0
$\begingroup$

I want to know if there exists a polynomial $ P(z, x_1,x_2,\ldots,x_n)$ over the rationals such that the set $$ Z_P = \{z | \exists x_1,\ldots,x_n. P(z, x_1,x_2,\ldots,x_n) = 0 \} \subsetneq \mathbb Q $$ is infinite but with no accumulation points.

Ideally I would like $Z_P = \mathbb N$ or $Z_P = \mathbb Z$, but a infinite discrete subset of the rationals is enough.

The context is to encode nonlinear integer arithmetic in nonlinear rational arithmetic, which is also the reason I need $\mathbb Q$ instead of an otherwise more reasonable $\mathbb R$.

Improve this question
$\endgroup$
3
  • $\begingroup$ With $z,x_j$ where (in $\mathbb Q,\mathbb R,\dots$)? $\endgroup$
    – fedja
    Jun 12, 2019 at 17:57
  • $\begingroup$ all of them should be in $\mathbb Q$ $\endgroup$
    – afiori
    Jun 12, 2019 at 21:04
  • 1
    $\begingroup$ Getting $Z_P = \mathbb{Z}$ is a well-known open problem (existential definition of $\mathbb{Z}$ in $\mathbb{Q}$) and possibly $Z_P = \mathbb{N}$ is no easier, but I am not an expert. See the paper of Koenigsmann in the Annals 2016 for the latest partial results on this. $\endgroup$
    – Felipe Voloch
    Jun 13, 2019 at 1:55

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.