Defining Valuation Rings - PDF Free Download

East Carolina University, Greenville, North Carolina, USA June 8, 2018

Outline 1 What? Valuations and Valuation Rings Definability Questions in Number Theory 2 Why? Some Questions and Answers Becoming More Ambitious 3 Other Applications of Definitions of Valuation Rings Infinite Algebraic Extensions of Q Function Fields Using Definitions of Valuations over Function Fields of Positive Characteristic Using Definitions of Valuations over Function Fields of Characteristic 0. 4 How?

Product Formula Fields Some of the objects we consider: number fields, i.e. finite extensions of Q; function fields of transcendence degree 1 over a finite field; function fields of transcendence degree 1 over arbitrary constant fields. The last class includes function fields of curves over number fields or algebraic closure of Q.

non-archimedean Discrete Valuations Let K be a product formula field. Let v : K Z { } be a map satisfying the following conditions: 1 For all x, y K \ {0}, we have that v(xy) = v(x) + v(y). 2 v(1) = 0. 3 v(0) =. 4 v(x + y) min(v(x), v(y)). 5 If v(x) < v(y), then v(x + y) = v(x). 6 If K is a function field of transcendence degree 1 over a constant field k, then v k is trivial, i.e. for every non-zero constant c k, we have that v(c) = 0.

Valuation Rings Let K be a product formula field and let v be a discrete valuation on K. Then the valuation ring of v is the ring R v = {x K v(x) 0}. The unique maximal ideal p = {x K v(x) > 0} of R p is often identified with the valuation itself. The field R v /p is the residue field of the valuation. In the case of a function field, the residue field is always a finite extension of the field of constants. If the extension is of degree 1, we say that the valuation (or the prime) is of degree one. If K is an infinite algebraic extension of a product formula field, we will define R v, to be the integral closure of R v in K (i.e. the ring of all elements of K satisfying monic irreducible polynomials with coefficients in R v.)

The Questions Let K be a product formula field or its infinite algebraic extension K. 1 Is there a polynomial f v (T, X 1,..., X M ) K[T, X 1,..., X m ] such that for any element t K the sentence E 1 X 1 K... E m X m K : f (t, X 1,..., X m ) = 0 is true if and only if t R v, v M for some set M of valuations of K? 2 Is there a polynomial f (T, X 1,..., X M ) K[T, X 1,..., X m ] such that for any element t K the sentence E 1 X 1 K... E m X m K : f v, (t, X 1,..., X m ) = 0 is true if and only if t R v,? Here E i stands for one of the quantifiers: or.

The Language of Rings The Classical Language of Rings The language of rings is the language where we are allowed to use only the following symbols to form sentences together with logical connectives, and the universal and existential quantifiers: L R = (0, 1, +, ). Occasionally we augment the language by adding names of some specific elements of the ring. Generally speaking a formula of the language is an expression of the form: E 1 X 1... E m X m P(X 1,..., X m, T 1,..., T s ), where, as before, E i stands for one of the quantifiers: or, and P(X 1,..., X m, T 1,..., T s ) is a polynomial over a given ring. If all variables are within the range of some quantifier, the formula becomes a sentence that is either true or false.

Two General Problems The General Definability Problems for Countable Rings 1 Is there an algorithm to determine whether a sentence of the language is true over a specific ring? (The answer will vary with the ring of course.) 2 What subsets of a given ring can we define using formulas of the language? The quantifiers may range over the whole ring or a subset of the ring. On many occasions we will require that only existential or only universal quantifiers are used in the formulas.

The Origin Is there an algorithm which can determine whether or not an arbitrary polynomial equation in several variables has solutions in integers? Using modern terms one can ask if there exists a program taking coefficients of a polynomial equation as input and producing yes or no answer to the question Are there integer solutions?. This problem became known as Hilbert s Tenth Problem

The First-Order Theory of Z is Undecidable J.B. Rosser, Extensions of Some Theorems of Gödel and Church, J. Symb Logic, vol. 1, (1936), 87-91. Grundlagen der Mathematik, by David Hilbert and Paul Bernays, 1934-39, Springer In other words, there is no algorithm to determine whether a sentence of the form E 1 X 1... E m X m P(X 1,..., X m ), is true when the variables are ranging over Z. (As above, E i is either or.)

Number Field Results of Julia Robinson Theorem (1949) Z is definable by a first-order formula over Q. Thus the first-order theory of Q (in the language of rings) is undecidable.

Number Field Results of Julia Robinson Theorem (1949) Z is definable by a first-order formula over Q. Thus the first-order theory of Q (in the language of rings) is undecidable. Theorem (1959) If K is a number field, then Z is definable over K by a first-order formula. Thus the first-order theory of K (in the language of rings) is undecidable.

The Answer to Hilbert s Question This question of Hilbert was answered negatively (with the final piece in place in 1970) in the work of Martin Davis, Hilary Putnam, Julia Robinson and Yuri Matiyasevich. Actually a much stronger result was proved. It was shown that the recursively enumerable subsets of Z are the same as the Diophantine sets.

Recursive, Recursively Enumerable Sets. Definition (Recursive Sets) A subset of integers is called recursive or computable if there is an algorithm to determine membership in the set.

Recursive, Recursively Enumerable Sets. Definition (Recursive Sets) A subset of integers is called recursive or computable if there is an algorithm to determine membership in the set. Definition (Recursively Enumerable Sets) A subset of integers is called recursively (or computably) enumerable if there is an algorithm to list the elements of the set.

Diophantine Sets: a Number-Theoretic Definition For an integral domain R, a subset A R m is called Diophantine over R if there exists a polynomial p(t 1,... T m, X 1,..., X k ) with coefficients in R such that for any element (t 1,..., t m ) R m we have that x 1,..., x k R : p(t 1,..., t m, x 1,..., x k ) = 0 (t 1,..., t m ) A. In this case we call p(t 1,..., T m, X 1,..., X k ) a Diophantine definition of A over R.

Undecidable Diophantine Sets Theorem (MDRP) There are undecidable Diophantine sets over Z.

Undecidable Diophantine Sets Theorem (MDRP) There are undecidable Diophantine sets over Z. Corollary HTP is undecidable or positive existential theory of Z is undecidable.

Undecidable Diophantine Sets Theorem (MDRP) There are undecidable Diophantine sets over Z. Corollary HTP is undecidable or positive existential theory of Z is undecidable. Proof. Let f (t, x) be a Diophantine definition of an undecidable Diophantine set. If HTP is decidable, then for each t Z we can determine if the polynomial equation f (t, x) = 0 has solutions in Z. However, this process would also determine whether t is an element of our set, contradicting the fact that the set was undecidable.

Some Properties of Diophantine Sets and Definitions over Subrings of Algebraic Extensions of Q Intersections and unions of Diophantine sets are Diophantine (unions always, intersection over not algebraically closed fields). One = finitely many (not algebraically closed fields)

A General Question A Question about an Arbitrary Recursive Ring R Is there an algorithm, which if given an arbitrary polynomial equation in several variables with coefficients in R, can determine whether this equation has solutions in R? The most prominent open questions are probably the decidability of HTP for R = Q and R equal to the ring of integers of an arbitrary number field.

Undecidability of HTP over Q Implies Undecidability of HTP for Z Indeed, suppose we knew how to determine whether solutions exist over Z. Let Q(x 1,..., x k ) be a polynomial with rational coefficients. Then x 1,..., x k Q : Q(x 1,..., x k ) = 0 y 1,..., y k, z 1,..., z k Z : Q( y 1 z 1,..., y k z k ) = 0 z 1... z k 0. So decidability of HTP over Z would imply the decidability of HTP over Q.

Using Diophantine Definitions to Solve the Problem Lemma Let R be a recursive ring containing Z and such that Z has a Diophantine definition p(t, X ) over R. Then HTP is not decidable over R.

Using Diophantine Definitions to Solve the Problem Lemma Let R be a recursive ring containing Z and such that Z has a Diophantine definition p(t, X ) over R. Then HTP is not decidable over R. Proof. Let h(t 1,..., T l ) be a polynomial with rational integer coefficients and consider the following system of equations. h(t 1,..., T l ) = 0 p(t 1, X 1 ) = 0... p(t l, X l ) = 0 It is easy to see that h(t 1,..., T l ) = 0 has solutions in Z iff (1) has solutions in R. Thus if HTP is decidable over R, it is decidable over Z. [1]

HTP for Q and Valuation Rings Non-Archimedean Valuations of Q Every non-archimedean valuation of Q can be associated with a prime number p Z. For any integer z Z, let ord p z be the highest power of p dividing z. Given a non-zero element x = a b with a, b Z, we can let v p (x) = ord p x = ord p a ord p b. The valuation ring R p corresponding to this valuation is the set of all elements of Q with a reduced denominator not divisible by p. Reformulating the problem of defining Z inside Q Now we can view the problem of defining Z inside Q by a Diophantine definition as a problem of defining existentially p P R p, where P is the set of all rational prime numbers.

A Conjecture of Barry Mazur The Conjecture on the Topology of Rational Points Let V be any variety over Q. Then the topological closure of V (Q) in V (R) possesses at most a finite number of connected components. A Nasty Consequence There is no Diophantine definition of Z over Q. Unfortunately, as it stands now, the reformulation of the of the problem in terms of valuations does not give us anything beyond the connection between defining valuation rings and defining Z.

More Rings Definable through Valuations inside Q If we choose a subset S P, and let W = P \ S, then we can consider a ring p W R p = Z[S 1 ] and the problem of existentially defining Z[S 1 ] over Q, as well as the problem of defining Z over Z[S 1 ]. As of now we can construct the first definition if S is co-finite, and the second one, if S is finite. (These two Diophantine definitions are due to J. Robinson.)

First-order and Existential (Diophantine) Models Another way to show undecidablity of the first-order (existential) theory of a ring R is to construct a map n A n, where A = i=0 A i R is first-order (existentially) definable over R, the collection {A i, i Z 0 } is a partition of A, the set + = {(x, y, z) R 3 : x A i, y A j, z A i+j } and the set = {(x, y, z) R 3 : x A i, y A j, z A ij } are first-order (existentially) definable.

Discrete Valuations of a Number Field Let K be a number field, and let O K be the ring of integers of K, i.e. the integral closure of Z in K. Then we identify each valuation of K with a prime ideal p of O K, and for x 0 we set v p (x) = ord p x. Here, if x O K, then ord p x is the non-negative integer n such that x p n \ p n+1. If x = a b, a, b O K \ {0}, then ord p x = ord p a ord p b.

Using Elliptic Curves to Generate Integers Theorem (Poonen, 2002) Let K be a number field, and let E be an elliptic curve over K of positive rank. We choose a model of E placing the neutral element of E at the origin. Then there exists G P K such that P K \ G is finite, and the following assertions are true for every p G. There exists a point P E(K) of infinite order such that: 1 P ker(red p ), 2 P has homogeneous coordinates of the form (x 1 : y 1 : 1), where ord p x 1 > 0, ord p y 1 > 0. We let ord p P = min(ord p x 1, ord p y 1 ), 3 for every positive integer n, we have that ord p [n]p ord p P, 4 for every positive integer l Z >0, there exists m Z such that ord p [m]p > l, 5 for every positive integer n, we have that ord p ( x([n]p) x(p) n) ord p P.

Using Finitely Generated Elliptic Curves and Valuation Rings Theorem (S., 2017) Let K be an infinite algebraic extension of K, and assume E(K) = E(K ). For convenience, we will also assume that K /K is normal. Now let x K be such that the following sentence is true. x(q) x(p) y K P, Q E(K ) : x R p,. y Then x K. Further, if x Z >0, then the sentence is true for x.

Some Proof Ideas Given x K, consider a field M, the Galois closure of K(x) in K. Let ˆx be any conjugate of x over K, and let y K. Now if then x(q) x(p) x y x(q) x(p) ˆx y R p,, [2] R p,. [3] Let q be any prime above p in M. From [2] and [3], we can conclude that ord q (x ˆx) ord q y. [4] Since y can be an arbitrary element of K, the only way [4] can be satisfied for all y K is for ˆx = x. Since ˆx is an arbitrary conjugate of x over K, we can conclude that x K.

Positive Integers Satisfy the Sentence To prove that positive integers satisfy the sentence y K P, Q E(K ) : x(q) x(p) x y R p,, let x = n Z >0, and assume without loss of generality that we are given y K. Choose a point P, as above, with ord p P > ord p y, and let Q = [n]p. In this case, ( ) x(q) ord p x(p) x > ord p y, and x(q) x(p) x y R p R p,.

The First-Order Theory of K is Undecidable We have constructed a first-order definition over K of a subset of K containing all positive integers. Since all elements of K can be written down as linear combinations of some basis elements over Q, and all elements of Q are ratios of integers, we have, in fact constructed a first-order definition of K over K. We can now use the result of J. Robinson stating that the first-order theory of any number field is undecidable to reach the desired conclusion.

Fields where Assumptions on Elliptic Curves and Valuation Rings are Satisfied By a theorem of B. Mazur and K. Rubin (2017), there are infinitely many infinite algebraic extensions of Q with finitely generated elliptic curves of positive rank, and where by a theorem of S., the integral closure of a valuation ring of a number field is definable. However, we do not have a definitive description of a class of infinite algebraic extensions of Q, where our assumptions hold.

Valuations of Function Fields Trivial on Constant Fields Valuations of Rational Fields Let k be a field, and let t be transcendental over k. All valuations of k(t), except for one, can be identified with prime ideals of k[t] using the order function. The remaining valuation, attached to the degrees of polynomials, corresponds to a prime ideal generated by 1 t in k[ 1 t ]. Valuations in Algebraic Extensions Let K/k(t) be a finite extension. Let O K be the integral closure of k[t] in K, and let O K 1 be the integral closure of k[ 1 t ] in K. Then all valuations of K correspond to the prime ideals of O K and O K 1 via the order function.

Different Ways of Defining Multiplication Addition and Divisibility. (J. Robinson, 1949) Let L be a first-order language generated by {0, 1, +, }. Then multiplication is first-order definable in this language over Z 0. Adding Division with p-th Powers. (T. Pheidas, 1987) Let L be a first-order language generated by {0, 1, +,, p }, where p is a prime number, and x p y is equivalent to x = yp s, s Z 0. Then multiplication is definable existentially in this language over Z 0.

Modeling Z 0 over a Function Field K. Let P = {(x, y) K 2 : s Z 0 y = x ps }, and let p be a prime of K. If P is first-order definable over K, then the map n t pn for a fixed element t of K gives rise to a first-order model of Z 0. If P and R p are both existentially definable over K, then the map n {x K ord p x = n} gives rise to an existential model of Z over K. Theorem (K. Eisenträger, S., 2017) 1 The first-order theory in the language of rings of any function field of positive characteristic is undecidable. 2 The existential theory in the language of rings of any function field of positive characteristic not containing the algebraic closure of a finite field is undecidable.

Some Open Problems about Valuations in Positive Characteristic Define R p over a function field K with a constant field k, not algebraic over a finite field. Define R p over a function field K with a constant field k containing the algebraic closure of a finite field.

Elliptic Curves of Rank One over Function Fields of Characteristic 0. Let K be a function field of characteristic 0, and let E be an elliptic curve of rank 1 over K. We again choose a model of E placing the neutral element at the origin. Let p be a prime of K such that E has a good reduction at p, and there exists a point P ker(red p ) of infinite order. Let (x : y : 1) be the homogenous coordinates of P, and let (x n : y n : 1) be the homogeneous coordinates of [n]p. Then, as before, we have that ord p ( xn x 1 n) ord p P. If R p is existentially (first-order) definable over K, then the map n {x K ord p x = n} for n 0, gives rise to an existential (first-order) model of Z.

Existence of Elliptic curves of rank 1 over Function Fields of Characteristic 0 Theorem (Moret-Bailly, 2005) Let K be a function field of characteristic 0. Then there exists an elliptic curve of rank 1 defined over K.

Valuations over Function Fields of Characteristic 0 Theorem (H. Kim and Fred Roush, 1995) Let k be a field such that it is formally real (i.e. -1 is not a sum of squares), or can be embedded into a finite extension of Q p. Let p be a prime of degree one. Then R p is existentially definable over k(t). Extensions This theorem was partially extended to finite algebraic extensions of k(t) by K. Eisenträger and L. Moret-Bailly. They independently constructed an existential definition of a subset of R p. This partial extension of Kim and Roush result allowed Eisenträger (2007) and Moret-Bailly (2005) to show that HTP was undecidable over finite extensions of k(t). S. [2018] constructed an existential definition of a valuation ring for an arbitrary prime of a finite extension of k(t).

Open problems 1 Describe constant fields k such that a function field over k has a definition (existential or first-order) of an arbitrary valuation ring. 2 In particular, determine whether one can define valuation rings when k is algebraically closed. 3 In the case the valuation rings are not definable, find another method for showing that the corresponding function fields have undecidable existential or first-order theory. 4 In particular, determine the status of theories of Q(t) and C(t).

From Poles to Divisibility of Order at a Prime Let q be a rational number, let K be a product formula field, let p be a prime of K, let a K be such that ord p a = 1. Now let x K, and observe that ord p ( xq a + 1 a ) 0 mod q if and only if q ord p x 0. Indeed, suppose ord p x 0, then ord p ( xq a ) 1, and ord p ( xq a + 1 x a ) = min(ord q q p a, ord p 1 1 a ) = ord q p a = q. Suppose q now that ord p x < 0. Then ord p ( xq a ) = q ord 1 p x 1 < q = ord p a. q

Using Quadratic Forms Consider a quadratic form x1 2 bx 2 2 cx 3 2 + bcx 4 2. This form is anisotropic (i.e. the only solutions to x1 2 bx 2 2 cx 3 2 + bcx 4 2 = 0 are x 1 = x 2 = x 3 = x 4 = 0) if and only if there exists a prime p such that ord p b is odd, and c is not a square mod p (or vice versa). So let s fix a prime p and c not equivalent to a square modulo p. Let ord p a = 1 and consider the equation x 2 1 bx 2 2 cx 2 3 + bcx 2 4 = 0, where b = xq a + 1 a q. If this equation has non-trivial solutions, then ord p b is even, and ord p x 0.

Satisfying the Equation When x is Integral at p The main tool is Hasse-Minkowski local-global principle. It states that the form x 2 1 bx 2 2 cx 2 3 + bcx 2 4 is anisotropic globally if and only if it is anisotropic locally, i.e. over K q the q-adic completion of K with respect to q, at some prime q of K.

From Quadratic Forms to Norm Equations We can rewrite the equation as or x 2 1 bx 2 2 cx 2 3 + bcx 2 4 = 0 c = x 1 2 bx 2 2 x3 2 bx 4 2, c = N K( b)/k (y), where y K( b). If we consider norm equations instead of quadratic forms, we are not restricted to extensions of degree 2. This is useful if one works over a global field of characteristic 2, or if one works over an infinite extension of Q where the local degrees of factors of p are divisible by arbitrarily high powers of 2. R. Rumely (1980) was the first person to use norm equations for this purpose. Instead of Hasse-Minkowski theorem, we use Hasse norm principle to make sure our equations are satisfied when c does not have a negative order at p.