Diophantine sets of polynomials over number fields

Let R be a recursive subring of a number field. We show that recursively enumerable sets are diophantine for the polynomial ring R[Z].


Introduction
Let R be a recursive subring of a number field. In this paper, we show that recursively enumerable (r.e.) subsets of R[Z] k are diophantine.
For any recursively stable integral domain, one can easily see that every diophantine set is recursively enumerable (see the end of section 1.1). However, the converse problem -are recursively enumerable sets diophantine? -is much more difficult.
The intuition of a recursive ring is a ring in which we can effectively compute, it is a ring whose elements can be represented by a computer. The recursive presentation θ gives every element of R a "code", such that, given the codes of X and Y , we can compute the code of X + Y and of XY . If we have two different recursive presentations θ 1 and θ 2 , then an element X of R has two "codes" θ 1 (X) and θ 2 (X). A ring is recursively stable if and only if θ 2 (X) can be effectively computed from θ 1 (X).
Definition. Let R be a recursively stable ring with a recursive presentation θ : R → N. Then a subset S ⊆ R k is called recursively enumerable if and only if θ ⊗k (S) is an r.e. subset of N k .
Intuitively, we can still think of r.e. subsets of R k as sets which can be printed by an algorithm (possibly running infinitely long). The requirement that R is recursively stable implies that the definition of r.e. subsets of R k does not depend on the choice of θ. One can prove (see [7]) that every field which is finitely generated over its prime field is recursively stable. Furthermore, a recursive integral domain with a recursively stable fraction field is automatically recursively stable. Since we assumed that R was recursive we have that R is recursively stable, hence R[Z] is recursively stable. To construct an example of a ring which is not recursive, consider any non-r.e. subset S of N. Now take the localization of Z where the n-th prime number is inverted if and only if n ∈ S. This is a non-recursive subring of Q.
Definition. Let R be an integral domain and S a subset of R k . Then S is called diophantine if there exists a polynomial p(a 1 , . . . , a k , x 1 , . . . , x n ) with coefficients in R such that S = {(a 1 , . . . , a k ) ∈ R k | p(a 1 , . . . , a k , x 1 , . . . , x n ) = 0 for some x 1 , . . . , x n ∈ R}.
(1) The polynomial p is called a diophantine definition of S. A function f : When dealing with decidability questions (analogues of Hilbert's Tenth Problem) it often makes sense to restrict the coefficients of the polynomial p to a subring of R. This is certainly necessary if R is uncountable. However, if we want to prove that r.e. sets are diophantine, then every singleton in R needs to be diophantine. Therefore, we might as well assume that we take all of R as ring of coefficients.
If R is a recursively stable ring, then every diophantine set is recursively enumerable. To see this, consider a diophantine set S defined as in (1). Construct an algorithm which tries all possible values (a 1 , . . . , a k , x 1 , . . . , x n ) ∈ R k+n and evaluates p(a 1 , . . . , a k , x 1 , . . . , x n ). Whenever is zero is found, it prints (a 1 , . . . , a k ). This algorithm will print exactly the set S.

Overview
Let K be a number field and let R be a subring of K with fraction field K. In order to prove that r.e. sets are diophantine for R[Z], the main result is the following from section 3: Theorem. Let R be a noetherian integral domain of characteristic zero such that the degree function To prove this, we first show that the set of polynomials in R[Z] which divide some Z u − 1 is diophantine. This is done using a Pell equation, similarly to the definition of powers of Z in [2], Section 4. A polynomial F dividing Z u − 1, normalised such that F (0) = 1, has coefficients in Z if and only if F (h) ∈ Z for a sufficiently large number h (depending only on the degree of F ). In this way, we diophantinely define the polynomials in Z[Z] dividing some Z u − 1. We call these the root-of-unity polynomials. This set is Z- At several points in the proof above we need a diophantine definition of the degree function deg : R[Z] \ {0} → Z. We give such a diophantine definition in section 4. We apply a result by Kim and Roush who showed in [8] that diophantine equations over L(Z) are undecidable if L is contained in a finite extension of Q p for some p ≥ 3. They showed undecidability by giving a diophantine definition of the discrete valuation ring L[Z] (Z) . Since "negative degree" is a discrete valuation, the same method gives a diophantine definition of "degree" in R[Z].

Special polynomials
In this section, we state some properties of the Chebyshev polynomials X n and Y n and cyclotomic polynomials Φ n . We also define root-of-unity polynomials. Everything in this section concerns only the ring Z[Z].

Chebyshev polynomials
Definition 1. Let n ∈ Z and define polynomials X n , Y n ∈ Z[Z] using the following equality: , this definition makes sense for negative n.
The degree of X n is |n|; the degree of Y n is |n| − 1 for n = 0, while Y 0 = 0.
In the literature, X n is called the n-th Chebyshev polynomial of the first kind and Y n+1 is called the n-th Chebyshev polynomial of the second kind (such that the n-th Chebyshev polynomials have degree n for n ≥ 0).
The couples (X n , Y n ) satisfy the Pell equation X 2 − (Z 2 − 1)Y 2 = 1. Conversely, we have: Proof. See [5], Lemma 2.1. Since X −n = X n and Y −n = −Y n , we do not need to put ± in front of Y n (T ).
The Chebyshev polynomials also satisfy the following identity: Proposition 3. In Q(Z), the following equality holds for all n ∈ Z: Proof. This easily follows from (2).

Cyclotomic and root-of-unity polynomials
Let Φ n ∈ Z[Z] denote the n-th cyclotomic polynomial.
Proof. If µ denotes the Möbius function, then we have Since n ≥ 2, we have a|n µ(a) = 0 and we can multiply by 1 = a|n (−1) µ(a) to get: Now we evaluate this product modulo Z 2d .
If n/a ≥ 2d then (1 − Z n/a ) µ(a) is congruent to 1 (mod Z 2d ). The same happens if a is not squarefree since in this case µ(a) = 0. The only squarefree a dividing n such that n/a < 2d equals a = n/d. So we have If k is even, then µ(n/d) = 1 and we have the desired result. If k is odd, Corollary 5. Let d ∈ N and s ∈ {−1, 1}. Then there exist infinitely many If r is any squarefree number coprime to m, then it follows from Proposition 4 that Φ rm (Z) is congruent to 1 ± Z d (mod Z 2d ), where the sign of Z d is determined by the parity of the number of factors in rm. Now the statement clearly follows. Definition 6. We call a polynomial F ∈ Z[Z] a root-of-unity polynomial if it satisfies one of the following three equivalent conditions: 3. F (0) = ±1, F is squarefree and all the zeros of F are roots of unity.
Let C denote the set of all root-of-unity polynomials, and let C + denote those with constant term equal to 1.
If we are working in the Z-adic topology, then "F ≡ M (mod Z d )" means that M is an approximation of F with a precision of Z d . Since the units of Z[[Z]] are exactly the power series F with F (0) = ±1, the proposition can be rephrased as follows: the set of root-of-unity polynomials is Proof. Since the set C is invariant under changing sign, we may assume without loss of generality that F (0) = 1.
The proof will be done by induction on d, which means that we will construct better and better approximations of F . For d = 1, we can take M = 1.
If c happens to be zero, then we can take M = M 0 .
First consider the case c > 0. By Corollary 5, we can find an We can iterate this procedure. Set The case c < 0 is analogous, the only difference is that we need to multiply with polynomials which are congruent to 1 − Z d (mod Z d+1 ).

Defining polynomials with integer coefficients
Throughout this section, R is a noetherian integral domain of characteristic zero such that the degree function

Divisors of
We give a diophantine definition of the divisors of Z u − 1, without requiring that they have coefficients in Z. For technical reasons, we first restrict ourselves to polynomials of degree at least 3.
Since X n (1) = 1, the condition X ≡ 1 (mod Z + S − 2) forces the sign of X to be positive. The formula Y = 0 is equivalent to n = 0.
In the last part of formula (6), we are working modulo G = 1 − ZS. But this means that S ≡ Z −1 (mod G). So, that formula becomes equivalent to Using Proposition 3, this is equivalent to Z n ≡ 1 (mod G). Without loss of generality, we may assume that n ≥ 0 (otherwise multiply both sides by Z −n ). Then we can rewrite Z n ≡ 1 (mod G) as G | Z n − 1.
Proposition 9. In R[Z], the set of all polynomials dividing Z u − 1 for some u > 0 is diophantine.
Proof. Let F be an element of R[Z]. We claim that F divides some Z u − 1 if and only if If formula (7) is satisfied, 1, F ). Then G will divide Z 3u − 1. Since F divides Z u − 1, its constant coefficient must be a unit, therefore G can be chosen to have G(0) = 1.

Root-of-unity polynomials
Now we have a diophantine definition of the divisors of Z u − 1, but we only want those divisors with integer coefficients. We take care of this using the following proposition, which was inspired by [4] and [16].
Proposition 10. Let K be a number field and O its ring of integers. Let F ∈ O[Z] be a polynomial satisfying F (0) ∈ {−1, 1} whose zeros (over an algebraic closure) are all roots of unity. If F (2 deg F + 1) is an integer, then every coefficient of F is an integer.
Proof. By changing sign if necessary, we may assume without loss of generality that F (0) = 1. Let d be the degree of F and write where α i ∈ O. Note that α d = 0 and α 0 = 1.
. Now assume that d ≥ 1. Over an algebraic closure, F can be factored as where every ζ i is a root of unity. We see that F (0) = α d (−1) d d i=1 ζ i . This must be equal to 1, therefore α d is also a root of unity. Write σ d,i for the i-th elementary symmetric polynomial in d variables. Since σ d,i has d i terms, it follows that α i = α d · σ d,i (ζ 1 , . . . , ζ d ) is the sum of d i roots of unity.
Let | · | be an archimedean absolute value on K (i.e. an absolute value coming from an embedding K → C). 1. The degree of G is at most d.

G(2 d + 1) is an integer.
3. |γ i | ≤ 2 d−1 for every coefficient γ i of G and every archimedean absolute value on K.
Clearly, the elements of Z[Z] having degree at most d and coefficients in the There are (2 d + 1) d+1 such polynomials. We claim that these are the only elements of G d . Since F is in G d , this claim implies the proposition.
To prove the claim, take any G in G d and write G = d i=0 γ i Z i (where we allow γ d = 0). We have the following bound for all h ∈ Z with h > 1: The coefficients of G and H have absolute value at most 2 d−1 , therefore |δ i | ≤ 2 d . Since δ e ∈ O is integral over Z, we have |δ e | p ≤ 1 for every nonarchimedean (p-adic) absolute value on K. From the product formula for absolute values it follows that |δ e | ≥ 1 for some archimedean absolute value on K. If we take such an absolute value, then (10) implies the following contradiction: Consider again the set G d . We just showed that G(h) cannot take the same value for two different elements G in G d . Since G(h) ∈ Z by definition of G d and |G(h)| ≤ (h d+1 − 1)/2, it follows that G d has at most h d+1 elements. But we already know that there are h d+1 elements in Taking Propositions 9 and 10 together, we can now prove: Proof. The R[Z]-divisors of Z u − 1 are diophantine by Proposition 9. If we take only those polynomials with F (0) = ±1, they satisfy the conditions of Proposition 10 with K = Q(ζ u ) where ζ u is a primitive u-th root of unity.

All polynomials with integer coefficients
Proposition 11 gives us a diophantine definition of C, which is a subset of Z[Z].
To define all of Z[Z] in R[Z], we use Proposition 7. By taking remainders of the elements of C after Euclidean division by Z d , we get all elements of Z[Z] with constant coefficient 1 or −1. We don't actually need that the set of powers of Z is diophantine, we can divide by elements of C + 1, which contains the powers of Z. In order for Euclidean division to be diophantine, we need "degree" to be diophantine. To get all elements of Z[Z], we just need to add an integer to the polynomials we get as remainders.
Assume that X is indeed in Z Z − 1 | D prevents D from being constant (note that 0 is not an element of C). Since all elements of C are monic up to sign, D + 1 is also. Formula (13) says that R is the remainder of the Euclidean division of M by D + 1, therefore R ∈ Z[Z]. Since C ∈ Z, it also follows that X ∈ Z[Z].

Diophantine definition of degree
We start with a lemma which shows that defining the degree function R[Z] \ {0} → Z is equivalent to defining a certain "weak" degree equality relation.
Since δ is diophantine and Y (1) = d is equivalent to Z − 1 | Y − d, this formula is clearly diophantine.
Assume that (15) is satisfied. Since Y n (1) = n for any n ∈ Z, the subformula " by Proposition 2. In particular, X is an element of Z[Z] of degree d. By the assumptions on δ, this implies that deg(F ) = d.
Conversely, if the degree of F equals d, then we set X = X d (Z) and Y = Y d (Z). This satisfies (15).
As in the Introducion, let K be a number field and R a subring of K with fraction field K.
To diophantinely define degree in R[Z], we use the fact that "negative degree" is a discrete valuation on K(Z). More precisely, if F, G ∈ R[Z], then v Z −1 (F/G) := deg(G) − deg(F ) defines a discrete valuation on K(Z). Therefore, the problem reduces to showing that the discrete valuation ring at Z −1 in K(Z) is diophantine. For this, we need certain quadratic forms used by Kim and Roush (see [8]) to prove undecidability for rational function fields over so-called p-adic fields with p odd. This undecidability has been generalised to arbitrary function fields over p-adic fields with p odd (see [10] or [6]).

Definition 14.
Let p be a prime number. A field K is called p-adic if K can be embedded in a finite extension of Q p .
It is clear from this definition that every number field is p-adic for every p.
For the rest of this section, we fix any odd prime p. Following the method by Kim and Roush, we need to work over a field satisfying Hypothesis (H).
Definition 15. Let L be a p-adic field with p odd and let v p be a discrete valuation on L extending the p-adic valuation on Q. We say that L satisfies Hypothesis (H) if and only if L contains elements α and π such that 1. v p (π) is odd and π is algebraic over Q.
2. α is a root of unity.
3. L contains a square root of −1.

Proposition 16 ([8], Proposition 8).
Let K be a p-adic field for an odd prime p. Then there exists a finite extension L of K which satisfies Hypothesis (H).
The next two propositions deal with certain quadratic forms. Our variable Z is the inverse of the variable t that Kim and Roush use.
Proposition 17 ([8], Proposition 7). Let L be any field of characteristic 0 and suppose that 1, −α 1, π is an anisotropic quadratic form over L. Let F ∈ L(Z) such that v Z −1 (F ) is non-negative and even. Then one of the following two is anisotropic over L(Z): Z, −αZ, −1, −F 1, π (16) Z, −αZ, −1, −αF 1, π . (17) The following proposition follows from [8]. However, here we use a reformulation by Eisenträger (see [6], Theorem 8.1). Note that the condition that G has algebraic coefficients is missing from Eisenträger's paper, but it is necessary and it does appear in Kim and Roush.
Proposition 18. Let L be a p-adic field satisfying Hypothesis (H) for elements α and π in L. Let U ⊆ L(Z) such that U ∩Q is dense in Q p 1 ×· · ·×Q pm for every finite set of rational primes {p 1 , . . . , p m }. Let G ∈ L(Z) such that v Z (G) = −2 and v Z −1 (G) = 1. Assume that G = G N (Z)/G D (Z) for polynomials G N and G D with coefficients algebraic over Q. Then there exist γ 3 , γ 5 ∈ U such that, if we let then the following quadratic forms are both isotropic over L(Z): The most natural choice for U would be U = L. However, for our applications, U needs to be diophantine in L(Z). In the article by Kim and Roush, U is a subset of L. However, since enlarging the set U only weakens the proposition, we can even take U in L(Z).
Taking these last two propositions together, we can prove the following: Proposition 19. Let L and U be as in Proposition 18 with the additional condition that every element A ∈ U has v Z −1 (A) ≥ 0. Let X ∈ L(Z) with algebraic coefficients and define G(Z) := (Z + Z 2 ) + X 3 Z 3 + Z 2 X 3 .
Then v Z −1 (X) ≥ 0 if and only if there exist γ 3 , γ 5 ∈ U such that the quadratic forms  Let α ∈ R such that K = Q(α) and let d := [K : Q]. Now any element X of R[Z] can be written as with X i in Z[Z] and y in Z \ {0}.

Now let S ⊆ R[Z]
be an r.e. set, we have to show that S is diophantine. To S we associate a set T ⊆ Z[Z] d+1 using (21): the set T has one tuple (X 0 , X 1 , . . . , X d−1 , y) ∈ Z[Z] d+1 for every X ∈ S. This tuple (X 0 , X 1 , . . . , X d−1 , y) is not unique but that is not a problem, we can for example try all possible tuples and take the first one which works for a given X. This way, we have a bijection between S and T . Moreover, the set T will also be r.e., since we can construct T from S using a recursive procedure. Since T is a subset of Z[Z] d+1 , it will be diophantine over R [Z]. Now it immediately follows that S is diophantine: X ∈ S ⇐⇒ ∃(X 0 , X 1 , . . . , X d−1 , y) ∈ T Xy = X 0 +X 1 α+· · ·+X d−1 α d−1 .
The argument for sets S ⊆ R[Z] k is very similar, using a set T ⊆ Z[Z] (d+1)k .