalgebra with the same fraction field. So what? The nice part is that in such cases $R$ has "finite height" in the sense of valuation theory, and there is a simple recursive procedure that constructs finite-height valuation rings in finitely many steps from valuation rings of smaller height. So if one has set up the appropriate inductive technique, it becomes possible to reduce one's problem to the height-1 case. Using "approximation in the completion" (which needs some work), it is often possible to then even reduce to the case of complete height-1 valuation rings. Now one can appeal to techniques from rigid-analytic geometry. Very nice: problems over general schemes can sometimes be reduced to problems in rigid-analytic geometry, thanks to the use of spaces of valuation rings with unrestricted height. </p> http://mathoverflow.net/questions/12279/what-are-examples-illustrating-the-usefulness-of-krull-i-e-rank-1-valuation/16197#16197 Answer by B A Sethuraman for What are examples illustrating the usefulness of Krull (i.e., rank > 1) valuations? B A Sethuraman 2010-02-23T22:14:16Z 2010-02-23T22:14:16Z <p>For a different flavor from the answers to date: valuations beyond the classical discrete valuations arising from absolute values have proved extremely useful in analysing structures of certain division algebras finite-dimensional over their centers. Division algebras (even when finite-dimensional over their centres) are notoriously difficult to get a handle on: for instance, it is almost always impossible to describe their subfields. However, when the centre $F$ is Henselian with respect to a general valuation (Henselian simply means that Hensel's lemma holds: this substitutes for the notion of completeness that is used in the classical case), then it is easy to see that the valuation on $F$ extends uniquely to $D$, and if further the division algebra $D$ is "tame" (i.e., the characteristic of $\overline{F}$ does not divide $[D:F]$), then one can use the valuations on $F$ and $D$ to obtain significant information about $D$. </p> <p>As an example: if further $D$ is totally ramified over $F$, i.e., $|V(D)/V(F)| = [D:F]$, where $V()$ stands for the value group, then it turns out that there is a nondegenerate alternating pairing from $V(D)/V(F) \times V(D)/V(F)$ to the group of $m$-th roots of unity in $\overline{F}$, where $m$ is the period of $D$ in the Brauer group of $F$. This pairing completely determines $D$: $D$ decomposes as a tensor product of "symbol algebras" (i.e., generalized quaternions), where the tensor factors correspond to a symplectic base for $V(D)/V(F)$. The isomorphism classes of $F$-subalgebras of $D$ are determined by subgroups of $V(D)/V(F)$, and the $F$-isomorphism classes of subfields of $D$ are determined by totally isotropic subgroups of $V(D)/V(F)$. This in turn completely determines tame division algebras over fields of the form $k((x_1))\cdots ((x_n))$ where $k$ is separably closed.</p> <p>For a survey see the paper: Wadsworth, A. R., Valuation theory on finite dimensional division algebras. Valuation theory and its applications, Vol. I (Saskatoon, SK, 1999), 385--449, Fields Inst. Commun., 32, Amer. Math. Soc., Providence, RI, 2002. (Math Reviews: MR1928379 (2003g:16023) )</p>