Let $S$ be the zero set in $\mathbb{R}^n$ of a polynomial with real coefficients. Let $B$ be the ring of polynomials, with INTEGER coefficients, that are bounded on $S$.
I would like to know how to get basic information about $B$: When is $B=\mathbb{Z}$? When is $B$ finitely generated? How do I find some comprehensible set of ring generators of $B$? Any suggestions or references would be appreciated.
Let $\bar{S}$ by the closure of S in $\mathbb{P}^n(\mathbb{R})$. If a polynomial with zero constant term is bounded on S, then its highest degree term vanishes on $S':=\bar{S} \setminus S \subset \mathbb{P}^{n-1}$. In particular, if S' is Zariski dense in $\mathbb{P}^{n-1}$ over Z, then B=Z (classically, sets with similar properties were called "generic"). On the other hand, S' could be defined over $\overline{\mathbb{Q}}$ even when S is not (e.g. $y^2=\pi x$).
Since obviously $B \neq \mathbb{Z}$ in the case in which $B \neq \mathbb{R}^n$ and B is defined over $\overline{\mathbb{Q}}$, it would be interesting to find an irreducible S not defined over $\overline{\mathbb{Q}}$ which is unbounded (i.e. $S' \neq \emptyset$) and for which $B \neq \mathbb{Z}$.
$|p(x)| < r$holds for every $x$ in $S$. How could $B=\mathbb{Z}$? For example, consider the line $y=\pi x$ in $\mathbb{R}^2$. If a non-constant polynomial $f(x,y)$ with integer coefficients was bounded on that line, then $f(x,\pi x)$ would reduce to a constant, which would imply that $pi$ is rational. So in this case $B=\mathbb{Z}$. $\endgroup$