12
$\begingroup$

Suppose $f(x_1,x_2,\dots)=\frac{P}{Q}$, where $P,Q$ are polynomials in several variables with integer coefficients that have the same degree. Let's denote by $S(f)$ the set of integers $n$ for which $f(x_1,x_2,\dots)=n$ is solvable in integers.

Which sets $S\subset \mathbb Z$ can be written as $S(f)$ for some $f$ as above?

For example we have, $S(\frac{x_1^2+x_2^2}{x_1x_2+1})=\lbrace -5,0,1,4,\dots,k^2,\dots\rbrace$.

This question is just a musing from playing around with variations to Hilbert's tenth problem. A more direct question would be: Is every Diophantine set representable as some $S(f)$?

$\endgroup$
1
  • 2
    $\begingroup$ The restriction to degree zero doesn't actually matter. If $P$ and $Q$ have different degrees you can always introduce a new variable $z$ and replace $P$ with $P + 4PQz$ and $Q$ with $Q+4PQz$. The new rational function will be degree $0$ and can only be an integer when $z=0$, when $P=0$, or when $P=Q\ne 0$. $\endgroup$
    – zeb
    Jan 29, 2013 at 4:14

1 Answer 1

8
$\begingroup$

A set $T \subseteq \mathbb{Z}$ can be written as $S(f)$ if and only if $T$ is effectively enumerable.

Proof: As in zeb's comment, the restriction to degree zero doesn't matter, and we consider rational functions of arbitrary degree.

Suppose $T$ is effectively enumerable. By the MRDP theorem choose a polynomial $f(\bar{z},x)$ such that $T$ is precisely all $x$ for which the equation $f(\bar{z},x)=0$ is solvable in integers $\bar{z}$. Then the rational function \begin{equation*} r(\bar{z},x):=x+\dfrac{f(\bar{z},x)^2}{1+f(\bar{z},x)^2} \end{equation*} has the value $x$ if $f(\bar{z},x)=0$, and otherwise is not an integer. Therefore $T=S(r)$.

Conversely, it is intuitively clear that every set of the form $S(r)$, with $r$ a rational function, is effectively enumerable.

$\endgroup$
1
  • $\begingroup$ Ah, nice! The degree zero condition was an attempt at avoiding easy ways of encoding the MRDP theorem. Not a successful one, it seems! :-) $\endgroup$ Jan 29, 2013 at 8:21

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.