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
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| Apr 15, 2010 at 11:45 | comment | added | damiano | Your example works for what I asked, even though I would have wanted an example in the n=2 case. Also, the ideal of all polynomials vanishing on the S you described is generated by the two linear terms in the equation (you do not care about non-reducedness, nor about complex points). In your case, the "trace at infinity" of S is contained in the vanishing set of y-sqrt(2)x and it is maybe $\mathbb{Z}$-dense in there. I would have wanted an example which is "genuinely" not defined over $\mathbb{Q}$: in some sense, the example you gave has one "side" defined over Q and one which isn't. | |
| Apr 15, 2010 at 11:14 | comment | added | Sidney Raffer | Yes! An example I gave (way below) maybe does the job: 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 assume you mean irreducible over R.) | |
| Apr 15, 2010 at 10:42 | history | edited | damiano | CC BY-SA 2.5 |
small omissions corrected
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| Apr 15, 2010 at 10:36 | history | edited | damiano | CC BY-SA 2.5 |
small omission corrected
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| Apr 15, 2010 at 10:28 | history | answered | damiano | CC BY-SA 2.5 |