How to see the ring of all polynomials (with integer coefficients) that are bounded on a given real algebraic set? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T12:24:04Z http://mathoverflow.net/feeds/question/21355 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/21355/how-to-see-the-ring-of-all-polynomials-with-integer-coefficients-that-are-boun How to see the ring of all polynomials (with integer coefficients) that are bounded on a given real algebraic set? SJR 2010-04-14T16:40:38Z 2010-04-15T10:42:13Z <p>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 $S$. </p> <p>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.</p> http://mathoverflow.net/questions/21355/how-to-see-the-ring-of-all-polynomials-with-integer-coefficients-that-are-boun/21377#21377 Answer by Wlog for How to see the ring of all polynomials (with integer coefficients) that are bounded on a given real algebraic set? Wlog 2010-04-14T19:25:20Z 2010-04-14T19:25:20Z <p>This is a comment. I might not be as comfortable with the notation, but if $B$ is a ring of polynomials, how can $B = \mathbb Z$? Is $B$ also in $\mathbb R^n$? Can you provide more details on why $B$ is always a ring, as opposed to just a set of polynomials? Finally, what do you mean for a polynomial to be bounded by a set?</p> http://mathoverflow.net/questions/21355/how-to-see-the-ring-of-all-polynomials-with-integer-coefficients-that-are-boun/21440#21440 Answer by damiano for How to see the ring of all polynomials (with integer coefficients) that are bounded on a given real algebraic set? damiano 2010-04-15T10:28:23Z 2010-04-15T10:42:13Z <p>Let $\bar{S}$ by the closure of <em>S</em> in $\mathbb{P}^n(\mathbb{R})$. If a polynomial with zero constant term is bounded on <em>S</em>, then its highest degree term vanishes on $S':=\bar{S} \setminus S \subset \mathbb{P}^{n-1}$. In particular, if <em>S'</em> is Zariski dense in $\mathbb{P}^{n-1}$ <strong>over Z</strong>, then <em>B=Z</em> (classically, sets with similar properties were called "generic"). On the other hand, <em>S'</em> could be defined over $\overline{\mathbb{Q}}$ even when <em>S</em> is not (e.g. $y^2=\pi x$).</p> <p>Since obviously $B \neq \mathbb{Z}$ in the case in which $B \neq \mathbb{R}^n$ and <em>B</em> is defined over $\overline{\mathbb{Q}}$, it would be interesting to find an irreducible <em>S</em> not defined over $\overline{\mathbb{Q}}$ which is unbounded (i.e. $S' \neq \emptyset$) and for which $B \neq \mathbb{Z}$.</p>