The Valuation Theory Home Page - Open Problems

Click here for a different list of open problems which was assembled for the Workshop on Valuation Theory in Positive Characteristic, which took place at the Mathematics Village, Sirince, Turkey, June 15 to 19, 2011.

Open Problem 1: Generalize Abhyankar's "Going Up" and "Coming Down" for local uniformization to arbitrary finite transcendence degree.

Possible applications: To local uniformization and resolution of singularities, in particular, in positive characteristic.

Posted by F.-V. Kuhlmann on February 4, 1999 

A valued field \((K,v)\) is called spherically complete if every nest of balls has a non-empty intersection. This holds if and only if every Pseudo-Cauchy sequence in \(K\) has a pseudo limit in \(K\). By the work of Kaplansky, this in turn holds if and only if the field is maximally valued, i.e., has no proper immediate extensions. Take a polynomial \(f\) with coefficients in \(K\). The following is known: 1) If \(f\) is a polynomial in one variable, then the image \(f(K)\) is spherically complete, just as a set with the ultrametric induced by the valuation \(v\). If f is an additive polynomial in several variables, then under a certain additional condition, \(f(K)\) is again spherically complete.

Open Problem 2: Prove or disprove that \(f(K)\) is spherically complete for all additive polynomials in several variables. More generally, prove or disprove that the same holds for all polynomials in several variables.

Possible applications: To the open problem whether the power series field \(Fp((t))\) over the finite field \(Fp\) with \(p\) elements (also called "field of formal Laurent series over \(Fp\)") has a decidable elementary theory.

Posted by F.-V. Kuhlmann on February 4, 1999

Open Problem 3: What are the subsets and what are the additive subgroups of \(F_p((t))\) which are definable in the language of valued rings (or fields)?

Possible applications: To the open problem whether the power series field \(F_p((t))\) has a decidable elementary theory.

Posted by F.-V. Kuhlmann on February 4, 1999

Open Problem 4: Is there a definable bijection between the \(p\)-adic integers and the p-adic integers without zero?

Posted by Luc Belair on September 1, 1999

Solution by Raf Cluckers: (postscript file)

Open Problem 5: Study the model theory of pairs of fields, i.e., one field embedded in another with the embedding not necessarily elementary, beyond the few known examples.

Here is some valuation theoretic background for this problem: Let \(K\) be the ring of Laurent series over a field \(k\) of characteristic 0, or the field of Witt vectors over a perfect field \(k\) of finite characteristic. Fix a proper subfield \(k'\) of \(k\), and let \(A\) be the subring of \(K\) obtained by taking the inverse image of \(k'\) under the residue map for the natural valuation. Then \(A\) is a henselian local ring which is not a valuation ring and the theory of the pair \((k',k)\) is a first-order invariant of \(A\).

Posted by Luc Belair on September 1, 1999

Open Problem 6: Characterize the subsets of an (ultra)metric space which have the following property: for every point in the space there is a (not necessarily unique!) closest point in the subset.

If the ultrametric space is the space underlying a valued field, we would also like to know whether this characterization is preserved under polynomial maps. (Cf. Problem 2.)

Posted by F.-V. Kuhlmann on March 4, 2000

Open Problem 7: This is an open problem about finite extensions of a valued field \((K,v)\) within its absolute inertia field (that is, the extension is unramified, the residue field extension is separable, and equality holds in the fundamental inequality for this extension: \(n = \sum e_i f_i)\). In short words, the question is whether the valuation ring of the extension can be obtained by adjoining finitely many henselian elements to the valuation ring of \((K,v)\). An element a is called henselian over \((K,v)\) if there is a polynomial \(f\) with coefficients in the valuation ring of \((K,v)\) such that \(v f '(a) = 0\). For the exact formulation of the problem and a detailed description of the background and of what is known in special cases, see
(dvi file) --- (postscript file)

Posted by F.-V. Kuhlmann on March 4, 2000

Partial solution by Peter Roquette: (dvi file) --- (postscript file)

Solution for valuations of finite rank by Lou van den Dries: (postscript file) --- (pdf file) 

Open Problem 8: Given any finitely many Abhyankar places on a function field \(F\) and finitely many elements in the intersection \(O\) of their valuation rings, is there a subring \(R\) of \(O\) with quotient field \(F\) such that \(R\) is regular with respect to all given places and contains the given elements?

Possible applications: Simultaneous local uniformization of finitely many Abhyankar places on a function field.

Posted by F.-V. Kuhlmann on May 18, 2000

Open Problem 9: Let \(n\) be a positive integer, \(K\) a field and \(R\) a subring of \(K\). Let \(n\) not be divisible by the characteristic of \(K\). Assume that \(R\) is local and that \(a_n \in R\) or \(a_-n \in R\) for each \(a \in K\). Is \(R\) a valuation ring?

Remark (F.-V. Kuhlmann): This does not hold if \(n>1\) is divisible by the characteristic \(p\) of \(K\). Because then, for all \(a\) in \(K\), \(a^n\) lies in the subfield \(K^p\). Take \(R=K^p\). If \(n\) is a power of \(p\) and \(K\) is infinite, then one can also take \(R\) to be any valuation ring of \(K^p\). In both cases, the conditions hold, but \(R\) is not a valuation ring of \(K\).

Posted by Andrew Cosnick on June 6, 2001

Solution: The answer is “yes” when \(n\) is prime to the characteristic of the residue field \(k\) of \(R\), provided \(k\) has at least \(n\) elements. There are counterexamples where \(n\) is not prime to the characteristic of \(k\) or \(k\) has less than \(n\) elements.

Solution by Peter Roquette: (dvi file) --- (postscript file)

Partial solution by Hagen Knaf: (postscript file) 

Open Problem 10: A field \(K\) is called large if it is existentially closed in \(K((t))\). Suppose that \(L|K\) is a field extension and that \(L\) admits a \(K\)-rational place, i.e., a place which is trivial on \(K\) and has \(K\) as its residue field. Does it follow that \(K\) is existentially closed in \(L\)?

We know that this is the case if \(K\) is perfect. But there are also large fields which are not perfect, since all PAC fields are large. We know that this is the case if \(K\) is perfect. But there are also large fields which are not perfect, since all PAC fields are large. It could also be proved if we knew that every rational place of an algebraic function field \(F|K\) admits local uniformization. So far, this is only known if the place is an Abhyankar place. If \(K\) is perfect, then the place admits local uniformization on a finite extension of \(F\), still having \(K\) as its residue field, so \(K\) is existentially closed in this extension field and hence in \(F\), too.

Do you have a problem (in valuation theory)? Send it to fvk@math.usask.ca!