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Hilbert's 10Th Problem for Solutions in a Subring of Q

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Hilbert's 10Th Problem for Solutions in a Subring of Q Hilbert’s 10th Problem for solutions in a subring of Q Agnieszka Peszek, Apoloniusz Tyszka To cite this version: Agnieszka Peszek, Apoloniusz Tyszka. Hilbert’s 10th Problem for solutions in a subring of Q. Jubilee Congress for the 100th anniversary of the Polish Mathematical Society, Sep 2019, Kraków, Poland. hal-01591775v3 HAL Id: hal-01591775 https://hal.archives-ouvertes.fr/hal-01591775v3 Submitted on 11 Sep 2019 HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche français ou étrangers, des laboratoires abroad, or from public or private research centers. publics ou privés. Hilbert’s 10th Problem for solutions in a subring of Q Agnieszka Peszek, Apoloniusz Tyszka Abstract Yuri Matiyasevich’s theorem states that the set of all Diophantine equations which have a solution in non-negative integers is not recursive. Craig Smorynski’s´ theorem states that the set of all Diophantine equations which have at most finitely many solutions in non-negative integers is not recursively enumerable. Let R be a subring of Q with or without 1. By H10(R), we denote the problem of whether there exists an algorithm which for any given Diophantine equation with integer coefficients, can decide whether or not the equation has a solution in R. We prove that a positive solution to H10(R) implies that the set of all Diophantine equations with a finite number of solutions in R is recursively enumerable. We show the converse implication for every infinite set R Q ⊆ such that there exist computable functions τ1; τ2 : N Z which satisfy ( n N τ2(n) , τ (n) ! 8 2 0) 1 : n N = R . This implication for R = N guarantees that Smorynski’s´ ^ τ2(n) 2 theorem follows from Matiyasevich’s theorem. Harvey Friedman conjectures that the set of all polynomials of several variables with integer coefficients that have a rational solution is not recursive. Harvey Friedman conjectures that the set of all polynomials of several variables with integer coefficients that have only finitely many rational solutions is not recursively enumerable. These conjectures are equivalent by our results for R = Q. 2010 Mathematics Subject Classification: 03D25, 11U05. Key words and phrases: Craig Smorynski’s´ theorem, Diophantine equation which has at most finitely many solutions, Hilbert’s 10th Problem for solutions in a subring of Q, Martin Davis’ theorem, recursive set, recursively enumerable set, Yuri Matiyasevich’s theorem. 1 Introduction and basic lemmas Yuri Matiyasevich’s theorem states that the set of all Diophantine equations which have a solution in non-negative integers is not recursive, see [3]. Martin Davis’ theorem states that the set of all Diophantine equations which have at most finitely many solutions in positive integers is not recursive, see [1]. Craig Smorynski’s´ theorem states that the set of all Diophantine equations which have at most finitely many solutions in non-negative integers is not recursively enumerable, see [4, p. 104, Corollary 1] and [5, p. 240]. Let denote the set of prime numbers, and let P = p ; q ; r ; p ; q ; r ; p ; q ; r ;::: ; P f 1 1 1 2 2 2 3 3 3 g where p1 < q1 < r1 < p2 < q2 < r2 < p3 < q3 < r3 < : : : 1 1 αi βi γi Lemma 1. For a non-negative integer x, let pi qi ri be the prime decomposition of i=1 · · x + 1. For every positive integer n, the mapping which sends x N to Q 2 β β ( 1)α1 1 ;:::; ( 1)αn n Qn − · γ + 1 − · γ + 1 2 1 n ! is a computable surjection from N onto Qn. Let s : N Qn denote the surjection defined in Lemma 1. n ! Lemma 2. For every infinite set R Q, a Diophantine equation D(x ;:::; x ) = 0 has no ⊆ 1 n solutions in x ;:::; x R if and only if the equation D(x ;:::; x ) + 0 x = 0 has at most 1 n 2 1 n · n+1 finitely many solutions in x ;:::; x R. 1 n+1 2 Let R be a subring of Q with or without 1. By H10(R), we denote the problem of whether there exists an algorithm which for any given Diophantine equation with integer coefficients, can decide whether or not the equation has a solution in R. 2 A positive solution to H10(R) implies that the set of all Diophantine equations with a finite number of solutions in R is recursively enumerable In the next three lemmas we assume that 0 ( R Q and r Z R for every r R. Every f g ⊆ · ⊆ 2 non-zero subring R of Q (with or without 1) satisfies these conditions. Lemma 3. There exists a non-zero integer m R. 2 Proof. There exist m; n Z 0 such that m R. Hence, m = m n (Z 0 ) R. 2 n f g n 2 n · 2 n f g \ Lemma 4. Let m (Z 0 ) R. We claim that for every b R, b , 0 if and only if the equation 2 n f g \ 2 4 2 2 y b m yi = 0 · − − = Xi 1 is solvable in y; y ; y ; y ; y R. 1 2 3 4 2 Proof. If b = 0, then for every y; y ; y ; y ; y R, 1 2 3 4 2 y b m2 y2 y2 y2 y2 = m2 y2 y2 y2 y2 6 m2 < 0 · − − 1 − 2 − 3 − 4 − − 1 − 2 − 3 − 4 − p If b , 0, then b = , where p N 0 and q Z 0 . In this case, we define y as m2 q and q 2 n f g 2 n f g · observe that m2 q = (m q) m R as m q R and m Z. Hence, · · · 2 · 2 2 p y b = (m2 q) = m2 p m2 (N 0 ) · · · q · 2 · n f g By Lagrange’s four-square theorem, there exist t ; t ; t ; t N such that 1 2 3 4 2 y b m2 · − = t2 + t2 + t2 + t2 m2 1 2 3 4 Therefore, y b m2 (m t )2 (m t )2 (m t )2 (m t )2 = 0; · − − · 1 − · 2 − · 3 − · 4 where m t ; m t ; m t ; m t R. · 1 · 2 · 3 · 4 2 2 Lemma 5. We can uniquely express every rational number r as r = r, where r Z, 2 r N 0 , and the integers r and r are relatively prime. If r R, then r R. 2 n f g 2 2 d d Proof. For every r R, r = r r r Z R. 2 d· 2 · ⊆ d Lemma 6. Let R be a non-zero subring of Q with or without 1. We claim that for every d T ;:::; T Rn and for every x ;:::; x R, the following product 0 k 2 1 n 2 n x r r 2 (1) i · i − i (r ;:::; r ) T ;:::; T i=1 1 n Y2 f 0 kg X d differs from 0 if and only if (x ;:::; xn) < T ;:::; Tk . Product (1) belongs to R. 1 f 0 g Proof. The last claim follows from Lemma 5. Lemma 7. Let R be a non-zero subring of Q (with or without 1) such that there exists an algorithm which for every (a; b) Z (Z 0 ) decides whether or not a R. Let ρ : Qn Rn 2 × n f g b 2 n ! denote the function which equals the identity on Rn and equals (0;:::; 0) outside Rn. We claim that for every positive integer n the function ρ s : N Rn is surjective and computable. n ◦ n ! Theorem 1. Let R be a non-zero subring of Q (with or without 1) such that Hilbert’s 10th Problem for solutions in R has a positive solution. We claim that the set of all Diophantine equations with a finite number of solutions in R is recursively enumerable. Proof. By Lemma 3, there exists a non-zero integer m R. For every (a; b) Z (Z 0 ), the 2 2 × n f g solvability in R of the equation b x a = 0 is decidable. Hence, for every (a; b) Z (Z 0 ) · − 2 × n f g we can decide whether or not a R. By Lemmas 4 and 6, the answer to the question b 2 in Flowchart 1 is positive if and only if the equation D(x1;:::; xn) = 0 is solvable in Rn θ(0); : : : ; θ(k) . Hence, by Lemma 7, the algorithm in Flowchart 1 halts if and only if n f g the equation D(x1;:::; xn) = 0 has at most finitely many solutions in R. 3 Start Input a Diophantine equation D x1,..., xn = 0 θ := ρn sn ◦ k := 0 Yes k := k + 1 2 Does the equation D (x1,..., xn)+ 2 n 2 4 y x r [r m2 y2 = 0 · i · i − i − − i i = 1 i = 1 r ,...,Yrn X X 1 ∈ θ(0), . , θ(k) { } is solvable in x1,..., xn, y, y1, y2, y3, y4 R ? ∈ No Print "The equation D x1,..., xn = 0 has at most finitely many solutions in R" Stop Flowchart 1 Theorem 1 remains true when R = 0 . The flowchart algorithm depends on f g m (Z 0 ) R. For a constructive proof of Theorem 1, we must compute an element of 2 n f g \ (Z 0 ) R.
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