Timeline for Real algebraic curves of $\mathbb R^2\sim\mathbb C$ containing all zeroes of infinitely many rational polynomials
Current License: CC BY-SA 2.5
7 events
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May 3, 2010 at 14:01 | comment | added | Roland Bacher | Yes, the elliptic curve $-x+x^3-2y^2+xy^2$ has this property in a non-trivial way (I did not check if it has infinitely many rational points, elliptic curves with infinitely may rational points have this property trivially of course). One can then apply the constructions mentionned above to get examples of higher genus. | |
Apr 30, 2010 at 19:13 | comment | added | Dror Speiser | @roland: can you please add an example of a hyperelliptic curve with the property? | |
Apr 30, 2010 at 16:28 | comment | added | damiano | I guess that I was answering the last question you mentioned in your post (and Dror seems to have done the same!). So you are only interested in curves having the property you mention, not the ones lacking, right? | |
Apr 30, 2010 at 14:55 | comment | added | Roland Bacher | No. I am asking for real curves in $\mathbb R^2$ containing all points $(x,y)$ for $z=x+iy$ root of a rational polynomial (and containing all such points for many polynomials). In particular, all points corresponding to roots of such polynomials have algebraic coordinates. A good example is the complex unit circle defined by $x^2+y^2-1$ where $z=x+iy\in\mathbb C$ which contains all roots of all cyclotomic polynomials. There are less trivial examples, eg. elliptic curves with only finitely many rational points. | |
Apr 30, 2010 at 13:56 | answer | added | Dror Speiser | timeline score: 3 | |
Apr 30, 2010 at 13:09 | comment | added | damiano | I am not sure that I understood the question correctly: is the curve with equation $y=\pi$ an example? The real and imaginary parts of the roots of a polynomial with rational coefficients are algebraic numbers, and the only imaginary parts of points on the curve above are transcendental. | |
Apr 30, 2010 at 12:08 | history | asked | Roland Bacher | CC BY-SA 2.5 |