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> Concentric circles, e.g. x^2 + y^2 = 1 and x^2 + y^2 = 4. They don't intersect at all, right? Ah, except you forgot to count points over the complex numbers,
by eafer 6y ago
> Concentric circles, e.g. x^2 + y^2 = 1 and x^2 + y^2 = 4. They don't intersect at all, right? Ah, except you forgot to count points over the complex numbers, where they do.
How is this possible? Are you saying that there are pairs of complex numbers (x,y) such that x^2 + y^2 = 1 = 4?
- alcolade 6y agoYou need to also count "points at infinity" in the "projective" plane. So the "projectivized" equations are actually X^2 + Y^2 = 1Z, and X^2 + Y^2 = 4Z. The intuition is similar to how 2 parallel lines will meet at the horizion (infinity).
- eafer 6y agoThanks, I think I got it.
- impendia 6y agoOops, brain fart. No, 1 is not equal to 4 even in Grothendieck's world. Scheme theory only gets you so far... A correct example of what I had in mind is x^2 + y^2 = 1 and (x - 2)^2 + y^2 = 1. That said, algebraic geometry takes care of this "bad example" also. This is done exactly as alcolade explained: consider "points at infinity" in projective space.