Hilbert's 10th problem

← Mathematics

Abstract. These are notes from a talk I gave on Hilbert's 10th problem, which asks if there exists an algorithm which determines whether a polynomial with integer coefficients has a solution in the integers. We follow the proof given in B. Poonen's expository article, [6], on how Turing machines, and specifically the Halting problem, are used to show Hilbert's 10th conjecture is false. We then mention generalizations and applications to Mazur's conjecture.

Turing machines
Hilbert's 10th problem
Hilbert's 10th problem for arbitrary rings
Mazur's conjectures
References

1. Turing machines

A Turing machine, introduced by A. Turing, [7], is a way to mathematically formalize the idea of an algorithm. We will not give the explicit definition here, as we will only need an informal notion for our purposes. A Turing machine can be thought of as a finite program having infinite memory, i.e., it is free of any physical memory constraints that a real computer has. It executes infinitely fast, meaning we are more concerned with whether a Turing machine halts, or stops after a finite number of steps, with respect to a given input rather than its run time.

Visually, a Turing machine $ T $ can be viewed as a string of code, which is equivalent to a finite piece of tape:

\[1, 0, 0, 1, 1, 1\]

Its memory $ M $ is likewise a piece of tape, but this time infinite, although at any point during the computation, it can only contain finitely many 1's:

\[\dots, 0, 0, 0, 1, 1, 0, 1\]

Thus, $ M $ is initialized to some infinite sequence containing finitely many 1's, and then $ T $ acts on $ M $ until it halts; it is also possible that the machine does not halt.

We can identify Turing machines with integers via any bijection between the natural numbers $ \mathbb{N} $ and the integers $ \mathbb{Z} $. Thus, we will write $ T \in \mathbb{Z} $. Likewise its initial input, since it only contains finitely many 1's, can also be identified with an integer $ a \in \mathbb{Z} $. We will denote inputting $ a $ to $ T $ by $ T(a) $. If our Turing machine halts or its memory converges to some infinite sequence, then we will also use $ T(a) $ to denote this value.

Let $ P \subseteq \mathbb{Z} $ be a subset of the possible inputs. We say that a Turing machine $ T $ solves the decision problem $ P $ if, for each $ a \in \mathbb{Z} $, $ T(a) $ halts with output 1 if $ a \in P $ and 0 if $ a \not \in P $. There are uncountably many possible decision problems but only countably many Turing machines, so not every decision problem has a solution; we call such decision problems unsolvable.

To shorten our terminology, we say a decision problem $ P $ is recursive if there exists a Turing machine solving it. We say $ P $ is listable if there exists a Turing machine $ T $ which outputs $ P $ if left running forever, i.e., $ T(0) = P $.

Proposition 1.1. Recursive implies listable.

Proof. Let $ T $ solve $ P $, and let $ Z $ be a Turing machine that prints the integers $ 0, -1, 1, -2, 2, \dots $. Then taking each output of $ Z $ and inputting it to $ T $ lists $ P $. Q.E.D.

The halting problem asks whether there exists a Turing machine $ H $ whose input is a pair $ (T, a) \subseteq \mathbb{Z} \times \mathbb{Z} $ (which we can, of course, identify with an integer), and whose output is whether $ T(a) $ halts. As it turns out, Turing showed that the halting problem is undecidable.

Theorem 1.2 ([7]). The decision problem is undecidable.

Proof. Suppose $ H $ exists. Then there would exist a Turing machine $ G $ such that $ G(a) $ halts if and only $ a(a) $ does not halt. But by setting $ a $ to be $ G $, we see that $ G(G) $ halts if and only if $ G(G) $ does not halt, a contradiction. Q.E.D.

Corollary 1.3. There exists a listable set $ L \subseteq \mathbb{Z} $ which is not recursive.

Proof. Set

\[L = \{2^T 3^a \mid T, a \in \mathbb{Z} \text{ and } T(a) \text{ halts} \}.\]

Then $ L $ is not recursive by theorem 1.2. It is, however, listable: for each $ N \in \mathbb{N} $ and $ T, a \in \mathbb{Z} $ with $ \vert T \vert, \vert a \vert \leq N $, print $ 2^T 3^a $ if $T(a) $ it halts within $ N $ steps. Q.E.D.

2. Hilbert's 10th problem

Hilbert's 10th problem asks if there exists an algorithm that determines whether a polynomial with integer coefficients has a solution in the integers. We can rewrite this using Turing machines as follows. Let $ P $ be the decision problem

\[P = \{ F \in \mathbb{Z}[X_1, \dots, X_n] \mid \exists a \in \mathbb{Z}^n \text{ s.t. } F(a) = 0 \};\]

we are interested in whether $ P $ is recursive.

We say a set $ S \subseteq \mathbb{Z}^n $ is diophantine if there exists a polynomial

\[F \in \mathbb{Z}[X_1, \dots, X_n, Y_1, \dots, Y_m],\]

where $ S $ is the set of points $ a \in \mathbb{Z}^n $ such that there exists a $ b \in \mathbb{Z}^m $ with $ F(a, b) = 0 $. If we let $ \mathcal{V}(F) \subseteq \mathbb{Z}^{n+m} $ denote the zero locus of $ F $, this is the same as saying

\[S = \{ (a_1, \dots, a_n) \mid a \in \mathcal{V}(F) \}.\]

The following result about diophantine sets was shown by Y. Matijasevǐc, [4], using the work of M. Davis, H. Putnam, and J. Robinson, [2].

Theorem 2.1. Let $ S \subseteq \mathbb{Z}^n $. Then $ S $ is listable (after identifying $ \mathbb{Z}^n $ with $ \mathbb{Z} $) if and only if it is diophantine.

Proof. In [2], it is shown that every listable set is exponential diophantine, i.e., the zero set of a function created from integers and variables using addition, multiplication, and exponentiation. It is then shown in [4] that the exponential relation $ y = C^x $ is, in fact, diophantine; this is done using sequences of integer solutions of the pell equation

\[X_1^2 - n X_2^2 = 0,\]

where $ n $ is not square. We will not go into too much detail here, and refer the reader to article [3] for a short-ish, complete proof. Q.E.D.

Corollary 2.2. Hilbert's 10th problem is undecidable.

Proof. Recall the set $ L $ given in corollary 1.3 is listable but not recursive. Hence, it is diophantine, and there exists a polynomial $ F \in \mathbb{Z}[X_1, Y_1, \dots, Y_m] $ such that

\[L = \{ a \in \mathbb{Z} \mid \exists b \in \mathbb{Z}^m \text{ s.t. } F(a, b) = 0 \}.\]

If Hilbert's 10th problem were true, then there would exist a Turing machine $ T $ such that, for each $ a \in \mathbb{Z} $, $ T(a) $ determines if $ F(a, X) $ has a solution $ X $. In particular, $ L $ would be recursive, a contradiction. Q.E.D.

3. Hilbert's 10th problem for arbitrary rings

A natural question is whether Hilbert's 10th problem is decidable for rings other than $ \mathbb{Z} $. In [6], the following chart is given:

Ring	Hilbert's 10th problem
$ \mathbb{Z} $	False
$ \mathbb{C} $	True
$ \mathbb{R} $	True
$ \mathbb{F}_q $	True
$ p $-adic fields	True
$ \mathbb{F}_q((t)) $	Unknown
Number fields	Unknown
$ \mathbb{Q} $	Unknown
Global funnction field	False
$ \mathbb{F}_q(t) $	False
$ \mathbb{C}(t) $	Unknown
$ \mathbb{C}(t, u) $	False
$ \mathbb{R}(t) $	False
$ \mathcal{O}_K $	Unknown

Here, $ K $ is a number field and $ \mathcal{O}_K $ its ring of integers. We do know the decidability of Hilbert's 10th problem for certain $ \mathcal{O}_K $ but not all. Indeed, theorem 14.1 in loc. cit. tells us it is false when…

$ K $ is totally real.
$ K $ is a quadratic extension of a totally real number field or of an imaginary quadratic.
$ K $ has exactly one conjugate pair of nonreal embeddings.
There exists an elliptic curve $ E $ over $ \mathbb{Q} $ such that the abelian groups $ E(\mathbb{Q}) $ and $ E(K) $ both have rank 1.
For every Galois prime-degree extension $ L/K $, there exists an elliptic curve $ E $ over $ K $ such that $ E(K) $ and $ E(L) $ have the same positive rank.

See Poonen's paper for the remaining details.

I do not believe there have not been any major changes to the table since its original writing. The recent survey [1] mentions progress towards the rational case, and they explain developing local-global methods.

4. Mazur's conjectures

In [5], for $ V $ a variety over $ \mathbb{Q} $, B. Mazur made a series of conjectures about the topological closure of the $ \mathbb{Q} $-valued points, $ V(\mathbb{Q}) $, in the $ \mathbb{R}$-valued points, $ V(\mathbb{R}) $. Let us $ \overline{V(\mathbb{Q})} $ the topological closure of $ V(\mathbb{Q}) $ in $ V(\mathbb{R}) $. Mazur's first conjecture was that, assuming $ V $ is smooth and $ V(\mathbb{Q}) $ Zariski dense in $ V $, $ \overline{V(\mathbb{Q})} $ is a finite union of connected components of $ V(\mathbb{R}) $; this is equivalent to the topological closure being open, and it was shown to be false. A subsequent conjecture of his is the following:

Mazur's conjecture. $\overline{V(\mathbb{Q})} $ has at most a finite number of components.

Let us replace "variety" with "algebraic subset" in his conjecture. Write $ X $ for an algebraic subset of $ \mathbb{Q} $. We say $ S \subseteq X(\mathbb{Q}) $ is diophantine over $ \mathbb{Q} $ if there exists regular morphism $ f \colon Y \to X $ such that $ S = f(Y(\mathbb{Q})) $.

Theorem 4.1. If Mazur's conjecture is true, then the closure of $ S $ in $ X(\mathbb{Q}) $ has at most finitely many components.

Proof. Indeed, if Mazur's conjecture were true, then for some $ n \in \mathbb{N} $, $ \overline{Y(\mathbb{Q})} $ would have $ n $ connected components. Since $ f $ is continuous, $ f(\overline{Y(\mathbb{Q})}) $ has at most $ n $ components. Because the closure of a finite union is the union of the closures, $ \overline{f(\overline{Y(\mathbb{Q})})} $ has at most $ n $ connected components, meaning

\[\overline{S} = \overline{f(Y(\mathbb{Q}))} = \overline{f(\overline{Y(\mathbb{Q})})}\]

has at most $ n $ connected components. Q.E.D.

Corollar 4.2. If $ \mathbb{Z} $ is diophantine over $ \mathbb{Q} $, then Mazur's conjecture is false.

5. References