HTP in Positive Characteristic

Transcription

1 HTP in Positive Characteristic Alexandra Shlapentokh East Carolina University November 2007

2 Table of Contents 1 A Brief History of Diophantine Undecidability over Number Fields The Original Problem Extensions to Number Fields 2 How Things are Done Diophantine Sets, Definitions and Models HTP over a Field vs. HTP over a Subring 3 A Brief History of HTP over Function Fields of Characteristic 0 Some Definitions Field Results Ring Results 4 HTP over Function Fields of Positive Characteristic A Longer History of Diophantine Undecidability for Function Fields of Positive Characteristic Multiplication through Addition and Divisibility P-th Powers over a Function Field Defining Order at a Prime 5 Back to the First-Order Theory A Short History of First-Order Decidability over Function Fields of Positive Characteristic P-th Powers are Enough

3 A Brief History of Diophantine Undecidability over Number Fields The Original Problem Hilbert s Question about Polynomial Equations Is there an algorithm which can determine whether or not an arbitrary polynomial equation in several variables has solutions in integers? This problem became known as Hilbert s Tenth Problem

4 A Brief History of Diophantine Undecidability over Number Fields The Original Problem The Answer This question was answered negatively (with the final piece in place in 1970) in the work of Martin Davis, Hilary Putnam, Julia Robinson and Yuri Matijasevich.

5 A Brief History of Diophantine Undecidability over Number Fields Extensions to Number 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? Arguably, the most important open problems in the area concern the Diophantine status of the ring of integers of an arbitrary number field and the Diophantine status of Q.

6 A Brief History of Diophantine Undecidability over Number Fields Extensions to Number Fields The Rings of Integers of Number Fields Theorem HTP is unsolvable over the rings of integers of the following fields: Extensions of degree 4, totally real number fields and their extensions of degree 2. (Denef, 1980 & Denef, Lipshitz, 1978) Note that these fields include all Abelian extensions. Number fields with exactly one pair of non-real embeddings (Pheidas, S. 1988) Any number field K such that there exists an elliptic curve E of positive rank defined over Q with [E(K) : E(Q)] <. (Poonen, S. 2003) Any number field K such that there exists an elliptic curve of rank 1 over K and an Abelian variety over Q keeping its rank over K. (Cornelissen, Pheidas, Zahidi, 2005)

7 A Brief History of Diophantine Undecidability over Number Fields Extensions to Number Fields The Rings in between the Ring of Integers and a Number Field A Ring in the Middle of a Number Field K Let V be a set of primes of a number field K. Then define O K,V = fx K : ord p x 0 p Vg. If V =, then O K,V = O K the ring of integers of K. If V contains all the primes of K, then O K,V = K. If V is finite, we call the ring small. If V is infinite, we call the ring big or large.

8 A Brief History of Diophantine Undecidability over Number Fields Extensions to Number Fields HTP over Small Subrings of Number Fields Theorem HTP is unsolvable over small subrings of Q. Theorem For any number field K, if HTP is unsolvable over O K, then HTP is unsolvable over any small subring of K. (Julia Robinson and others)

9 A Brief History of Diophantine Undecidability over Number Fields Extensions to Number Fields HTP over Large Subrings of Number Fields Theorem Let K be a number field satisfying one of the following conditions: K is a totally real field. K is an extension of degree 2 of a totally real field. There exists or an elliptic curve E defined over Q such that [E(K) : E(Q)] <. Let ε > 0 be given. Then there exists a set S of non-archimedean primes of K such that 1 The natural density of S is greater 1 [K : Q] ε. HTP is unsolvable over O K,S. (S. 2002, 2003, 2006)

10 A Brief History of Diophantine Undecidability over Number Fields Extensions to Number Fields HTP over Very Large Subrings of Number Fields Theorem Let K be a number field with a rank one elliptic curve. Then there exist recursive sets of K-primes T 1 and T 2, both of natural density zero and with an empty intersection, such that for any set S of primes of K containing T 1 and avoiding T 2, Hilbert s Tenth Problem is unsolvable over O K,S. (Poonen 2003: the case of K = Q; Poonen, S. 2005: the general case)

11 How Things are Done Diophantine Sets, Definitions and Models Diophantine Sets Diophantine Sets Let R be an integral domain. Then 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 m-tuple (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.

12 How Things are Done Diophantine Sets, Definitions and Models Diophantine Sets Other Descriptions Diophantine sets can also described as projections of algebraic sets or sets existentially definable in the language of rings.

13 How Things are Done Diophantine Sets, Definitions and Models Diophantine Subsets of Z MDRP Theorem The recursively enumerable subsets of Z are the same as the Diophantine subsets of Z. Corollary There are undecidable Diophantine subsets of Z.

14 How Things are Done Diophantine Sets, Definitions and Models Existence of Undecidable Diophantine Sets Implies No Algorithm Suppose A Z is an undecidable Diophantine set with a Diophantine definition P(T, X 1,..., X k ). Assume also that we have an algorithm to determine existence of integer solutions for polynomials. Now, let a Z >0 and observe that a A iff P(a, X 1,..., X K ) = 0 has solutions in Z k. So if can answer Hilbert s question effectively, we can determine the membership in A effectively.

15 How Things are Done Diophantine Sets, Definitions and Models Diophantine Models Definition Let R 1, R 2 be two recursive rings and let φ : R 1 R2 m, m Z >0 be an injective recursive map sending Diophantine sets of R1 k, k Z >0 to Diophantine sets of R2 k+m. Then φ is called a Diophantine model of R 1 over R 2. Remark If R 1 R 2 and φ is the inclusion map, then R 1 has a Diophantine definition over R 2. Conversely, if R 1 has a Diophantine definition over R 2, then R 2 has a Diophantine model of R 1 with φ being the inclusion map.

16 How Things are Done Diophantine Sets, Definitions and Models Diophantine Models and Diophantine Undecidability Proposition Suppose R 1 has undecidable Diophantine sets and R 2 has a Diophantine model of R 1. Then R 2 also has undecidable Diophantine sets. Corollary If R is a countable ring with a Diophantine model of Z, then R has undecidable Diophantine sets and therefore HTP is unsolvable over R. Remark Most of the known Diophantine undecidability results over algebraic extensions of Q are obtained by constructing a Diophantine definition of Z. However, there are notable exceptions to this pattern, e. g. Poonen s Theorem, where a Diophantine model which is not a Diophantine definition is constructed.

17 How Things are Done Diophantine Sets, Definitions and Models Other Types Of a Diophantine Model A Fat Model Let R be a recursive ring. Let φ : Z R be a recursive injection such that every Diophantine set of Z is mapped inside a Diophantine set of R. Then R has undecidable Diophantine sets and therefore HTP is unsolvable over R.

18 How Things are Done Diophantine Sets, Definitions and Models Other Types Of a Diophantine Model A Class Model Let R be a recursive ring. Let be a recursive Diophantine equivalence relation on R. Let φ be a recursive injective map from Z into the set of equivalence classes of R under such that the following subsets of R 3 are Diophantine: (x 1, x 2, x 3 ) Plus n 1, n 2 Z : x 1 φ(n 1 ), x 2 φ(n 2 ), x 3 φ(n 1 + n 2 ) and (x 1, x 2, x 3 ) Times n 1, n 2 Z : x 1 φ(n 1 ), x 2 φ(n 2 ), x 3 φ(n 1 n 2 ) Then HTP is unsolvable over R.

19 How Things are Done HTP over a Field vs. HTP over a Subring 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.

20 How Things are Done HTP over a Field vs. HTP over a Subring Diophantine Undecidability of the Field is a Stronger Statement Proposition Let K be any field. Let R be a subring of K such that K is the fraction field of R, the set of non-zero elements of R is existentially definable over R. Then Diophantine undecidability of K implies Diophantine undecidability of R.

21 A Brief History of HTP over Function Fields of Characteristic 0 Some Definitions Review: What is a Function Field? Definition (Function Fields) Let C be a field and let t 1,..., t k be algebraically independent over C. Let K be a finite extension of C(t 1,..., t k ). Let C K be the algebraic closure of C in K. Then K is called a function field in k variables over a constant field C K. Definition (Formally Real Field) A field is called formally real if 1 is not a sum of squares.

22 A Brief History of HTP over Function Fields of Characteristic 0 Field Results Field Results Theorem HTP is unsolvable over function fields of the following types: Over constant fields which are formally real or are subfields of a finite extension of Q p for some rational prime p. (Denef 1978, Kim and Roush 1995, Moret-Bailly 2006, Eisenträger 2007) Over C and of transcendence degree at least 2. (Kim and Roush 1992, Eisenträger 2004) Remark If the field is uncountable we have to adjust the statement of the problem

23 A Brief History of HTP over Function Fields of Characteristic 0 Field Results Proving Undecidability of HTP over Function Fields of Characteristic 0 The Main Tools An elliptic curves of rank 1 Diophantine definability of the valuation ring of a prime of the field: O p = fx K : ord p x 0g Theorem (Moret-Bailly, 2006) Let K be any function field of characteristic 0. Then there exists an elliptic curve of rank 1 defined over K. Let K be a function field such that for some prime p of K we have that O p is existentially definable over K. Then HTP is unsolvable over K.

24 A Brief History of HTP over Function Fields of Characteristic 0 Ring Results Big and Small Rings inside a Function Field Definition (Holomorphy Ring) Let K be a one variable function field. Let W be a set of primes of K. Then let O K,W = fx K : ord p x 0 p Wg Theorem (Moret-Bailly, S. work in progress) Let K be any function field of characteristic 0 and let W be any set of primes of K not containing all the primes of K. Then HTP is not solvable over O K,W.

25 HTP over Function Fields of Positive Characteristic A Longer History of Diophantine Undecidability for Function Fields of Positive Characteristic HTP over Rational Function Fields of Positive Characteristic Theorem HTP is unsolvable over the following fields: rational function fields over finite fields of characteristic greater than 2 (Pheidas, 1991); rational function fields over a constant field C, where C is a proper subfield of the algebraic closure of a finite field (Kim and Roush, 1992). rational function fields over finite fields of characteristic 2 (Videla, 1994).

26 HTP over Function Fields of Positive Characteristic A Longer History of Diophantine Undecidability for Function Fields of Positive Characteristic HTP over Algebraic Function Fields of Positive Characteristic Theorem HTP is unsolvable over the following fields: algebraic function fields over finite fields of characteristic greater than 2 (S. 1996); a field K = C(u, v) Z/p F, where p > 2, C is algebraic over Z/p and has an extension of degree p, u is transcendental over C, v is algebraic over C(u), and C(u, v) and F linearly disjoint over Z/p (S. 2000); K as above for p = 2 ( Eisenträger 2003) any field K finitely generated over Z/p (S. 2002) a field K = E Z/p F, where E is finitely generated over a field C algebraic over Z/p and with an extension of degree p, and E and F linearly disjoint over Z/p (S. 2003)

27 HTP over Function Fields of Positive Characteristic A Longer History of Diophantine Undecidability for Function Fields of Positive Characteristic The Main Unsolved Question A Problem Let C p be the algebraic closure of Z/p for some rational prime p. Show that the existential theory of a function field (or even a rational function field) over C p is undecidable.

28 HTP over Function Fields of Positive Characteristic Multiplication through Addition and Divisibility p-divisibility Definition Let x, y Z 0 and let p be a rational prime. Then we will say that xj p y if y = xp s, where s Z 0. Proposition (Pheidas 1987) Let p be a rational prime. Then multiplication is existentially definable in the system (Z 0, +, j p ).

29 HTP over Function Fields of Positive Characteristic Multiplication through Addition and Divisibility Proving Diophantine Undecidability Proposition Let K be a countable function field over a field of constants C of positive characteristic p. Let q be a prime of K. Suppose the following subsets of K are Diophantine over K: INT = fx K : ord q x 0g; p(k) = f(x, y) K 2 : y = x ps, s Z 0 g. Then HTP is unsolvable over K.

30 HTP over Function Fields of Positive Characteristic Multiplication through Addition and Divisibility Constructing a Model of (Z 0, +, j p ) Proof. Send n A n = fx K : ord q x = ng. Observe the following: For any x K we have that n : x A n ord q x 0 x x, y A n ord q y = 0, ord q x 0, ord q y 0 xy x A n, y A m, z A n+m ord q z = 0, ord q x 0, ord q y 0, ord q z 0 x A n, y A m, nj p m s Z 0, z K : y = z ps, ord q z x = 0, ord q x 0, ord q y 0, ord q z 0

31 HTP over Function Fields of Positive Characteristic P-th Powers over a Function Field The General Plan Notation Let C be a field of characteristic p > 0, let t be transcendental over C, let K be a finite separable extension of C(t). The Three Step Program 1 Define p-th powers of t. 2 Define p-th powers of a set of functions with simple zeros and poles. 3 Define p-th powers of arbitrary functions.

32 HTP over Function Fields of Positive Characteristic P-th Powers over a Function Field p-th Powers of t over Rational Function Field of Characteristic Greater Than 2 Lemma (Pheidas) Let C be a finite field of characteristic p > 2. Let t be transcendental over C. Then the equations below are satisfied with u, v, w C(t) if and only if for some s Z 0 we have that w = t ps. { w t = v p v 1 w 1 t = (1) up u Satisfiability is easy For any x K and any s Z 0 x ps x = (x p(s 1) +x p(s 2) +...+x) p (x p(s 1) +x p(s 2) +...+x) (2)

33 HTP over Function Fields of Positive Characteristic P-th Powers over a Function Field Constructing p-th powers of t We proceed in two steps. First we show that if w satisfies equation below, then it is a p-th power. { w t = v p v 1 w 1 t = (3) up u Second, we show that if w = w p 1 above: we can rewrite the equations { w1 t = (v p w p 1 ) + (w 1 v) = v p 1 v 1 1 w 1 1 t = up 1 w p w 1 u = u p 1 u 1 (4)

34 HTP over Function Fields of Positive Characteristic P-th Powers over a Function Field Why is w a p-th power? { w t = v p v 1 w 1 t = up u Since in a rational function the class number is one and we are working over a perfect field of constants, it is enough to show that the divisor of w is a p-th power of another divisor. Let p be the zero of t and let q be the pole of t. Then ord p t = 1 and ord q t = 1. If t is a prime of C(t) not equal to p or q and ord t w 0, then from the equations above ord t w 0 mod p. Further, if ord p w 0 mod p or ord q w 0 mod p then ord p w = 1 and ord q w = 1 and thus w = tz p. (5)

35 HTP over Function Fields of Positive Characteristic P-th Powers over a Function Field Why is w a p-th power? We can now rewrite w t as tz p t = t(z 1) p = t( f 1 f 2 ) p, where f 1, f 2 are relatively prime polynomials in t over C with f 2 monic. From the equation w t = u p u of our original system we now have that t( f 1 f 2 ) p = u p u. From this equations we see that every pole of u is either a pole of t or a zero of f 2. Also, p is not a pole of u, because the order of all poles of the right-hand side is divisible by p and the order of the left-hand side at p is not divisible by p. Further, if t fq, pg and ord t u < 0 then ord t u = ord t f 2. Hence f 2 u can have a pole at q only and thus is a polynomial in t.

36 HTP over Function Fields of Positive Characteristic P-th Powers over a Function Field Why is w a p-th power? Let s = f 2 u and rewrite the the equation above sf p 1 2 = s p tf p 1. Since p > 2, we have that p 1 2. Let c be a zero of f 2 and observe that ord c (s p tf p 1 ) > 1. Thus, ord c (s p tf p 1 ) > 0 or ord c (f p 1 ) > 0. But (f 1, f 2 ) = 1. So f 2 has no zeros and therefore is 1. Thus, w is a polynomial in t. Similarly, we can show that 1 w is a polynomial in 1 t. Hence, w = ctk. But if we consider our assumptions that ord p w = 1 and ord q w = 1 and the original equations, we conclude that w = t.

37 HTP over Function Fields of Positive Characteristic P-th Powers over a Function Field What happens in algebraic extensions? Plan We try to reduce the algebraic case to the rational case. Assumptions K/C(t) is a separable extension of degree p The zero (p) and the pole (q) of t remain inert in this extension. w C(t) and t still satisfy (1).

38 HTP over Function Fields of Positive Characteristic P-th Powers over a Function Field What happens in algebraic extensions? Claim N K/C(t) (w) is a p-th power. Proof. For any prime t p, t q we have that ord t w 0 mod p. Further, ord P N K/C(t) (w) = f ord p w, where f is the relative degree of p over C(t) and P is the prime below p in C(t). Since p is inert in the extension, f = p. So ord P N K/C(t) (w) 0 mod p. Similarly, we also have ord Q N K/C(t) (w) 0 mod p for the prime Q lying below q in C(t). Thus, the divisor of the norm is a p-th power of another divisor of C(t). As before, this is enough to conclude that the norm of w is a p-th power.

39 HTP over Function Fields of Positive Characteristic P-th Powers over a Function Field What happens in algebraic extensions? From Norms to Minimal Polynomials to Being a p-th Power Suppose we show that N K/C(t) (w a i ) is a p-th power in C(t) for i = 0,..., p, where a i C and are all distinct. (This of course implies that jcj must be bigger than p + 1.) If H(X ) = B 0 + B 1 X +... X p is the monic irreducible polynomial of w over C(t), then we show that H(a i ) = p j=1 (a i w j ) = z p i, a p-th power in C(t). (Here w = w 1,..., w p are the conjugates of w over C(t).) Thus we have a linear system 1, a 0,..., a p , a p,..., a p p B = z p 0... z p p We can solve this system for B 0, B 1,..., B p 1 and determine that all the coefficients are p-th powers. Now we can conclude that w is a p-th power.

40 HTP over Function Fields of Positive Characteristic P-th Powers over a Function Field Putting Things Together Lemma Let K/C(t) be a separable extension of degree p. Let a 0 = 0, a 1,..., a p, b 0 = 0,..., b p C, with all a i distinct. Let p i be the zero of t + b i and let q be the pole of all t + b i. Assume that q, p 0,..., p p lie above primes of C(t) inert in the extension K/C(t). Let w, t, u 0,..., u p, v 0,..., v p K satisfy the following equations. { (w ai ) (t b i ) = u p i u i, i = 0,..., p = v p i v, i = 0,..., p 1 w a i 1 t b i Then either w C(t) or w is a p-th power. Disclaimer Some (annoying) technical details are omitted.

41 HTP over Function Fields of Positive Characteristic P-th Powers over a Function Field Existence of the Required Rational Subfield Theorem Let K be a function field of positive characteristic p over a constant field C algebraic over a finite field. Assume C has an extension of degree p. Then for any r Z >0 there exists a finite extension Ĉ of C such that the field compositum ˆK = ĈK contains a non-constant element t satisfying the following conditions: 1 [ ˆK : Ĉ(t)] = pk, k Z 0 2 There exist a 0 = 0, a 1,..., a r+p k Ĉ such that the Ĉ(t) prime P i which is the zero of t + a i ˆK/Ĉ(t). is inert in the extension 3 The Ĉ(t) prime Q which is the pole of t is inert in the extension ˆK/Ĉ(t).

42 HTP over Function Fields of Positive Characteristic P-th Powers over a Function Field What happens when transcendence degree goes up? Part 1: p > 2 Definition Let u : A B be a morphism of abelian groups. We say that u is almost bijective if u is injective and Coker u is a finite p-group. Theorem (More-Bailly) Let C be a field of characteristic p > 2, and assume that C is of transcendence degree at least one over a finite field. Let K be a function field in one variable with constant field C, let E : y 2 = P(x) be an elliptic curve which is defined over a finite field contained in C. There exists a non-constant element t K such that the elliptic curve E given by E : P(t) y 2 = P(x) has the property that the natural homomorphism E(F (t)) E(K) induced by the inclusion F (t) K is almost bijective.

43 HTP over Function Fields of Positive Characteristic P-th Powers over a Function Field What happens when transcendence degree goes up? Part 1: p > 2 Lemma Let E, t be as above. Let q be the size of the finite field containing the coefficients of E. Let s = P(t). Then the point (t qm, s qm 1 ) E(F (t)).

44 HTP over Function Fields of Positive Characteristic P-th Powers over a Function Field What happens when transcendence degree goes up? Part 2: p = 2 We Know Less We can reduce the case of a finite transcendence degree of the constant field to the case of an algebraic constant field. (This means that the algebraic closure of the finite field in the given field of constants must have an extension of degree 2.) We can also handle some cases of infinite transcendence degree. The general situation is not clear.

45 HTP over Function Fields of Positive Characteristic Defining Order at a Prime A Property of Order at a Prime Let K be a function field and let x, y K. Let p be a prime of K. Assume that ord p x < ord p y. Then ord p (x + y) = ord p x. Now let a K be such that ord p a = 1 and consider ord p (a q + ax q ) = { ordp ax q 1 0 mod q if ord p x < 0, ord p a q 0 mod q, if ord p x 0.

46 HTP over Function Fields of Positive Characteristic Defining Order at a Prime Inert Primes and Norms Lemma Let K/F be a function field extension, p a prime of F inert in the extension K/F, and consider the following equation: N K/F (z) = y. For any y F there exists z K satisfying this equation only if ord p y 0 mod [K : F ].

47 HTP over Function Fields of Positive Characteristic Defining Order at a Prime Satisfying the Norm Equation Let F be a function field over a finite field of constants C of characteristic p > 0. Let q p be a rational prime. Let p be a prime of F. Let b C be such that it is not a q-th power modulo p. Let K = F ( q b). Observe that K/F is not ramified.

48 HTP over Function Fields of Positive Characteristic Defining Order at a Prime Satisfying the Norm Equation To begin with we consider N K/F (z) = (a q + ax q ). Observe that for any F -prime t not occurring in the divisor of a, all the poles of y = a q + ax q are of order divisible by q. Our goal is to make sure that for all primes t p we have that ord t y 0 mod q.

49 HTP over Function Fields of Positive Characteristic Defining Order at a Prime Satisfying the Norm Equation To this end we find a finite extension F of degree q over F where all the primes occurring in the divisor of y not equal to p ramify completely and p splits completely into distinct factors if it is a pole of x. (We pretty much take the q-th root of y and possibly make some adjustments for primes occurring in the divisor of a. ) Then we consider the norm equation N KF /F (z) = (aq + ax q ). Now Hasse Norm Principle and the fact that the extension is unramified should assure that norm equation can be satisfied as long as p is not a pole of x.

50 Back to the First-Order Theory A Short History of First-Order Decidability over Function Fields of Positive Characteristic What Do We Know about the First-Order Theory Theorem The first-order theory is undecidable for the following fields: rational function fields over perfect field of constants (Cherlin, 1984), function fields over algebraically closed fields of positive characteristic (Duret, 1986 ), rational function fields over any field of constants of characteristic greater than 5 (Pheidas, 2004), any function field of characteristic greater than 2 (Eisenträger, S. 2007), any function field over a field of constants algebraic over a finite field (Eisenträger, S. 2007).

51 Back to the First-Order Theory P-th Powers are Enough Multiplication via Divisibility and Addition Proposition (Julia Robinson) Multiplication is definable in Z 0, +, j. Proposition Let K/C(t) be a finite extension with C of characteristic p > 0. Assume p(k, t) = fx K : s 0, x = t ps g is first order definable. Then the following sets are first-order definable: B(K, t) := f(t ps, x ps, x), s Z >0, x Kg C(K, t) := ft pa, t pb, t pa+b, a, b > 0g Theorem Assume that t has a simple pole or a simple zero. Suppose that the set p(k, t) is first-order definable in K. Then Z >0, +, j has a model over K.

52 Back to the First-Order Theory P-th Powers are Enough Proof. We map s > 0 to t ps. Then s2 s = s 1 + s 2 (t ps1, t ps2, t ps ) C(K, t). Further s 1 if and only if (p s 1 1) (p s 2 1) if and only if there exists x K such that since at least one pole or zero of t is simple. Hence s 1 s2 if and only if x ps 1 1 = t ps 2 1, (6) x, y K ( (t ps1, y, x) B(K, t) y/x = t ps2 /t ). The result now follows from the fact that the sets p(k, t), B(K, t) and C(K, t) are all first-order definable in K.