Hilbert's Tenth Problem over Q

Hilbert's 10th problem was to find a procedure which, given any polynomial $p \in \mathbb{Z}[x_1,\ldots,x_n]$, would decide if $p$ has an integral root, i.e. a tuple $(a_1, \ldots, a_n) \in \mathbb{Z}$ such that $p(a_1, \ldots, a_n) = 0$. However, the MRDP (Matiyasevich, Robinson, Davis, and Putnam) Theorem states that there is no such computable procedure.

Given any countable ring $R$ with a computable presentation, one can ask a version of Hilbert's 10th problem over $R$. Namely, is the following problem decidable: given a polynomial $p \in R[x_1, \ldots, x_n]$, determine whether $p$ has a root in $R$? The answer to this question is known for several rings and open for others:

• For $R = \overline{\mathbb{Q}}$, the algebraic closure of the rationals, the answer is yes, i.e. the problem is decidable. This follows from quantifier elimination for algebraically closed fields.
• As stated above, for $R = \mathbb{Z}$ the answer is no. This was proved by Matiyasevich [Mat70], building on work by Davis, Putnam and Robinson [DPR61].
• Alpöge, Bhargava, Ho, and Shnidman [ABHS26] proved that if $R$ is the ring of integers of any number field then the answer is no.

OpenAI has announced a proof that Hilbert's 10th problem over $\mathbb{Q}$ is undecidable. See here.

This question is known to be equivalent to the question of whether there is a decision procedure for the problem of determining whether a variety over $\mathbb{Q}$ has a rational point.

To give a negative answer to the question, it is enough to show that $\mathbb{Z}$ is diophantine over $\mathbb{Q}$, i.e. that there is a polynomial $p(x, y_1, \ldots, y_n)$ over $\mathbb{Q}$ such that $a \in \mathbb{Q}$ is in $\mathbb{Z}$ if and only if $p(a, y_1, \ldots, y_n)$ has a rational root.