Welcome to MathOverflow

Question

7

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.

flag	asked Apr 14 2010 at 16:40 SJR 3,082●1●6●16

2

Someone care to explain the vote to close? This seems like a perfectly reasonable question to me. – Qiaochu Yuan Apr 14 2010 at 17:37

Some random comments. (a) this is "not a proper question", like I always say: B=Z precisely when it's Z, for example. But on the other hand it feels like it has potential. As you say as a comment in Wlog's answer, if S={y=pi.x} then B=Z. But it's not true that if S is not defined over Q-bar then B=Z: for example if S={y=pi} then f(x,pi) had better be constant so B=Z[y]. Is B always finitely-generated? That's a question. Do you know a counterexample? Aah---if S={y=0} then f(x,0) had better be constant so B might well be Z[y,xy,x^2y,...]---is that right? And that's not f.g.. – Kevin Buzzard Apr 14 2010 at 20:08

In fact here's a better example: if S={x^2+y^2=pi} then certainly S isn't defined over Q-bar but B is everything. – Kevin Buzzard Apr 14 2010 at 20:40

@Kevin: Why is question not proper? Because I'm speaking as if the embedding of $\mathbb{Z}$ in $\mathbb{Z}[x]$ is an inclusion? Isn't this a common and convenient and utterly harmless manner of speaking? Please consider editing out your "not proper" comment, unless you have other objections. I would hate to loose some potential comment or answer. – SJR Apr 14 2010 at 20:56

There are some things that seem odd in the formulation of the question. If you are want the polynomials in B to have integer coefficients, surely you also want to the polynomial defining S to have integer coefficients? Also, if g and h have the same restriction to S, are they the same or different as elements of B? I assumed they were the same; if you want them to be different then your ring will almost never be finitely generated because of examples like S={x=0}. – David Speyer Apr 14 2010 at 21:31

show 6 more comments

damiano · Answer 1 · 2010-04-15 10:42:13Z UTC

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 $B \neq \mathbb{Z}$.

Wlog · Answer 2 · 2010-04-14 19:25:20Z UTC

-1

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?

answered Apr 14 2010 at 19:25

Wlog
67●4

	@Wlog: the question is clear. B is "the ring of polynomials with integer coefficients that happen to have property X" which I think is perfectly clear. If no non-constant polynomials have property X then B is the integers. Being "bounded on S" means that there is a constant C such that \|p(s)\|<=C for all s in S. – Kevin Buzzard Apr 14 2010 at 19:42
	@Wlog: A polynomial $p$ is "bounded" on a subset $S$ of $\mathbb{R}^n$ if for some real $r$, the inequality $\|p(x)\| < r$ 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 $B=\mathbb{Z}$. – SJR Apr 14 2010 at 19:46
	@Kevin: We posted simultaneously. – SJR Apr 14 2010 at 19:55

How to see the ring of all polynomials (with integer coefficients) that are bounded on a given real algebraic set?

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

2 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.