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HTP in Positive Characteristic


Alexandra Shlapentokh

East Carolina University

November 2007


Table of Contents1 A Brief History of Diophantine Undecidability over Number

FieldsThe Original ProblemExtensions to Number Fields

2 How Things are DoneDiophantine Sets, Definitions and ModelsHTP over a Field vs. HTP over a Subring

3 A Brief History of HTP over Function Fields ofCharacteristic 0

Some DefinitionsField ResultsRing Results

4 HTP over Function Fields of Positive CharacteristicA Longer History of Diophantine Undecidability for FunctionFields of Positive CharacteristicMultiplication through Addition and DivisibilityP-th Powers over a Function FieldDefining Order at a Prime

5 Back to the First-Order TheoryA Short History of First-Order Decidability over Function Fieldsof Positive CharacteristicP-th Powers are Enough


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 anarbitrary polynomial equation in several variables has solutions inintegers?

This problem became known as Hilbert’s Tenth Problem



The Original Problem

The Answer

This question was answered negatively (with the final piece inplace in 1970) in the work of Martin Davis, Hilary Putnam, JuliaRobinson and Yuri Matijasevich.



Extensions to Number Fields

A General Question

A Question about an Arbitrary Recursive Ring R

Is there an algorithm, which if given an arbitrary polynomialequation in several variables with coefficients in R, can determinewhether this equation has solutions in R?

Arguably, the most important open problems in the area concernthe Diophantine status of the ring of integers of an arbitrarynumber field and the Diophantine status of Q.




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 theirextensions 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 ofpositive rank defined over Q with [E (K ) : E (Q)] <∞. (Poonen,S. 2003)

Any number field K such that there exists an elliptic curve ofrank 1 over K and an Abelian variety over Q keeping its rank overK . (Cornelissen, Pheidas, Zahidi, 2005)




The Rings in between the Ring of Integers and aNumber 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

OK ,V = fx ∈ K : ordp x ≥ 0 ∀p 6∈ Vg.

If V = ∅, then OK ,V = OK – the ring of integers of K . If Vcontains all the primes of K , then OK ,V = K . If V is finite, we callthe ring small. If V is infinite, we call the ring big or large.




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 OK , then HTPis unsolvable over any small subring of K .

(Julia Robinson and others)




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-archimedeanprimes of K such that

The natural density of S is greater 1− 1

[K : Q]− ε.

HTP is unsolvable over OK ,S .

(S. 2002, 2003, 2006)




HTP over Very Large Subrings of Number Fields

Theorem

Let K be a number field with a rank one elliptic curve. Then thereexist recursive sets of K -primes T1 and T2, both of natural densityzero and with an empty intersection, such that for any set S ofprimes of K containing T1 and avoiding T2, Hilbert’s TenthProblem is unsolvable over OK ,S . (Poonen 2003: the case ofK = Q; Poonen, S. 2005: the general case)


How Things are Done

Diophantine Sets, Definitions and Models

Diophantine Sets

Diophantine Sets

Let R be an integral domain. Then a subset A ⊂ Rm is calledDiophantine over R if there exists a polynomialp(T1, . . . ,Tm,X1, . . . ,Xk) with coefficients in R such that for anym-tuple (t1, . . . , tm) ∈ Rm we have that

∃x1, . . . , xk ∈ R : p(t1, . . . , tm, x1, . . . , xk) = 0

m

(t1, . . . , tm) ∈ A.

In this case we call p(T1, . . . ,Tm,X1, . . . ,Xk) a Diophantinedefinition of A over R.


How Things are Done


Diophantine Sets

Other Descriptions

Diophantine sets can also described as projections of algebraic setsor sets existentially definable in the language of rings.


How Things are Done


Diophantine Subsets of Z

MDRP Theorem

The recursively enumerable subsets of Z are the same as theDiophantine subsets of Z.

Corollary

There are undecidable Diophantine subsets of Z.


How Things are Done


Existence of Undecidable Diophantine SetsImplies No Algorithm

Suppose A ⊂ Z is an undecidable Diophantine set with aDiophantine definition P(T ,X1, . . . ,Xk). Assume also that wehave an algorithm to determine existence of integer solutions forpolynomials. Now, let a ∈ Z>0 and observe that a ∈ A iffP(a,X1, . . . ,XK ) = 0 has solutions in Zk . So if can answerHilbert’s question effectively, we can determine the membership inA effectively.


How Things are Done


Diophantine Models

Definition

Let R1,R2 be two recursive rings and let φ : R1 −→ Rm2 ,m ∈ Z>0

be an injective recursive map sending Diophantine sets ofRk

1 , k ∈ Z>0 to Diophantine sets of Rk+m2 . Then φ is called a

Diophantine model of R1 over R2.

Remark

If R1 ⊂ R2 and φ is the inclusion map, then R1 has a Diophantinedefinition over R2. Conversely, if R1 has a Diophantine definitionover R2, then R2 has a Diophantine model of R1 with φ being theinclusion map.


How Things are Done


Diophantine Models and DiophantineUndecidability

Proposition

Suppose R1 has undecidable Diophantine sets and R2 has aDiophantine model of R1. Then R2 also has undecidable Diophantinesets.

Corollary

If R is a countable ring with a Diophantine model of Z, then R hasundecidable Diophantine sets and therefore HTP is unsolvable overR.

Remark

Most of the known Diophantine undecidability results over algebraicextensions of Q are obtained by constructing a Diophantine definitionof Z. However, there are notable exceptions to this pattern, e. g.Poonen’s Theorem, where a Diophantine model which is not aDiophantine definition is constructed.


How Things are Done


Other Types Of a Diophantine Model

A Fat Model

Let R be a recursive ring. Let φ : Z −→ R be a recursive injectionsuch that every Diophantine set of Z is mapped inside aDiophantine set of R. Then R has undecidable Diophantine setsand therefore HTP is unsolvable over R.


How Things are Done


Other Types Of a Diophantine Model

A Class Model

Let R be a recursive ring. Let ∼ be a recursive Diophantineequivalence relation on R. Let φ be a recursive injective map from Zinto the set of equivalence classes of R under ∼ such that thefollowing subsets of R3 are Diophantine:

(x1, x2, x3) ∈ Plus

~

�

∃n1, n2 ∈ Z : x1 ∈ φ(n1), x2 ∈ φ(n2), x3 ∈ φ(n1 + n2)

and(x1, x2, x3) ∈ Times

~

�

∃n1, n2 ∈ Z : x1 ∈ φ(n1), x2 ∈ φ(n2), x3 ∈ φ(n1n2)

Then HTP is unsolvable over R.


How Things are Done

HTP over a Field vs. HTP over a Subring

Undecidability of HTP over Q ImpliesUndecidability of HTP for Z

Indeed, suppose we knew how to determine whether solutions existover Z. Let Q(x1, . . . , xk) be a polynomial with rationalcoefficients. Then

∃x1, . . . , xk ∈ Q : Q(x1, . . . , xk) = 0

~

�

∃y1, . . . , yk , z1, . . . , zk ∈ Z : Q(y1

z1, . . . ,

yk

zk) = 0 ∧ z1 . . . zk 6= 0.

So decidability of HTP over Z would imply the decidability of HTPover Q.


How Things are Done

HTP over a Field vs. HTP over a Subring

Diophantine Undecidability of the Field is aStronger 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 definableover R.

Then Diophantine undecidability of K implies Diophantineundecidability of R.


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 t1, . . . , tk be algebraically independentover C . Let K be a finite extension of C (t1, . . . , tk). Let CK bethe algebraic closure of C in K . Then K is called a function fieldin k variables over a constant field CK .

Definition (Formally Real Field)

A field is called formally real if −1 is not a sum of squares.



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 ofa finite extension of Qp for some rational prime p. (Denef1978, Kim and Roush 1995, Moret-Bailly 2006, Eisentrager2007)

Over C and of transcendence degree at least 2. (Kim andRoush 1992, Eisentrager 2004)

Remark

If the field is uncountable we have to adjust the statement of theproblem



Field Results

Proving Undecidability of HTP over FunctionFields of Characteristic 0

The Main Tools

An elliptic curves of rank 1

Diophantine definability of the valuation ring of a prime of thefield:

Op = fx ∈ K : ordp x ≥ 0g

Theorem (Moret-Bailly, 2006)

Let K be any function field of characteristic 0. Then thereexists an elliptic curve of rank 1 defined over K .

Let K be a function field such that for some prime p of K wehave that Op is existentially definable over K . Then HTP isunsolvable over K .



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 ofK . Then let

OK ,W = fx ∈ K : ordp x ≥ 0∀p 6∈ Wg

Theorem (Moret-Bailly, S. work in progress)

Let K be any function field of characteristic 0 and let W be anyset of primes of K not containing all the primes of K . Then HTPis not solvable over OK ,W .


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 PositiveCharacteristic

Theorem

HTP is unsolvable over the following fields:

rational function fields over finite fields of characteristicgreater than 2 (Pheidas, 1991);

rational function fields over a constant field C , where C is aproper subfield of the algebraic closure of a finite field (Kimand Roush, 1992).

rational function fields over finite fields of characteristic 2(Videla, 1994).




HTP over Algebraic Function Fields of PositiveCharacteristic

Theorem

HTP is unsolvable over the following fields:

algebraic function fields over finite fields of characteristic greaterthan 2 (S. 1996);a field K = C (u, v)⊗Z/p F , where p > 2, C is algebraic overZ/p and has an extension of degree p, u is transcendental overC , v is algebraic over C (u), and C (u, v) and F linearly disjointover Z/p (S. 2000);

K as above for p = 2 ( Eisentrager 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 fieldC algebraic over Z/p and with an extension of degree p, and Eand F linearly disjoint over Z/p (S. 2003)




The Main Unsolved Question

A Problem

Let Cp be the algebraic closure of Z/p for some rational prime p.Show that the existential theory of a function field (or even arational function field) over Cp is undecidable.



Multiplication through Addition and Divisibility

p-divisibility

Definition

Let x , y ∈ Z6=0 and let p be a rational prime. Then we will say thatx jpy if y = xps , where s ∈ Z≥0.

Proposition (Pheidas 1987)

Let p be a rational prime. Then multiplication is existentiallydefinable in the system (Z≥0,+, jp).




Proving Diophantine Undecidability

Proposition

Let K be a countable function field over a field of constants C ofpositive characteristic p. Let q be a prime of K . Suppose thefollowing subsets of K are Diophantine over K :

INT = fx ∈ K : ordq x ≥ 0g;

p(K ) = f(x , y) ∈ K 2 : y = xps, s ∈ Z≥0g.

Then HTP is unsolvable over K .




Constructing a Model of (Z≥0,+, jp)

Proof.

Send n −→ An = fx ∈ K : ordq x = ng. Observe the following:

For any x ∈ K we have that ∃n : x ∈ An ⇔ ordq x ≥ 0

x , y ∈ An ⇔ ordqxy = 0, ordq x ≥ 0, ordq y ≥ 0

x ∈ An, y ∈ Am, z ∈ An+m ⇔ ordqxyz = 0, ordq x ≥

0, ordq y ≥ 0, ordq z ≥ 0

x ∈ An, y ∈ Am, njpm⇔ ∃s ∈ Z≥0,∃z ∈ K : y =zps, ordq

zx = 0, ordq x ≥ 0, ordq y ≥ 0, ordq z ≥ 0



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 andpoles.

3 Define p-th powers of arbitrary functions.




p-th Powers of t over Rational Function Field ofCharacteristic Greater Than 2

Lemma (Pheidas)

Let C be a finite field of characteristic p > 2. Let t betranscendental over C . Then the equations below are satisfied withu, v ,w ∈ C (t) if and only if for some s ∈ Z≥0 we have thatw = tps

.{

w − t = v p − v1w −

1t = up − u

(1)

Satisfiability is easy

For any x ∈ K and any s ∈ Z≥0

xps−x = (xp(s−1)+xp(s−2)

+. . .+x)p−(xp(s−1)+xp(s−2)

+. . .+x) (2)




Constructing p-th powers of t

We proceed in two steps. First we show that if w satisfies equationbelow, then it is a p-th power.

{

w − t = v p − v1w −

1t = up − u

(3)

Second, we show that if w = wp1 we can rewrite the equations

above:{

w1 − t = (v p − wp1 ) + (w1 − v) = v

p1 − v1

1w1− 1

t = up − 1w

p1

+ 1w1− u = u

p1 − u1

(4)




Why is w a p-th power?

{

w − t = v p − v1w −

1t = up − u

(5)

Since in a rational function the class number is one and we areworking over a perfect field of constants, it is enough to show thatthe divisor of w is a p-th power of another divisor. Let p be thezero of t and let q be the pole of t. Then ordp t = 1 andordq t = −1. If t is a prime of C (t) not equal to p or q andordt w 6= 0, then from the equations above ordt w ≡ 0 mod p.Further, if ordp w 6≡ 0 mod p or ordq w 6≡ 0 mod p thenordp w = 1 and ordq w = −1 and thus w = tzp.





We can now rewrite w − t as

tzp − t = t(z − 1)p = t(f1

f2)p,

where f1, f2 are relatively prime polynomials in t over C with f2

monic. From the equation w − t = up − u of our original systemwe now have that

t(f1

f2)p = up − u.

From this equations we see that every pole of u is either a pole oft or a zero of f2. Also, p is not a pole of u, because the order of allpoles of the right-hand side is divisible by p and the order of theleft-hand side at p is not divisible by p. Further, if t 6∈ fq, pg andordt u < 0 then ordt u = ordt f2. Hence f2u can have a pole at q

only and thus is a polynomial in t.





Let s = f2u and rewrite the the equation above

sfp−1

2 = sp − tfp

1 .

Since p > 2, we have that p − 1 ≥ 2. Let c be a zero of f2 andobserve that

ordc(sp − tfp

1 ) > 1.

Thus,ordc(sp − tf

p1 )′ > 0 or ordc(f

p1 ) > 0.

But (f1, f2) = 1. So f2 has no zeros and therefore is 1. Thus, w isa polynomial in t. Similarly, we can show that 1

w is a polynomial in1t . Hence, w = ctk . But if we consider our assumptions thatordp w = 1 and ordq w = −1 and the original equations, weconclude that w = t.




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 thisextension.

w 6∈ C (t) and t still satisfy (1).





Claim

NK/C(t)(w) is a p-th power.

Proof.

For any prime t 6= p, t 6= q we have that ordt w ≡ 0 mod p.Further, ordP NK/C(t)(w) = f ordp w , where f is the relativedegree 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 ordP NK/C(t)(w) ≡ 0 mod p.Similarly, we also have ordQ NK/C(t)(w) ≡ 0 mod p for the primeQ lying below q in C (t). Thus, the divisor of the norm is a p-thpower of another divisor of C (t). As before, this is enough toconclude that the norm of w is a p-th power.





From Norms to Minimal Polynomials to Being a p-th Power

Suppose we show that NK/C(t)(w − ai ) is a p-th power in C (t) fori = 0, . . . , p, where ai ∈ C and are all distinct. (This of courseimplies that jC j must be bigger than p + 1.) IfH(X ) = B0 + B1X + . . .X p is the monic irreducible polynomial of wover C (t), then we show that −H(ai ) =

∏pj=1(ai − wj ) = z

pi , a p-th

power in C (t). (Here w = w1, . . . ,wp are the conjugates of w overC (t).) Thus we have a linear system

1, a0, . . . , ap0

. . .1, ap, . . . , a

pp

B0

. . .1

=

zp0

. . .z

pp

We can solve this system for B0,B1, . . . ,Bp−1 and determine that allthe coefficients are p-th powers. Now we can conclude that w is ap-th power.




Putting Things Together

Lemma

Let K/C (t) be a separable extension of degree p. Leta0 = 0, a1, . . . , ap, b0 = 0, . . . , bp ∈ C , with all ai distinct. Let pi

be the zero of t + bi and let q be the pole of all t + bi . Assumethat q, p0, . . . , pp lie above primes of C (t) inert in the extensionK/C (t). Let w , t, u0, . . . , up, v0, . . . , vp ∈ K satisfy the followingequations.

{

(w − ai )− (t − bi ) = upi − ui , i = 0, . . . , p

1w−ai

− 1t−bi

= vpi − v , i = 0, . . . , p

Then either w ∈ C (t) or w is a p-th power.

Disclaimer

Some (annoying) technical details are omitted.




Existence of the Required Rational Subfield

Theorem

Let K be a function field of positive characteristic p over aconstant field C algebraic over a finite field. Assume C has anextension of degree p. Then for any r ∈ Z>0 there exists a finiteextension C of C such that the field compositum K = C K containsa non-constant element t satisfying the following conditions:

1 [K : C (t)] = pk , k ∈ Z≥0

2 There exist a0 = 0, a1, . . . , ar+pk ∈ C such that the C (t)prime Pi which is the zero of t + ai is inert in the extensionK/C (t).

3 The C (t) prime Q which is the pole of t is inert in theextension K/C (t).




What happens when transcendence degree goesup? Part 1: p > 2

Definition

Let u : A→ B be a morphism of abelian groups. We say that u isalmost 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 oftranscendence degree at least one over a finite field. Let K be afunction 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 curveE 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.




What happens when transcendence degree goesup? Part 1: p > 2

Lemma

Let E , t be as above. Let q be the size of the finite field containingthe coefficients of E . Let s =

√

P(t). Then the point(tqm

, sqm−1) ∈ E(F (t)).




What happens when transcendence degree goesup? Part 2: p = 2

We Know Less

We can reduce the case of a finite transcendence degree ofthe constant field to the case of an algebraic constant field.(This means that the algebraic closure of the finite field in thegiven field of constants must have an extension of degree 2.)

We can also handle some cases of infinite transcendencedegree.

The general situation is not clear.



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 ordp x < ordp y . Then ordp(x + y) = ordp x . Now leta ∈ K be such that ordp a = −1 and consider

ordp(aq + axq) =

{

ordp axq ≡ −1 6≡ 0 mod q if ordp x < 0,ordp aq ≡ 0 mod q, if ordp x ≥ 0.




Inert Primes and Norms

Lemma

Let K/F be a function field extension, p – a prime of F inert inthe extension K/F , and consider the following equation:

NK/F (z) = y .

For any y ∈ F there exists z ∈ K satisfying this equation only ifordp y ≡ 0 mod [K : F ].




Satisfying the Norm Equation

Let F be a function field over a finite field of constants C ofcharacteristic p > 0.

Let q 6= 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.





To begin with we consider

NK/F (z) = (aq + axq).

Observe that for any F -prime t not occurring in the divisor of a, allthe poles of y = aq + axq are of order divisible by q. Our goal is tomake sure that for all primes t 6= p we have that ordt y ≡ 0 mod q.





To this end we find a finite extension F ′ of degree q over F whereall the primes occurring in the divisor of y not equal to p ramifycompletely and p splits completely into distinct factors if it is apole of x . (We pretty much “take the q-th root of” 1 + 1

y andpossibly make some adjustments for primes occurring in the divisorof a. ) Then we consider the norm equation

NKF ′/F ′(z) = (aq + axq).

Now Hasse Norm Principle and the fact that the extension isunramified should assure that norm equation can be satisfied aslong as p is not a pole of x .


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 positivecharacteristic (Duret, 1986 ),

rational function fields over any field of constants ofcharacteristic greater than 5 (Pheidas, 2004),

any function field of characteristic greater than 2 (Eisentrager,S. 2007),

any function field over a field of constants algebraic over afinite field (Eisentrager, S. 2007).



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 = tpsg is first order definable.Then the following sets are first-order definable:

B(K , t) := f(tps, xps

, x), s ∈ Z>0, x ∈ Kg

C(K , t) := ftpa, tpb

, tpa+b, a, b > 0g

Theorem

Assume that t has a simple pole or a simple zero. Suppose that theset p(K , t) is first-order definable in K . Then 〈Z>0,+, j〉 has amodel over K .



P-th Powers are Enough

Proof.

We map s > 0 to tps. Then

s = s1 + s2 ⇔ (tps1 , tps2 , tps) ∈ C(K , t). Further s1

∣

∣

∣s2 if and only if

(ps1 − 1)∣

∣

∣(ps2 − 1) if and only if there exists x ∈ K such that

xps1−1 = tps2−1, (6)

since at least one pole or zero of t is simple.

Hence s1

∣

∣

∣s2 if and only if

∃ x , y ∈ K(

(tps1, y , x) ∈ B(K , t) ∧ y/x = tps2

/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 .

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