Timeline for Algebraic independence of two polynomials in $\mathbb{C}[x, y]$ and Combinatorial Nullstellensatz.
Current License: CC BY-SA 3.0
7 events
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| Sep 12, 2011 at 15:09 | comment | added | Albertas | Thank you for the answer. Using that projections of constructible sets are constructible, I could prove that the converse is true. | |
| Jul 15, 2011 at 21:26 | comment | added | Andreas Blass |
$S$ is the image of the variety $\{(x,y,z,w):z=f(x,y), w=g(x,y)\}$ under projection to the last two components. Projections of constructible sets are constructible (Tarski-Chevalley theorem).
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| Jul 15, 2011 at 17:47 | comment | added | Albertas | Thank you; I have corrected the definition; I do not see why $S$ is constructible. | |
| Jul 15, 2011 at 17:32 | history | edited | Albertas | CC BY-SA 3.0 |
added 68 characters in body
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| Jul 14, 2011 at 0:03 | comment | added | Andreas Blass | The assumption of algebraic independence tells you that $S$ doesn't lie on a curve. Being a constructible set, won't $S$ then have dimension 2 and contain a whole neighborhood of some point (and therefore product sets of any desired width up to the cardinal of the continuum)? | |
| Jul 13, 2011 at 18:32 | comment | added | Andreas Blass |
You defined $S$ as a certain pair of sets, but I guess you intended a set of pairs, $\{(f(x,y),g(x,y)):(x,y)\in\mathbb C^2\}$.
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| Jul 13, 2011 at 16:01 | history | asked | Albertas | CC BY-SA 3.0 |