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arXiv:0903.1442v2 [math.LO] 24 May 2009

Henson and Rubel’s Theorem for Zilber’s

Pseudoexponentiation

Ahuva C. Shkop

May 25, 2009

Abstract

In 1984, Henson and Rubel ([2]) proved the following theorem: If

p(x1,...,xn) is an exponential polynomial with coefficients in C with

no zeroes in C, then p(x1,...,xn) = eg(x1,...,xn)for some exponential

polynomial g(x1,...,xn) over C. In this paper, I will prove the analog

of this theorem for Zilber’s Pseudoexponentiation directly from the

axioms. Furthermore, this proof relies only on the existential closed-

ness axiom without any reference to Schanuel’s conjecture.

1 Introduction

In [7], Zilber constructed an exponential field, Zilber’s Pseudoexponentiation,

of size continuum that satisfies many special properties. Schanuel’s conjec-

ture is true in this field and every definable set is countable or co-countable

(quasiminimality). It is still unknown whether Pseudoexponentiation is iso-

morphic to complex exponentiation.

In [2], Henson and Rubel prove that the only exponential polynomials

with no zeros are of the form exp(g) where g is some exponential polynomial.

Although this seems to be a question in exponential algebra, this proof uses

Nevanlinna theory.

The goal of this paper is to prove the following theorem:

Theorem 1. Let p(x1,...,xn) be an exponential polynomial with coefficients

in Zilber’s Pseudoexponentiation K. If p ?= exp(g(x1,...,xn)) for any expo-

nential polynomial g(x1,...,xn), then p has a root in K.

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D’Aquino, Macintyre, and Terzo have also explored this problem and offer

an alternate proof of this theorem in [1]. We will use purely algebraic tech-

niques and give a proof directly from the axioms. This proof uses only basic

exponential algebra and is entirely independent of Schanuel’s conjecture.

We will begin with the following definitions.

Definition 2. In this paper, a (total) E-ring is a Q-algebra R with no zero

divisors, together with a homomorphism exp : ?R,+? → ?R∗,·?.

A partial E-ring is a Q-algebra R with no zero divisors, together with a Q-

linear subspace A(R) of R and a homomorphism exp : ?A(R),+? → ?R∗,·?.

A(R) is then the domain of exp.

An E-field is an E-ring which is a field.

We say S is a partial E-ring extension of R if R and S are partial E-ring,

R ⊂ S, and for all r ∈ A(R), expS(r) = expR(r).

We now set some conventions. Let K be any algebraically closed field and

α ∈ N. Throughout this paper, a variety is a (possibly reducible) Zariski

closed subset of Gα(K) := Kα× (K∗)αor some projection of Gα(K). We

will use the notation ¯ y for a finite tuple y1,...,ym, and we will write exp(¯ y)

instead of exp(y1),...,exp(yn). Similarly for a subset S of an E-ring, exp(S) is

the exponential image of S. We write tdA(¯b) to be the transcendence degree

of the field generated by (A,¯b) over the field generated by A.

To prove Theorem 1, we recall that Zilber’s field, which we will call K,

satisfies the following axiom:

Axiom 3. If a variety V ⊆ Gα(K) is irreducible, rotund, and free, then there

are infinitely many ¯ x such that (¯ x,exp(¯ x)) ∈ V .

The definitions of a rotund variety and a free variety will be given later in

the paper. The outline of the proof is as follows:

1. Given an exponential polynomial p(¯ x), we construct a variety Vpsatis-

fying

∀¯ a(∃¯b(¯ a,¯b,exp(¯ a),exp(¯b)) ∈ Vp ⇐⇒ p(¯ a) = 0).

2. We reduce to the case where Vpis irreducible and free.

3. We prove that if p(¯ x) ?= exp(g(¯ x)), then Vpis rotund.

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2 Constructing Vp

Recall the following construction of K[X]E, the exponential polynomial ring

over an E-field K on the set of indeterminates X: (see [6],[3])

If R is a partial E-ring, we can construct R′, a partial E-ring extension

of R, with the following properties:

• The domain of the exponential map in R′is precisely R.

• If for i = 1,...,n, yi / ∈ A(R), then tdR(expR′(¯ y)) in R′will be exactly

the Q-linear dimension of ¯ y over A(R).

• R′is generated as a ring by R ∪ exp(R).

For K an E-field and X a set of indeterminates, let K[X] be the partial

E-ring where A(K[X]) = K. Then the exponential polynomial ring over K,

K[X]E, is simply the union of the chain

K[X] = R0֒→ R1֒→ R2֒→ R3֒→ R4֒→ ···

where Rn+1= R′

n.

This construction yields a natural notion of height.

Definition 4. For p an exponential polynomial and n ∈ N, the height(p) = n

if and only if p ∈ Rnand p / ∈ Rn−1.

Example 5. The exponential polynomial p(x1,x2) = exp(exp(x1

in C[x1,x2]Ehas height 2.

2+x2

2))+x3

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We now have the background necessary to begin the construction of Vp.

Let K be an algebraically closed E-field of characteristic 0 and p an

exponential polynomial with coefficients in K.

Definition 6. We will call a set T of exponential polynomials a decomposi-

tion of p if it is a minimal set of exponential polynomials such that:

• ∃t1,...,tk ∈ T : p ∈ K[¯ x,exp(t1),...,exp(tk)], the subring of K[¯ x]E

generated by ¯ x,exp(t1),...,exp(tk).

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• ti∈ T ⇒ ∃t1,...,tl∈ T : ti∈ K[¯ x,exp(t1),...,exp(tl)].

• There is an L ∈ Z∗such thatx1

L,...,xn

Lare in T.

We will call elements of T T-bricks.

Consider the parallel between exponential polynomials and terms in the

language L = {+,−,·,0,1,exp} ∪ {ck : k ∈ K}. This parallel extends to

subterms and T-bricks. Considering this parallel, notice that every T-brick

can be written as a polynomial in ¯ x and the exponential image of the T-

bricks of lower height. Furthermore, all decompositions are finite. To satisfy

the third bullet consider the following: While there are several terms which

correspond to the same polynomial, we can choose one such term and take

the least common multiple of the denominators of the rational coefficients of

all the elements of ¯ x which appear in the term.

Example 7. Consider p(x1,x2) = exp(exp(x1

{x1

2)} is a decomposition of p. Notice that

not in the decomposition since exp(x1

in the decomposition to satisfy the third bullet.

2+ x2

2)) + x3

1. Then T =

x1

2+ x2

2). We need

2,x2

2,x2

2,exp(x1

2+ x2

2is

x2

2+ x2

2) = exp(x1

2)exp(x2

2

Definition 8. We say that a decomposition T is a refined decomposition if

T is Q−linearly independent over K.

Lemma 9. Given a decomposition T, we can form a refined decomposition

T′.

Proof. We induct on the size of T. Clearly, if the decomposition is empty, it

is refined. Suppose T is not refined, and |T| = m and assume the claim for

decompositions of size less than m. Suppose t ∈ T is a Q−linear combination

over K of other T-bricks. That is, for all i ∈ I ⊂ {1,...,m}, t ?= tiand

t = a +

?

i∈I

ai

biti

for some ai,bi∈ Z,bi?= 0,a ∈ K, and the least common multiple of the biis L.

(L > 1 since otherwise T is not minimal and thus not a decomposition.) Then

after replacing each ti,i ∈ I with1

p. exp(t) is now a polynomial in the variables exp(1

I})\({t} ∪ {ti: i ∈ I}) contains a smaller decomposition and by induction,

we can find a refined decomposition T′of p.

Lti, this set will contain a decomposition of

Lti) SoˆT = (T ∪{ti

L: i ∈

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Remark 10. To simplify notation, let ¯ x′= (x1

invertible change of variables, we may and do assume L = 1.

L,...,xn

L). By making this

We now set T0to be a refined decomposition of p, and α = |T0|. Further-

more, we order the T0-bricks in order of height, i.e., height(ti) ≤ height(tj)

for i ≤ j. For convenience, we let the first n elements of T0be x1,...,xn. So

T0= {¯ x} ∪ {ti: n + 1 ≤ i ≤ α}.

We now name the polynomials which witness T0being a decomposition.

For each n + 1 < i < α, let pi∈ K[¯ x,y1,...,yi−1] be such that

pi(¯ x,exp(¯ x),exp(tn+1),...,exp(ti−1)) = ti

Let p∗∈ K[¯ x, ¯ y] be such that p∗(¯ x,exp(¯ x,tn+1,...,tα)) = p.

Let Vp⊆ Gα= Kα× (K∗)αbe the variety given as follows:

(x1,...,xn,wn+1,...,wα,y1,...,yα) = (¯ x, ¯ w, ¯ y) ∈ Vp

⇐⇒ (

α?

i=n+1

wi= pi(¯ x, ¯ y)) ∧ p∗(¯ x, ¯ y) = 0

Please note the indexing. We will maintain this indexing for coordinates

of points in the variety as well.

Proposition 11. For any ¯ a ∈ K

∃¯b((¯ a,¯b,exp(¯ a),exp(¯b)) ∈ Vp) ⇐⇒ p(¯ a) = 0

Proof.

∃¯b((¯ a,¯b,exp(¯ a),exp(¯b)) ∈ Vp) ⇐⇒

∃¯b(∀bi∈¯b(bi= pi(¯ a,exp(¯ x′),exp(bn+1,...,bi−1))) ∧ p∗(¯ a,exp(¯ a),exp(¯b)) = 0)

⇐⇒ ∃¯b(∀bi∈¯b(bi= ti(¯ a)) ∧ p∗(¯ a,exp(¯ a),exp(¯b)) = 0)

Since p∗(¯ a,exp(¯ a),exp(tn+1(¯ a)),...,exp(tα(¯ a))) = p(¯ a), this is if and only

if p(¯ a) = 0.

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