Timeline for How to see the ring of all polynomials (with integer coefficients) that are bounded on a given real algebraic set?
Current License: CC BY-SA 2.5
17 events
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| Apr 15, 2010 at 10:31 | comment | added | Sidney Raffer | @David, Kevin: Here's an example where S is not defined over Z: Fix algebraically independent reals r and s, and take S to be the subset of R^3 defined by (y-2^(1/2)x+r)^2+(z-8^(1/2)rx+s)^2=0. Then the polynomial y^2-2x^2+z is bounded on S. I don't see any systematic way to deduce all results like this! | |
| Apr 15, 2010 at 10:28 | answer | added | damiano | timeline score: 1 | |
| Apr 15, 2010 at 7:43 | comment | added | Kevin Buzzard | @David: don't my examples convince you that S not being defined over Z can still give a wealth of interesting possibilities? | |
| Apr 15, 2010 at 7:42 | comment | added | Kevin Buzzard | @anon: it's "not a proper question" because it is not of the form "prove this", it's of the form "give me necessary and sufficient conditions for this", so a logically valid but unhelpful answer would be "B=Z iff B=Z". | |
| Apr 15, 2010 at 3:30 | comment | added | Sidney Raffer | Are you thinking that if $S$ is defined by a polynomial equation with integer coefficients, then all bounded polynomials on $S$ are constant? This is not true -- For example if $S$ is defined by the polynomial equation equivalent to $y-x=\frac{1}{1+x^2}$, then $y-x$ is bounded but not constant on $S$. The situation in $>2$ variables seems harder, since one can't just write out and analyse finitely many Puiseux series representing the branches at infinity of $S$. | |
| Apr 15, 2010 at 1:00 | comment | added | David E Speyer | For that you need them to be nonconstant as functions on B. You can always find a polynomial which is nonconstant on R^n and bounded on S -- namely, the polynomial that defines S! | |
| Apr 14, 2010 at 22:43 | comment | added | Sidney Raffer | @ David Speyer I'd be happy to have an answer in the case that $S$ is defined by polynomials with integer coefficients. By $B$ I mean polynomials, formal objects, not functions on $S$..... And yes, non-finite-generation would seem to be typical, but I can't pin down exact conditions on $S$. Some background: Finding non-constant elements of $B$ is the crux of Runge's Theorem on Diophantine equations (cf Sprindzuk's Classical Diophantine Equations, Ch 1.) I want to generalize this theorem to equations of more than 2 variables. | |
| Apr 14, 2010 at 21:31 | comment | added | David E Speyer |
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}.
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| Apr 14, 2010 at 20:56 | comment | added | Sidney Raffer | @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. | |
| Apr 14, 2010 at 20:40 | comment | added | Kevin Buzzard | 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. | |
| Apr 14, 2010 at 20:08 | comment | added | Kevin Buzzard | 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.. | |
| Apr 14, 2010 at 19:55 | comment | added | Sidney Raffer | @Kevin: We posted simultaneously. | |
| Apr 14, 2010 at 19:46 | comment | added | Sidney Raffer |
@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 $B=\mathbb{Z}$.
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| Apr 14, 2010 at 19:42 | comment | added | Kevin Buzzard | @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. | |
| Apr 14, 2010 at 19:25 | comment | added | Wlog | 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? | |
| Apr 14, 2010 at 17:37 | comment | added | Qiaochu Yuan | Someone care to explain the vote to close? This seems like a perfectly reasonable question to me. | |
| Apr 14, 2010 at 16:40 | history | asked | Sidney Raffer | CC BY-SA 2.5 |