Timeline for Projections of real algebraic curves
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 20, 2012 at 0:57 | vote | accept | Marina Iliopoulou | ||
| Jan 20, 2012 at 0:52 | answer | added | Lyosha | timeline score: 3 | |
| Jan 19, 2012 at 17:41 | comment | added | Lucas Kaufmann | You can only ensure that the projection is semi-algebraic. See en.wikipedia.org/wiki/Tarski%E2%80%93Seidenberg_theorem | |
| Jan 19, 2012 at 17:06 | answer | added | Marina Iliopoulou | timeline score: 2 | |
| Jan 19, 2012 at 15:36 | comment | added | Marina Iliopoulou | Thank you! In the case of the line, the projection on a generic plane is still a line (there is only one plane the projection on which is not a line). | |
| Jan 19, 2012 at 14:52 | comment | added | J.C. Ottem | Well, if the curve is given by $x=y=0$ and you are projecting onto the $(x,y)$-plane, then you certainly don't get a curve. If however, $\gamma$ is not a line, you can consider the elimination ideal $(p_1,\ldots,p_k)\cap \mathbb{k}[x,y]$ which defines the image of the projection | |
| Jan 19, 2012 at 13:36 | history | asked | Marina Iliopoulou | CC BY-SA 3.0 |